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a A «
LIBRARY
JC^HNS HOPKINS UNIVERSITY

A
TEEATISE
ON
PLANE AND SPHERICAL
TRIGONOMETRY.
BY
WILLIAM GHAUVENET,
PBOPESSOa OP MiTHEKATICS AND A8TK0N0MT IN •WASHINGTON UNIVERSIir,
SAINT LOUIS.
iinti ^Yiiin,
PHILADELPHIA:
J. B. LIPPINOOTT & 00.
1878.

Q./\ 6-^1
C f
Entered according to Act of Congress, in tcs year 1860, by WILLIAM CHACTBHET, in the
Clerk's Office of the District Court of the Eastern District of Pennsylvania.

PEEFACE.
I HAVE in this treatise endeavored to arrange a course of
trigonometrical study sufficiently extensive to enable the
student to comprehend readily any applications of trigonometry
he may meet with in the works of the best modem
mathematicians. With this object, some topics have been
introduced which are not usually found in works devoted
specially to this subject.
Among those topics, the most important is the solution
of the general spherical triangle, or the triangle whose sides
and angles are not limited, according to the usual practice,
to values less than 180° The advantage of introducing
such triangles into astronomical investigations is sufficiently
shown in the applications made of them in the works of
BESSEL and other German mathematicians; and especially
in the Tliearia Motus Corporum Oodestiwn of GAUSS, who
was the first to suggest their employment.
The subject of Finite Differences of triangles, plane and
spherical, occupies a large space in CagnoH's treatise, but
has not been admitted into more recent works. It here
occupies only a few pages, but no important result of

4 PKEFACE.
Cagnoli's Table has been omitted, while a number of tHe
formulae are much simpler than the corresponding ones
given by him.
Although my plan embraces a much more extensive
course than is contained in the text-books commonly used,
I have studiously kept in view the wants of academic and
collegiate classes; and have so arranged the work that a
selection of subjects of immediate importance may be readily
made. The more elementary portions are printed in a larger
type, and are intended to form, independently of the matter
in the smaller type, a connected treatise which may be
studied as though it were in a separate volume.
Those who may afterwards wish to extend their knowledge
will appreciate the advantage of having the higher
departments of the subject treated in connection with those
fundamental ones to which they are most intimately re
lated. W. C.
XT. S. NAVAL AOADEMT,
Annapolis, Md., May 1, 1860
NOTE TO THE FOURTH EDITION.
IN this edition, besides a number of minor changes, and the correction of somt
typographical errors, a very important modification has been made in the solution of
the equation tan a; = i^ tan y by series (p. 145), which was given in former editions
in the usual form as stated by all writers on trigonometry. This form was discovered
to lack generality, and consequently to fail in certain applications, in consequence
of the omission of the arbitrary term njr now introduced. Several subsequent
investigations, depending on this, have in like manner been rectified.
TJ. S. NATAL AOADBMT, April 1, 1864.

CONTENTS.
PAET I.
PLANE TEIGONOMETRT.
CHAPTEE L
FAOB
MEASUEBS OF ANGLES AKD ABOS 9
CHAPTEB, XL
SiNBS, TANGENTS, AND SECANTS. FUNDAMENTAL FOEMUL^ 14
CHAPTER XXI.
TEIQONOMETKIO FUNCTIONS OE ANGULAE MAGNITUDE IN GENBKAI 22
Sine and Tangent of a Small Angle or Arc
SO
CHAPTER XV.
OENEEAL FOEMUXiE 31
Formulae for Multiple Angles 36
Relations of Three Angles 38
Inverse Trigonometric Functions 41
CHAPTER V.
TEXGOSOMBTEIO TABLES 43
Elementary Method of Constructing the Trigonometric Table 47
CHAPTER VX.
HoiTJTioN OS PLANE RIGHT TEIANGIES 51
Additional Formulse for Right Triangles 54
CHAPTER VIX.
FoEMULiE FOE THE SOLUTION Or PLANE OBLIQUE TEIANGLBS 57
A2 6

6 CONTENTS.
CHAPTER VIXX.
Pict
SOLUTION OP PLANE OBLIQUE TEIANQLES
Area of a Plane Triangle 74
G4
CHAPTER IX.
MISCELLANEOUS PEOBLEMS RELATING TO PLANE TEIANQLES 75
CHAPTER X.
SOLUTION or CEBTAIN TEIOONOMETEIO EQUATIONS AND OE NUMEEIOAL EQUA­
TIONS or THE SECOND AND THIRD DEGEEES 85
CHAPTER SI.
DUEEEENCES AND DiFEEEESTIALS OF THE TEIGONOMETEIC FUNCTIONS 101
CHAPTER XII.
DOTEEENCES AND DiEEEEENTIALS OFPLANE TEIANGLES 105
CHAPTER XIII.
TKIGONOMETEIC SEEIES. DEVELOPMENTS OE THE FUNCTIONS OF AN ANGLE IN
TEEMS OE THE AEO, AND EECIPEOOALLY 115
Computation of Natural Sines and Cosines by Series 116
Computation of the Ratio of the Circumference of a Circle to its Diameter 120
Computation of Logarithmic Sines and Cosines
V22
CHAPTEB XXV.
EXPONENTIAL POEMUL^. TEINOMIAL OE QUADEATIO FACTOBS 127
CHAPTER XV.
1'BIGONOMETEIO SERIES CONTINUED. MULTIPLE ANGLES 135
Development of the Sine and Cosine of the Multiple Angle in a Series of
Ascending Powers of the Cosine of the Simple Angle 137
Development of the Sine and Cosine of the Multiple Angle in a Series of
Ascending Powers of the Sine of the Simple Angle 139
Development of the Sine and Cosine of the Multiple Angle in a Serie's of
Ascending Powers of the Tangent of the Simple Angle 140
Development of any power of the Cosine of the Simple Angle in a Series
of Sines or Cosines of the Multiple Angles, the Cosine of the Simple
Angle being positive 141
Development of any power of the Cosine of the Simple Angle in a Series
of Sines or Cosines of the Multiple Angles, the Cosine of the Simple Angle
being negative 142
Development of any power of the Sine of the Simple Angle in a Series of
Sines or Cosines of the Multiple Angles 144
Certain Equations developed in Series of Multiple Angles 145

CONTENTS. 7
PAET II.
SPHEEICAL TRIGONOMETRY.
CHAPTER L
P^^^
GENEEAL FoEMULiE 149
Gauss's Theorem 161
Additional Formulse 162
Deduction of the Formulse of Plane Triangles from those of Spherical
Triangles 160
CHAPTER II.
SOLUTION OE SPHEEIOAL RIGHT TEIANGLES 167
Additional Formulae for the Solution of Spherical Eight Triangles....- 176
Qaadrantal and Isosceles Triangles 177
CHAPTER III.
SOLUTION OE SPHEEIOAL OBLIQUE TEIANQLES 178
Solution of Spherical Oblique Triangles by means of a Perpendicular 206
Computation of Spherical Formulse by the Gaussian Table 211
CHAPTER XV.
SOLUTION OE THE GENEEAI SPHEEIOAL TEIASQLE 214
Note upon Gauss's Equations 227
CHAPTER V.
AEEA OE A SPHEEIOAL TEIANQLE 229
CHAPTER VL
DlEEEEENOES AND DiFFEEENTIALS OP SPHEEICAL TEIANQLES 232
CHAPTER VIL
APPEOXIMATB SOLUTION OP SPHEEIOAL TEIANQLES IN CEBTAIN CASES 241
Legendre's Theorem 244
CHAPTER VIIL
MISCELLANEOUS PEOBLEMS OE SPHEEIOAL TEIGONOMETET 240

PAET I.
PLANE TEIGONOMETET.
CHAPTER L
MEASURES OF ANGLES AND AECS.
1. TRIGONOMETRY is that branch of Mathematics which treats
of methods of subjecting angles and triangles to numerical computation.
2. PLANE TRIGONOMETRY treats of methods of computing plane
angles and triangles.
It embraces the investigation of the relations of angles in general,
a branch of the science not necessarily connected with the
elementary solution of triangles, and which has been distinguished
as the Angular Analysig.
3. By the solution of a triangle, in trigonometry, is meant the
computation of unknown parts of the triangle from given ones.
The triangle has six parts; three angles and three sides. It is
shown in geometry, that when any three of these parts are given,
provided one of them is a side, the triangle may be constructed,
and the unknown parts found by mechanical measurement.
In the same cases, by trigonometry, we compute the unknown
parts from the three given ones, without resorting to construction
and measurement: a method of inferior accuracy, on account of the
unavoidable imperfections of the instruments employed, and the
difficulty of distinguishing with the eye the smallest subdivisions of
lines and angles.
But here also the case is excluded in which the three angles are
given without a side, because there may be an indefinite number o*
plane triangles, whose angles are equal to the same three given ones, as
2 0

10 PLANE TRIGONOMETRY.
in Fig. 1. the triangles ABC, A'B' C\
&e.. In this case, all these triangles
are similar, and their sides are proportional
; or the ratio oi AB to A C
is equal to the ratio of A'B'to A! 0',
ko.; so that the ratios of the sides to
each other are fixed or determinate, although the absolute lengths
of these sides are indeterminate.
4. Now, in order to subject a triangle to computation, we must
first express the sides and angles by numbers. For this purpose
proper units of measure must be adopted.
The unit of measure for the sides of plane triangles is a straight
line, as an inch, a foot, a mile, &c.; and the number expressing a
side is the number of units of the adopted kind that the side contains.
5. The units by which angles are expressed are, the degree,
minute, and second; distinguished by the characters " ' "
A degree is an angle equal to -g'^ of a right angle; or a degree is
jjj; of the whole angular space about a point, or ^J^j of four right
Tig. 2.
A'
angles. Thus, Fig. 2, if the angular space about
0 is divided into 360 equal parts, of which A OB
is one, then A OB is one degree. The right angle
will be expressed by 90°; two right angles by
180°, and the whole angular space about a point
by 360°.
A minute is an angle equal to g'^ of a degree.
Therefore, 1° = 60'; and a right angle =90 X 60'= .5400'.
A second is an angle equal to g'j of a minute. Therefore, 1' =
60": 1° = 60x60" = 3600"; and a right angle = 90 X 60 x 60"
= 324000"
Angles less than seconds are sometimes expressed by thirds,
fourths, fifths, &c., marked '" '^ '', &c.; a third being g';; of a second;
a fourth, g'j of a third; &c. But the more convenient method is to
express them as decimal parts of a second; thus ^ of a right angle
will be either
or more conveniently
12° 61' 25" 42'" SI"', &c.
12° 51' 25".714, &c.
6. The above division of angles is called sexagesimal, from the divisor 60 esnployed
m the subdivision of the degree. The centesimal division, however, would be preferable
in all cases, but cannot now be generally introduced without, at the same time,
changing the arrangement of all our tables, the graduation of astronomical and

MEASURES OF AECS. 11
other instruments, charts, &o. Nevertheless, the attempt has been made in France,
and several standard works exist in the French language, in which it is employed
throughout.
In the centesimal or French division, the right angle is divided into 100 degrees;
the degree into 100 minutes ; the minute into 100 seconds, &c. The reduction of
these denominations from one to the other requires only a change in the position
of the decimal, point; thus, in this system 00° 75' 84"-8 is the same as 607584"'b
or 60°'75848 or 0'-C075848, the symbol q denoting a quadrant or right angle.
To convert centesimal into sexagesimal degrees, since 100° dec. = 90° sex. deduct one
tenth from the number of centesimal degrees.
EXAMPLE. Eequired the number of sex. degrees in 85° 47' 43" dec.
85°-4743 cent.
Deduct JV = 8 -54743
76°-92687 sex. degrees and dec. parts.
55'-6122
36"-782
or
76° 55' 36"-732 sexagesimal.
To convert sexagesimal into centesimal degrees, since we must take y of the WK.,
divide by 9 and move the decimal point one place to the right.
EXAMPLE. Eequired the number of centesimal degrees in 76° 55' 36"-732 sex.
Eeducing the minutes and seconds to the decimal of a degree, we have
76°-92687 ses.
y> of which is 85°-4743 cent,
or
85° 47' 43" centesimal.
To distinguish the degrees of the centesimal from those of the sexagesimal division,
the former are frequently called grades, and are denoted by the character »
instead of °; thus the preceding angle would be 85* 47' 43".
MEASURES OF ARCS.
7. Since the angles at the center of a circle are proportional to
the arcs of the circumference intercepted between their sides, these
arcs may be taken as the measures of the angles, and we may express
both the arc and the angle by the number of units of arc intercepted
on the circumference.
The units of, arc are also the degree, minute, and second. They
are the arcs which subtend angles of a degree, a minute, and a
second, respectively, at the Center. A'degree of arc is thus always
alt) of the circumference, whatever the radius of the circle may be;
and we o'btain the same numerical expression of
an angle, whether we refer it directly to the angu-"
lar unit, or to the corresponding unit of arc. The
right angle A OA', Fig. 3, and its measure, the
quadrant A A', are therefore both expressed by
90°; the semicircumference by 180°, and the
whole circumference by 360°.
"'~~^^
8. The radius of the circle employed in measuring anirles is then

12 PLANE TRXGONOMETBT.
arbitrary, and we may assume for it such a value as will most simplify
our calculations. This value is unity ; that is, the linear unit
employed in expressing the sides of our triangles, or other lines
considered. This value wUl be generally used throughout this
treatise.
9. To find the length of an arc of a given number of degrees,
minutes, &c.
The semi-circumference of a circle whose radius is unity is known
to be 3-14159265; or, the radius being B, the semi-circumference
IS 3-14159265 i2. Hence
When.B = l
Arc 180° = 3-14159265 72 =3-14159265
u 1° = 0-017453293 i? =0-017453293
a v =0-0002908882i? =0-0002908882
u I" = 0-000004848137 i? =0-000004848137
An arc x therefore, in the circle whose radius is unity, being expressed
in degrees, or minutes, or seconds, we find its length by the
formulse
Arc a; = 0-017453293 2!°
= 0-0002908882 a;'
= 0-000004848137 a;".
As these factors for finding the length of an arc are often used,
it is convenient to have their logarithms prepared.* Thus
Arc X = [8-2418774] x°
= [6-4637261] a;'
= [4-6855749] a;"
m which the rectangular brackets are used to express that the logarithm
of the factor is given instead of the factor itself.
EXAMPLE. What is the length of the arc a; = 38° 17'48", the
radius being = 1.
.38° 17'48" =137868" log. 5-1394635
Log. factor for seconds 4-6855749
a; = 0-6684031 log. x 9-8250384
10. To find the number of degrees, ^c. in an arc equal to the
radius, '
We have, from the preceding article,
* The logarithms in the examples of this work wiU be taken from Stanley's Tables,
(published in New Haven, by Durrie and Peck,) the best tables of seven-figure
logarithms yet published in this country.

MEASURES OF ARCS. 13
180°
^ = 3-44169265 =^^°-2^^^^^S
= 3437'-74677 = 206264' -806
11. The angle at the center measured by an arc equal to the radius, is often
taken as the unit of angular measure, as this angle will be of an invariable magnitude,
whatever is the length of the radius. If x is the number of such units in a
given angle, the number of degrees, &o., in it will be found by multiplying by the
value of the radius in degrees, &o., found in the preceding article. Thus,
x° = xR° = 57°-2957795a; = [1-7581226] x
x' =xB' = 3437'-74677 a; = [3-5362739] a:
x" = x U" = 206264"-806 x = [5-3144251] x
Beciprocally, the angle being given in degrees, &o., we reduce it to the unit radius,
by dividing by -8°, E, or R", thus,
_ x" _x; _x!'
which is evidently the same as multiplying by the faotSrs of Art. 9.
It appears, then,.that an angle is expressed in the unit of this article by the
length of the arc which measures the angle in the circle whose radius is unity
Hence, an angle thus expressed is said to be given in arc. If we put (as is usual)
ir = 3-14159265 • - •
K is the circular measure of two right angles, or it is the expression of two right
angles in arc. In trigonometry it is therefore common to employ rr to denote an
7r
angular magnitude of 180°; — a right angle ; 2 v four right angles, &o.
12. The complement of an angle or arc is the remainder obtained
by subtracting'the angle or arc from 90°.
The supplement of an angle or arc is the remainder obtained by
subtracting the angle or arc from 180°.
Thus the complement of 30° is 60°; the supplement of 30° is
150°.
Two angles or .arcs are complements of each other when their
sum is 90°. They are supplements of each other when their sum is
180°
13. According to these definitions, the complement of an arc
that exceeds 90° is negative. Thus the complement of 120° is
90° - -120° = — 30°. In like manner the supplement of 200° la
ISO"^ -200° = -20°.

14 PLANE TRIGONOMETRY.
CHAPTER n.
SINES, TANGENTS, AND SECANTS.
FUNDAMENTAL FORMULA.
14. HAVING expressed the sides and angles of triangles by numbers,
we are next to find such relations between them as shall enable
us to combine these two diiferent species of quantity in computation.
As every oblique triangle may be resolved into two right triangles
by dropping a perpendicular from one of the angles upon the opposite
side, the solution of all triangles is readily made to depend upon
Fig. 4. that of right triangles. Let us therefore
consider a series of right triangles, ABQ,
AB'Q', AB"Q", &c., Fig. 4, which have
a common angle A. The angles at B,
B', B", being also equal, the triangles are
0' a" similar; and by geometry
BO:AB = B'O': AB' = B"0": AB"
or by the definitions of ratio and proportion,
BO _ B^'_ B"G"
AB~ AB'~ AB"
In like manner it follows that
and
BQ B'O' B"0"
AQ A 0' AO"
AB _ AB' _ AB"
AQ" AQ' AQ"
Hence it appears that the ratios of the sides to each other are the
same in all right triangles having the' same acute angle; and,
therefore, if these ratios are known in any one of these triangles,
they will be known in all of them.
These ratios, then, depending on the value of the angle alone,
without regard to the absolute lengths of the sides, may be considered
as indices of the angle, and have received special names, as follows:
15. The SINE of the angle is the quotient of the opposite side
divided by the hypotenuse.

SINES, TANGENTS, AND SECANTS. 1,5
Thus, in the right triangle ABQ, Fig. 5, "«•»•
1 we designate the sides by the small letters
a, b, c, we shall have, (whatever the absolute
length of the sides)
, a • T> b
sm A = —, sm B = —
c 0
16. The TANGENT of the angle is the quotient of the opposite side
divided by the adjacent side.
Thus tan A=-, tan 5 = -
0 a
17. The SECANT of the angle is the quotient of the hypotenuse
divided by the adjacent side.
Thus secA = -^, sec^ = -
0 a
18. The COSINE, COTANGENT, and COSECANT of an angle, are respeotively
the SINE, TANGENT, and SECANT of the complement of the
angle.
Since the sum of the two acute angles of a right triangle is one
right angle, or 90°, they are, by Art. 12, complements of each other;
therefore, according to the preceding definitions, we shall have
sia A = cos B =
a
c
tan A = cot B =
a
1
sec A = cosec B =
e
1
cos J. = sin .B = —
c
cot A = tan B = —
a
cosec >4 = sec ^ = — a
19. Since—is the reciprocal of —, it follows from the first and
a
c
last of these equations, that the sine and cosecant of the same angle
are reciprocals; and from the other equations, also, that the cosine
and secant, the tangent and cotangent are reciprocals. That is,
sin A = r cosec A = -
cosec A
sin A
COS A = r
sec A
sec A •
COS A
(2)
1 . ^ 1
tan A = cot A r cot A- tan A
4T more briefly,
sin A cosec J. = cos J. sec J. = tan JL cot J. = 1 (3)
(1)

iQ
PLANE TRXGONOMETBT.
SINES, &C. OF ARCS.
"r
D
0
Fig. 6.
\ B'
2"
20. The sine, tangent, and secant of
an arc are respectively the sine, tangent,
and secant of the angle at the center
measured by that are. Thus, Fig. 6,
BQ_
sin AJS = sin^O.B =
OB
The sine of an arc, therefore, does not depend upon the absolute
length of the arc, but upon the ratio of the arc to the whole circumference,
(Art. 7.) It follows that the relations (2) and (3) are also
applicable when A expresses an arc.
21. If the radius = 1, all the trigonometric functions above defined
may be represented in or about the circle by straight lines.
Representing the arc AB, or angle AOB, by x, we have, when OA
= 0B = 1,
sma;
^^^?fi^BO
= OB
AT AT .^
tan x = YTJ = ^f~ =-a.jt
OT
sec X = OA =
OT
-^=0T
and from the arc A'B = 90° — a; we find in the same way
coax = BI>=00
cot X = A'T'
cosec a; = OT'
Therefore, in the circle whose radius is unity, the sine of an aro,
or of the angle at the center measured by that arc, is the perpendicular
let fall from one extremity of the arc upon the diameter
passing through the other extremity.
The trigonometric tangent is that part of the tangent drawn at one
extremity of the arc, which is intercepted between that extremity and
the diameter (produced) passing through the other extremity.
The secant is that part of the produced diameter which is inter'
cepted between the center and the tangent.
The cosine is the distance from the center to the foot of the sine.
In a circle of any other radius than unity, the trigonometric

FUNDAMENTAL FORMULAE. [7
functions of an arc will be equal to the lines drawn as above, divided
by that radius.
The properties here stated have heretofore been used by most
writers upon trigonometry as definitions, but without limiting the radius
to unity; and it is evidently from this mode of viewing these
functions that they have derived their names.
22. Besides the functions already defined, others have been occasionally employed
to facilitate particular calculations, as the versed sine, ^Mhich. in the circle is the
portion of the diameter intercepted between the extremity of the ai-c and the foot
of the sine; thus. Fig. 6, the versed sine of A £ is A O, or the radius being = 1,
versin a; = 1 — cos x (4)
by means of which formula we may always substitute versed sines for cosines, and
reciprocally.
The coversfd sinb (covers.) is the versed sine of the complement, and suversed sine
(suvers.) is the versed sine of the supplement.
The chords of arcs have also been used, and may be substituted for sines by the
formula
ch 2: = 2 sin J 3: (5)
which is evident from Fig. 6, where if the arc B B' ^=x, we have chord B B' =..
2BG=2sinAB.
23. From what has now been stated, the student will perceive that
angles are to be subjected to computation by means of the quantities
sine, cosine, &c., commonly designated by the comprehensive
term trigonometric functions.* It becomes necessary, therefore, for
the computer to know the values of these functions for any given
value of the angle. The trigonometric tables, contain these values
for every minute, and sometimes for every second, from 0° to 90° ;
and with these tables all the numerical computations of trigonometry
are carried on. In practice, then, we are not required to compute
the functions themselves, and we shall therefore defer the methods
for that purpose to a subsequent part of this work, and proceed
at once with the investigation of the formulse and methods by which
these tables are rendered available.
FUNDAMENTAL FORMULA.
24. Q-iven the sine of an angle, to find the cosine.
From the right triangle ABQ, Fig. 7, we **^have
by geometry a^ + 5^ = 0^
Dividing by c^, this equation becomes
a^ b^ ,
c^ e^
* Also trigonometric lines, from the properties explained in Art. 21
R
B2

18 PLANE TRIGONOMETRY.
or, by the definitions of sine and cosine (1),
sin^ J. 4-cosM = 1 (6j
in which the notation sin^ A signifies " the square of the sine of A."
From this formula, if the sine is given, we find
cos^ A = l — sin^ A = {1 + sin A) (1 — sin A)
coaA = ^{l- sin^ A) = s/ [(1 -f sin A) (1 - sin AJ] (7)
and if the cosine is given, we find
sin JL = N/ (1 - cos'^ ^) = N/ [(1 + cos A) (1 - cos A)} (S)
25. (riven the sine and cosine of an angle, to find the tangent.
By (1) we have
also
therefore
tan^ = T
0
sinu4
cos J.
a _ b
c ' c
sin^
a
"b
tan J. =
cos.
And since the cotangent is the reciprocal of the tangent,
. cos J.
cot A = —; f
sm^
26. Q-iven the tangent of an angle, to find the secant.
The right triangle ABQ, Fig. 7, gives
e^ = J^ -f a^
Dividing by 5% this becomes
= l-fc'
¥~^ • b^
or, by the definitions of secant and tangent (1),
secM = l -f tan^J.
This formula applied to the complement of A gives
Kg- 8.
(9)
(10)
(11;
cosec^ J. = 1 + cot^ A
(12)
27. The preceding formulse are also
directly obtained from Fig. 8. If the
angle AOB, or the arc AB, be denoted
by X, the right triangle OBQ, gives
BQ^ + OQ^=OB^
or remembering that the radius is unity,
by Art. 21,
sin'' X + cos^ a; = 1 (13)

FUNDAMENTAL POEMULjE. 19
The triangle O^C gives by the definition. Art. 16,
BQ
tan A OB = -^
sin a;
or tana; = (14)
cos X
'
Since the angle BOD is the complement of BOQ, ta.nBOB =
cot a;, and the triangle BOB gives
„^^ BB OQ
tan502>=^ = -^
cos X , ,
01 cot X = —. (15;
sm a;
^ '
In a similar manner the triangles AOT, A! OT' give
sec^ 2; = 1 + tan^ x (16)
cosec* a; = 1 -f cot^ x (17)
28. The following equations are easily demonstrated by combining (13), (14),
(15), (16), (17), and employing the property of the reciprocals (2). They .are of
frequent use.
1 . tan X cos x
3in3;= == tan a; cos a; =—— :^ n81
cosec X sec x cot x ^ •
1 , . cot X oin -y
^ cot X sm X =
sec X • cosec x tan x ^ '
sina; = v'(l — cos" a;), cos2: = ,^/(l — sin'^) (20)
seca; = Y/(1-1-tan'a;), cosec i = ^ (1-(-cot'2:) (21)
tan a: = .^ (sec'a; — 1), cot a: = ^ (cosec'a;—1) (22)
tan X 1
sma;: V(l+.tan'2:) v'(l +cot'a:)
cot 2; 1
(23)
''°^^~,/(l + cotf'2:)~"v'(l + tan'a:)
sin 2; -1/(1 — cos'a:)
(24)
tana; = ^—^ = ^-^ ^
Y/ (1 — sm' X) cos X
(25)
cos a; »/(l — sin'2;)
cota;=^ ^-^— - ^ (26)
,y{l — cos'a:) sma: '^ '
29. To find the sine, ^c. of 30° and 60°.
In Fig. 8, let the arc AB = 30°, and BB' = 2AB = 60° By
Art. 21, sin JL^ = BQ, and by geometry the chord of 60°, or of onesixth
of the circumference, is equal to the radius = 1; therefore
2sin30° = 2 5C = ^.B' = l
whence
sm 30° = 1 = cos 60°
'27)'

20
PLANE TEXGONOMETRY.
dhd by (7)
cos 30° = x/[(l + J) (1-i)] = v/(i X i)
whence cos 30° = J x/ 3 = sin 60°
Then, by (9) and (2),
(28)
sin 30° i 1
tan 30° =
= cot 60°
cos 30° iv^3 ,/3
(29)
1 _
cot 30° =
tan 30° ^
1
sec 30° =
cos 30°
v" 3 = tan 60°
:6Q°
(30)
(31)
1
cosec 30° = = 2 = sec 60°
•.sin30° ~ " ~"^^""
^^^^
30. To find the sine, ^c. of 45° Since 45° is the complement of
45°, we have
sin 45° = cos 45°
whence by (13), putting x = 45°,
sin2 45° -f 008^45° = 2 sin^" 45° = 2 cos^ 45° = 1
sin^ 45° = cos^ 45° = J
sin 45° -•: cos 45° = >/ i = i v/ 2
(33)
sin 45°
tan45°^ : cot45° = 1
(34)
cos 45°
1
sec 45° = cosec 45° •
sin 45 -o = ^/2 (35)
These values are readily verified in the circle.
Fig. 9, where OA TA' is a square described upon
the radius. The diagonal OT bisects the right
angle, whence J-OT = 45°, and tan 45° == AT
= OJ. = 1; cot 45° = ^' T = 1; sin 45° = BQ
= Oa = cos45°, &c.
31. The sines and cosines of two angles being given, to find
the sine and cosine of the sum, and the sine and cosine of the difference
of those angles.
Kg 10
Fig. 11 Let the two angles he AOB
cB
y and BOQ, Figs. 10 and 11.
B
/
f-iE ^
\y
At any point B in the line
B
/"; f
\^ 0 B draw B G perp. to O'B.
Draw BA and C B perp. to
/I- DA L_
'
" o^— AD OA, and B E perp. to QB.

FUNDAMENTAL FOEMUL^. 21
Then the triangles B QB and BOA are mutually equiangular,
the three sides of the one being perp. to the three sides of the other
respectively; therefore the angle B QB= AOB.
Let
Then, Fig, 10,
Fig. 11,
and in
Fig. 10, sin {x + y)
Fig. 11, sin [x — y)-
and in both figures
Again in
Fig. 10,
Fig. 11,
x = AOB=BQB
y=^BOQ
x+y = QOB
x-y= QOB
QB BA+QE BA
~ Q.0~ QO ~,'Q0'^
QB BA - QE BA
QO QO QO
QE
QO
QE
QO
BA BA ' M) .
^ ^ X-^ = sm a; cos y
QO ~
QB QE QB
QO ~
"QB x ^ =cosa; smy
which being substituted in the above expressions of sin {x + y) and
sin (x — y) give
sin [x + y) =sin x cos y + cos x sin y
(36)
sin {x — y) == sin x cos y — cos x sin y (37)
and in both figures.
therefore
, , ^ OB OA-EB OA
cos {x + y)~ Q^- Q^ - Q^
OB OA + EB OA
cos{x-y)-^^- 0(7 ~0Q'^
OA OA OB
OQ = -OB''-OQ =
EB EB BQ .
OQ^W'OQ^''''''''''^
'"'''''''y
EB
' OQ
EB
00
cos (x -i- y) = cos a; cos a/ — sin x sin y (38)
cos {x — y) = cos a; cos ^ + sin a; sin^ (39)
and (36), (37), (38), and (39) are the required formula.
These maybe considered as the fundamental formulse of the trigonometric
analysis, and will form the basis of our subsequent investigations.
They are equally applicable to arcs represented by x and
y (Art. 20).

•i'i
PLANE TEIGONOMETEY
CHAPTER III.
TRIGONOMETRIC FUNCTIONS OF ANGULAR MAGNITUDE IN GENERAL.
32. THE definitions of sine, &c. given in the preceding chaptei
apply only to acute angles, since the angle is there assumed to be one
of the oblique angles of a right triangle. But we shall now take a
more general view of angular magnitude and of the functions by
means of which it is subjected to computation.
If, Fig. 12, we suppose the line OA to revolve
from the position OA to OA' in the direction of
the arc AA' (or from right to left), it will describe
\B an angular magnitude of 90°; when it ai-rives at
OA" it will have described an angular magnitude
of 180°; at OA'", 270°; and at OA again, 360°.
If it now continue its revolution, when it arrives
at OA' again, it will have described an angular magnitude of
360° + 90°, or 450° ; and thus Ave may readily conceive of an angular
magnitude of any number of degrees. In like manner we may have
arcs equal to or greater than one, two, or more circumferences.
To obtain trigonometric functions for angles and arcs thus generally
considered, we shall avail ourselves of the fundamental formulae
established in the preceding chapter; first deducing their values
analytically, and then explaining their geometrical signification.
33. To find the sine, ^c. of 0° and 90°. In (37) and (39)
let X — y; the first members become sin [x — x) = sin 0°, and
cos (x —' x) = cos 0° ; and by (13) they are reduced to
sin 0° = sin x cos x — cos a; sin a; = 0
cos 0° = cos^ X -f sin^ x =1
and smce 0° and 90° are complements of each other. Art. 12,
from which by (9) and (2)
sin 0° = cos 90° = 0 (40)
cos 0° = sin 90° = 1 (41)
sin 0° 0
tan 0° = cot 90° = ^ = r = 0 (42i
cos 0° 1 ^ •

FUNCTIONS OF ANGULAR MAGNITUDE. 23
cotO°= tan90°=J^, =!-=«, ^43)
sec 0° = cosec 90° = —^ = ^ = 1 (44)
cos 0° 1 ^ '
cosec 0° = sec 90° = ~ = 1 _ •„ UK^
sin 0° 0 ~ °° (*^)
34. To find the sine, ^c. of 180°. In (36) and (38) let
z=.-y = 90° ; these equations become by means of the preceding
values
whence by (9) and (2)
sin 180° =1x0 + 0x1 = 0 (46)
cos 180° = 0 x 0 - l x l = - l (47)
0 1
tan 180° = —^ = 0 cot 180° = ^ = oo (48)
sec 180° = —^ = - 1 cosec 180° = ^ = oo (49)
35. To find the sine, ^c. of 270°. In (36) and (38) let
x =180°, y = 90°, then
sin270°=0x0+(-l)xl = - l (50)
cos 270° = (-l)x0-0xl = 0 (51)
tan 270° = ^ = 00 cot 270° = — = Q (52)
sec 270° = - = 00 cosec 270° = J- = — 1 (53)
36. To find the sine, ^c. of 360°. In (36) and (38) let
x=y = 180°; then
sin360° = 0 x(-l) + (-l) x0 = 0 (54)
cos360°=(-l)x(-l)-0 xO = l (55)
the same values as for 0°, whence it follows that all the trig, functions
of 360° are the same as those of 0°.
The same process continued will give for 450° (= 360" + 90°),
the same trig, functions as those of 90°; for 540° the same functions
as for 180°, &c.

24 PLANE TEIGONOMETEr.
37. The preceding values now furnish us at once with the values
of the functions for all possible values of the angle. In (36) and
(38) let X = 0°, they are reduced to
sin y = sin 0° cos y -f cos 0° sin y — siny
cosy = cos 0° cos«/ — sin0° sin y = cosy
which are simply identical equations, and reveal no new property.
But if in (37) and (39) we put x = 0°, we have, after substituting
the functions of 0°,
whence by (9) and (2)
sin (—«/) = — sin y cos (— «/) = cosy (56)
, . sin (— w) — sin y .^„^
tan (— w) = —7—H = ^ = — tan«/ (57)
"^ ^^ cos{ — y) cosy
. COS(— W) COSW . ,;.QV
cot(—«) = -^—^= .^- = -coty (58)
^ ^' sm{ — y) —smy "
sec (— y) = 7 N = = sec y (59)
^ ^^ cos('^.?/) cosy " ^ '
cosec (— ,y) = -^—, c = =— = — cosec y (60)
^ ''' sm (— 2/) — sm y ^ ^ '
or, the sin., tan., cot., and cosec. of the negative of an angle are the
negative of the sin., tan., cot., and cosec. of the angle itself; and the
cos. and sec. are the same as those of the angle itself.
38. In (37) and (39) let a; = 90° ; we find after reduction
sin (90° —y) = cos y
cos (90° — «/) = sin y
which agree with the definition of cosine, but give no new relations.
But in (36) and (38) let x = 90°, we find
sin (90° + 2/) = cos y, cos (90° -f «/) = — sin ?/ (61)
whence by (9) and (2),
tan (90° + 2/) = — cot 2/ cot (90° +«/) = — tan y (62)
sec (90°+ «/) = —cosec y cosec (90°+«/) = sec y (63)
or, the sin. and cosec. of an angle are equal to the cos. and sec. of the
excess of the angle above 90° ; and the cos., tan., cot., and ges. are
equal to the negatives of the sin., cot, tan., and cosec. of the excess
of the angle alove 90°.

FUNCTIONS OF ANGULAR MAGNITUDE. 25
69. In (37j and-(39) let x = 180° ; we find
sin (180° - ?/) = sin y cos (180° —y) = -cosy (64;
tan (180° — y) = — tan y cot (180° — y) = — coty (65)
sec(180° — y) = — secy cosec (180° — y) = cosecy (66)
or, the sin. and cosec. of the supplement of an angle are the same as
those of the angle itself; and the cos., tan., cot., and sec. are the
negative of those of the angle itself.
40. If y is acute (that is, less than 90°), all its trig, functions are
positive ; and since its supplement 180° — yis obtuse (that is, greater
than 90°), it follows from the preceding article, that the sin. and
cosec. of an obtuse angle are positive, while its cos., tan., cot., and
sec. are negative.
41. In (36) and (38) let x = 180° ; we find
sin (180° +y) = — siuy cos (180° + t/) = - cos j/ - (67)
tan (180° +y) = tan y cot (180° + y) = cot y (68)
sec (180° -h y) = —secy cosec (180° + y) — — cosee«/ (69)
by means of which, if y is acute, we obtain the values of the sines,
&c. of angles between 180° and 270°.
42. In (37) and (39) let x = 270° ; we find
sin (270° -y) = -cosy cos (270° —y) = -smy (70)
'tan(270°-?/) = cot?/ cot (270° — ?/) = tan t/ (71)
sec (270°—«/) =—cosec?/ cosec (270°— «/)=—sec?/ (72)
43. In (36) and (38) let x = 270° ; we find
sin (270° +y) = -cosy cos (270° + y) = sin y (73)
tan (270° +y) = -coty cot (270° + ?/) = - tan y (74)
sec (270° +y) = cosec y cosec (270° + ?/) = - sec.y (75)
44. In (37) and (39) let x = 360° ; we find
sin (360° -y) = -siny cos (360° - y) = cos y (76)
tan (360°-?/) = -tan y cot (360°-?/) = -cot j/ y11)
sec (360° —y) = secy cosec (360° -«/)=- cosecy (78;
or the functions of 360° — y are the same as those of — ?/ (Art. 37).
45. In (36) and (38) let x = 360°; we find
sin(360°+ ?/)=siny cos (360° + ?/) = cos?/ (79i
9r, the functions of an angle which exceeds 360° are the same as
those of the excess above 360°

ye
PLANE TF.XGONOMETRT.
It follows that the functions of 720° + y are the same as those
of 360° + y, and therefore the same as those of y; and in like
manner for an angle which exceeds any multiple of 360°.
46. Since y — 90° is the negative of 90° — y, we obtain from
Art. 37,
sin (?/ — 90°) = — sin (90° — ?/)-= — cos ?/
cos {y — 90°) = cos (90° —y) = sin y
(80)
whence also tan., &c.; and in the same manner we may find the functions
of ?/ - 180°, y - 270°, y - 360°, &o.
47. We shall now give the geometrical interpretation of the preceding
results.
In Fig. 13, let the radius revolve from the
position OA to OA!, OA!', kc, as in Art. 32,
thus describing a continuously increasing angular
magnitude; or, which is equivalent, let
the arc commencing at A increase continuously
to AB, AA!, AB', kc. Then the changes
in the values of the several trigonometric lines
may be traced as follows.
1st. TAesme being, by Art. 21, the perpendicular from one extremity
of the arc upon the diameter drawn through the other extremity,
we shall have sin AB = BQ, sin AB' = B' Q', sin A A!' B" = B" 0',
sin AA"B'" = B'"Q, and if we make
we have
AB = A"B' = A"B" = AB'" = y
siny = BQ
sin (180° -y) = B' Q'
sin (180° +y)= B" Q'
sin (360° - ?/) = B'" Q
The lines BQ, B' Q', B" Q', B'" Q, however, represent only the
numerical values of the sines, and are here equal. But the results
above obtained from our formulae enable us to distinguish between
them by means of their algebraic signs. Thus, by (64), (67), (76),
sin (180° — y) = sin y
sin (180* +y) = — ainy
sin (360° — y) = — Biay
so that the sines from 0° to 180° are positive, while those from 180°
to 360° are negative; or the sines which are above the diameter
A A" are positive, while those which are below this diameter are

FUNCTIONS OF ANGULAR MAGNITUDE. 27
negative; or still more generally, the sines that have opposite directions,
with reference to the fixed diameter from which they are
measured, have opposite signs.
2d. The cosine being, by Art. 21, the distance from the center to
the foot of the sine, we have
but by (64), (67), (76),
cos y •= 0 Q
cos (180°-?/) = OC"
cos (180° -\-y)=
cos (360° ~y)=
OQ'
OQ
cos (180° — ?/) = — cosy
cos (180° + y)= — cosy
cos (360° — ?/) = cos y
so that the cosines on the right of the diameter A' A'" are positive,
while those on the left of this diameter are negative; or rather the
cosines that have opposite directions, with reference to the diameter
from which they are measured, have opposite signs.
We have here only exhibited a well-known principle in the application
of analysis to geometry, viz.: that all lines measured in opposite
directions from a fixed line have opposite signs.
To interpret the results (56), it is only necessary to observe that
a negative arc will be one reckoned from A towards B'", or in the
opposite direction to that of the positive arc, so that
sin A B'" = sin (- ?/) = B'"
cos A B " = COS ( — y) = OQ — cos y
Q=-BQ=-siny
as in (56).
The same principle applies to the tangents, but it will be simpler
in practice to obtain their signs (as also those of the secants), analytically,
from those of the sine and cosine, as has been already shown.
It will be sufiicient to bear in mind the following table, which is alsc
expressed by Fig. 13.
1st QUAD.
2d QUAD.
3d QUAD.
4th QUAD.
SINE
+
+
—
CCSINB
+
-f

•IS
PLANE TRIGONOMETRY.
48. The particular values of the sine and
cosine at A, A', A", kc, or sin. and cos. of
0°, 90°, 180°, &c., may also be found by
Fig. 13, upon the same principles; but this
we leave to the student.
49. GENERAL REMARK.—In the demon,
strati m of the fundamental formulse for
sin [x ± y), and cos [x ± y), Art. 31, the
A"'
angles x, y and x^y were all taken less than 90° and positive.
In this chapter these formulae have been applied to angles of any
magnitude, and the resulting functions have been shown to take
opposite signs when the lines representing them take opposite directions.
It follows that, in deducing trigonometric formulse from
geometrical figures, we need not embarrass our demonstrations with
the consideration of the various cases of the problem, or of the
various values of the angles of the figure. The formula deduced
from any supposed position of the lines of the figure will be of
general application, provided in the practical application of this formula
to the particular cases, we observe those values and signs of the
trigonometric functions which have now been determined.
50. The results of this chapter may be expressed hy a few general forniulfB.
From (70) it appears that all the trigonometric functions return to the same values
after one or more complete revolutions of 300°. If we represent the semi-circumference,
or two right angles, hy a-(Art. 11), and let n = any whole number or zero,
we shall have
sin 4 n — := 0
cos 4 re — = 1
(81)
sin (4 re + 1) -n- = 1
cos(4re4-l).l- = 0
(82)
sin(4«+2) j = 0
cos(4re+2).^ = —1
(83)
whence
sin (4 n + 3) .^ :
cos (4 re + 3) — = 0
(84)
tan 4 n — = 0 tan (4 n -f- 1) — = 00
tan (4 n -f 2) • tan (4 re-f- 3) — = oo
or the tan. of the even multiples ^,f-;-- = 0, and of the odd multiples =oo , so that
ive may write more simplytan
2 re ^ = 0 tan(2n-f 1) — = co (85)

FUNCTIONS OF ANGULAR MAGNITUDE. 29
In these formulaB we have only to give n one of the values 0, 1, 2, 3, 4, &c., to
obtain the functions of any given multiple of the right angle. Thus, we lind
Bin 450° = sin 5 -1 == sin ^4 -{- 1) ^ = 1 by making n = 1, in (82).
Since the subtraction of 8 re— from the arc will not change tUo functions, lii«
above formulse are also true when re is a negative whole number.
51. In a similar manner we obtain
sin 4 re — + y = siny cos [4 re .^ + y | = cos t/. (86;
sin 1^(4 re + 1) ^ + yj = cos y cos [|(4 re -f- 1) -^ -f- j/l = — sin y (87)
sin |^(4re + 2)|- + 2/j = — siny cos r(4re + 2)|l + yj = —cosj/ (88)
sin {i n-\- S) ^-\-y = — cos y cos (4 re -)- 3) — + y == sin y (8&)
tan [^2 re |- -f 2/] = tan j; tan [^(2 re + 1) ^ + j-J = — cot y (90)
in which n may be any whole number, positive oi negative, and y any angle, positive
or negative.
52. A still more concise form may be given to the formulse of the two preceding
articles, as follows : re being, as before, any whole number, positive or negative.
sin 2 re-|^ = 0 cos 2 re ^ = (— 1)" (91)
sin (2 re -f 1) ^=(—1)» cos (2 re + 1) ^ = 0 (92)
sin 12 re y-f-y | = (—l)"siny cos 12 re y + 2/ = (—1)" cos?/ (93)
B'n[(2n-1 l)|. + y]=(—1)»C0S2/ cos[(2re+l) | + ;/] = —J-lj" sinj (91)
and from these (85) and (90) may be directly deduced.
53. We have seen that an angle being given, there is but one corresponding sine.
On the other hand, a sine being given, there is an indefini'te number of angles corresponding
; for if a denote the given sine, and y any corresponding angle, then a is
also the sine of all the angles
TT — y, 2^+2/, Srr — y, &.O.
— TT — y, —lTT-\-y, — 3 !r — y, &c.
or in general
o2
a z= sin y = sin [ rea- + ( — 1)" 2/] (95)

80 PLANE TRIGONOMETRY.
In like manner if a is a given cosine, and y any corresponding angle,
a = cos y = cos (2 rejr ± y) (96)
and if « is a given tangent corresponding to the angle y,
a = tan y = tan [mr -\- y)
(97_'
SINE AND TANGENT OF A SMALL ANGLE OR ARC.
54. When the angle AOB=x, Fig. 14, -s
very small, the sine and tangent are very nearly
equal to the arc AB, which measures the angle,
\ii the radius being unity ; and the cosine and secant
are nearly equal to 0A=1 (Art. 21). Therefore,
to find the sine or tangent of a very small
angle approximately, we have only to find the
ength of the arc by Art. 9 ; thus
,sin 1" = arc 1" = 0-000004848137
log. sin 1" = 4-6855749
and X being a small angle, or arc, expressed in seconds,
If X is expressed in minutes,
sin X = tan a; = a; sin 1" (98)
sin X = tan a; = a: sin 1' (99)
If X expresses the length of the arc, the radius being unity,
sin X = tan x = x (100)
The employment of these approximate values must be governed by
the degree of accuracy required in a particular application. It is
found, for example, that they are sufiiciently accurate when the
nearest second only is required in our results, provided the angle
does not much exceed 1°.
55. If X and y are any two small angles, it follows from the preceding
article that
sin x-.siny = x sin 1" •.ys\aV' = x:y
that is, tlie sines (or tangents) of small angles are proportional to. the
angles themselves. The application of this theorem, however liko
that of the preceding, must depend upon the accuracy required in
the problem in which it is employed.*
•* For a full discussion of tlie limits under which this theorem may be employed,
.see a pnper, by the author of this work, in the Astronomical .lournal, fCambridee'
Mass.1 Vol i p. ?1.
^ '

GENERAL FORMULA 81
CHAPTER rV.
GENERAL FORMULiB.
56. WE have already obtained four fundamental equations, (36),
(37), (38), and (39), involving two angles, x and y. From these we
shall now deduce a number of formulse, either required in the subsequent
parts of this work, or of general utility in the applications
of trigonometry.
57. The sum and difierence of the equations (36) and (37) are
sin (a; -f" 2/) + ^^"^ (^ — 2/) ^^ 2 sin x cos y
(1^1)
sin {x -\- y) — sin (x — ?/) =; 2 cos x sin y (102)
and the sum and difi'erence of (38) and (39) are
68. If we put
cos (a; + ?/) + cos (x — «/) = 2 cos x cos y
cos {x -\- y\ — cos (a; — ?/) = — 2 sin a; sin y
X -\- y =^ x'
«^—!y = y'
whence 2 a; = a;' -f ?/', x=\{x' -\-y')
equation (101) will become
2?/ = a;'-?/', y^W~y')
sin x' + sin ?/' = 2 sin \ {x' + y') cos J {x' — y')
(1^3)
(10^)
and (102), (103), and (104) admit of a similar transformation. But
since x' and ?/' admit of all varieties of value, we may omit the
accents and apply the formulas to any two angles x and y; we have
thus
-• sin a; + sin ?/ = 2 sm 1 (a; + ^ cos J fe — ^ (105)
sin a; — sin ?/ = 2 cos i (a; + y) sin i (a; — ?/) (1061
cos a; -f cos ?/ = 2 cos J (a; + y) cos Hx — y) _, (107)
cos a; - cos ?/ = - 2 sin J (x + y) sin \ {x - y) (108)
Each of these equations may be enunciated as a theorem; thus

32 PLANE TRIGONOMETRY.
(105) expresses that "the sum of the sines of any two angles la
equal to twice the sine of half the suln of the angles multiplied by
the cosine of half their difi'erence."
These formulse are of frequent use (especially in computations
performed by logarithms), in transforming a sum or difierence into
a product.
69. Dividing (105) by (106), we have by (14) and (15)
or by (2)
sin a; -+- sin ?/ , _ , , , , , , ,
-. r-^ = tan *(x + y) cot i (x — y)
smx— smy ^\ ' ai 2 v J)
sin X -f sin y _ tan \{x +y)
sin X — •m\y tan \{x — y)
and from (107) and (108) we find in the same manner
We find also
a09)
cos a; — cosy
cos a; cosy
= — tani {x + y) tan|(a; — ?/) (HO)
sin a;
cos a;
+
+
+
+
+
sin X — siny
sin?/
= tani(a; + ?/) (Ill)
cosy
sin a; — siny
= tan^(a;-?/) (112)
cos a; cosy
sin X siny
= — cot |(a; — ?/) 1113)
cos a; — cosy
cos a; — cosy
-cot|(a;-fy) (111)
60. Divide the equations (36), (37_), (38) and (39) by cos a cosy;
then by (14) we have
sin (x + y)
-^ -' = tan X + tan y (li 6)
cos X cosy ^ V "y
sin Ix — y)
-^^j^t.nx-Uny (116)
cos (x -\- y)
^ ^ = 1 — tan a; tan y (117)
cos a; cos?/ ^ V-^-^V
cos {x.— y)

GENERAL FORMUL.*;.
61. Divide (36), (37), (38) and (39) by sin a; siny; and by sin x
cos y; then
sin (x ± ?/)
. ^ . -^^ = cot y ± cot X (119)
sm X smy "^ ^ '
cos (x ± ?/)
-V ^ . -^^ = cot a; cot w qp 1 (120)
sm X smy ^ ^^ ^ •'
sin (a; ± ?/)
"^'^ r?7
• 1 ± cot a; tan y (121)
sm a; cos" ^ ^ •'
cos (a; ± y)
^-^ ^ = cot a; q= tan y (122;
sm X cos y ^ '
62. Divide (115) by (117), and (116) by (118); then by (14)
SJi
/ s tana;—^ tan?/ rt^ix
tan (a; — ?A = -— i-—^ (124)
^ '^' 14- tan .-Han?/ ^ '
by which, when the tangents of two angles are given, we may compute
the tangent of their sum or difierence. To find the cotangent
of the sum or difi'erence when the cotangents of the angles are given,
divide (120) by (119),
^, , , cot ?/cot a; =p 1 ,-ioc\
cot (x±:y) = —^—;—^^^- (125)
^ "^^ cot?/±cota; '
63. Dividing (115) by (116), and (117) by (118j, (or from the
equations of Art. 61), we have
sin (a; + ?/) tan a;-j-tan ?/ cot?/-+cota; /loei
sin (x — y) tan x — tan y cot y — cot x
cos(x+y) 1 —tana;tan?/ cota;cot?/—1
cos {x—y)~l-jr tan x tan t/ ~ cot a; cot ?/ -fl
(127)
04. Formulas for secants are obtained from those for cosines by means of (2) j
thus we find
sec (xdcy) = - cos s cos y ::p sm a; sm 2/
and multiplying numerator and denominator by sec x sec y,
I ,' \ sees sec 2/ -mo.
sec (a: ± J/) = -—— -f- (128)
1 q:; tan x tan y

34 PLANE TRIGONOMETRY.
Also since
sec X rfc sec y -
1 , 1 cos y ± cos X
cos a: ~^ cos 2/ cos a; cosy
we find by (107) and (108)
sec X -f- sec y
2 cos J (x 4- y) cos J [x — y)
cos a; cos y
(129)
sec X — sec y
In the same manner from (105) and (106)
2 sin H* + y) sin i(x—y)
cos X cos 2/
(ISO,
, 2 sin J (a; -|- y) cos i (x — 2^)
cosec X 4- cosec y = ^ ^ T ' . ^-^ — (131)
sm X sm y ^ '
2 cos i (a; -|- y) sin i (x — y)
cosec X — cosec y = ^-^—-. —. ^-^ =^ (132)
" - sin X sm y ^ '
These formulse, although generally omitted in treatises on trigonometry, will be
found useful in a subsequent part of this work.
66. The product of (36) and (37), and of (38) and (39), are
sin {x -+- y) sin {x —«/) = sin^ x cos^y — cos'^ x ain^y
cos [x + y) cos {x — y) = cos^a; coa^ y — sin^ x ain^y
By (13) we have cos^ a; = 1 — sin'- x and cos^ y = 1 — sin^ y,
which substituted in the preceding equations, give
sin (x + y) sin [x — y) = sin^ x •— sin^y = coa^y — cos^ x (133)
cos {x + y) cos (x — y) = cos^ x — sin^ ?/ = cos^y — sin^ a; (134)
66. In (36), (38) and (123), let ?/ = a;, we find
-i sin 2 a; = 2 sin a: cos x (l^^j
cos 2 a; = cos^ x — sin^ x (136)
_
2 tan X
by which the functions of the double angle may be found from those
of the simple angle.
67. To find the functions of the half angle from those of the
whole angle, we have, from (13) and (136),
the sum and difi'erence of which are
cos^ X + sin^ a; = 1
cos^ X -^ sin^ X =.cos 2 x
2 cos^ a; =• 1 -f cos 2 a;
2 sin^ a; = 1 — cos 2 x

GENERAL FOEMULIE. 35
As these express the relations of an angle 2 x and its half x, their
meaning will not be changed by writing x and J x instead of 2 a;
and X; whence
2 cos^ J a; = 1-f cos a; (138)
2sin^ia; = l-cosa; (139)
the quotient of which is
1 —cosa;
68. The faUowing may be proposed as exercises.
_ 2 tan J a; 2
1 4" tan' J X cot \x-\- tan J x
(141)
2 tan \x 1
tan a; = = . ^ , = —-j ^, (142)
1 — tan' \x cot J a: — tan \x ^ '
tan' J a; 4- 2 cot a; tan J a; — 1 = 0 (143)
tan' \x — 2 cosec x tan J a; 4- 1 = 0
1 — tan' \ X
1 4- tan" \ X
(•'•^^)
(145)
1 i 1 — cosa; ,, ,„,
tan \ X = cosec a; — cot x = ; (146)
'• sm a; ^ •'
cot i a; = cosec a: 4- cot x = —V (147)
'' ' sm a; ^ •"
. , 1-4-sin a: — cosa; /IAOS
tan lx = —-J (148)
1 -f- sm a; 4- cos a; ,
69 Several useful formulse result from the preceding, by introducing
45° or 30°. If a; = 45° in (36), (37), (38), and (39), we
have, by (33),
cos ?/ d= sin y
sin (45° ± y) = cos (45° =F y) = — .^ (^"^^^
whence
cos ?/ ± sin ?/
tan (45° ± v) = cot (45° =F ?/) = ^ (150)
^ ^'^ ^ •'^ cosy =p smy ^ ^
in which either the upper signs must be taken throughout, or the
lower signs throughout.
If we divide the numerator and denominator of (150) by cos y or
sin?/,
1 rb tan y cot ?/ d- 1

d&
PLANE TEIGONOMETEY.
From this, by (67),
tan?/ — 1
tan (y - 45°) = ^ '' . ^ 152)
^•^ ' tan y + 1
70. Again, let a; = 90° ± ?/ in (138), (139) (140), and^(146),
sin (45° ± i'y) = cos (45° ^ i y) = J (l-^^^ (153)
tan(45°=t=j2.)=J(^fS|)
(^^^^
From the last we find
tan (45° ± * y) = L^^l^ = -^f- (155>
^ - •'•' cos y 1 ::^ sm y ^
tan (45° + i y) + tan (45° _ J j') = ^ = 2 sec i/ (150)
tan (45° 4- J j/) _ tan (45° _ J j,) = IfifLZ = 2 tan y (157)
the quotient of which is
tan (45° 4-} 2^)-tan (45°-^ 2,) _
tan (45° + i y) + tan (45° - i y) "'" ^ ^'°°''
71. In (101), (102), (103) and (104), let a; = 30° ; then by (27)
and (28)
sin(30° 4-?/)4-sin(30°-?/) = cos?/ (159)
sin (30° 4- y)- sin (30° - ?/) = sin?/ x/ 3 (160)
cos (30° 4- ?/) 4- cos (30° - ?/) = cos ?/ >/ 3 (161)
cos(30°4-?/)-cos,(30° —?/) = -siny (162)
and in a similar manner we may introduce 60° ; but it is unnecessary
to extend these substitutions, as they involve no difiiculty, and
can be made as occasion demands.
FoKMUL^ FOK
MULTIPLE ANGLES.
72/ From (101) and (102) we have
sin (y -{- x) = 2 sin y cos a; — sin (y — x)
sin {y -[- x) = 2 cos y einx -]- sin (y — x)
in which let j^ = (m — 1) a;; then
sin mx = 2 sin (rei — 1) a; cos x — sin (m — 2) a; (163)
sin mx = 2 cos (m — 1) a: sin x -{- sin (m — 2) a; '164)
which are the general formulse for computing the sine of any multiple mx, from the
lower multiples (m — 1) a: and (m — 2) x, and the simple angle x.

FOEMULiE FOR MULTIPLE ANGLES. 37
If we make m successively 1, 2, 3, 4, &c., these formulge give
sin X == sin x = sin x
sin 2 a; = 2 sin x cos x =2 cos x sin x
sin 3 a; = 2 sin 2 a; cos x — sin a; =2 cos 2 a:, sin a; 4- sin a;
sin 4 a; ^ 2 sin'3 x cos a: — sin 2 a: = 2 cos 3 x sin x -{• sin 2 a:
&c.
&c.
73. From (103) and (104)
cos (2/ 4" ^) = 2 cos y cos a; — cos (y — x)
cos (2/ 4" 3;) = — 2 sin y sin x -\- cos (y — x)
which, if we put y = (m — 1) 2:, become
cos mx 1= 2 cos (m — 1) a: cos x — cos (m — 2) a; {^^^)
cos mx = — 2 sin (m — 1) a; sin x -{• cos (m — 2) a; (166)
If m is taken successively equal to 1, 2, 3, 4, &o.
cos 2: = cos X = cos X
cos 2 2; = 2 cos X cos x — 1 = — 2 sin x sin x -{• X
cos 3 a: = 2 cos 2 a; cos x — cos x = — 2 sin 2 x sin a; -)- cos x
cos 4 a: = 2 cos 3 x cos a; — cos 2 a; = — 2 sin 3 a; sin a; 4- cos 2 x
&c.
&c.
74. In (123) let 2^ = (m — 1) a;; then
tan (m — 1) a; -4- tan x
,-ir.r,s
tan mx = ^ -,—'-^ (167)
whence
1 — tan (m — 1) a; tan x
„ tan 2 X •\- tan x
tan X = tan 2; tan 3 a::— 1 — tan 2 X tan x
2 tan 2; , . tan 3 2; -i- tan x
tan 2 a; = ;—-— tan 4 a; ^ = fc—7 •
1 — tan' X 1 — tan 3 x tan x
&e.
75. If in the expression for sin 3 x, Art. 72, we substitutt, the value of sin 2 x,
we find
sin 3 a: = 4 sin x cos' x — sin x
by which we find the sine of the multiple directly from the functions of the simple
angle. If this be substituted in the expression for sin 4 x, the latter will also be
expressed in terms of the simple angle. By these successive substitutions we easily
nbcain the following tables:
sin X = sin x
sin 2 a; = 2 sin x cos x
sin 3 a; = 4 sin x cos' x — sin x
sin 4 a; =: 8 sin x cos" x — 4 sin a; cos x
&o.
76. cos X = cos X
cos 2 a; = 2 cos' x — 1
cos 3 2: = 4 cos' X — 3 cos x
cos 4 a; = 8 cos* x — 8 cos' a; 4" 1
&c

38 PLANE TRIGONOMETRY.
77. If in these equations we substitute for cos' x = 1 — sin' z they become
sin X = sin a:
sin 2 a; ^ 2 sin 2: ^ (1 — sin' a;)
sin 3 a; = 3 sin x — 4 sin' x
sin 4 a; = (4 sin a; — 8 sin' x) ,y (1'— sin' x\
&o.
78. cos a; =: ^ (1 — sin' x)
cos 2 a: = 1 — 2 sin' x
cos 3 a; = (1—4 sin' a:) .^ (1 — sin' a;)
COS 4 a; ^ 1 — 8 sin' x-\- ^ sin* x
&c.
From the preceding tables it appears that the cosine of the multiple angle may
always be expressed rationally in terms of the cosine of the simple angle ; but that
the sine of only the odd multiples and the cosine of only the even multiples can be
expressed rationally in terms of the sine of the simple angle.
79. By successive substitutions we find from the formulse of Art. 74.
tan
X =: tan x
„ 2 tan X
tan 2 2; = - tan' X
tan 8 X :
tan 4 a; =:
3 tan X — tan' x
1—3 tan' X
4 tan X — 4 tan" x
1 — 6 tan' X 4- tan' x
80. The preceding results are but particular applications of general formulse to
be given hereafter, (Chapter XV.) They are introduced here for the convenience
of reference in elementary applications. The powers of the sine or cosine of the
simple angle may also be expressed in the multiples of the angle: but they are most
readily obtained from the general formulas of Chapter XV.
RELATIONS OF THHEE AUQLES.
81. Let X, y, and e be any three angles ; we have, by (36) and (38),
sin (a; 4- y + ^) = sin (x •\- y) cos 2 4- cos (a; -j- J/) sin z
^ sin X cos y cos z -\- cos x sin y cos z
-[- COS X cos y sin z — sin a; sin 2/ sin z (168)
cos {x-{- y -\- z) = cos {x 4- y) cos z — sin {x -f- y) sin z
= COS X cos y cos z — sin a; sin y cos z
— sin X cos y sin z — cos x sin y sin z (169)
and in the same way we may develop the sines and cosines oi x -\- y — z, 2; —y-\-Z;
&c. ; but we may find these directly from (168) and (169) by changing the sign of z,
y, &c., and observing (56).

RELATIONS OF THREE ANGLES. 39
The quotient of (168) divided by (169) gives, after dividing the numerator ana
denominator by cos x cos y cos z,
tan (x + y + z) = ^an a; 4- tan 2/ 4- tan z — tan x tan y tan z ^^
1 — tan X tan y — tan x tan z — tan y tan z '
82. Let X, y and z be any three angles, and from the equations
let cos z be eliminated; we find
sin (x — z) = sin x cos z — cos x sin z
sin (2/ — z) = sin 2/ cos z — cos 2/ sin z
sin y sin (x — z) — sin x sin (y — z) = sin z (sin a; cos y — cos x sin y)
If sin z is elicunated, we find
:= sin z sin (a; —'• y)
cos y sin (2; — z) — cos x sin (2/ — z) = cos z sin (a; — y)
These equations may be more elegantly expressed, as follows:
sin X sin (y — z) 4- sin y sin (z — a:) 4- sin z sin (a; — y) = 0 (171)
cos X sin (2/ — 2) 4- cos 1/ sin (z — a;) 4- cos z sin (a: —• 2/) = 0 (172)
A number of similar relations may be deduced from these by substituting 90° ± z,
&c., for X, &c.
83. Let
v = i (x-^y + z)
we have by (104)
2 sin V sin (u — a;) = cos x — cos (2 v — x) = cos x — cos (y -[- z)
2 sin (« — 2/) sin (v — z) = cos (y — z) — cos (2 v — y — z)
the product of which is
= cos (y — z) — cos X
4 sin V sin (« — x) sin (« — y) sin (v — z) = cos x [00s (y — 2) 4- cos (y -\- z)]
Reducing the second member by (103) and (134) ;
— cos' X — cos (y 4- 2) cos {y — z)
4 sin V sin (v — x) sin (» — ?/) sin (v — z) = 2 cos a; cos y cos z — cos' x
In the same manner we find
— cos' y — cos' z 4- 1 (173)
4 cos V cos (» — a:) cos (« — y) cos (» — z) = 2 cos a; cos 2/ cos z -j- cos' a;
4- cos' y + cos' z — 1 (174)
84. The following may be proposed as exercises.
sm a; 4-sin 2/4-sin z — sin (x-^-y-\-z) ^ 4 sin J (a; 4-2/) sin J (x-\-z') sin ^ (y-^z) (175)
eoB X •[• cosy-}-cos z-{-aoa(x-\-y-\-z) =4cos J(a;4-2/)oos-J(a;4-z) cos J (y 4-2) (176)
sin (2; 4- y -t- z)
tan x 4- tan y 4- tan z — tan a: tan y tan z = '^ (177)
' " ' cos a; cos y cos z ^ '
cos (2; 4- '/ 4~ ^)
cot a; 4- cot y 4- cot z — cot x cot 2/ cot z = ^ ;—'•—r—- CHSl
^^ ' T^ " sm a: sm y sm z

40 PLANE TRIGONOMETRY.
4 [si n (a;-f 3/4-z) -,-2 sin x sin y sin z]' = 4 [sin [x-\-y) cos z -f- cos (a;—y) sin zj'
= [1 — cos (2 X 4- 2 y)-\ (14- cos 2 z) 4- [1 4- cos (2 a; — 2 t/)] (1 — cos 2z)
4- 2 (sin 2 z -\- sin 2 J/) sin 2 z
(179)
= 2 (14-siu 2 a; sin 2 2/4-sin 2 a: sin 2 z 4" sin 2 y sin 2 z — cos 2 a; cos 2 y cos 2 z) _
85. Let the sum of three angles x, y and z be K-, or a multiple of TT, that is, an even
multiple of — a condition which is expressed by the equation
2:4-2,4-z = 2n.^ (180)
then, tan (x-{- y .-\- z) = 0, and the first member of (170) being thus reduced to
zero, the numerator of the second number must be zero, or
tan X -\- tan y -{- tan z = tan z tan y tan z (181)
an equation, it must be remembered, that is true only under the condition (180).
Since x, y and z may be selected in an infinite variety of ways so as to satisfy (180),
it follows from (181) that there is an infinite number of solutions of the problem,
" to find three numbers whose sum is equal to their product."
Let the sum of three angles z, y and z be — or an odd multiple of -^ ; that is, let
a;4-2/ + z=(2«4-l).| (182)
then, tan (z -\- y -\- z) = cx, and the denominator of (170) must be zero, or
tan X tan y -{- tan x tan z -f- tan y tan z = 1
which, divided by tan x tan y tan z, gives
cot X 4" cot y 4" cot z = cot x cot y cot z (183)
a relation that holds only under the condition (182).
86. Let
x-\-y-\-z = n7r = 2n-^ (184)
We have by (93) and (91)
cos {x -\- y — z) = cos (n ir — 2 z) = (— 1)" cos 2 z
ops {x — y-\- z) = COS (n 9r — 2y) = (— 1)" cos 2 y
cos (y -\- z — z) = cos (nrr — 2 a:) = (— 1)" cos 2 x
cos (y -\- z -{- x) = cos njr = (— 1)»
the sums of the first two and of the second two are by (103)
2 cos X cos (2/ — z) = (— 1)» (cos 2 z 4- cos 2 y)
2 cos X cos (y -f z) = (— 1)" (cos 2 a; -f 1)
and the sum and differende of these equations are
4 cos X cos y cos z = _(— 1)" (cos 2 z 4- cos 2 y 4. cos 2 a; -{- 1)
4 cos X sin y sin z = (— 1)" (cos 2 z 4- cos 2 y — cos 2 a; — 1)
or ±4 cos X cos y cos z = cos 2 a: -^r cos 2 y 4- cos 2 z 4- 1 (185)
± 4 cos a; sin y sin z = — cos 2 2: -|- cos 2 y 4- cos 2 z — 1 (186)
'he upper sign being taken when n in (184) is even, the lower when n is odd.

INVERSE TRIGONOMETRIC FUNCTIONS. 41
In the same manner we obtain
rp 4 sin a; sin 2/ sin z = sin 2 a: 4- sin 2 y 4- sin 2 z (187)
=p 4 sin a; cos y cos z = — sin 2 2; 4- sin 2 y 4- sin 2 z (188)
the signs being taken as .above.
Again, let
2;4.y4-z= (2«4-l)^ (189;
we shall find by the same process
=t 4 sin a: sin y sin z ^ cos 2 a: 4- cos 2y -\- cos 2 z — 1 (190)
± 4 sin 2: cos y cos z = — cos 2 2: 4- cos 2 y -|- cos 2 z -|- 1 (191)
± 4 cos X cosy cos z = sin 2 2: 4- sin 2 y 4- sin 2 z (192)
rb 4 cos X sin y sin z = — sin 2 a: 4- sin 2 y -{- sin 2 z (193)
-p or — according as n in (189) is even or odd.
87. If
INVERSE TRIGONOMETRIC FUNCTIONS.
y = sva. X
y is an explicit function of x, and, since x and y are mutually dependent,
X is an implicit function of y; but to express x in the
form of an explicit function of y, we write*
X = sin ~' y
which is read, "a; equal to the angle (or arc) whose sine is y," and
X is called the inverse functioti of y, or of sine x.
In like manner tan"''?/ is "the angle or arc whose tangent is
y," kc.
88. Many of the formulas already given may be conveniently expressed with the
aid of this not.ation. Thus, by (16),
or if we put y = tan x
X = sec ~' Y^ (1 4- tan' x)
tan-' y = sec-' ^/ (1 4- y')
* This notation was suggested by the use of the negative exponents in algebra.
If we have y = nx, we also have x =z n~^ y, where y is a function of x, and x is
the corresponding inverse function of y. The latter equation might be read " x is
a quantity which multiplied by n gives y." It may be necessary to caution the beginner
against the error of supposing that sin"' y is equivalent to —;——.
For a general view it the nature of inverse functions, see Peirce's Difi". Calo.
Arts. 13, et seq.
6 B2

i2
PLANE TRIGONOMETRY.
And in the same way tho formulse of Art. 28 give
sin —' y = cosec —' — =: cos —'

TRIGONOMETRIC TABLES. 43
CHAPTER V.
TRIGONOMETRIC TABLES.
90. BEFORE prot :'eding to the numerical computation of triangles
and to other applications of the preceding formulse, the student should
make himself acquainted with the arrangement of, and the mode of
consulting, the trigonometric tables. We shall here speak of those
points only that are common to all tables, but it will be necessary
to consult also the explanations that are always prefixed to a table
in order to understand any peculiarity that may attach to it. We
suppose also that he is acquainted with the nature and use of the
common tables of logarithms of numbers.
There are two principal trigonometric tables;* the first, called the
Table of Natural Sines, ^c, contains simply the numerical values
of the sines, tangents, &c. for each given value of the angle; the
* The most convenient seven-figure tables yet published in this country are Sta?iley's,
already mentioned, p. 12. Attached to these are also five-figure tables, and a
table of anti-logarithms.
Computers, engaged in extensive and varied calculations, generally provide themselves
not only with tables of seven figures, but also with those of six, of five, and
even of four figures—the selection and use of a particular table in any case being
determined by the degree of precision sought for in the results. We might, indeed,
employ seven-figure, or even ten-figure tables in all cases, and reject the final figures
of our results, when a lower degree of approximation is thought sufficient; but it is
clearly a loss of time and labor to employ other figures besides those which are ne
cessary in arriving at the proposed degree of precision.
The best six-figure tables are to be found in Bremiker's N'ova Tabula Berolinensis,
(Berlin. 1852,) which are distinguished for simplicity of arrangement, as well as accuracy.
Bowditch's five-figure tables, in his Epitome of Navigation, are valuable on account
of their undoubted accuracy.
Four-figure tables are to be found in various collections, as for instance, in Schumacher's
Il'dlfstafeln, (edited by Warnstorff.)
Of the German seven-figure tables we may cite those of Vega, of which Bremiker's
edition is the best: of the English, Taylor's, Hutton's, Babbage's, Shortrede's; and
if the French, Callet's, Bagay's, Borda's. T.aylor's, Shortrede's, and Bagay's give
the log. functions to every second of the quadrant; Borda's give the functions corresponding
to the centesimal division of angles, (Art. 6.)
For computations requiring more than seven figui'es recourse must be had to the
ten-figure tables of Vlacq, Thesaurus Logarithmorum Completus, edited by Vega,
'Leipzig, 1794.)

44 PLANE TRIGONOMETRY.
second, called the Table of Bogarithmic Sines, ^c, contains the lo>
garithras of the numbers in the first table. As the greater part of
the computations of trigonometry are carried on by logarithms, the
latter table is by far the most useful.
TABLE OF NATURAL SINES,
91. The arrangement of this table will be understood from a simple
inspection. It contains the sines, &c. of angles between zero
and 90°, generally for every minute, and the functions of angles
consisting of a number of degrees, riinutes, and seconds, have to be
found by interpolations similar in their nature to those that are required
in using tables of logarithms of numbers. This interpolation
is based upon the supposition that the diiferences of the sines, &c.,
are proportional to the differences of the angles; and this proportion,
though theoretically inexact, gives, in general, a sufiicient approximation,
provided the diflerences of the angles of the table are
sufficiently small. When the greatest accuracy is desired, the tables
should give the angles to every second, or at least to every 10", and
the sines, &c., should be given to at least seven decimal places.
92. As every angle between 45° and 90° is the complement of
another between 45° and 0°, every sine of an angle less than 45°
is the cosine of another greater than 45° ; every tangent is a cotangent,
&c.; hence the angles at the top of the tables generally extend
only to 45°, and the same functions answer for the remaining 45°,
by giving them at the bottom of the table the names of the complemental
functions.
93. As the sines, &c., pass through all their possible numerical
values, while the angle varies from 0° to 90°, the tables are not extended
beyond 90°; but we easily deduce the functions of all other
angles by the principles of Chap. III.
For the functiohs'of an angle between 90° and 180°, we may taks
the same functions of its supplement, observing to prefix the proper
algebraic sign. Art. 39. Thus, from Hutton's Tables we fini
sin 140°, 16' = sin 39° 44' = 0-6392153
cos 140° 16' = - cos 39° 44' = - 0-7690278
tan 140° 16' = - tan 39° 44' = - 0-8-311992
cot 140° 16' = - cot 39° 44' = - 1-2030810
sec 140° 16' = - sec 39° 44 = - 1-3003431
cosec 140° 16' ==- cosec 39° 44 = 1-5644181

TABLE OF TRIGONOMETRIC SINES. 45
remembering that in the 2d quadrant all the functions are negative
except the sine, and its reciprocal, the cosecant.
Or, we may (Ai't. 38) deduct 90° from the given angle and take
from the table the complemental functions of the remainder, prefixing
the signs as before; thus
sin 140° 16' =
cos 50°a6' = &c.
cos 140° 16' = - sin 50° 16' = &c.
which is the better practical method, as the subtraction of 90° may
be performed mentally.
94. For angles between 180° and 270°, we deduct 180° and take
the same functions of the remainder, prefixing the signs that belong
to the 3d quadrant. Art. 41; thus
sin 220° 26' = - sin 40° 26'
cos 220° 26' = - cos 40° 26'
tan 220° 26' = 4- tan_40° 26' &c.
95. For angles between 270° and 360°, we may deduct 270° and
take the complemental functions of the remainder, prefixing the sio-ns
that belong to the 4th quadrant, Art. 43; thus
sin 331° 27' = — cos 61° 27'
cos 331° 27' = 4- sin 61° 27'
tan 331° 27' = -
cot 61° 27' &c.
Or we may take the same functions of the difference between the
angle and 360°, Art. 44, observing the signs.
96. Above 360° we deduct 360°, and take the same functions
with their signs, Art. 45; and if the angle exceeds 720°, 1080°, &c.,
we deduct 720°, 1080°, &c. ; thus
cos 840° 45' = cos 120° 45° = — sin 30° 45'
tan 1372° 13' = tan 292° 13' = - cot 22° 13'
TABLE OF LOGARITHMIC SINES, &;C.
97. In this table we find the logarithms of the numbers in the
Table of Natural Sines arranged in precisely the same manner; it
will therefore require but little additional explanation.
As the sines and cosines are all less than unity (being by their
definitions proper fractions), their logarithms properly have negative
indices; but these are,avoided in the usual manner by increasing the

4b
PLANE TEIGONOMETEY.
index by 10, so that we find the index 9 instead of — 1, 8 instQp,d
of — 2, &c. The tangents under 45° being also less than unity,
the indices of their logs, are also increased by 10.
In some tables, to preserve uniformity, the indices of the logs, of
all the functions are increased by 10, so that the log. secants and
cosecants are always greater than 10. In using these tables, we ha^ e
the general rule that for each log. function added in forming a sum
we must deduct 10 from that sum.
98. Since
we have
sec A
1
cos A
log sec -4. = — log cos A
or since the tabular log. cos. is increased by 10,
log sec A = 10 — log cos A
that is, the log. secant is the arithmetical complement of the tabular
log. cosine. For a like reason log. cosec. is the ar. co. of the log.
sm.; and log. cot. is the ar. co. of the log. tan.
Also since
sin J.
tan A = J
cos^
log ten A = log sin A — log cos A
by which property, together with the preceding, we may obtain by
subtraction only, the log. tan. cot. sec. and cosec. from a table containing
only the log. sin. and cos.
99. When the natural sines, &c. are negative, we shall in this
work indicate it by prefixing the negative sign also to their logarithms
;* thus we shall write
cos 140° 16' = - 0-7690278
and log cos 140° 16' = - 9-8859420
* strictly speaking, negative numbers have no logarithms, but in practice, the
multiplication, division, &c.-of numbers is performed without reference to their signs,
1. e. as if they were all positive, and the sign of the result is then deduced from the
signs of the factors according to the rules of algebra. We employ logarithms simply
to effect the first of these operations, i. a. the multiplication, division, &c. of
the numbers considered as positive; and to facilitate the second operadon, or the
determination of the sign, we prefix to the logs, the signs which are prefixed to the
numbers to which they belong.

CONSTEUCTION OF TEIGONOMETEIC TABLES. 47
A^the logs, of all the trig, functions are positive (being rendered
BO by the addition of 10 where necessary), it will easily be remembered
that the sign in the latter case is not that of the logarithm,
but of the number to which it belongs.
ELEMENTARY METHOD OF CONSTRUCTING THE TRIGONOMETRIC
TABLE.
100. By dividing TT = 3-1415926 by the number of seconds in
180°, we found (Art. 9) the length of the arc 1", and (Art. 54), the
sine of 1", which is sensibly equal to the arc. In the same manner
we find, by dividing by 10800,
and by (7)
sin 1' = 0-0002908882
cos 1' = v^ (1 - sin^ 1') = ^/ [(1 4- sin 1') (1 - sin 1')]
= V (1-0002908882 x-9997091118)
or performing the arithmetical operations
cos 1' = 0-9999999577
Then by (101) and (103)
sin (a; 4- ?/) = 2 sin x cos y — sin [x — y)
cos (a; -f ?/) = 2 cos x cos y — cos {x — y)
in which we can suppose y to be constantly equal to 1' and x to become
successively 1', 2', 3', &c. Thus, first substituting y = V,
sin (a; 4- 1') = 2 sin x cos 1' — sin {x — 1')
cos (a; -f 1') = 2 cos x cos 1' — cos {x — 1')
themif x = 1', 2', 3', &c., we find for the sines
sin 2' = 2 cos 1' sin 1' - sin 0' = 0-0005817764
sin 3' = 2 cos 1' sin 2' - sin 1' = 0-0008726646
sin 4' = 2 cos 1' sin 3' - sin 2' = 0-0011635526
sin 5' = 2 cos V sin 4' - sin 3' = 0-0014544407
&c.
&c.
BT^ for the cosines
cos 2' = 2 cos 1' cos r - cos 0' = 0-9999998308
cos 3' = 2 cos 1' cos 2' - cos 1' = 0-9999996193
cos 4' = 2 cos 1' cos 3' - cos 2' = 0-9999993232
cos 5' = 2 cos 1' cos 4' - cos 3' = 0-9999989425
&c.
kc.

48 PLANE TRIGONOMETRY.
The whole difBculty in this operation consists in the multiplication
of each successive sine or cosine by 2 cos 1' = 1.9999999154; but
.this multiplication is much shortened by observing that
so that if we put
2 cos r = 1-9999999154 = 2 - -0000000846
m = -0000000846
we have 2 cos 1' = 2 — wi and therefore
sin 2' = 2 sin 1' — sin 0' — m sin 1'
sin S' = 2 sin 2' — sin V — m sin 2'
sin 4' = 2 sin 3' — sin 2' — 7n sin 3'
&c.
cos 2' = 2 cos 1' — cos 0' — m cos 1'
cos 3' = 2 cos 2' — cos V — m cos 2'
cos 4' = 2 cos 3' — cos 2' — m cos 3'
&c.
which are computed with great facility.
101. It is not necessary, however, to continue this process beyond
30° ; for by (159) and (162) we have
sin (30° + y) = cosy — sin (30° — y)
cos (30° 4- ?/) = cos (30° — y) — sin y
so that the table is continued above 30° by the simple subtraction
of the sines and cosines under 30° previously found. Thus, making
y successively 1', 2', 3', &c.
sin 30° r = cos 1' - sin 29° 59'
sin 30° 2' = cos 2' - sin 29° 58'
sin 30° 3' = cos 3' - sin 29° 57'
&c.
cos 30° r = cos 29° 59' - sin 1'
J
cos 30° 2' = cos 29° 58' - sin 2
cos 30° 3' = cos 29° 57' - sin 3'
&c.
This last process requires to be continued only to 45° since the
sines and cosines of the angles above 45° will be respectively the
cosines and sines of their complements below 45°.

CONSTRUCTION OF TRIGONOMETRIC TABLES. 49
102. The tangents and cotangents may be found from the sin&s
and cosines by the formulsB
sin X
cos X
tan X = cot X = -:
cos X
sm X
and the secants and cosecants by the formulse
sec X =
1 1
cos X
cosec x
sm X
103. In so extended a computation as the construction of the entire
table, it is necessary to verify the accuracy of the work from
time to time, by separate and independent calculations. By means
of (138) and (139) we can find from the cosine of an angle the sine
and cosine of its half; hence from the cos. 45° = >/ J we can find
sin. and cos. of 22° 30', and from these the sin. and cos. of 11° 15
by the same formulse; and from cos. 30° = J \/ 3 we can find sin.
and COS. of 15°, 7° 30', and 3° 45'. If these agree with those found
by the first process, the whole work may be considered as correct.
104. There are various other angles whose functions can be expressed under finite
forms more or less Smple, and may therefore be employed for the purpose of verification.
Let x = 18°; then 3 a; 4- 2 a; ^ 90° and cos 3 a; = sin 2 a;, whence, by Art. 70,
4 cos' X — 3 COS X = cos 3 2; = 2 sin x cos x
4 cos' X — 3 = 2 sin a;
4 (1 — sin' a;) — 3 = 2 sin 2;
sin' a: 4- J sin a;
= J
which equation of the 2d degree being resolved, gives sin x =
sin 18° = cos 72° =
^ ^ " ^
whence
cos 18° = sin 72° = N/aO + ^v"^)
4
From these by (138) and (139), we find the sine and cosine of 9° and 36°; then
by (37) and (39) those of 36° — 30° = 6°, whence those of 3°; after which it
will be easy to form a table of the exact values of the sines and cosines for every
3° of the quadrant.* These expressions, however, are not of much use, directly,
in tho construction of tables, as we have much better methods; but they lead to a
formula of verification which is of some importance. We findeos36°
= ^^Aii
* A table of this kind is given by Cagnoli in his Trigonometrie.
7 E

60 PLANE TRIGONOMETRY.
And by (102)
sin (72° -f y) — sin (72° — y) = 2 cos 72° sin y = ^ ~ sin j,
sin (36° -f y) — sin (36° — y) = 2 cos 86° sin y = "^ "^ sin y
the difference of these equations gives
sin (36° -f y) — sin (36° — y) = sin (72° 4- y) — sin (72° — y) 4" sin y
which is Buler'a formula of verification. By giving y any value at pleasure, the correctness
of five sines of the tables is examined. By substituting 90° — y for y in
this formula it is easily reduced to the following
sin (90° — y) 4- sin (18° 4- y) 4- sin (18° —y) = sin (54°-}- y) 4- sin (54° — y;
which is known as Legendre's formula, though not essentially different from Euler's.
105. The method that has here been given for computing the trigonometric table,
though simple in principle is nevertheless sufficiently operose. The method by infinite
series, to be given hereafter, will bo found to be much more rapid and simple
in practice.

SOLUTION OF PLANE RIGHT TEIANGLES. 51
CHAPTER VI.
SOLUTION OF PLANE RIGHT TRIANGLES.
106. IN order to solve a plane right triangle, two parts in addition
to the right angle must be given, one of which must be a side,
The solution is effected directly by means of our Mg. 15.
definitions of sine, &c., which are expressed by
the equations (1). As three of the six functions
are only the reciprocals of the other three, we
shall base the solutions upon the foUowino- three ;
(Fig. 15):
, a , b , a
svn A = — cos A = — tan A = -r
0 c 0
Since each of these equations expresses a relation between three
parts-^-an angle and two sides—it follows that in order to apply them,
or in order to solve the triangle trigonometrically, there must be
given two of these parts ; and that of the three parts considered,
one must be an angle while the other two are sides. Thus, if an
angle and side are given, the third part sought must be a side; but
if two sides are given, the third part sought must be an angle.
In every instance the choice of the proper equation will be determined
by the precept,—employ that trigonometric function of the
angle which is equal to the ratio of the two sides considered.
107. CASE. I. G-iven the hypotenuse and one angle, or c and A.
To find a. We consider a, c and A; and since the ratio of a and
c is given by the sine, we have
sin A = — whence a = c sin A (195)
0
To find b. Considering b, c and A, we have the ratio of b and c
expressed by the cosine, or
b
cos A = — whence b = c cos A (19^)
e
To find B. We have 5 = 90° - A.

52 PLANE TRIGONOMETRY.
Tho required quantities a and b being equal to the product of two
factors, the computation is conveniently performed by the addition
of the logarithms of these factors.
EXAMPLES.
1. Given c = 672-3412, A = 35° 16' 25"; to find the other parts.
By (195) By (196)
e = 672-3412 log 2-8276897 log 2-8275897
A = 35° 16' 25" log sin 9-7615382 log cos 9-9119049
a = 388-2647 log* 2-5891279 5 = 548-9018 log* 2-7394946
Ans. a = 388-2647
b = 548-9018
B = 54° 43' 35"
2. Given a = 42567-2, B = 87° 49' 10"; find the other parts.
Ans. a = 1619-626
b = 42536-37
A= 2° 10' 50"
108. CASE II. Given the hypotenuse and one side, or c and b.
To solve this case trigonometrically, we must first find an angle.
To find A. We have
cos A = ^ (197)
To find a.
We have, by the preceding case,
sin J. = — a = c sin J. (198)
c
But a may be found by geometry from the equation
a^ -j-52 = c2 whence a^ = c^ — h^
a = ^ {c^- b') = v' [(e 4- b) {c - 6)] (199)
EXAMPLES.
1. Given c = 672-3412, b = 548-9018; find A and a.
By (197) By (198)
b = 548-9018 log 2-7394946
c = 672-3412 log 2-8275897 log 2-8275897
A = 35° 16'25" log cos 9-9119049 log sin 9-7615382
a = 388-2647 log 2-6891279
* Ten is rejected from each of these indices because the logarithms of the sine
• and cosine in the table are ten too great. Art. 97

SOLUTION OF PLANE RIGHT TRLANGLES. 53
By (199)
c= 672-3412
b = 548-9018
c + b = 1221-2430 log 3-0868021
c - b= 123-4394 log 2-09145-38
2)5-1782559
a = 388-2647 log 2-5891279
Ans. A =.36°16'25"
B = 64°43'35"
a = 388-2647
2. Given c = -092357, b = -058018; find a.
Ans. a = -071859
109. CASE III. Given an angle and its adjacent side, or A and b.
To find a. We have
tan A = -r whence a = b tan A. (200^
To find e.
We have
cos A = — whence, by (2), e = -r = b sec A (201)
or directly from the secant
see A — -J- whence c = b sec A
b
EXAMPLES.
1. Given A = 88° 59', 5 = 2-234875 : find the other parts.
Ans. a = 126-9365
c = 125-9563
5 = 1° r
2. Given B = 60°, a = 10; find e. (See Art. 29).
Ans. c = 20.
110. CASE IV. Given an angle and its opposite side, or A and a.
To find e. We have
sin JL = — c = -.—T = a cosec A (202)
c sm J. ^ '
To find b. We have
tan ^ = 4 5 =^-^ = a cot JL (203^
6 tan A

64 PLANE TEIGONOMETEY.
EXAMPLES.
1. Given A = 25° 18'48", a = -085623; find 5.
Ans. b = -1810278
2. Given B = 39° 17'5", 6 = -01; find c.
Ans. c = -0157934
111. CASE V. Given the two sides, or a and b.
To find A and B. We have
To find c. We have
tan A = cotB = j (204)
sin ^ = — c = --. 7 = a cosec A (205)
0 sm A ^ '
We may also find c directly by geometry, from
e^ = a^ 4- 6^ whence e = -y [a^ + b^)
but this is not readily computed by logarithms.
EXAMPLES.
1. Given a = 30, 5 = 40; find c. Ans. c = 60
2. Given a = 8-678912, b = 2-463878; find A and c.
Ans. J[= 74°9'4"-l
c = 9-021875
ADDITIONAL FoEMCLiE FOE. EIGHT
TEIANQLES.
112. By inspecting the tables it will be seen that when the angles are very small,
the cosines differ very little from each other ; consequently a small angle cannot be
found with very great accuracy from its cosine. For a similar reason an angle that
is nearly 90° cannot be accurately computed from its sine. It is therefore desirable,
when a reqiiired angle is small, to find it by its sine, and when near 90° by its cosine,
or in either case by its tangent or cotangent; and for this purpose special formula!
.are sometimes necessary. We shall deduce several such formulse, from which
one adapted to a particular case may be selected.
113. From (197) we find, by (139^
., . n . « . . ^ ^ c — b
1 — cos yl = 2 sm' A ^ = 1 =
c c
= Jte)
(206)
which may be used instead of (197), when A is small, that is when b is nearly equal
to c. It glTes also
c-'b = 2c sin' J A (207)
by which c — b may be accurately found when A is small

ADDITIONAL FORMULA FOE EIGHT TRIANGLES, 55
Also from (197), by (140)
-M = jG-^-^)=JC^)="-^ (20S)
114 From the equation
we deduce by (153) and (154)
Bin A = —
c
Bin(4.5°±J^)=J(^'') (209)
a
and from tan J4 = -7- we find by (151),
cos (45° ± M) = J ( ^ ) (210)
tan(45°±J^) = J(i0) (211)
115. By (136) we have
tan (45° dt= A) = l^!^ (212)
o zp. a
cos 2 A ^ cos' A — sin' A = 4'—a'
(vhich, since 2.4=.4-f-90° — .8 = 90° — (B — A), gives
By (135)
Bin(^-^)=i^-^/-'') (213)
cos (.B — ^) = sin 2 ^ = 2 sin ^ cos .4 = - ^ (214)
and from (213) and (214)
tan(^-^)=i^ + |)i^) (215)
by which B — ^ is found with great accuracy when b and a are nearly equal.
EXAMPLE. Given

66 PLANE TRIGONOMETRY.
rapid change of these diJerences. We avoid the use of these large difl'erences, and
gain somewhat in accuracy, by employing the approximate value of sin. J A given
by (98), whence
sin M = J ^ Bin r, iA = ?^^Ai
sin 1"
Thus we have found above log sin J J4 = 7-9160547
Art. 54, log sin 1" = 4-6855749
J ^ = 1700"-12 = 28' 20"-12
log J A ='3'-2304798
But to obtain J A with the utmost precision, recourse must be had to the following
process, which is constantly employed in observatories, and wherever sma; 1 angles
are to be computed with extreme accuracy. Special tables are prepared containing
for every minute from 0° to 2° the logarithms of
sin X
tan x
which do not vary rapidly, and may therefore be taken with accuracy from the
tables. Then we have
sin a; = a;, sin a:
tan a: = a;.
X
tana:
z
Bin X
tana;
ANjtable of this Hnd will be found on page 156 of Stanley's Tables, where the
notation used is
q r= log sin x,
n = log x
and therefore in the column marked j—n we find the log . Thus in the above
example we have found log sin J ^ = j = 7-9160547
and from the table - q—n = 4-6855700
J .4 = 1700"-14 = 28' 20"-14 log J ^ = „ = 3-2304847
which is the true value ot ^ A within 0"-01.
Stanley's Table contains also the values of
, tan X
log = q — « (? = log tiin ^> n = log 2:1
'"S ilTS = ? 4- « (? = log cosec X, n = log x)
^°^ te^ = ? 4- » (? = log cot X, n = log x)
the use of which may easily be inferred from the example just given.

FOEMULiE FOE OBLIQUE TRIANGLES. 57
CHAPTER VII.
FORMULA FOR THE SOLUTION OF PLANE OBLIQUE TRIANGLES
116. As every oblique triangle may be resolved into two right
triangles by a perpendicular from one of the angles upon the
opposite side, we are enabled to deduce all the formulse for their solution
from those of the preceding chapter.
117. The sides of a plane triangle are proportional to the sines
of their opposite angles.
Denote the angles of the triangle ABQ,
Pig. 16.
Fig. 16, by A, B and Q, and the sides opposite
these angles respectively by a, b and a.
From G draw GP perp. to AB and put
QB = p. Then in the right triangles A G.P, B QB, we have,
by (195)
whence
p = b sm. A,
5 sin ^ = a sin B
which, converted into a proportion, gives
p = a ain B
a : b = sin A : sin B
and in the same way we may prove that
a : c = ain A : sin Q
b : c — ain B : sin G
and these three proportions may be written as one, thus :
or thus,
a -.b : c = sin A : sin B : sin. G
a l e
sin A sin B sin G
When the perpendicular falls without the
triangle. Fig. 17, the angle GBP is the supplerafent
of B, but by Art. 39, it has the same
8in«, so that the triangle GBP gives A
^ = a sin QB P = a sin J5
Fig. 17.
(216)
(217)
(218)

fiS
PLANE TRIGONOMETRF.
the same as was found from Fig. 16. The proposition i^ therefore
general in its application.*
118. The sum of any two sides of a plane triangle is to their difference
as the tangent of half the sum of the opposite angles is to
the tangent of half their difference.
For, by the preceding article,
a : b = sin A : sin B
whence, by composition and division,
a -h b : a — b = ain A -{- sin B : ain A — sin B
But from (109) ii x = A, y = B yfe obtain the proportion
sin J. 4- sin 5: sin A — sinB = tan J (J. 4- JB) : tan ^{A — B)
which, compared with the above, gives
a 4- 6 : a - 6 = tan i (J. 4- 5) : tan J (^ - ^) (219)
This may also be written
a 4- S _ tan i{A+B)
b tan ^{A-B)
(220)
and we may infer the same relation between b, c, B, Q and a, c,
A, Q.
119. The square of any side of a triangle is equal to the sum of
the squares of the other two sides diminished by twice the rectangle
of these sides multiplied by the cosine of their included angle.
In the triangle ABQ, Figs. 16 and 17,*
we have either
^ BP = c-AP or BP=AP-o
but in both cases
But the triangle A GP gives by (196)
BP^ = AP^ + c^ — 2cx
AP
Adding QP^ to both members, we find
AP
a2 = 52 + c^-2c
=b cos A
xAP
* The consideration of Fig. 17 was not strictly necessary according to the principle
stated in Art. 49. It may, however, be useful for the student to verify thai
principle when convenient.

FORMULAE FOR OBLIQUE TRIANGLES. 59
which substituted in the preceding equation gives
as was to be proved.
a^ = 53 -f c'- - 2bc cos A (221)
We have in the same way
53 = a2 4- c2 _ 2 -xc cos B (222)
c^ = a" -{•b^-2ab cos (?> (223)
120. The same result is obtained from the following equations (which are evident
from Fig. 16, where c = AP-\- BB}
From the first of these
whence
a =z b cos G -{- c cos B "I
i = e cos yl 4- fflcos C I (224)
c = a cos B -\- b cos A
c cos B ^ a — b cos 0

60 PLANE TRIGONOMETRY.
This may be simplified by representing the half sum of the three
sides by s, or by putting
a-f5-fc = 2s
whence
a-b + c = a-i-b-i-c-2b = 2s-2b = 2{s-b)
a-^b — c=a + b-i-c-2c = 2s-2c = 2{s — c)
which substituted in (226) give
sin-.^^(-^H^--) (227)
In the same manner we should find from (222) and (223)
si,. X 5 = (i^^liHf), 3in^i(7=^^^=4r^- (228)
122. Add both members of (226) to unity; then
But by (138)
, 2bc-hh^ + c^-a^ (b-hcY-a^
l + cosA= 26^ = 2bc
1 4-cos J. = 2cos^i^
Also, (5 4- e)' - a' = (5 4- e 4- a) (5 4- (? — a)
therefore
crf-A
cos 2 ^ -
(5+c4-a)(54-.-a)
4 j^
Substituting g in the numerator as in the preceding article
and therefore also
cos^i^=lii_Z^ (229)
cos^B = t(l^, eosHa=ii^) (230)
123. Dividing (227) by (229), we have, by (14)
and in the same manner
tanHg = (^-;H«-^),
,,,.^a^is-a){s-h)
s{s — b^ ^ 8{s — c)
(232)

FORMULA FOR OBLIQUE TRIANGLES. 61
124. The preceding formulse are sufficient for the resolution of all the usual cases
of plane triangles ; but there are others that are occasionally useful. From (218)
we find, by (105), (106) and (135),
a-Jf- b _ sin 4 4- sin B _ sin j (A -\- B) cos ^{A — B)
c sin C sin J C cos | 0
a — 4 sin .4 —• sin B cos^J (4 4"'.^) sin J (4 — B)
c sin C sin J C cos J Q
But since .4 4- J? 4- C = 180°,
.4 4- i? = 180° — C, J {A -f .B) = 90° — J C
sin J (.4 4- 5) = cos J 0,
by means of which we find
cos J {A -f ^) = sin J C7
a-\-b _ cos i {A — B) _ cos } (A ~ B)
c sin J (7 cos \ [A -\- B)
a~b _ sin J (.4 — ^) _ sin J [A — B)
c cos J C sin J (^ 4- -^)
(233)
(234)
The quotient of (233) divided by (234) gives (220).,
125. Adding unity to both members of (233), or subtracting it, we have, oy
(103) and (104)
a-\- b -{. c _ cos I [A-{- B) -\- cos j- [A — B) _ 2 cos J ^ cos J i?
c cos I (4 4- JB) sin J 0
a -\- b — c cos^ [A — B) — cos ^ [A-\- B) 2 sin |^ ^ sin j- B
c cos J [A -\- B) sin J C
Similarly from (234) we find
c-f a —4 _ sin i (4 4- ^) 4- sin j (A — B) _ 2 sin J ^ cos J 5
c sin J (4. -\- B) cos | 0
_c —^4-_4 _ sin ^ (4 4- -B) — sin | (4 — B) _ 2 cos ^ A sin \ B
c sin J (.4 •}- B) cos J G
Substituting s =: J (a 4- 5 -j- ")> these equations become
'
c
cos J A cos J B
sin J C
(2S5)
a — c sin ^ A sin J i?
c sin J C
(236)
3
t
— 4
C
—
c
a
sin
M cos J
cos iG
B
cos J 4 sin J B
cos ^ G
F
(237)
C2S«\

62 PLANE TEIGONOMETEY.
From these equations we can deduce immediately (227) &o. ; for example exchanging
c for 4 in (236), we have
« — 4 ^_ sin ^ A sin ^ (7
4 sin J 5
which, multipUed by (236) gives (227).
126. Four times the product of (227) and (229) is by (135)
whence
Bin' ^ = ^j^ . » (J — a){s — 4) (» — c)
BinA = -^,/ls{s — a)(s-b){3 — c)] (239)
Exchanging A for B and C successively, this gives also
sin B = —^/la (« — a) {a — b) (« — c)] (240)
In these equations put*
sin

FOEMULiE FOE OBLIQUE TELINGLES. 63
128 The following equations are added as exercises.
tan i A s — 4 ,^ ,„ s — e
?--55 = , tan A A tan i B = •
ta.ni B s — a ^ ^ s
sin [A — B) (g 4- 4) (a — 4)
sin {A-\- B)~ c'
tan J A tan J B cot J C =
cot J ^ 4- cot } i? 4- cot J C = -^
{s-cy
A'
sin i ^ sin i .B sin i C = —;—-
abca
Ks
cos J J1 COS i JS cos i C = —;—
^ ^ ^ abc
tan J .4 tan J i? tan J C = —
£^ z= « (« — a) (i— 4) (s — c)
= i [44'(T- - (a' - 4' -

tt4
PLANE TRIGONOMETRY.
Kg. 18.
CHAPTER VIIL
SOLUTION OF PLANE OBLIQUE TRIANGLES.
129. CASE I. Given two angles and one
side, or A, B and a. Fig. 18.
To find the third angle. We have
C=180°-(J.4-.B)
To find b and c. We apply the theorem of Art. 117, and state
the proportions thus: the sine of the angle opposite the given side ia
to the sine of the angle opposite the required side, as thi' given side
is to the required side. Thus we have
whence
and
whence
ainA:sinB
= a:b
asmB
0 = —-.—T- = aamB cosec :A (244)
smA
sin A: sin Q = a: c
a sin (7 . ~
c = —f—T— = asmG cosec : A (245)
am. A
EXAMPLES.
1. Given A = 50° 38' 62", B = 60° 7' 25"
to find G, b and c.
A = 50° 38' 52"
B = 60° 7' 25"
G = 69° 13' 43"
a = 412-6708
A + B = 110° 46' 17"
G = 69° 13' 43"
By (244).
log cosec 0-1116730
log sin 9-9380702
log 2-6166037
log b 2-6653469
h = 462-7505
and a = 412-6708,
By (245).
log cosec 0-1116730
log sin 9-9708129
log 2-6156037
log c 2-6980896
c = 498-9875

SOLUTION OP OBLIQUE TRIANGLES. 65
2. Given A = 100° 16' 35", B = 25° 16' 13", and b = 29-167
find a and c.
Ans. a = 67-22857
c = 55-59178
130. CASE I. Given A, B and a. Second solution. We find C = 180° — (4 -f B) ;
then, by (233) and (234)
, , coal (B — G) COS cos J i{B~G) (.B —. G) „, ,
4 4- c = a . 1\„. J = a . sin ^-^T—i—'- J J4 (24B
' COsiiB-\-G) sin i/( y '
sin J (B — G)
which give the sum and difi'erence of the required sides ; adding half the difference
to half the sum, we find the greater side, and subtracting half the difference from
half the sum, we find the less side.
131. CASE I. Given A, B and a. Third Solution. When A and B are nearly equal,
and great accuracy is desired, we may compute the difference between a and 4 ; for
we have, from (244),
a — 4 = a
a sin B sin A — sin B
sin A ' ain A
^ _ J ^ 2 g cos i (^ 4- B; sinijA- B)
sin A ^"
EXAMPLE. Given A = 35° 40' 12"-3, B = 35° 37' 48"-6, and a = 26246-948.
A = 35° 40' 12"-3
B = 3-5° 37' 48"-6
i [A -{• B)=^ 35° 39' 0"-5
i{A — B)= 0° 1' ll"-9
ffl = 26246-948
a — b= 25-499
4 = 26221-449
log cosec 0-2342442
log 2 0-3010300
log cos 9-9098720
log sin 6-5423038
log 4-4190788
log 1-4065288
0-23424
0-30103
9-90987
6-54230
4-41908
1-40652
One of the advantages of this process is, that a —-4 may be found with sufficient
accuracy with five-figure tables, as in the second column of logarithms above. If a
had been given to ten figures instead of eight, we should still have been able with
the seven-figure logs, to find a — 4 to seven figures, and therefore 4 to ten figures,
which could not be done by the ordinary methods without ten-figure tables.
132. CASE II. Given two sides and an angle opposite one of
them, or a, b and A.
To find B. To find the angle opposite the other given side, we
apply Art. 117, and state the proportion thus : the side opposite the
given angle is to the side opposite the required angle as the sine of
the given angle is to the sine of the required angle. Thus, with the
present data, we have
, . .1 . T. , . _ 5 sin ^ ^,
a: 0 = sin ^ : sm B whence sm B = (249)
9 F2

66 PLANE TRIGONOMETRY.
To find G. •\5Pe have. C= 180° -{A + B)
To find c,.. Having found Q, we now have the data of Case I,
therefore,, by (245)
c = 0! sin C cosec A (250)
Fig. 19.
133. It is shown in geometry that when two
sides and an angle opposite one of them are
given, there may be constructed two triangles,
as in Fig. 19, whenever the given angle is ^.
acute and the given side opposite to it is less
• than the other given side. In one of them, the required angle B is
acute, and in the other it is obtuse, and the two values are supplements
of each other ; for
B = BB'Q^ 180°-AB' G
These two values of B are given in the trigonometric solution by
the consideration that sin B found by (249) is at once the sine of an
acute angle, and the sine of its supplement. Art. 39.
In general, when an angle is determined
only by its sine,it admits of two values, supplements
of each other, unless the conditions of
the problem' are such as to exclude one of these
values. In the present case, the obtuse value
of B is excluded when a is greater than b, and
there is but one triangle whether A is acute or
obtuse, as in Fig. 20.
134. If the given parts were such that
a = b ain A
a would be equal to the perpendicular from Q upon the side e, and we
should have but one solution, namely, a right triangle, B and its
supplement both being 90°.
135. If the given parts were such that
a 90°.
136. When there, are two solutions, represent the two values of .B
by B' and B", then the two values of G will be
G' ^ 180° -{A + B') = 180° -B' -A = B"-A (251)
G" = 180° - {A 4- B") = 1%0°-B"-A = B' -A (252)

and the two values of e will be
SOLUTION OF OBLIQUE TRIANGLES. 67
c' = a sin Q' cosec A c'' = a sin Q" cosec A (253)
EXAMPLES.
1. Given a = 31-23879, b = 49-00117 and A = 32° 18'; find B, Q
and G.
a = 31-23879
b = 49-00117
A= 32° 18'
B'=
66°66'56".3
5" = 123° 3' 8".7
G'= 90° 46' 3"-7
C"=
24°38'56".3
ar. CO. log 8-5063068
log 1-6902064
log sin 9-7278277
log sin 9-9233399
log sin 9-9999627
log cosec A 0-2721723
log a 1-4946942
log sin 9-6201962
0-2721723
1-4946942
log c' 1-7668292 log c" 1-3870627
c' = 58-46601 c" = 24-38163
Ans. B = 56° 56'56"-3 ^ C-B^ 123° 3' 3"-7
(7= 90° 45' 3"-7 lori C= 24°88'56"-3
c = 58-45601 J t c = 24-38163
2. Given a = -061234, h = -042356, ^ = 55° ; find B, G and c.
Ans. B = 42° 37' 32"-7
Q= 82° 22'27"-3
c= -06199202
3. Given a = -042356, b = -051234, ^ = 55° ; find B, G and c
Ans. B = 82°
WS5"-1
(7 = 42° 46'24"-3 )-or <
c = -03610331
f 5 = 97°45'24"-3
Q = 27° 14' 35"-7
c = -02366993
4. Given a = 40, 5 = 50, A = 63° 7'48"-4; find B.
Ans. B = 90°.
5. Given a = 40, 6 = 60, J. = 60° ; solve the triangle.
Ans. Impossible.
6. Given b = 40, c = 50, i? = 100° ; solve the triangle.
Ans. Impossible,

68 PLANE TRIGONOMETRY-.
fig 21. _ 137. CASE II. Given a, 4 arid ^. Second Solution. W«
may solve, separately, the two right triangles AJPG, BPG,
Fig. 21, which is a convenient method when there are
two solutions. We first find B by (249) ; then we have
^P = 4cos4, .BP= a cos JB
and

SOLUTION OF OBLIQUE TEIANGLES. 69
2. Given a = 31-0005, b = 15-1101, Q = 10° 15'; find A and B.
Ans. A = 160°17'13".7
B = 9° 27'W.3
8. Given a = 2-4*878, b = 9-021876 and Q = 74° 9'4"-2; find
A and B.
Ans. A = 15°50'55"-8
B = 90° 0' 0"
4. Given b = 15-1101, c= 31-0005, A = 10° 15'; find 5 and Q.
Ans. B = 9°27'46"-3
G = 160° 17' 13"-7
139. Having found A and B as above, the most convenient mode of finding c is, bj
(!i33) or (234), which give
.= (g+4)i^!H£±4) (2.56)
' cos J (4. •—• B)
^ '
,, sin l(A-\- B)
c = (a — b) -.~iy.~^~-J (257)
^ ' Bin J (^ — B) ^ ^
for we have, from the process of finding A and B, the log. of g 4" ^> or of a — 4,
and the values of J (^ -f- B) and J (.4 — P), so that we have only two new logs, to
find, which are taken out at the same opening of the tables with the tangents of
i{A-JfB)s.ndi(A—B).
140. CASE III. Given a, b and Q. Second Solution. When a
and b are given by their logarithms, which occurs when they are
deduced by a logarithmic process from other data (as, for example,
in the computation of a series of triangles in a survey), we proceed
as follows. Let x be an auxiliary angle, such that
tan a: = -I" (258)
an assumption always admissible, since a tangent may have any
value from 0 to oo .
We deduce
tan X — 1 a — b
tan a; -f 1 a -f 6
or by (152)
tan (a; - 46°) = ^ ^
ivhich substituted in (254) gives
tan 1 (.1 - .B) == tan (a; - 45°) tan ^{A-^B) (259)

70 PLANE TRIGONOMETRY.
We find X from (258) and employ its value in (259). As this
method does not require the preparation of a -f J and a — 5, it is
quite a? short in practice as (254).
EXAMPLE.
Given log a = 8-7950941, log b = 8-3706056, and (7= 110° 32'.
(Same as Ex. 1. Art. 138.)
log a = 8-7950941
log 5 = 8-3706056
X = 69° 22' 46"-8 log tan 0-4244885
X - 45° = 24° 22' 46"-8 log tan 9-6562825
J (JL 4- B) = 34° 44' log tan 9-8409174
^{A-B)-= 17° 26' 32"-9 log tan 9-4971999
141. CASE III. Given a, 4, and G. Third Solution. To express .4 or .B directly
in terms of the data, we have, from (218) and (224)
the quotient of which is
and in the same manner
c sin A =zz a sin 0
c cos .4=4 — a cos G
. a sin C /„»„,
tan A = J 7, (260)
4 — a cos G ^ ''
tan 5 = - ^ - ^ (261)
a — 4 COB O
'• ^
142. CASE III. Given a, 4 and G. Fourth Solution. To find c directly from the
data, we have, by (223)
o" = a» 4- 4' — 2 a4 cos C
which, however, is not adapted for logarithmic computation.
follows. Substitute by (139)
then
cos (7 = 1 — 2 sin' J G
c* = o' 4- 4' — 2 a4 -f 4 ai sin' J C
= (a — by+4: ab sin' | G
It may be adapted as
, IS / /i I 4 a4 sin' i G\
(a-4)'
(262)
Let X be an auxiliary angle, such that
tan'I :
4 04 sin' } G
(a-4)'
2 sin i G ,—r ,„,.„,
tan X = _J ^ab (263)

SOLUTION OF OBLIQUE TRIANGLES. 7]
then the radical in the above equation becomes ^/(l 4- tan' x) = sec 2;; tierefore,
c = (a — 4) sec x (264)
143. Wo may also adapt (223) for logarithmic computation by means of (138)
which gives
whence
Let
cos C = — 1 4- 2 cos' J O
c* = a« 4- 4' 4- 2 a4 — 4 a4 cos' J G
1
, , , , / / . , 4 a4 cos' i C\
(265)
2 cos i (7 ^—J- _„..,
sin a; ^ -A—. .^ ab (266)
(2 -{- 4
then the radical becomes v'' (1 — sin' z) = cos a:; therefore,
c = (a 4- *) cos X (267)
144. It is to be observed, that the supposition (263) is always possille, since a
tangent may have any value between 0 and co , and therefore an angle z may always
be found having any given number as its tangent. As the greatest value of a
sine is unity, it is not so obvious that the supposition (266) is always possible; but
whatever the values of a and 4
therefore
(a—4)'^0
{a 4- 4)'^ 4 a4
4^

72 PLANE TRIGONOMETRY.
145. CASE IV. Given the three sides, or a, b and c.
To find A, B and G. We have from (227) and (228),
[s—b){s—c)
sin i J. - J (^F^) be
si.ii>-j(fc^4^')
sin
j

SOLUTION OF OBLIQUE TRIANGLES. 73
EXAMPLES.
1. Given a = 10, 5 = 12, c = 14; find the angles.
By (268).
a = 10 arCO 1 9-0000000 ar co 1 9- 0000000
5 = 12 ar col 8-9208188 ar col 8-9208188
c = 14 arCO 1 8-8538720 ar co 1 8-8538720
2s=36
s = 18
8 —a= 8
8-6= 6
s— c= 4
log sines
i J. = 22° 12'27"-6
log 0-7781513
log 0-6020600
log 0-9030900
log 0-6020600
2)9-1549021 2)9-3590220
^A 9-5774510
log 0-9030900
log 0-7781518
2)9-6020601
rB 9-6795110 iG 9-801030G
i.B = 28°33'39"-0 J C=39°13'63"-6
J. = 44°24'55"-2 5 = 57° 7'18"-0 (7= 78° 27'47"-0
Verification. A-\-B-h G = 180°
2. Given a = -8706, b = -0916, c = -7902; find the angles.
Ans. A = 149° 49' 0".4
B= 3° 1'66".2
C= 27° 9' 3".4
3. Given a = .6123864, b = -3538971, c = -3090507 ; find Q.
Ans. C = 36°18'10"-2
146. The computation by (270), when all the angles are required, will be much
facilitated by the introduction of an auxiliary quantity*
(s — a) [a — 4) (« — e)
•=JC
(271)
from which we find by (270)
tan J .4 = ,
-' a — a
tan J B :
T
tan i (7 =
(272ti
•* This quantity T is the radius of the inscribed circle. See (289)
'0 G

74 PLANE TRIGONOMLTRY.
ExAMPLi. Given a = 6053, 4 = 4082, c = 7068. We find
a = 8601-5
ar. CO . log 6-0654258
» — a = 2548-5
log 3-4062846
« — 4 = 4519-5
log 3-6550904
« — c = 1533-5
log 3-1856838
J J = 29° 20'54"-47
^ .B = 17° 35' 31"-70
J C7 = 43° 3' 33"-83
rerifieation. 90° 0' 0"-00
Hg. 22.
the difference of which is
log?-
r
log tan i A •= log —
r
log tan J P — log ^ J
log tan J C = log
T
2)6-3124846
= 3.1562423
= 9-7499577
= 9-5011519
= 9-9705585
147. The case where the three sides are given Is sometimes
solved as follows. From G, Fig. 22, draw C?
perp. to c. Then
_ ^C" = ^P'4- CP', PC" = PP'4-CP'
AG' — BG'^ AP' — BP'
or (.4C4-.BC7) (AG—BG) = {AP-\-BP) (AP—BP)
and if .4P — BP = d, this equation gives
^^(5+a)J4-a)
^2^3^
Then, since ^P-f PP = c, and ^P —PP =

PROBLEMS RELATING TO PLANE TRIANGLES,
CHAPTER IX.
raSCELLANEOUS
PROBLEMS RELATING TO PLANE TEIANGLES.
149. In a given plane triangle, to find the perpendicular from one of the angles upoi.
the opposite side.
\,eip be the perpendicular from G upon c. We have
or by (239) and (278),
^ == 4 sin ^ (279)
P = ^-yi^{^-a){a-b){s-c)-]=^-^ (280)
the expression for^ in terms of the three sides, where a z= }^ {a-\- b -\- e) and A'is
the area of the triangle.
If we substitute in (279) sin J1 = — sin C, it becomes
or, if we substitute the value

76 PLANE TEIGONOMETEY.
161 To find the radius of the circle circumscribed about a plane triangle.
fig.^
The center 0 of the circle. Fig. 23, lies in the perpen
dicular erected from the middle point of one of the sides,
as A B. Let the radius = P. We have, by geometry,
and in the triangle
AOB = iAOB = C
AOB,
AB c
sin ADD =z sin G = AO 2R
whence
R = 2 sin G
(284)
Substituting the value of sin G from (241),
B
a4c
abc
• '4:,y\a{s — a){a — b){s — c)-] ~~ iK
(285)
From (229) and (230), we easily find
Ks
cos i- A cos i B cos i G = -5—
a4c
irhioh combined with (285) gives
^
f
4 cos J A cos J P cos J G
(286)
152. To find the radius of the circle inscribed in a plane triangle.
Kg. 24.
The required center 0, Fig. 24, is in the intersection
of the three lines bisecting the angles,
E/^ \ /V^
and each of the perpendiculars 0 D, 0 E, OF, is
equal to the required radius := r. The value of
0 P in terms of ^ P = c, OAB = \A, - d
^^ OB A = J Bis by (282)
sin J .4 sin J P. sin J A sin \ B
^ — ^- sin \(A + B) ~ " ' cos J C
(287)
This is reduced by means of (236) to
Substituting the value of tan J G,
J- = (a — c) tan J C
r = J/ {s-a)(s-b) (s-c)\ ^ X
(288)
(289)
This is reduced by means of (231) and (232) to
7 = s tan J A tan J P tan J C
(290;

PROBLEMS RELATING TO PLANE TRIANGLES. 77
153. Besides the inscribed circle, strictly so called, there are three other circles
that touch the three sides, (or sides produced), and are exterior to the triangle, iis
in Fig. 25 These have been named escribed circles. Their centers are frund
Tig. 25.
geometrically, by bisecting the exterior angles PCC", GBB', &o. Designate the
centers of the circles lying within the angles A, B, and G respectively, by 0', 0".
and 0'", and their radii by /," r", r'". We find the perpendiculars from 0', &« ,
upon B 0, &c by (282) to be
r"= 4
cos J B cos J G
' sin J (P4- G)
cos J A cos J 0
Bin i{A+0)
= 4.
cos J A cos J B
sinJ(^4-P) -
By means of (235) we reduce those values to
T' =1 a tan J 4, r" = a tan J P,
Substituting the values of tan J ^, &o.
cos i B cos J O
cos J A
cos J .4 cos J (7
cos J P
cos i ^ cos J P
cos J (7
r'"
JT
VV «— a / s — a
][s — a) (s — e) \ __ K
s — / s — 4
-a) (s —4) \ K
/ » — c
a tan J (7
(291)
(292)
(293)

78 PLANE TEIGONOMETEY.
Also, by means of (236) applied successively to a, 4 and c, we may reduce (291)
to the following:
r* z= (« — a) tan J A cot J B cot J G
t-" = (s — 4) cot J ^ tan J P cot J C • (294)
T"' = (s — c) cot J ^ cot J P tan J C
154. Relations between the radii of the circumscribed, inscribed, and three escribed circles
of the preceding article, and the three perpendiculars from the angles upon the opposite
sides.
The four equations of (289) and (293) give
•r' r"
K*
s(s — a) (s — 4) (
Dividing this successively by r', r", &c.
/ /' jj'i r r" r"
- = :^ = ^•
c) A^'
= {s-af
rr'r'"
r" = (»-*)' ——- = {a- ef
Again, we have, (Art. 127),
tan J A tan J P 4- tan J A tan i G + tan J P tan J C = 1
and substituting in this the value of the tangents from (292)
r' r" +r' r"' + r" r"' = s' = '''""''".'
(295)
r296)
From (292) we find
tan IA
tan J P
14-1-1-1 = 1
r' ^ r" ^ r'" r
r" tan J ^ z= r' tan J P
(297)
from which it follows that in Fig. 25, the distances A D and P D', are equal (P, D'
being the points of contact of the circles 0', 0" with .4 P produced), and therefore
BD^^Ajy. Other curious geometrical properties may be traced with the aid of
our equations.
From (284),
a 4 c
R = . 4 sin J ^ cos J -4 4 sin J P cos J P 4 sin J C cos J C
which combined with (287) and (291) give, by Art. 127,
r
-g- = 4 sin J .4 sin J P sin J C7 = cos .4 -j- cos ..> -f cos (7 — 1
^ = 4 sin J .4 cos J P COB J (7 = — cos .4 -f- cos P 4- cos (7 4- 1
r"
^ = 4 cos J ^ sin J P cos J C = cos .4 — cos B+ coa C+l
r'"
^ = 4 cos J ^ cos J B. sin J C = cos ^ -f cos P — cos C -f- 1
(298)

PROBLEMS EELATING TO PLANE TEIANGLES.
Changing the signs of the first of these equations, the sum of the four is
r' + T" + r"'—r
= 4, R = i{r' + r" + r'" — r)
B
79
(299)
Finally, if p', p", p'" denote the three perpendiculars from the angles upon the
Sides a, 4, c respectively we have by (283), (289) and (297) the following relation :
1+L+L=L+L+L
f
r"'
(299*)
155.. To find the distance between the centers of the circumscribed and inscribed circles.*
Let P, Fig. 26, be the center of the circumscribed, Kg. 26.
and 0, that of the inscribed circle. Put P 0 ^ D.
By Arts 151 and 152,
whence
and by (221)
P^P = 90° —C, OAB = iA
FAO = 90° — G—lA = i{B—G)
POf = PA^+ 0A^ — 2PA . OA cos PAO
Z)' = if 4- ^
sin' J A
2g?-cos^(P— G)
sin J A
By (298)
4 Rr sin J P sin J (7
sin' I A sin J 4
therefore
D
; if-
2Prcos^ {B+ G)
sin J A
P'
if —2P?- (300)
156. Let P 0' =^ If, Fig. 26, 0' being the center of the escribed circle lying
within the angle A. If r* = radius of this circle, we have, as in the preceding
wkicle
2 Er' cos 1{B— G)
Zl" = if 4- -.•
sin' I A sin J .4
4 P/ cos J P cos J C
sin' J A "~ sin J ^
7)" = if -f 2 P/
i)'" = if 4- 2 Rr"
(801
D"" = P" 4- 2 Rr"'
Hymers' Trig. Appendix, Art. 58.

80 PLANE TEIGONOMETEY.
the expressions for the distances of the centers of the three escribed circles from
that of the circumscribed circle.
The sum of (300) and (301) gives by (299)
p» + B" + P"' + JD"" = 12 R'^ (302)
157. Given two sides of a plane triangle and the difference of their opposite angles, (or
a, 4, and A — P), to solve the triangle.
We have i (A + B) directly from (220), which also solves the case where two
angles and the sum or difference of two sides are given.
158. Given the angles and the sum of the sides, (or -4, P, G, ar.d a -j- 4 -j- c = 2 s).
By (235)
_ sin i G
' cos J A cos J P
ind a and 4 are found by similar formulae.
159. Given one angle, the opposite side, and th' aum of the squares of the other two
ndes, (or C, e, and a' -|- 4' = e').
In the identical equations
(a 4- 4)' = e' 4- 2 a4, (a — 4)' = e" — 2 a4
substitute the value of 2 a4 given by (223), namely,
.,» ..'
2a4 = cos 0
we find (a4-6)' = e'4 -, (a_4)' = «' —! 1.
' cos C ^ ' cos C
which determine a -j- 4, and a — 4, and therefore a and 4.
To compute these equations by logarithms, let
e' — c' (e 4- c) (e — e) ,„„„,
S = ?r = Vr (303
' cos G cos O ^ '
then (a 4-4)'= e'-f-^', {a — by = e'— g'
that is a -|- 4 is the hypotenuse of a right triangle whose sides are e and g; and
a — 4 is one side of a right triangle whose other side is g, and whose hypotenuse is t.
Let the angle opposite g be denoted by x in the first triangle and by x' in the second,
then by the formulaj of right triangles
tan X = —
a 4" * i
sinx'=^ a —• b = e cos j/
(304J
so that the problem is solved by logarithms by finding log g from (303) and employing
its value in (304).
The above may serve as an example of ° geometrical method of introducing the
auxiliary quantities, which is occasionally useful. The analytical process in the
present instance is similar to that of Art. 143; thus
«'+* = ^JO + f)
a-4 = .J(l-f)

PEOBLEMS EELATING TO PLANE TEIANGLES. 81
therefore if tan a; = — we have /(l-|-^)=secs
and if sin x' = — we have /(l — — j=:Cosx'
whence the same formula as before.
160. Given an angle, its opposite side, and the difference of the squares of the other two
sides, (or G, c, and a' — 4' r^ /').
We have by multiplying (233) by (234)
sin {A — B) _ a'— 4' _/'
Bin G ~ ? ~ 1
sin {A — B)== 1- sin 0
whence A — P, and since A + B = 180° — G, the angles are determined There
will be two solutions given by sin [A — P) except where the obtuse value of ^ — 7i
is greater than A + B.
161. Given the three perpendiculars from the three angles upon the opposite sides.
Denote the perps. upon a, 4 and c respectively by a', b' and c', and let
«" = 4, ^" = ^, - = i
a 0 c
If i = 2 area of the triangle
and therefore
aa' = 44' = cc' T^TC
a ^ a" h, 4 r^ V'lc, c = c"k
Substituting these values of a, 4 and c in (225), (227,) &c.
J"'4.c"'_a"' .,, . (s"_4")(«" —c")
cos A = — \ ,„ „ , sm' J 4 = '- ^, i, &c.
2 4 a 4 c
in which 2 s" = a" 4- 4" 4- c".
162. Given the radii of the circumscribed and inscribed circles, and the perpendicular
from one of the angles upon the opposite side, to solve the triangle.
Let c be the side to which the perpendicidar (p) is drawn. We have found forP, /
and^ the expressions.
P = 2 sin (7 ~ 2 sin (.4 4- P) — 4 sin i (_A + B) cos J (4 -f P)
'•="•
sin I A sin I B
sinj(^4-P)
11
sin A sin P
•P = " • Sn(^4-P)

82
PLANE TIUGONOMF.TET.
Eliminating c, we have
—^ =•= sin A sin B
2R
(m)
r
4 cos J (^ 4- P) sin J ^ sin JP
(n)
from which two equations A and P are to be found.
(n) becomes
Developing cos i {A + B),
— = 4 sin J ^ cos J ^ sin J P cos J P — 4 sin' J ^ sin' J B
P
= sin j4 sin P — 4 sin' J ^ sin' J P
which subtracted from (m) gives
^~f r = 4 sin' J ^ sin' J P
(»)
•Dividing the square of [m) by (o), we find
2P(^ —2r)
= 4 cos' J ul cos' J P
whence
sinMsinJ^ = J f ^ ' ' ) =
JO —2r
2^\2B{p — 2r)1
cos J^ COS ^B = P
2,/\2R{p-2r)-\
Tlie difference and sum of these two equations give
COB \{A + B)
\/[2R(p-^r)}
COS l(A — B) = v/ [2 P (i> - 2 r)-]
(305)
which determine J (-4 -f- P) and i {A — P) and therefore A and P.
are then found by the formula
c = 2 P sin C
The sides
Fig. 27. 163. In a given plane triangle ABO, Fig. 27, to find a
point P ^Hch that the three lines dr

PROBLEMS RELATING TO PLANE TRIANGLES.
In the triangles APG,BPG,y^c
from which
BG:
4 sin z a sin y
sin ^ sin t
have
sm z a sin /S
sin y 4 ' sin a.
sin x+ sin y tan ^(z + y) m+1
sin X — sin y tan J (x — y) m — 1
tan I {z — y)
m —-1
m4- 1
To compute this equation by logarithms, let
tan •
a sin ,8
4 sin a
tan i {x + y)
then by (152), tan ^{x — y) = tan {y — 45°) tan i{x + y)
eo that the angles x and y are found by (306) and (307).
164. The following problems are proposed as exercises.
In a plane triangle AB G —
1. Given c, the perp. upon c = p and a -f- 4 = m.
Bin' X :
4^'
[m + c) [m — c)
tan J C =
2pc
{m + c) (in — c)
2. Given c, the perp. upon c ^ p, and a — 4 = m.
tan' X
4j>'
(c -j- «) (c — n)
tan J C =
• 4 = c cos X
a 4" 6 = " sec s
(c 4" ") (" — '*)
• 2^0
83
(307)
8. Given C, e, and a4 = g'.
tan z •= — cos J C
a 4" ^ = " sec 3
2 (7
sin z' = — sin J C
a — 4 = c cos a;'
4. Given C, the perp. from G =: p, and a -f 6 = m.
tan 2 == — tan 1 C c = m tan i a;
5. Given G, the perp. from C = p, and a — b = n.
tan X = — cot J C c = ncotlx
P
6. Given c, C, and a 4- * = '"-
cos J (,4 — P) = — sm J C a — 4 = c
sm h (4 — P)
cos } C

84 PLANE TRIGONOMETRY.
7. Given c. A, and a+ b = m.
tan i B ^ —,— cot i A
^ m+ c ^
8. Given a 4- ^ = '"> the perp. upon c = p, and the difterenoe of the segments
of c = (f.
tan H^ 4- ^) = J [_-.,,._ ^,) .„,, _ ^, _ ^^. J
2 »(?
tanH^-^)=^;;^
or with an auxiliary angle
4i,'
(m 4"

SOLUTION OF TRIGONOMETRIC EQUATIONS. 85
CHAPTER X.
SOLUTION OF CERTAIN TRIGONOJIETEIC EQUATIONS AND OF NU­
MERICAL EQUATIONS OP THE SECOND AND THIRD DEGREES.
166. THE solution of a prohlem in -which the unknown quantity
IS an angle, often depends upon that of one or more equations, involving
different functions of the angle, which cannot he reduced by
merely algebraic transformations. We shall select a few simple examples
of such equations from among those that most frequently
occur in astronomy.
167. To find zfrom the equation
sin (ct -f 2) — m sin z (309")
in which a, and m are given. "We have, by (119),
sin ( ci -t- 2) = sin a sin z (cot z 4- cot oi)
which becomes identical with (309) by taking
whence the required solution
If the proposed equation were
we should find
sin «, (cot z 4- cot oi) =m
cotz = -. cot 01, (310)
sm «.
sin [ot, — z) = m sin z (311)
cot 2 = —; 4-cot« (312)
sin«
Unless z is limited by the nature of the problem in which these
equations are employed, there -will be an indefinite number of solutions
; for all the angles a, 2 4- 180°, z 4- 360°, z + 540°, &c., in
general all the angles s 4- w TT have the same cotangent. [See
(68), (79).J In most cases, however, we consider only the first two
of these solutions, taking the values of z always less than 360°
H

g6
PLANE TRIGONOMETRY.
Similar remarks apply in all cases where an angle is determined
by a single trigonometric function; but if the prohlem is such as to
give the values of two functions of the required angle, as the sine and
cosine, the solution is entirely determinate under 360°, since there
cannot be two difi'erent angles less than 360° that have the same
sine and cosine.
168. The solution of the preceding article requires the use of a
table of natural cotangents ; to obtain a formula adapted for logarithmic
computation entirely,^ we deduce from (309) the following
sin (a 4- 2) 4- sin 2 _ m 4- 1
sin (* 4- 2) — sin 2 m — 1
But by (109), if a; =» 4- 2, 2/ = 2, we have
which substituted above, gives
sin {ot, -\- 2) -f sin 2 _ tan (24-2"*)
sin (a 4- 2) — sin 2 tan J a,
, / , 1 X m-\- 1
tan iz + *a,) = =- tan i «
^ ^ ' TO — 1 ^
which determines 2 -f J a, whence 2 is found by deducting f a,.
The computation of this equation is facilitated in most cases by
introducing an auxiliary angle, such that
tan (p = m
an assumption always admissible, since while the angle varies from
0 to 90° the tangent varies from 0 to oo, so that an angle cp may
always be found having any given number as its tangent.
We have then by (152),
TO -f 1 tan (p -f 1
T = . ! 1 = cot (

SOLUTION OP TRIGONOMETRIC EQUATIONS.
hj
169. To find z from the equation
We deduce
tan (o4 -f 2) = TO tan 2 (316)
tan («. -f 2) -f tan 2 TO 4- 1
tan (a 4- 2) — tan 2 TO — 1
80 that by (126) and (152) the solution is
tan (p = TO, sin (» 4- 2 2) = cot (cp — 46°) sin

88 PLANE TRIGONOMETRY.
172. The preceding examples wiU sufiice to indicate the method to be followed
with all the equations of the following table. The solutions of the equations involving
cosines may be obtained from those involving sines, by exchanging z for 90° ± z,
or a for 90° rt a..
Logarithmic solutions of the first four will be obtained by imitating the process of
Art. 171.
EQUATIONS.
SOLUTIONS.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
sin {a. ± z) sin z = nt
cos (a. rb z) cos z = m
sin (a. ± z) cos z ==. m
cos (a ± z) sin z = m
sin (i ±z) = m sin z
cos {dL rbz) = m cos z
sin {a.:±iz) •= m cos z
cos (a ± z) :^ OT sin z
tan (« ± z) tanz = m
tan (rt rtz) = m tanz
cos (a ± 2 z) = cos :t =p 2 m
cos (a ± 2 z) = 2 m — cos a.
sin (a ± 2 z) = 2 m •— sin a
sin (a ±2z) = sina±2 7?s
tan tp =m, tan (z ± J a) = cot ((6 q= 45°) tan J a
tan0 = m, tan (^^a±z) = tan(45° —

SOLUTION OF TRIGONOMETRIC EQUATIONS. 89
l78. To find z from the equation
sin (a -|- z) = m sin (/S + z).
Put z' = ,S -j- z, a' = a — jj, then this equation becomes
sin (a' + z') = m sin z'
which is of the form (309) and may be solved by (309*) or (311) ; then z = z' — /3.
In the same manner equations of this form, involving cosines or tangents, may be
reduced to those of the preceding table.
174. To find k and z from the equations
We have, by division.
& sin 2 = TO
h cos z = n
m
tan 2 = —
n
y (320,
I
which gives two values of z, one less, the other greater than 180°;
whence, also, two values of h from either of the equations
TO
sm 2
The solution becomes entirely determinate (2 not exceeding 360°)
as follows :
1st. When the sign of h is given. Eor if k is positive, sin 2 has
the sign of TO, and cos 2 the sign of n, and 2 must be taken in the
quadrant denoted by these signs. If k is negative, the signs of sin 2
and cos 2 are the opposite to those of TO and n, and 2 must be taken
accordingly.
2d. • When z is restricted by either the condition 2 < 180°, or
2 > 180°. For under either of these conditions the tangent gives
but one solution. If 2 < 180°, k has the sign of TO ; and if 2 > 180°
k has the opposite sign to that of TO.
3d. When 2 is restricted to acute values, positive or negative. For
under this condition a positive tangent will give 2 between 0° and
-r 90° ; and a negative tangent, between 0° and — 90° ; and k will
always have the sign of n.
It follows that TO and n being any given numbers whatever, we
may always satisfy the conditions expressed by (320), 1st, by a positive
number k and an angle 2 between 0° and 360° ;' 2d, by a number
Ic (unrestricted as to sign) and an angle 2 < 180° ; 3d, by a
number k (unrestricted as to sign) and an angle 2 > 180°; and 4th,
by a number k, and an.angle 2 in the 1st or 4th quadrant.
12 H2

90 PLANE TRIGONOMETRY.
EXAMPLE.
To find k and 2 from (320), {k being a positive number), whn
,„ = _ 0-3076268, n =-{• 0-4278735.
A; sin 2 - 0-3076258
A; cos 2 4- 0-4278736
{a) log & sin 2 - 9-4880228
{b) log A cos 2 4- 9-6313147
(a) - (6) 'log tan 2 - 9-8567081
z 324° 17' 6"-6
(c) log sin 2 - 9-7662280
(a) - (e) log k 4- 9-7217948
k 4- 0-5269808
Upon this problem and the deductions we have made from it, rests
the method of introducing the auxiliary angles required in solving
many of the formulse of spherical trigonometry. It is applicable
to any equation that can be reduced to the form of that solved in
the following article.
175. To solve the equation
m cos z + n ainz = q
{2>2V
TO, n and q being given.
The fij-st member will be reduced to the form h sin (cp 4- 2") by as
suming k and cp such that
whence
k sin cp = TO, /c cos (p = w (322)
k sin

SOLUTION OF TRIGONOMETRIC EQUATIONS.
If we restrict cp to values less than 180°, (as we may do according
to the last article), we may find it by the equation
yi
and then
TO
tan (J) = —
n
sin (cp 4- 2) = — sin $ = — cos

92 PLANE TRIGONOMETRY.
176. To sohe the equation
a sin (a 4- z) 4- 4 sin (/S -j- z) 4- c sin (>. 4- z) 4- &c. = q (825)
Developing by (36) and putting
this becomes
a sin a 4" * ^i'^ Z' 4" '^ ^'"^ > "t" ^''- ^^ "'
a cos a 4" * "OS /? 4- " ''°^ y "^ ^"^ = "
m cos z 4- »» sin z = J
which is solved in the preceding article. The same process applies if any or all of
the terms contain cosines.
177. To find k and zfrom the equations
k sin (a 4- z) = m
4 sin [0+ z) :^ n
The sum and difference of these equations are, by (105) and (106),
whence
2 k sin [J (a 4- ^) 4- ^] cos J (a — /?) = m + n
2 k cos [J (a 4- ;8) 4- z] sin f (a — 13) = m — n
m -j- n
2 k sin [J (o 4- /?) 4- z] =
cos J (o — ^)
2 k cos [J (a 4- /J) 4- z] =
m — n
Bin J (a — ;S)
(326)
(327)
from which 2 i and J (a -f- /3) -j- « are determined by Art. 174. The logs, of the
second members of these equations should be computed separately, for the purpose
of readily discovering the signs of the sine and cosine in the first members. The
Bolution is determinate (according to Art. 174) when the sign of k is given.
From (327) we find, by division.
tan [J (. 4- /?) 4- z] = ^-±^ tan i{a-fs) (328)
which requires a less number of logs, than the separate computation of (327), but
we are obliged to refer to (327) to determine (by an inspection of the second members)
the signs of the sine and cosine.
If we assume
tan * =
we may compute (328) by tho formula
tan [J (a 4- ;3) 4- z] = tan (45° 4- 0) tan J (a — /?)
- (329)

SOLUTION OF TRIGONOMETRIC EQUATIONS.
EXAMPLE.
In (326) given a = 200°, ^ = 140°, m = — 0-42345 and « = — 0-20123, to
find z and k, k being positive.
(")
(a) - (4)
(a) -
(c)
By (35!7).
m + n
m — »
i (" 4-,?)
J (" - ^)
log (m 4- ")
log cos J (ii — /3)
log 2 A sin [J (a 4- /3) 4- z]
log (m — n)
log sin J (a — ^)
log 2 i cos [J (a + 0) + z]
log tan [J (a + /l) + z]
H-4-/2 4-2)
^
log sin [J (a -f /3) 4- z]
log 2 k
2 k
k
— 0-62468
— 0-22222
170°
30°
— 9-7956576
4- 9-9375306
— 9-8581270
— 9-3467831
4- 9-6989700
— 9-6478131
4- 0-2103140
238°21'38"-6
68°21'38"-6
— 9-9301171
4- 9-9280099
0-8472467
0-4236234
178. A more general solution of (326) is the following.* Let j. be any angle as
sumed at pleasure, and in (171) let
a; = o4-z, y = /3-}-z, z' = 3-4-z
(distinguishing the z of (171) by an accent); then we shall find
sin (a —/S) sin (j- 4- z) = sin (a — y) sin (^ 4- z) — sin (^ — y) sin (a -|- z)
In this let y (whose Talue is arbitrary) be exchanged for j. 4- 90° ; then
sin (a— 0) cos (y + z) = — cos (a — y) sin (/S + z) + cos {0 — y) sin (a 4- z)
Multiplying these equations by k and substituting m and n from (326)
k sin (a — 0) sin (j. -f z) = m sin [y — 0) — n sin {y — a)
k sin [a — 0) cos (y + z) =i m cos {y — B) — n cos {y — o)
which (y being assumed at pleasure), determine k .ind y -\- z.
If we take j. = 0, we find
•— OT sin 5 4- ?! sin u
tan z = -—-J
m cos 0 — n cos "
If } = a,
k sin (a -f- z) ^ TO 1
, , , > mco3(a—/?)-B I (381)
h cos (a -f z) = r^-^ '-^—. j ^ -'
^ ' ' sin (a — B)
li y = 0, we have a similar result.
If J- ^ J (a 4- (3) we obtain the solution of the preceding article.
If A is required, without first finding z, we have, by adding the squares of (330)
k sin (o — /?) = .^ [m' 4- n' — 2 m m cos (? — 0)] (332),
-* GAUSS. Theoria Motus Corporum Ccclestium, Art. To.
(330)

94 PLANE TRIGONOMETRY.
J 79. To find k and zfrom the equations
k coa [a+ z) = m )
kcoa{0+z) = n J ^ '
These are reduced to the form (320) by substituting 90° + % and 90° -f /? for a and /?.
We find, however, by a process similar to that of Art. 177,
2kBi.l_U^ + 0) +
'-\=^f^'0)
2.cos[J(a4-/J)4-^] = - ^ P y )
tan A =
cot [J {a + 0) + z] = tan (45° 4- 0) tan ^ (

EQUATIONS OF THE SECOND AND THIRD DEGREES. 95
182 In like manner, if the proposed equation is
ive assume
whence
n cos (d -j- z) = m cos (/? + z)
k cos (a 4- z) = TO
k cos [0 + z) = n
and k and z are found by Arts. 179, 180. As the sign of k (in this and the preceding
article) may be arbitrarily assumed, there will be two solutions.
NuMEEicAL EQUATIONS OF THE SECOND AND THIRD
DEGREES.
183. To solve the equation
x'' + .px+ q = 0 (338)
when q ia easenfially positive, and p either positive or negative.
We have from (144), exchanging x for ip,
tan' } * — 2 cosec * tan J * -f 1 = 0 (339)
and (338) may be reduced to this form by substituting
X = z .yq
In which we may take the radical only with the positive sign, since we may assume
X and z to have the same sign. We thus reduce (338) to
which compared with (339) gives
P
— 2 cosec 0 = -^—, z = tan J ^
V?
or sin 0 = ^^, x = y/qt&n J ^ (340)
p
which gives two values of (p less than 360° and consequently two values of x. If
e be the least of these two values of

96 PLANE TRIGONOMETRY.
184. To solve the equation
x' + px — q^Q (842)
when — q is essentially negative, p being either positive or negative.
We have, by (143)
tan' \'p + 2 cot

EQUATIONS OF THE SECOND AND THIRD DEGREES. 97
BO tliat, by the theory of equations, «, and t^ are the two roots of an equation of the
a^
second degree in which the absolute term is — -H= and the coefiioient of the second
tern is 4; that is, they are the roots of the equation
e+bt — "^ = 0 (m\
If then we find the two roots «, and 4 of (m) by the preceding methods we shau
have
x = y + z = ^i, + ^t^
(n)
It will be necessary to consider the sign of a in the equation (OT).
1st. When a is positive, (m) comes under the form (342) and the solution bv (344)
gives
and by (n)
x = J~[^ta.ni 9 — ^001^9^
and if we assume
this becomes, by (142)
tan ^ Ip = ^ tan J 9
a; = - 2 j - | cot^i
Collecting these results we have, for the solution of (345), ivhen a is positive,
2 /"a^
tan 9 = -T- v/ 97' ^^^ J 'P = -^ tan J 9
2 / -^ cot (
• (346)
in which the radicals
/ —- and / — are to be considered positive, and 9 is to be
taken < 90° with the sign of the tangent. But two of the three values of ^ tan J 9
being imaginary, the given equation has but one real root.*-
•^ If »• represent the real value of ^ tan J 9, and a„ a^ the two imaginary root* of
unity, the real value of x ia
and the imaginary values are
'' =J Y (.'"^ -^) "' = JT (- - k)
or since a, a, = 1
13 I

98 PLANE TRIGONOMETRY.
2d. When a is negative and — 4 a' < 27 4'. Equation (m) becomes
and is of the form (338); its roots are therefore found by (341) which gives
2 I a» / 4a»
4'
8ine=-^J-2= = -J-274
/ «"" / ^
cot J 9
ana by (M) a: = / ^ f .^ tan J 9-}-.^ cot J 9 j
or if we put, as before, tan J 0 = -^ tan J 9, the solution of (345), when a is negative, ia
2 a"
sin 9 = —-- / —27 tan J0 = i^tanj I
(347)
cosec 5*
which gives one real root, (the other two being imaginary, as above), when sin 9 ia
possible, i. e. when — 4 a" < 27 4'.-*
Substituting the values of a, and a,
-14-V/-3 - 1 - ^ - 3
a. = ft, _ 2
and also r = tan J ^ — = cot J $
J
r
a
-^ (cot 0 4- cosec ^ y/ — 3)
;~a~
a:,. = I -^ (cot 0 — cosec ip .^ — 3)
flr finally, x, being the real root, the imaginary roots are
^, = — -2 Y~ sec ^ v' — 1
a:. = — — 4
Y~ ^^"'P'/ — ^
• * The two imaginary roots will be found, by a process similar to that employed
in the preceding note, to be
X^ = 2 ^ COB^ v/ —1
a:, a:, ^ 3
a;, = — -g- 4- —|— cos 0 v' — 1
in whioh a;, is the ro^l root found ty (347).

EQUATIONS OF THE SECOND AND THIRD DEGREES. 99
3d. When a is negative and — 4 a' > 27 4'. In this case sin 9, in (347), is impossible
and the preceding solution fails. This is the irreducible case of Cardan's rule, the
roots appearing under imaginary forms, although it is known that they are all three
real. It is, however, readily solved trigonometrically.
In Art. 77, putting ^ for x, we have
sin'

100 PLANE TRIGONOMETRY.
so that tnere are but three different values of sin

DIFFERENCES AND DIFFERENTIALS. 101
CHAPTER XI.
DIFFERENCES AND DIFFERENTIALS OF THE TRIGONOMETRIC
FUNCTIONS.
186. IN the applications of trigonometry, it is often required to
compute a function of one angle from that of an angle which differs
from the first by a small quantity. In such cases it is generally
most convenient to compute the difference of the two functions,
which may be applied to either to obtain the other.
187. To find the increment of the sine or cosine of an angle, corresponding
to a given increment of the angle.
Let the angle x be increased by A x, (this notation signifying difference,
or increment of as), and let the corresponding difference or
increment of the sine be expressed by A sin x and of the cosine by
A cos a;; we have, by this notation,
and by (106) and (108)
A sin a; = sin (a; -f A a;) — sin x
A cos X = cos (a; 4- A a;) — cos x
A sin a; = 2 cos (a; 4- J A x) sin J A a; (350)
A cos a; = — 2 sin (a; -1- J A a;) sin |- A a; (351)
which are the required formulse.
We here consider the difference always as an increment, i. e. an
increase, and give it the positive (algebraic) sign; its essential sign
may, however, be negative, and it will then be in fact a decrement.
Thus, in (351) the second member will be negative so long as
a; 90°, and < 270°
188. To find the increment of the tangent and cotangent. We have
and by (116) and (119)
A tan X = tan (a; 4- A a;) — tan x
A cot a; = cot (a; -f A a;) — cot x
sin A a; / , . \ • ^ /orroN
A tan a; = T—r^—\ = sec (a; 4-A a;) sec a; sm A a; (651)
cos (a; 4- A a;) cos a; ^ '
A cot a; = s—^— = — cosec ix 4- A a;) cosec a; sin A a; (363)
sin (a; 4-A a;) sma;
'
i2

102 PLANE TRIGONOMETRY.
189, Tb find the increment of the secant and cosecant. We have
or by (130) and (132)
h sec X =
sec [x + ixx) —• sec x
A cosec z = cosec (a: 4- ^ ^) — cosec x
2 sin (a: 4- j- A x) sin } A a;
A sec a; = ^—,—;——4 — (354)
cos (x + t^x) cos z ^ '
— 2 COS (a: 4- J A a:) sin J A. a;
A cosec z = ;—^7—-7-^—r-A = (855^
sm (a; -f. A a:) sm x
190. To find the increments of the squares of the trigonometric functions corresponding
to a given increment of the angles
We have
whence by (133)
A sin' X = sin' (a: 4- A a:) — sin' x
= cos' X — cos' (x + AX)
A sin' X = — A cos' a; = sin (2 a; 4- A a;) sin A a! (356)
From (115), (116), and (119) we deduce
, . , . sin (x +y) sin (z — y)
tan' X — tan' y = ^ ^^' \ ^^
cos' X COS y
whence
cot' :. - cot' y =
^^^(?^+l)^^±=l)
sin' X sin' y
, „ sin (2 a; -4- A a:) sin A a; ,„,_,
A tan' z = \ , I ( , (357)
cos' (a; 4- A a;) cos' z ^ '
— sin (2 a; 4-A a;) sin A a;
A cot' X = —. , ) / / . , (358)
sm' (a; 4- A a;) sin' x ^ '
From (16) we have
the difference of which gives
and in the same manner, from (17),
sec' (a; 4- A a;) = tan' (a; 4- A a;) 4- 1
sec' z = tan' x+\
A sec' a: = A tan* x (359)
A cosec' a; = A cot' x (360)
and the values of A tan' x, A cot' x, may be substituted in (359) and (360).
191. When the increment of an angle, or arc, is infinitely small,
it is called the differential of the angle, or arc; and the corresponding
increments of the trigonometric functions are the differentials of
these functions.
The differential of x is denoted by dx; of sin a; by d! sin x, kc.

DIFFERENCES AND DIFFERENTIALS. 103
192. To find the differentials of the trigonometric functions from
the differential of the angle.
Let the angle x and its increment Aa; be expressed in the unit of
Art. 11; or, which is equivalent, let x and Aa; be the arcs which
measure the angle and its increment in the circle whose radius = 1.
It is evident that the less the arc, the more nearly does it coincide
with its sine or tangent; therefore, when Aa; is infinitely small, or
becomes dx,
sin dx = dx sin ^dx = ^ dx
This may be demonstrated more rigorously thus.
infinitely small, we have cos dx = 1, whence
sindx , - ,
5— = cos dx= 1
tan dx
sin dx= tan dx
When dx is
but the arc cannot be less than the sine, nor greater than the tangent,
and therefore
dx = sin dx = tan dx
Again, when Aa; is infinitely small, or becomes dx, we must, according
to the principles of the differential calculus, reject it when
connected with finite quantities by the signs -f or —; thus we must
substitute a; for a; 4- dx, or for a; 4- J dx.
Upon these principles we find the differentials directly from the
finite differences (350), (351), (352), (353), (354) and (355) as follows:
d sinx = cos x dx (361)
cZ cos a; = — sin x dx (362)
di&n X = sec^ a; c?a; = (1 -f tan^ x) dx (363)
dcotx= — cosec^ xdx = — {!-{• cot^ a;) dx (364)
d sec X = tan x sec x dx (365)
d cosec X = — cot x cosec x dx (366)
193. In the same manner the equations (356), (357), (358), (359) and (360) give
d sin' X = — d cos' x = sinix dx (367)
d tan' a; = i sec' a: = ?^^4^ dx (368)
cos z
=
2 sin a; 2 tan a; .
=— dx = —-^— aa; (369)
COB x cos'X
— sin 2 a; , .....
dcot' x= d cosec' x = —-, dx (61O)
sin z
=.-=l^dx=^^l^dx (371)

104 PLANE TRIGONOMETRY.
194. Although the equations (361), (362), (363), (364), (365) and
(366), are rigorously true only when dx is infinitely small, they may
be used when dx is a finite difference, instead of the equations, (350),
(351), (352), (353), (364) and (365), provided dx is sufficiently small
to be considered equal to its sine without sensible error, and is also
very small in comparison with x. This is very frequently the case in
practice, and the differential equations are then preferred on account
of their simplicity. It is only necessary to observe that dx must
be expressed in arc, i. e. in terms of the unit radius ; if it is given
in seconds, it may be reduced to arc by Art. 9.
195. To find the differential of an angle from the differentials of
its famotions.
Fiom (361) we have
dx = ^^^ (372)
cos a; ^ '
but it is convenient in this case to employ the notation of inverse
functions. Art. 87. Thus, iS y = sin x, x = sin ~^y, and the preceding
equation becomes
dain-^y^^^f^_^^^ (378)
In the same manner from (362), &c., we find
~'y
_
—^y
^/(l-
f)
(374)
dtan' -ly:
dy
1+y^
(375)
dcot~ 'y-
— dy
1 + y'
(376)
'y =
dy
y'^ {y^-
•1)
(377)
d cosec"" 'y =
— dy
y^{y^-
1)
(378)

DIFFERENCES AND DIFFERENTIALS OF PLANE TRIANGLES. 105
CHAPTER XIL
DIFFERENCES AND DIFFERENTIALS OF PLANE TRIANGLES.
196. IN trigonometrical investigations it is Mg.29.
often necessary to determine the effect of a
small change in one of the data, upon the computed
parts. Thus, Fig. 29, ii A, AB and
A G, of the plane triangle ABQ, are the data,
and AQ is subject to an error of QQ', the required parts will be
subject to errors which are respectively, the differences between
A QB and AG'B, AB Gand AB Q', B Qand B G' In the same
figure, the data may be supposed to be A, A B and AB G, and the
angle ABQ may be regarded as subject to the error QB G' which
produces the corresponding errors in the remaining parts. In the same
manner, the data may be A, A B, and A QB, A GB being variable,
or. A, A B, and B G,B G being variable. In all these instances, A
and A B are constant, while the remaining four parts are variable, and
may be considered as receiving, simultaneously, certain increments
which are related to each other. We propose, then, to solve the
general problem :
Bfi a plane triangle, any two parts being constant, and the rest
variable, to determine the relations between the increments of the
variable parts.
It is evident that the solution of this problem resolves itself into
an investigation of the differences of two triangles which have two
parts in common. We shall consider the several cases successively ;
distinguishing the triangle formed from the given one by the application
of the increments, as the derived triangle.
197. CASE I. A andc constant. The six parts of the given triangle,
ABQ, Fig. 29, being A, B, Q, a, b, e, those of the derived triangle
formed by varying all but A and c, are A, B -h AB, C 4- A (7,
a -f Aa, b -f Ab, and c. In these two triangles we have
A-i-B-i-Q= 180°
A + B + AB+Q-^AG= 180°
whence AB+AQ==0, AB = - AO (379)
14

106 PLANE TRIGONOMETRY.
Also in the two triangles we have
a = c sin A cosec G
a -{• Aa = c sin A cosec {G 4- AC)
the half difference of which by (355) is
[m)
(w)
- _ e sin A cos {G-h i AQ) ain ^ AG , .
i^''-- sinCsin(C+AC) ^^'
jAa _ j-Aa _ acos((7-f|AC7)
sin IAB sin I AG sin(C4- AG)
(380)
The half sum of (m) and («) by (131) is
'^+ ^^"'
_ csin J.sin((74-i-AC)cos JAC
sinCsin((74-A(7)
which combined with {p) gives
jAa _ ^ Aa _ a-f jAa
tan J A5 "" tan i AC "~ tan ((7 4- |A6')
(381)
From (260) we have
c sin ^
tan C = b— ccosJ.
whence
J — ccos J. = e sin J. cot G
b -^ Ab — c coa A = c ain J cot (C 4- AQ)
the difference of which by (353) is
, _ csin.4 sin A(7
sinCsin(C4-A0)
therefore
Ab Ab a
sin AB sin AC sin (C 4- AC)
(382)

DIFFERENCES OF PLANE TRUNQLES. 107
This equation gives by (135)
_ J A5 _ a
sin 1 AC cos 1 AC " siir(CTXC)
and dividing (380) by this
Aa _ cos(C4-iAC)
A6 cos i AC
(383)
It is to be observed that the increments (or half increments) of
the angles must be deduced from their sines or tangents, since it is
only by these functions that a small angle can be accurately determined.
Moreover, a small arc being nearly equal to its sine or tangent,
the equations (380), (381) and (382) express very nearly the
ratios of the increments of the sides to the increments of the angles,
or rather to those increments reduced to arc by Art. 9, or Art. 54.
198. CASE II. A and a constant. We have as in the preceding
case AB = — A C; and in the two triangles •
b ain A = a sin B
(5 4- *A b) sin A = a sin {B 4- AB)
the difference and sum of which give
I Ab ain A = a cos (-S 4- J A.B) sin J AB
(5 4- i Ab) sin A = a sin {B -\- I AB) cos J AB
whence by di-vision
lAb ^ _ i-A5 ^ b-^^Ab
tan 1 AB tan J AC ~ tan (-B 4- i AB)
{p)
(384)
In the same way
hAe I'-C _ e4-iAc
taniAC tanjA^ tan (C 4-i AC)
(385)
From the equations
c sin A = a sin C
{c 4- Ac) sinA =• a sin (C 4- AC)
we find I Ac sin A = a cos (C 4- J AC) sin | AC {q)

log
PLANE TRIGONOMETRY.
which combined with the equation [p) gives, since sin JAC=— sinjA5,
From [p) we also have
A6 cos {B-{-l AB) ,Qgg.
Ac ~ cos (C 4- i AC) ^ >
jAb ^ _^A5 5 cos (.g 4-i A.g)
sin J A.B ~ sin J AC sin 5
(387)
which, when A 5 is to be found from AB, is more convenient than
(384). In the same way from [q)
I Ac _ _ I Ac ^ ccoa{Q+ I AG) .^
sin i AC ~" sin A C^B sin C ^
2
199. CASE III. 5 and c constant. We have
c sin B — b sin C
c sin (^ 4- A£) = & sin (C 4- AC)
the sum and difference of which give
c sin (^ 4- J AB) COS J A5 = J sin (C 4-. J AC) COS J AC {p)
ccos {B + l A-B) sin J A-S = 6 cos (C 4- J AC) sin i AC (?)
the quotient of these gives
By (224) we have
tan I AJ5 ^ tan {B + ^ nB) . „.
tan J AC tan (C 4-J AC) ^ '
a = b cos C 4- c cos 5
a 4- Aa = 6 cos (C4- AC) 4- c cos [B +
the sum and difference of which give
AB)
a4-iAa = Jcos(C4-JAC)COS JAC4-CCOS(5 -\-^AB) cos IAB
— jAa= Jsin(C4-i-AC)sinjAC4- csin{B -\- ^AB) sin^AS
These expressions are reduced by [p) and {q) to
a-f jAa=ccos(.B4-jA.B)cosi-A.ScotiAC(tan|A.B 4-tan|AC) (r)
— J Aa=c sin (.B 4- J A.B) cos J A.B (tan J A.B 4- tan J AC) (»)
and by division
I Aa _ a -f J A«
cot (-B f J A5)
(390)

DIFFERENCES OP PLANE TRIANGLES 109
In the same way we have
iAa
tan JA5
a-hiAa
cot (C 4- i AC)
(391)
Since the sum of the three angles is constant,
AJ. 4- A-B 4- AC = 0
i(A5 4- AC) = -iAA
therefore by (116)
tan
1 * r. ! 4. T ^rt sinl(A.B4-AC)
J A5 4- tan J AC = -——^ -—-^
cos i AB cos i AC
sin J
AA
cos J A5COS J AC
(9
which substituted in (s) gives
lAa _ esin{B-hi AB)
sin J AA cos J AC
(892)
and in the same manner
_i_Aa
sin J AA
_ 5sin(C4-jAC)
COS J A.B
(393)
Substituting [t) in (r) we find
sinjAC_
sin J AA
ecos(.B4-JA.B)
a -f J A a
(394)
siniA.B 5cos(C-f-i AC)
whence also -.—z -r = , , ,
sin J AA
a-\- i Aa
By differencing the equation
(395)
we find instead of (392) and (393)
a^ = J2 4- e^ — 2 5c cos J.
I Afl _bc sin (J. 4- J AA)
sin J A J ~ a-h^Aa
K
(396)

no
PLANE TRIGONOMETRY.
200. CASE IV. A and B constant. We have
smB
b = -;—T a
sin .4
5 4- A6 = -.—T (a + Aa)
sin.A
sin 5
whence Ab = ^^^ Aa
In this case the third angle is also constant and there are but
three variables related by the equation
_.^« = ^ = _ ^ (397)
sin A sm.B sm U
This case is not strictly included in the general problem as stated
in Art. 196, since the two triangles have not two parts in common.
201. The second members of the equations (380), (381), (382),
(383), (384), (385), (386), (387), (388), (389), (390), (391), (392),
(393), (394), (395), (396), involve the increments themselves, -which
are the quantities sought. It is therefore necessary, in many cases,
to solve these equations by successive approximations.
For a first approximation we consider the increments in the second
member to be = 0, employing ^ for ^ 4- ^AB, kc, and taking
cos ^ AB = 1, &c. This will evidently produce but a slight error
so long as the increments are small as compared with the entire
parts of the triangle. We then obtain a. second approximation, by
recomputing the equation in its complete form, employing in the
second members the approximate values of the incremeijts. With
these second values we may, in the same way, obtain a third approximation,
&c. Theoretically, it requires an infinite number of sucli
approximations to arrive at a perfect result; but in practice, the
tenths or hundredths of seconds being the limits of accuracy, it is
rare that more than a second approximation is necessary.
It is also to be observed that in computing the values of small
quantities such as the increments in question, we may employ logarithms
of only four or five decimal places and take the angles to the
nearest minute. This is in fact one of the chief advantages of computing
by differential formulse, rather than by the direct formulae
applied to each of the two triangles successively.

DIFFERENTIAL VARIATIONS OF PLANE TRIANGLES.
Ill
EXAMPLE.
In a plane triangle whose parts are
A = 68° 41' 48"-9 B = 35° 11' 3".4 C = 86° 7' 7 '-7
a = 6053 b = 4082 e = 7068
let A and a be constant while b is diminished by 50-5; to find the
change in the angle B.
We have in this case Ab = — 50-5; and by (387)
i Ab sin B
sin i AB = , / T, , , „,-
^ 5 cos (.B 4- i A.B)
1ST APPROX.
2D APPROX.
iA5
5
B
lAB
B+IAB
log 1 AS
ar. CO. log. b
log sin B
ar. CO. 1. COS {B -\- ^ AB)
log. sin J A^
lAB
- 26-25
4082
36° 11'
0
36° 11'
- 1-4028
6-3891
9-7606
0-0876
- 7-6396
-15'0"
35° 11'
-15'
34° 56'
1
\ - 7-6520
i
0-0863
- 7-6383
-14' 56"-8
It is evident that changing the angle -B 4- i- AB by only three seconds
would not affect the fourth place of its cosine; a third approximation
is therefore unnecessary, and we have finally A.B = — 29'53"-6.
As the log. sines of small angles do not vary proportionally with the
angles, it will conduce to accuracy to employ the methods explained
in Art. 115.
DIFFERENTIAL VARIATIONS OF PLANE TRIANGLES.
202. The equations (380), (381), (382), (383), (384), (385), (386),
(387), (388), (389), (390), (391), (392), (393), (394), (39.5), (396) and
(397) become differential by making the increments infinitely small,
that is, by omitting the increments -svhen connected with finite quanti-

112 PLANE TRIGONOMETRY.
ties by the signs 4- or —, and substituting the increment itself for its
sine or tangent, and unity for its cosine, (Art. 192.) The character
d must also be substituted for A. These changes being made, we
easily deduce the following differential relations.
CASE I. A and c constant.
dB = - dQ
da da
a cot C
JB^ ~ 'dQ ^
db
dB~
db
dQ
da
db^
a
sin C
cos C
> (398)
CASE II.
A and a constant.
db
dB
dB =
db
' dQ
- dO
bcoiB
dc
iQ
dc
'dB~
e cot C
(399)
db
dc ~
COS.B
cos C
CASE
III.
b and c constant.
dA-irdB-\-dQ = = 0
dB tan 5
tanC
da
—T7I ^^ — '^ tan B.
da
dA'
cainB = bsin C
da
= — a tan C
dB
> (400)
dC c .^ dB
-5-r = -cos B, -Tj—r. =
dA a ' dA
cos C

DIFFERENTIAL VARIATIONS OF PLANE TRIANGLES. 113
CASE IV. The angles. A, B, G, constant,
da db dc
sin A sin B sin C
(401)
203. These differential relations are often employed when the increments
are very small, instead of the equations of finite differences.
We have already seen that the equation of differences often requires
to be solved by successive approximations, the first approximation
being in fact obtained by employing the corresponding differential
equation. In all cases therefore where a second approximation in the
use of finite differences could not alter the result of the first, it is
plain that the differential equation is sufficiently accurate.
The increments of the angles must generally be expressed in arc.
Thus if dB is given in seconds we must divide it by B" = 206264".8,
or substitute dB sin 1" for dB.
dA
But in such fractions as —j^, this substitution is evidently unnodB
cessary provided the two increments are always expressed in the
same unit, as minutes, seconds, &c.
EXAMPLE.
In a plane triangle whose parts are
J. = 58°41'48"-9 5 = 36°ir3"-4 C = 86° 7' 7"-7
a = 6053 b = 4082 c = 7068
suppose b and c to be constant and the angle A to receive the increment
dA = 20"-6 ; find da and dQ.
From (400) we have
da — dA sin 1" c sin B
^ — dA ccosB
aU =
a
log dA 1-3139 log (- dA) - 1-3139
log sin 1" 4-6856 log c 3-8493
lege 3-8493 log cos.B 9-9124
log sin B 9-7606 ar. co. log a 6-2180
log da 9-6094 log cZC - 1-2936
da 0.407 dQ - 19".7
16 B:2

114 PLANE TRIGONOMETRY.
204. The error of employing the differentials in any case may be determined approximately
by developing the equation of finite differences and comparing it with
the corresponding differential equation. We shall select a simple example.
We have from (387) and its corresponding differential equation in (399)
, , b cos (B + i^B) . . _
i A6 = ^.^—W^ sm J AB
•^
sin B
Ab = b cot BAB
the first of -which when developed gives
sinl"
. , „ 2bsin (B+iAB) . , _ . , „
i Ab — b cot Bain iAB \ „ '- sin J AB sm J AB
2 -^ sm .o
or substituting sin J AB = J A5 sin 1", sin J AB = J AJS sin 1", and also B for
B + i AB inthe second term, which will affect so small a term but slightly,
Ab = bcotBAB
sin 1" — — (A5 sin 1")'
Comparing this -with the differential equation above, the error of employing the
latter is approximately
-|-(ABBinl")'
which for AB = 1° is — -000015 b.
It appears from this example that the error is expressed by a term involving the
square of the increment; and if we develop aU the equations of finite differences we
shall find that they differ from the corresponding differential equations by terms involving
the squares and higher powers of the increment. Hence, employing the differentials
instead of the finite differences amounts to neglecting the terms involving the squares
and higher powers of the increments.
205. The differential relations above obtained could have been deduced more directly
from the formulse of plane triangles by differentiation, employing the values
of the differentials given in Art. 192. Thus in CASE I, A and « being constant, if
we differentiate the equation
a = c sin ^ cosec O
we have da ^ c sin A rf cosec (7
= — c sin A cot G cosec G dG
= — a cot GdG
%a in (398)
The studsnt may exercise himself by deducing the other relations of (398), (399)
and (400) in. a similar manner.

TRIGONOMETRIC SERIES.
^15
CHAPTER XIII.
TRIGONOMETRIC SERIES. DEVELOPMENTS OF THE FUNCTIONS OF iiN
ARC IN TERMS OF THE ARC, AND RECIPROCALLY.*
206. THE investigation of trigonometric series is most readily
carried on with the aid of a few elementary principles of the Differential
Calculus. All that will be required here will be no more than
is generally given in the first chapter of a treatise on that subject,
namely, the differentiation of simple algebraic functions, and Taylor's
Theorem. We shall employ the following expression of this theorem :
fiy+h)=fy^^.]L+'^Jl.tt.+^lJy..Jt_+^^, (402)
i\!}-r ) ''"^ dy 1 dy^ 1-2 ^ dy^ 1-2-3 ^ ^ '
in which fy denotes what f {y + K) becomes when A = 0 and
d "ft/ cZ^ "Fa
—^—, ' „ , &c., are the successive differential coefficients, or dedy
dy^ '
rivatives of fy.
207. To develop sin x and cos x in terms of x.
We shall first develop sin [y -\- x) and cos [y 4- x) by (402). By
(361j and (362), if
fy = siny
, d.fy d sin y
we have -5-^ = —^—^ = cos y
dy dy
d^.fy _ d cos y
dy^ dy sin y
d\fy dsvny ^
dy^ dy "
d*.fy
dy*
d cos y
dy = am.y
* The leading results of this Chapter being of very general utility and const
application are printed in the larger type, but as they are not referred to in the ?
sequent large print of this work, and moreover require a limited acquaintance w
the Differential Calculus, the student can omit them at the first perusal, .and p
directly to Part II

119 PLANE TRIGONOMETRY.
SO that the values of the coefficients of the series (402) recur in the
order -J- siny, -f cosy, — siny, — cosy, and therefore/(?/ -f- a;) =
sin(«/4-a;) = sin«/4- cosy-j sin?/—-^ cosi/j^^Tg 4-&c. (40;?)
If we commence with
fy = cosy
the coefficients will recur in the order -{- cos y, — sin y, — cos y,
-f siny, and (402) will give
3? OJ 3"
cos (?/ 4- a;) = coay — siny -^ cosy-y^ 4- sinyj-^ +kc. (404)
If now we put y = 0 in (403) and (404), siny = 0, coay = 1, thtalternate
terms of the series vanish, and we have
or Q' 2/ Cu
sin a; = y - ^ ^ + 1.2.3.4.5 " 1.2-3-4-6-6-7 + ^°-
(^*^^)
cos a; = 1 - -j:2~ + rMl " 1-2-3-4-5-6 + ^"^
(4°^)
It may be observed that (406) can be deduced from (405) by differentiation.
208. The series (405) and (406) are directly available for the construction
of the trigonometric table. For this purpose x in the series
must be expressed in arc, since (361) and (362), upon which the pro
ceding demonstration rests, require x to be in arc. Art. 9.
Find COS. 10°.
EXAMPLE.
Reducing 10° to arc, by Art. 9, we have
a; = 10 X -01746329 = -1746329
and computing separately the positive and negative terms of (406),
1 = 1- - - ^ = - -01523086
=
1-2-3-4
-00003866
""""""""
- T ^ =- - -00000004
l__g
1-00003866 - -01623090
- -01523090
cos 10°= .98480776
agreeing with the tables, which give -9848078. The student may,
for practice, verify any other sine or cosine of his table.

TRIGONOMETRIC SERIES. 117
209. To develop tan x in terms of x.
Representing the coefficients in the series (405) and (406) by letters, we have
X — a,z' + a, z' — a, a;' -^- &c.
tana; =
1 •— a^z' + a,x* — a, x' + &c.
(407>
1 1
In which
1-2 a. = &o.
1-2-3
If we perform the division of the numerator by the denominator, we perceive that
the result will be a series containing only odd powers of a;, and commencing with the
term x. But as the law for the successive formation of the coefficients is not easily
shown in this way, we shall resort to the following process. Assume the series to be
tan z = c,x + c,x' + a, x' + &c.
and differentiate it; we find, by (363), after dividing by dx,
1 4- tan' a: = c, -f 3 c, a;' 4- 5 o, a:* 4- &o.
. or, since from the division of (407) we know that c, = 1,
tan' X z= Sc,z''+ 5c,z' +7 c,z' + 9c,2^ + &c.
The square of (m) is
tan' X = Ci Ct z^ + c, c,
which compared with (n) gives
+ c, c.
C, -^ c, c.
z' + c, c,
+ c, C-,
(c, c, + c, c.)
A' 4- c, f,
+ c, c,
+ c, e,
+ c,c^
x' + &c.
4-&C.
4- &c.
4- &c.
(m)
(n)
c, = Y (c, c, + c,c,+ c. e,)
c, = -g- (c, c, + c,c^+ c, c, + c, e,)
&c. &c.
where the law of derivation is obvious. We have preserved the factor c„ althongh
it is equal to unity, in order to render this law more apparent.
Since the first and last terms of these expressions are equal, as also the terras
equally distant from them, we may write them as follows :
c. = 1
Ci = Y (c. "•)
«. = -r (2 «a ".)
c, = y (2 c, e, 4- c, c.)
Vs = -- (2 c, c, -f 2 c, c,)
c„ = jy (2 c, c. 4- 2 c, c, 4- c, c.)
&c.
&c.

118
PLANE TRIGONOMETRY.
in which form any coefficient c,„ + , when n is even, is expressed by y terms all ol
ffl 4- 1
whose coefficients are = 2 ; and when n is odd, by —^—terms all of whose coefficients
are 2 except the last, which is 1.-*
If we now substitute the value of c, = 1, and deduce the numerical values of the
coefficients successively, we shall find
, x- , 2 a;' 17 a;' 62 x' 1382 a:"
,tan x-^z + -^ + - ^ + -gSTgy 4" -3i75:7;:9- + W^^T^ll + *"*• ^^^^>
210. To develop cot z in terms of z.
If we invert (407) we have
1 — a,a;' 4- a,a;' — &o. .....
cot X = ^ .T . J— (409)
X — a, x' 4- "•i^ — ""'-
and the first term of the actual division is —, the second term — (a, — a,) z, and
the succeeding terms evidently involve only the odd powers of x. Therefore let
cot X = d.x — d,x' — d.x' — &o. (o)
X
The coefficients cannot be determined by the method of the preceding article in
consequence of the negative exponent in the first term; but they are directly deducible
from those of the series for tan x. We have by (142)
tan a; := cot X — 2 cot 2 x
(p)
Now the series (o) being true for any value of x wUl give cot 2 x by substituting 2 z
for X, whence
2 cot 2 X = — — 2' rf, X — 2' rf,x» — 2' d,x' —- &o.
X
Subtracting this from (o) we have by (p)
tan X = (2' — 1) (7, X 4- (2* — 1) d,x' + {2' — 1) d, z' + &c.
Designating the coefficients of (408) by c„ c„ c„ &c. we have also
tan X = c^j} + c^x' + c,-, x' + &o

Substituting the values from (408)
TRIGONOMETRIC SEE,tES. 119
.-. = 1 c. = - c--. = -3-&o.
and reducing the coefficients to their simplest forms, we find the series (o) to be
1 a; x= 2z' x' 2x"
cot X = 1- . A,c C410)
X 3 3'-5 3'-5-7 3»-5'-7 3'-5-7-9-ll ^
211. By a process similar to that of Art. 209, but which we leave to the student,
we find
x' 5 X* 61 x^ 277 x"
secx = i4--4-—4-2j:g^4-2^T35:f 4-&0. (4ii)
And from (408) and (410) by means of the formula
cosec X = J (cot J a; 4- t^^i J ^)
•we find
1 z 7 x' 31 x' 127 x'
cosecx = -4- —4-25:3^4-2^^31:5:^-^ 2^:51:5^:7+ ^'=-
212. yo develop sin~^ ^ m ierms c/«/. (See Art. 87).
Let x = sin~^ y (or sin x = y); then by (373)
(^^^^
d^-
V{l-y')~^^~y'
Developing the second member by the Binomial Theorem,
| = l + i^^ + 2-3^^ + g|/-^&c. {m)
As this contains only even powers of y, the series from which it
would be obtained by differentiation must contain only odd powers
of y; therefore, let
X = a^ 4- a^^ -f a^^ + a^y'' 4- kc.
{n)
There will be no term independent of y if we limit x to values between
0 and db 90°, for then when «/ = 0 we must also have x — 0.*
Differentiating, we have
J
= «! 4- 3 a3?/^ 4- 5 a,y' + 7 ay 4- kc.
which compared with (m) gives
1 1*3 1"3'6
ai=l 3 33 = 2 5fl!j=2q '''ay = 2:^&c.
•*• The series (413) obtained under this limitation expresses but one of the values
of sin-^y, but if we denote the series by s, we shall have by (95) the following expression,
including all the values,
sin~' y = n ?r.+ (—1)"3
n being an integer, positive or negative, or zero.

120 PLANE TRIGONOMETRY,
therefore (n) becomes
a; = sin-2, = y 4 - | . 4 4 - g |-4-1|| • f + &c. (413)
It is unnecessary to develop cos~'^ since we have
cos '«/ = -o
Sin '«/
213. To develop ta.n~^y. Let x = tan^^y, then by (376)
'^ = {l + y^)-^ = l-y^+y*-tf+kc.
ay
from which we infer, as in the preceding problem, that the required
series contains only odd powers of y; therefore let
then
x = a^y-h a^y^ 4- a^ + ay 4- &c.
-T- = «! 4- 3 ay 4- 5 ay 4- 7 ay 4- &c.
ay
which, compared with (m), gives
ai=l 3a3 = — 1 5a5 = l 7a, = — l&c.
so that the series is
x = ta,n-^y = y — iy^-{-iy^—^y''-\-kc. (414)
214. To compute the ratio (= TT) of the circumference of a circle
to its diameter.
We have heretofore assumed this ratio to be known from geometry,
where it is found by means of circumscribed and inscribed polygons
which are made to differ from the circle by as small a quantity as we
please; but (414) enables us to express its value in a series. We
have tan -2" = 1) therefore if we make ^ = 1 in (414) we have
{m)
(n)
But this series converges too slowly to be of any use. To obtain a
rapidly converging series y must be a small fraction. We might em-
TT 1 . . .
ploy tan •— = —^-Q- (Art. 29), but in consequence of the radical, it i?

TRIGONOMETRIC SERIES.
12I
simpler to resolve -^ into two or more arcs whose tangents are known,
and to compute the value of each of these arcs by the series.
effect this let
then by (123)
To
-^ = tan ^ ^ 4- tan~'i{' (416 >
, ^ ., t-i-t'
tan -p = 1 =
4 1-tt'
whence *'=T+1 (417)
from which, assuming any value of t at pleasure, the corresponding
yalue of t' is found.
If we take i = 1, -we find if' = J; therefore by (416) and (414)
-— = tan ^ 1 -f tan"" 'J
A few terms of these series give
-J = -4636476 4- -3217506 = -7853982
TT = 3-14169
more accurately vr = 3-14159 26536 89793
1 3
If we take t =-^, we find *' = "r. but the above supposition is
evidently the best adapted for rendering both series sufficiently convergsnt.*
215. To resolve sin z and cos x into factors.
The series (405) shows that x is a factor of sin x, and gives
sinx = x(l-.^4-j-^_&c.)
(p)
16
* See NOTE at the end of th.s chapter, p. 124.

122 PLANE TRIGONOMETRY.
and the factors of the series within the parenthesis must evidently be of the form
x^
1-J (?)
A being a constant, but having a different value in each factor. The required factors
must be such as to reduce the second member of [p) to zero whenever the first
member is zero. Now sin x is zero for the value x = 0, whence x is a factor as already
seen, and also for x = rtnjr, « being any integer; therefore the general
valuj of (j) is
whence
A
.4 = n'sj-'
which, substituted in [q), gives as the general factor
l--i^
n'jr'
Making n successively = 1, 2, 3, &c., the equation {p) becomes therefore
^^^=^(^-l4)(^-2-^'-)0-3S^)---
^''')
The factors of cos x in (406) must also be of the form (j); but cos x is zero for
x=':ii^2n+\)
-^, n being any integer or zero, and the general value of (q) is
(27»4-l)V_Q
.4.2' ""
whence
A = -—"X^
which, substituted in (j), gives the general factor
2' x'
•'~(2«4-l)'5r'
Making n successively = 0, 1, 2, 3, &c., we have
216. Logarithmic ainea and cosines. By means of (419) and (420) the logarithmic
sines and cosines of the tables are readily computed.
^ 7r ,
Put X = m —, then
and taking the logarithms
TT mv 1. m'\ /. m'\ /, m'\
«nm-= — (1-2^) (^1-J,) (^1-gj) .:.
COS.-^=(l--)(l-g,)(l--)...
log sin . ^ = log |1 4- log m + log ^1 _ ^^ 4- log ^1 — ^^ 4. ..
logCOS -^ = log (1 - y,) 4- log (1 - ^,) 4- log (1 - ^,) 4- -

Developing these logs, by the known formula
TRIGONOMETRIC SERIES. 123
log(l — n)= — M(n+in''+in'+
&o.)
(inwliioh M = modulus of common logs.) and arranging according to the po-nersol
m, we have
log sin
= 108-0-4- log '»
•'"•T(2-'+T'+T' + ^''-)
.vf(i, 4-14-14-.0.)
•&o.
.EfL + ^+^ 6,4-^0.)
3 V2'
log cos - ^ = - m'. -(^.j, + - 4- _ + &c. )
—•I(|.-^|.+i+-)
-"^ •¥(-?+ 3-'+5-.+ *

124 PLANE TRIGONOMETRY,
Compute log sin 9°. We have
and therefore by (421)
EXAMPIE.
m X 90° = 9° m = — logm = — I
log sin 9° = 10-19611 98770 — 1-
— 0-00178 5%45
— 0-00000 14689
— 0-00000 00023
= 10-19611 98770 — 1-00178 74357
log sin 9°= 9-19433 24413
217. If in (419) we put x = -^, we have
«^'^f = ^=YO-i)0-i)0-i)-
_ fl- /2' — 1\ /4' — 1 \ /6' —1\
"~ 2 \ 2' / \ 4' / \ 6' / •"
_ ^ (2-l)(2 4-l)(4-l)(4-hl)(6-l)(64-l)...
" ~ 2 2 2 4 4 C 6 . . .
i j - 2 2 4 4 6 6
whence .^ = - . - . - . - . - . - . . . (423
which is Wallia'a expression of jr.
NOTE to page 121. Computation of ir. Many other series besides those of Art.
214, may be given for computing rr. One method of obtaining them is to resolve
tin"' t and tan~' t into two others, and thus make J A- to depend upon three or more
arcs. From (194) we easily deduce
1 1 n
tan-' — = tan-' —; 1- tan-' -5-j p- (a)
m m+ n rrir+ mn + \
1 1 • n
tan—' — = tan-' tan—' —^ J-T- • (V\
m m — n m —mn + \ ^
in which m being given, « may be assumed at pleasure. The numerators of the
ft-actions in the last terms will reduce to unity when m' -|- 1 is divisible by ra; ii
therefore we assume n and p so as to satisfy the condition
we shall have
np =. rt^ +\
tan—' — = tan—' ;— -f- tan—' —; 1 ol
m m + n m+ p
tan—' — =: tan—' tan—' • lel
m m — n p — m '
(c)

TRIGONOlVfETRIC SERIES. 125
For example, let m = B; then m' 4- 1 = 10 = 1 x 10 = 2 x 5, so that we may
take n = 1, J) = 10; or » = 2, ^ = 5, whence by (d) and (e)
tan—' — = tan-' — + tan"' —
O 4 Id
= tan—' -^ — tan-' —
Substituting in (418)
= tan~' — 4- tan—' -5-
~ = tan-' -^ + tan-' j + tan"' y^
= 2 tan"' — tan—' —
= 2 tan- i 4- tan-' i (/)
= tan-' y 4- tan-' — 4- tan— -^ (g)
The equation (/) was employed by CLAUSEN of Germany, in computing it- to 200
decimal places, and (g) was employed by DASB, also of Germany, in computing TT to
the same number of figures. These computations were carried on independently of
each other, and the results when communicated to SCHUMACHER, (who gives them in
the Astronomische Nachrichten, No. 589), were fotfnd to agree to the last figure
They prove the value previously found by Mr. Rutherford to be erroneous beyond
the 150th figure.
By means of the formulse (a), (6), (c), (d) and (c) we may again subdivide the
arcs as often as we please. Thus, it is easy to deduce
|- = 2 tan— -- 4- tan— -=-4-2 tan"' —
= 2 tan~' —- 4- tan—' —- -I- tan"' -— + tan-' -j7--r-
5 ' 4 ' 7 268
= 4 tan- i - tan- 1 4- tan"' i 4- tan- ^
= ^ *''""' T - *^'^~" W + '"""' 2l8
= 4 t a n - i - t a n - ^
which last is known as Machin'a formula. In deducing it we have reduced the difference
of two arcs to a single aro by means of formula (a).
Another method is, to find by trial, or otherwise, an arc a multiple of which is
nearly equal to -7-, and whose cotangent is a whole number; and then deduce the

i26
PLANE TRIGONOMETRY.
difference between this multiple and -j-.
Thus it is known (from the trigonometho
tables) that cot 11° 15' = 5 nearly; therefore by the last formula of Art. 79, putting
1
tan X = -—,
and by (194)
therefore
as was foimd above.
1 120
4 tan— -=- = tan— -—^
o 119
, 120 a- ,120 , , 1
*""" W - T = *"" IT^ - *''•'"' 1 = "^'^ 239
!!- 1
-r- = 4 tan—'
4 5
tan~
If'we resolve tan-' -—^ by means of (c), (d) and (e), we have m = 239,
m* -|- 1 = 57122 = 2-13' = np, which offers several suppositions for n and p; if
we take n = 13' = 169 and^ = 2-13" = 338, we find by (e)
^ =
4tan-l-tan-^4-tan-l
whicii was employed by Rutherford.
If we take n = 1, p = 57122, we find by {d)
^ =
4tau-i_tan-^-tan--^

EXPONENTIAL FORMUL.^. 127
CHAPTER XIV.
EXPONENTIAL FORMULiE.
TRINOMIAL OR QUADRATIC FACTORS.
218. To demonstrate Buler's formulae
cos a; = i(e='v'-i4-e-'^i^-') (424)
sina; = ^-^-—J-(e^>'—— e-«v'-i) (425)
in which e is the Naperian base of logarithms, or,
It is shown in the theory of logarithms that
wtcve for brevity we write
^^ = ^ + (!) + |)+($+($+^-
(1) = 1 (2) = 1-2 (3) = 1-2-3, kc.
We have by (406) and (406), employing the above notation,
^ a;^ ^ a;^ a;« .
cos a;=l-p-^ 4-^-^4-&c.
m
^^'^^=(l)-(3) + (5)-(7) + ^°the
terms of which are the same as those of (426), but with alternate
signs. If the signs in these two series were all positive, the sum of
the two would be equal to (426); and it is evident that we shall make
them positive by substituting
x^= —z^ or a; = a v' — 1
which gives
cosa; = l4-p^4-p^4-^4-&c,
. z^ s*" z'
Bin a; = s %/ — 1 (1 4- TgT -F Tgs -F 7=v 4- &c. ^

128 PLANE TRIGONOMETRY.
whence
But
z , ^ , z" z'
sm a; = 7^ 4- TQS 4- 7F\ 4- 7=v 4- &c.
^_1 — (1) ' (3)-(5)-(7)
,osx-^-^-^ainx^l-\-^^ + y^+^^ + kc.=^
therefore
cosa; —i/—1 sina; =e-'^''-i (427)
If in this equation we substitute — x for x, we have, by (56),
cosa; 4-^/ —1 sina; = e='''-i (428;
The sum and difference of these equations are
2cosa; = e'"»'-i4-e^"'»'~' (429)
2-^/-lsina; = e'"t'-i —e-*/-' (430)
whence (424) and (425).
219. The quotient of (430) divided by (429) is
220. If we put
y = e''^-'' = cos a; 4- %/ — 1 sin a; (432)
we have y~^ — e~=° i'"' = cos a; — i/ — 1 sin a; (433)
and (429) and (430) become
2 cos a; = y -f «/-' (434)
2V — Isinx =-y —y-^ (435)
If mx be substituted for x in these formulse, we have
ym = ^r^xv-•L = COS mx-\- >/ — 1 siu mx (436)
y-^ = e-m=^ V-i= COS ma; — v^ — 1 sin mx (437)
2 cos Twa; = t/ 4- ^z- (438)
2 s/ — i sm ???a: = J/ — y- (439)

EXPONENTIAL FORMULA. 129
221. Moivre's Formula. The value of y from (432), compared
with (436) gives
(cos X -\- \/ — 1 sin a;) = cos mx 4- >/ — 1 sin mx (440)
which is Moivre's Formula. It shows that the involution of the expression
cos a: -f N/ — 1 sin a; is effected by the multiplication of
the angle.
Again, if we multiply (432) by
we have
cos a:' 4- -v/ — lsina;'= e^'»'-^i
(cos a; -f v' — 1 sin a;) (cos a;' 4- -v/ — 1 sinx') = ei'^'^'^V- i
= coa ix-{• x')-\- \/ — 1 sin [x -f x')
which shows that factors of this form are multiplied by the addition
of the angles.
We have also
[cosx-'t\/—lsina;)(eosa;—\/—lsina;)=cos^a;4-sin^a;=e°= 1 (441)
222. General form of Moivre's Formula. As long as m is an integer, both members
p
of (440) can have but one value ; but if ??» = —• the first member becomes
v_
(cos X -f- y/ — 1 sin x)« = -v^ (cos z + ,y — 1 sin z)'
which has q different values* in consequence of the radical of the degree q, while
the second member
P
P
cos •=— x+ .y — Isin — X ( i)
S 3
has but one value.
In order that both members may have the same generality, as should be the case
with every analytical expression, it is necessary to suppose that we take for the arc
X not merely the arc less than the circumference which has the given sine and cosine,
but also all the arcs which have the same sine and cosine; that is, c denoting the
circumference, all the arcs
X, z + e, z + 2c, X + Sc, &c.
Now there is an infinite number of these arcs, but only q of them can give differentvalues
to (a); for all the values of the arc in (a) will be
Zx, ^z+^A' ^ x + ^ , .P.z+^J^=^,
q ' g ^ q' g 5 ? g
^x + l^, ^x+ ^1+^)P\ &c. &0.
g g g 2-
-* That is g values real and imaginary; thus it is shown in algebra that ^^u' = + a
and —a; ;ya'= + a, a( ^ ) and a( ^ )>
ya'=
4- a, — ff, 4- «^/ — 1, — a v^ — 1; &c.
1-

130 PLANE TRIGONOMETRY.
But —x+ .Ss— z= — X + pe has the same sign and cosine as — x;
g g g i
£- x+ '•" "*"—i-£-_ = (—x4-—)4-P

TRINOMIAL OR QUADRATIC FACTORS. 131
TRINOMIAL OR QuAnitATio FACTORS.
225. To find the quadratic {trinomial) factors of the expression z' — 2 z "• cos if -j- 1 ;
m being integral.
By (438) and (484) we have
y"» — 2 y"* cos mx -f 1 =0
y^ — 2y cos x 4-1 = 0
Therefore if we ^nt y = z, mx = 2n v + •p, or x = ^^'^ + 'i> ^^
m
2' —2z'»cos^-|-1 = 0 (448)
3' — 2z co8""'^+^-f-l = 0 (449)
As these two equations exist at the same time, they have common roots, and the
second is therefore a divisor or factor of the first; but this factor has m values in
consequence of the m values of cos "'^ "^ ^ (Art. 222), found by making
B = 0, 1, 2, 3 . . . OT — 1. Therefore the m quadratic factors of (448) are all ex
pressed by (449), and we have
sara_2z'" cos * -f 1 = (^ z' — 2z cos — -I- l)
\ m l
x(^'-2zcos ^^+^ -t 1)
V m l
x{z^-2zcos±^t
+i)
X
/ . n 2 (m — 1) v+ ip , , \
X (^z'-2zcos -^^ ^—H^ + lj (450)
226. To obtain the simple factors of (448), we have only to find the two simple
factors of each of the quadratic factors in (450), or to find the two factors of the
general quadratic (449). Now, by the theory of equations, if z, and z^ are the two
roots of (449), the first member is equal to
but we have by (432)
(Z — 0,) (Z — Zj)
1 , 1 . 2n7r+

132 PLANE TRIGONOMETRY
EXAMPLES.
1. Find the quadratic and simple factors of
z* —2z'4- 1
Eere m = 2, 2 cos * = 2, cos ^ = 1, t = 0; and by (450),
by (451)
s« _ 2 z' -f 1 = (z' — 2 z cos 0 4- 1) (z' — 2 2 cos fl- 4- 1)
= (z._2z4-l)(z'4-2z4-l)
= [z — (cos 0 4- v' — 1 sin 0)]
X [z — (cos 0 — v^ — 1 sin 0)]
X [z — (cos fl- 4- N/ — Is '-)]
X [z — (cos fl- —.,/ — 1 sin fl-)]
= (z_l)(z_l)(z4-l)(z4-l)
2. Find the factors of z* -f- 2 z' 4- 1- Here m = 2, 2 cos # = — 2, 0 = A-, and
z'4-2z'4-l = (z'4-l)(z'4-l)
3. Find the factors of z* — z' 4- 1.
= (z-v/-l)(z4-^/-l)(^-v'-l)(^4-^/-l)
z* — z' -t- 1 = (z' — 2 z cos 30° 4- 1) (z' 4- 2z cos 30° 4- 1)
= (z' — z v^ 3 4- 1) (z' 4- z V 3 4- 1)
= (z-i^S-i^-l)(z-.i^3 + i^-^
X (z4-iV34-Jv/-1)
4. Find the factors of z" — 2z° -f- 1.
(^4-Jv'S-Jv'-i)
2«_2z'4- 1 = (z'—2z4- 1) (z'-f z4- 1) (z'4--z4- 1)
= (z-l)'(z4-J4-Jv'-3)'(z4-J-J^-3)'
227. To find the quadratic factors of ^ — 1 when m is odd.
In (450) let ^ = 0, it becomes
(z" — 1)' = (z — 1)' X ( z' — 2 z cos ~ + l)
X ( z' — 2 z cos -i^ 4- l)
\ m l
X
x(z'-2zcos^^"'-^)"4-l) (452)
Now m being odd, m — 1 is even, and the number of trinomial factors in (452),
exclusive of (z — 1)', is even ; but
2(m —l)fl- / 2n-\
i —^ '-— = cos 12 B- ) = cos
m \ m J
2;
so that the first and last of these factors are equal. In the same manner it is shown
that any two of these factors equally distant from the first and last are equal,

TRINOMIAL OR QUADRATIC FACTORS.
13ii
Bierefore, imiting these equal factors and extracting the square root of both members,
we have, ichen m is odd,
z" — 1 = (z — 1) X ( z' — 2 z cos -i^ 4- 1)
X (z^ — 2zaoa — + l)
\ m /
X
X (z'-2zcosi!^^1^4-l) (453)
228. To find the quadratic factors of i^ — 1, when m is even.
When m is even, m — 1 is odd, the number of factors in (452), exclusive of (z — 1)',
is odd, and the middle factor will not combine with any other. This factor is the
( __ I and contains
and is therefore equal to
2(_
cos = cos fl- = 1
m
z" 4- 2 z 4- 1 = (z 4- 1)»
BO that uniting the remaining factors, and extracting the square root, we have,
when m is even.
. 1 = (z — 1) (z 4- 1) X ( z" — 2 z cos ^4-1)
x(z'-2zcos^4-l)
(^,._2zcos(^^4-l) (451)
229. To find the factors of z^ + 1, when m is odd.
In (450) let j) = A-, it gives
(z 4- 1)' = ( z' — 2 z COB -^ 4- 1)
X
(^z'-2zcos-^4-l)
/ „ . (2 m —l)fl- , A
X ( z^ — 2 z cos i '- \- 1 )
\ m /
and it is easily shown, as in the preceding articles, that the factors equally distant
from the first and last are equal, and that tlie middle term is z' -f- 2 z -j- 1 = (z -}- 1 )*'
M

134 PLANE TRIGONOMETRY.
Hence we find, when m is odd,
z" -}- 1 = (z 4- 1) X ( ^^ — - ^ "^os Vi "^ •^)
X ( z' — 2 z cos -^^ -1- 1 )
\ m J
230. To find the factors of if" + 1, when m is even.
The same process gives
(^,= _2zcos(^^^^4-l) (455)
.4-l =
(z'-2zcoB^4-l)
X (z' —2rco8^4-l)
X (,._2zcos(!!^^4-l) (456)
231. The simple factors of (453) and (454) are obtained from (451) by putting
^ = 0, and those of (455) and (456) by putting ip = JT. There will be found pairs
of equal factors as in the preceding articles, but all the different simple factors will
be found by taking only the positive sign of the radical ,y — 1.
232. Any function of the form z"" —2p z"" 4- ? ay also be resolved into quadratic
factors. It is only necessary to reduce it to one of the preceding forms. By
i-esolving the equation
we shall find from its two values of z"
2«m _ 2j3 z" -)- J = 0 (457)
z*"' —2i>z»4-j= (z- —(;>4-^^' —?)) X {^"'— (P — s/p'- g))
and if we put the absolute term in one of these factors = ±01'" (according to its
sign) it becomes
= a'" ( ^ ± 1 ) = a"* (z ± 1)
in which z = az', and the factors of this last expression may be found by one of
the preceding articles.
If, however, the values of z" in (457) are imaginary, i. e. if p" < q, this method
fails to discover the real quadratic factors, and we must proceed as follows. Put
q = o*", then the proposed function becomes
„^/^_i^ !!: + i) = ,.(^_i^.z'»4-i)
\ a"" a"' o"" ^ / \ a"' ' /
P
in wliich z = az'; and since in the present casep < a", -jj is a proper fractio:
and we may put — = cos (p, which reduces the given function to the form (450).

MULTIPLE ANGLES.
135
CHAPTER XV.
TRIGONOMETRIC SERIES CONTINUED.
MULTIPLE ANGLES.
233. THE true developments of sin mx and cos mz in series, when m is not restricted
to integral values, were first obtained by Poinsot, and form the subject of °
memoir read by him before the French Academy of Sciences, in 1823.* The following
problem is the basis of these investigations.
234. To develop (4 4- •v/'f — 1)"*, in a series of ascending powers of k. Let
and assume
z = (k+^lc" — !]
z = A„ + A,k + A^k^ +
Differentiating (o) and putting
we find
dz
A,k'
z' = m (A4-.y/i' —iV 'x(lthe
v/(*'-l)/ i)-
square of which gives
m'z'—(4' —l)z" = 0
Differentiating this and putting
we find, after dividing by z',
y. _ "^^
~" IT
m' z — kz! — (A' — l)z" = 0
Again, differentiating (6) twice, we find,
^ =A, + 2A,k+S A,7c'. + nA K'-' .
.4--in A"
v/(*'-l)
z" = 1.2 A, + 2.3 A,k+ 3.4 A^ 4' + (n—l) n A„ J»-^
Substituting in [d) the values of z, z', z", given by (6) and (e), we have
D = m'A + m'A,
J»
— -4,
4-1.2 A
4-2.3^,
A-f- m' ^
— 2^,
— 1.2 4,
4-3.44.
4-m'4„
— ra 4„
— (n —1)»4„
4-(»4-i)(«4-2)4,.+,
(a)
(A.
(«.'
(rf)
} (^
m which each of the coefiioients of the ppwei-s of k must be zero. To discover the
aw which governs these coefficients, it will suffice to examine that of the general
Uirm, or the coefScient of k", which is
(m' — «') A„+(n+ 1) [n + 2) 4„+, =0
whence
. m' —«' .
^»+. — — (j, + mn+2) ^"
•* See the published memoir, " Reeherches sur VAnalyse des Sections Angulaires,"
?aris, 1825

I'dH
PLANE TRIGONOMETRY.
so that frcm the first coefficient, A„ we find by making n = 0, 2, i, 6, &o.,
^'
TT^
_ m' —2' m' (m' — 2')
^4 34— ^' — r2"^34 "^°
_ '«° —4" ^ _ m'(m' —2') (m' —4')
•^' - 5^6~ ^' 1-2-3-4-5-6 ^°
and from the second coefBcient, 4„ we find by making « = 1, 3, 5, &o.
m" —1' ,
Therefore, if we put
m'-3' (m'-l')(m'-3')
^' f5~ '-• 2^34^5 ^'
. _ m' —5' _ (m' — 1') (m' — 3') (m' — 5°)
'~ 6-7 '~ 2-3-4-5-6-7
&c.
the equation (J) becomes
z = 4o^4-4,.K-'
and it only remains to find A„ and 4,. In (a), (6), (c) and (e), put & = 0; we find
z = (^ — l)"' = A„ z'= m (v'— 1) "•-'= 4.
Therefore we have, finally.
z = (J4-.^A;» —1)"'= (.y—l)''K+ (^ — l)''—mK' (458)
235. To develop (y/1 — h' + h .^/ — 1)"" in a series of ascending powers of h. We
nave
(^/l _ A' 4- A ^ - 1)- =(^ - !)•» (h + v'A' - 1)"
therefore by (458), exchanging k for h,
(.^ 1 _ A' 4- A ^ -1)-" = (v'-l )•» [(y - l) i? 4- (-v/ - l)"-' mS'-]
in which 5" and fl"' are what £• and K' become when h is put for k.
the imaginary factors in the second member, observing that
(v/ -1)" X (^/ D" = (v' l)"" = (1)?
Ooabining

MULTIPLE ANGLES. 137
(which must not be put equal to unity, since m may be a fraction, and unity has
imaginary roots,) and also that
(^_l)mX (V—l)'»-' = ^—1(,/ —l)-' X(y—l)"'-' = v/ —1(1)~
we have
in which
(v'l ^A"'4- h^—\)^ ={l)^ 11+ ,/ _ 1 (l)^m //' (459)
„ 1 »«' ,. . '«" (m' —2') ,
m—I
^'= A - f!l^ A'+ i!^!^i3ij^ A. - .c.
236. To develop the sine and cosine of the multiple angle in a series of ascending.powers
of the cosine of the simple angle.
When m is an integer, this problem requires us simply to develop sin mx and
. . P
cos mx in a series of powers of cos x ; but when m is a fraction = —, the Sngle mz
has q values which have the same sine and cosine, (Art. 222), if we consider x to
represent all the angles which have the same sine and cosine as the simple angle.
We shall therefore employ Moivre's Formula in its general form (442), or
(cos X 4- \/ — 1 sin x)"* =: cos m (2 »TT 4- ;;)
Putting k = cos x we have by (458) and (446),
(cosx4-^—Isinx) = (/c4-^A'—l)-"
= {^ — l)r-^E+ {^—l)'"-mK'
( m{in'+l)7r \ / m (in'+ 1) ,r\ _
/(m —l)(4n'4-l)fl--\ ^, , ^ , . /(m —l)(4?i'4-l)fl--\ ^,
-f cos (^^ '-^ !—'—].mK'+.^ — lsin[- ~ -~^ j.mK'
Comparing the real and imaginary terms of these two values of (cos x -{-y/— 1 sin x)*",
we have
^„ . , /m (47i'4-l)fl-\ /(m —1) (4«' + l)fl-\
cos OT (2 nfl--f-x) = cos( -^— '—'-— 1 . K+ cos I ^' '--^ •—'-—1 . m K'
• .o I N - /'"»(4«'4-l)^\ „, - /(m —l)(4«'4-l)fl-\ „,
Binm(2nfl--f-z) = sin( —i—g-!—'—] .K+ain[^

133 PLANE TRIGONOMETRY.
thsrefore these two angles can only differ by some multiple of 2 fl-, or we must
have
m(4«4-l)fl-_^m(4«'4-l)fl- ^^
2 2
whence ni (n — «') = n"
but m being a fraction —, and », n' numbers of the series 0, 1, 2,
cannot have m (n — «') equal to an integer n", unless it is zero ;•* therefore
n — «' = 0, n = n'
and the above developments are
. q — 1, we
/m(4n4-l)fl-\ ,^ , / (m —l)(4n4-l)flr\
cos m (2 «fl- 4- x)=cos (^ '' ^ ' ). S:+ cos (^ ^ ^g —^ )• ^^ (**^*')
/m(4»-|-l)fl-\ „ , . / (m —l)(4n4-l)fl-\
sin m (2 nfl-4- x)=sin (^ ^ ^ )- -Sr4- sin (^ ^ '-\ '—^ ). mK' (461)
in which
m' . , m' ()«' — 2') . .
Z- = 1 — ^^ cos' X 4 \.2,.s.i =08* X — &o.
m'—1' , , (m'— 1') (m'— 3') . ,
K' = COB X g^g— COB' X4- ^ 2^^^ " ~
It hence appears that, in general, it requires the combination of two series to express
the cosine and sine of a multiple angle in powers of the cosine of the simple
angle, when m is fractional.
237. When m is an integer, one of the terms of (460) and (461) will always become
zero, and we shall have but a single series to express the function of the multiple
angle. The first members become in all cases
cos (2 mn fl- -|- mx) = cos mx
sin (2 mn w + mx) = sin mx
and the second members vary according to the form of m. In (460), if
m = 4 m', COB mx = K
m = 4 m' -f- 1, cos mx = mK'
m ^ i m' + 2, cos mx = — K
m = i m' + S, cos mx ^ — mK'
and since when OT is even, the series K terminates, and when m is odd, the series K'
terminates, these four equations are all finite expressions, and will give the equation*
of Art, 76, by making OT = 1, 2, 3, &o.
In (461), if
OT = 4 m',
sin mx = — mK'
OT = 4 m' 4- 1, sin mx = K
m = i m' + 2, sin OTX = mK'
OT = 4 m' -f- 3, sin OTX = — K
- (462)
P
* Since — is supposed to be reduced to its lowest terms, p and q are prime to each
other , therefore, if =--^ is not zero, q must divide n — n'; which is impossible,
7
since the greatest value of either re or re' is g- — 1.

MULTIPLE ANGLES. 139
In these formulae, however, the series do not termitate, but by -differentiatin-g
(162) we find for
, = li ; = — m sin x (cos x •
m' 1'
OT = 4m'-f. 1, sin mx == sinx (1 ^-;j cos' x 4- &o.)
1-2
4 m' 4- 2, sin mx = m sin x (cos x -
4 m' 4" 3, sin mx = — sin x (1 —
2-3
2-3
• cos' X + &o.)
• cos' X + kc.)
-1'
•cos' X + &c.)
1-2
(463)
aU of which terminate and give the equations of Art. 75.
238. To develop the sine and cosine of the multiple angle in a series of ascending powert
•)/ the sine of the simple angle.
We take as before
(cos z + .^y — 1 sin x)"" = cos m (2 re3- 4- a:) 4- v^ — 1 sin rei (2 resr -j- x)
Putting A = sin X, we have, by (459) and (444),
(cos X 4- ^ — 1 sin x)-" = (v^ 1 — A' 4- A v* — 1)"
fn
ffl—1
= (!)•' H+^—l{l)~mH'
= cos m re' fl-. 11+ v^ — 1 sin m re' A- . S
+ y/— Icos (m— 1) re'fl-. mH' — sin (m — l)n'v .mil
Comparing the real and imaginary terms of these equations,
cos OT (2 ra^ 4- x) = cos mrirr . H— sin (m— 1) re'fl- . m H'
sin m (2 refl- -)- x) = sin mn'?r . 11+ cos (OT — 1) re'fl-. mil'
and to find what values of re and re' correspond, let x = 0, then A= sin x = 0, iif = 1,
ff' = 0, and we have
cos 2mmr = cos mn!TT
sin 2 OT refl- ^ sin mn'rr
from which we infer that 2 mnfl- = OTB'A-, or 2 re = re', and hence
cos m (2 ren- 4- 2;) = cos 2 OTre fl- . .ff — sin 2 (m — 1) re A- . mB'
sin m (2n-!r + z) = ain 2 mmr . II-^ coa 2 (m — 1) rear. mS'
in which m being a fraction = —, reis any number of the series 0,1, 2, 3,. .
and
M:
H'
1 — ^-^ sin' z 4-
OT* — 1'
2-3
' (m' — 2') sin* X — &c.
1-2-3-4
sin* X +
m' — 1') (m' — 3')
2-3-4-5
sin* X — &o
(464)
(465)
— 1;
239. When mis an integer, the first members of (464) and (^65) become cos mx
and sin mx; and the ooefacients of the second members jontain only multiples of
In-; therefore we have
cos mx = H
sin mx =: mil'
But the series .&• terminates only when m is even, and the series M' only when m li

140 PLANE TRIGONOMETRY.
odd, and we must also employ the derivatives of these equations to obtain finite expressions
in all cases ; thus we have also
sm mx = -
dll
mdz
cos mx = • dW
dx
Therefore differentiating the series i?and //', we shall have, when
2 m', cos mx : 1-2
OT = 2OT'-(-1, cosmxz= cos x (1-
sin' X 4" &o.
- (466)
. sin' X 4- &c.)
1-2
m = 2 m'.
sin OTX =: OT COB X (sin z
O / 1 1 • , • '
m = 2m + 1, sm mx = m (sm x
2-3
2-3
• sin' z -j- &c.)
sin' X + &c.)
(467)
all of which terminate, and give the equations of Arts. 77 and 78.
240. To develop the sine and cosine of the multiple angle in a aeries of ascending powert
of the tangent of the simple angle.
We have
coa m {2 nrr + x) + .y — 1 sin m (2 rea--f- z) = (cos z -J- \/ — 1 sin x)"
= cos"" x{\+^ — 1 tanx)"
Expanding by the Binomial Theorem, and putting
T= 1-
.^i-^-^) ,^^,^_^'n(m-l)[m-2)im-Z) ^^^, ^ _
1-2
-&a.
g" = OTtanz^'"^"'-^„H"'-^)
1-2-3
tan' X + &o.
we have
cos m (2 re fl-4-z) 4-y/—1 sin m (2 re B--f-z) = cos^z (?-}-^—1 T')
But the imaginary and real quantities are not yet distinctly separated in the second
member, for m being fractional cos" z has a number of imaginary values. If
we designate its real value by cos"" x, all its values are included in the expression
cos"" z(l)'" = cos'"x(cos2 OTre' T -|- v' — 1 sin 2 mn' it)
(vhich, substituted above for cos"" x gives
cosm(2refl-4-z) +.y—lsinm(2refl-4- x) rscos^z (cos 2 mre'fl-. T— sin 2 mre'r. T')
+ v'— 1 coB"'x (sin2mre'fl-. T + cos 2 mn'sr. 7")
Comparing the real and imaginary terms, we now have
cos OT (2refl- 4- x) = cos"" x (cos 2 mre's- T— sin 2 mrt-rr . T')
Bin OT (2 refl- 4- z) = cos"" x (sin 2 mn'v . T + coa 2 mn'-?r . T')
ind it is shown as in the preceding problems that re = re', whence
cos m (2refl- -|- x) = cos" z (cos 2mmr T— sin 2 OTBA- . 7") (468)
sin m (2 refl- 4- z) = cos"" x (sin 2 mrefl- T+ cos 2 mrefl-. 7") (469)

MULTIPLE ANGLES. 1-11
in which m being a fraction = .^, nis any number of the series, >, 1, 2, . . . ^ — I;
andCO3 x denotes only the real value of {/(aos'x)''.
241. By the division of (469) by (468)
, , tan 2mrefl-. 7"4- 7" ,,„„,
tan m (2refl-4-z) = ^5 5 ^ ^ (470)
* ' T —• tan zmuTr.T' ^ '
242. TFACT! m is an integer, both the series T and 7" terminate, and in all cases
cos 2 re.n fl- = 1, sin 2 mre fl- = 0 ; and (408), (469) and (470) give
cos mz = cos z. T (471)
sin mz = cos*" z. 7" (472)
7"
tan mz ^ -— (473)
which last expression embraces all the equations of Art. 79.*
243. Before the memoir of Poinsot, developments were given for the multiple arcs
m series of descending powers of the sine or cosine of the simple arc; but he has
shown that these developments are impossible, except when m is integral, and in this
case the series are the same as the preceding, with the terms written in inverse
order.
244. To develop any power of the cosine of the simple angle in a series of sines or cosines
of the multiple angles, the cosine of the simple angle being positive.
If y = cos z 4- -v/ — 1 sin z, we have, by (434) and the Binomial Theorem,
(2 cos z)-" = (y + y-')"" — y"" + my"^-' + "!:i^ L j,-. _j_ &c.
and by Moivre's Formula,
y" = cos m (2 re fl- 4" z) 4- -v/ — 1 sin m (2 re A- 4- z)
OT2/"^'=:mcos (m — 2) (2ren-4-z) 4- »" v''—1 sin (m — 2){2n-!r + z)
' ^ t ^ y ^ = . ^iapn COS (re^-4) (2n,r+z)+'!L(^^_i si„ (m-4)(2re,r4-x)
Therefore, if we put
we have
&c.
Pj„„_jj. = cos m(2njr + z) + m cos (m — 2) (2 refl- 4- z) -|- &c.
.P'anir^j = sin m (2 re fl- 4- z) 4- "* sin (OT — 2) (2n?r + x) + &c.
&c.
(2 cosz)" = P,„„^4-^ —1P',„,^ (a)
- Now m being a fraction (2 cos z)"" has imaginary values, but ivhen cos x is positive,
it will have at least one real positive value, and then (2 cos z) being understood to
lenote only this real value, all the values are included in the formula
(2 cos x)" X (1)*" = (2 cos x)'" (cos 2 mn'a- 4- \/ — 1 sin 2 mre'a-)
-* Although the formulse for multiple angles require, in general, the combination ol
two series when m is not an integer, yet there are certain cases, even when m is a
fraction, in which one or the other of the series will disappear. iSee the memoir of
I'omfot, cited at the beginning of this chapter.

142" PLANE TRIGONOMETRY.
Therefore we have
(2 cosx) (cos 2 mn' Tr+ ,/ — 1 sin 2 mn' ir) = i'dm + i -\- >/ — ^ 7*ann-+»
Cr.mparinf the real and imaginary terms,
(2 cos x)'" cos 2mn'ir = Pa„„+x
(2 cos z)"' sin 2 mn'ir = P',„ „^i
and to find the corresponding values of re and re', let z = 0, then (2 cos z)'" = 2"",
and tie series become
and in the same way
Therefore our formulje become
P,„, = cos 2 mre flr (1-|- OT 4- " »"" 4" ^°-)
= cos 2 mrefl- (1 + 1)""
= 2"" cos 2 mre ir
P',„,r = 2»" sin 2 mre T
2"" cos 2 mre' A- = 2*" cos 2 mre n-
2" sin 2 OTre'fl- = 2 sin 2 mresand
as in former cases, it is shown that re =: re', so that we have finally
(2 cos x)-" = -^''"^-'-'' (474)
cos 2 mre A-
(2 cos z)-" = - ."^''''--^ - (475)
^ ' sm 2 mre fl- ^ '
From this it appears that the real and positive value of (2 cos z)"'may be expressed
either by a series of cosines or by one of sines of the multiple angles, and by
comparing (474) and (475), we have the following constant relation between these
series.
P',„„+j, sin 2 mre fl-
-Pinir+x coa2mnir
245. If re = 0, (474) gives
(2 cos x)'"=Pj =; cos mx+m COS (m — 2) x -] ^—= — cos (OT — 4) x -j- &c. (476)
which may be employed as the general development of the real value of (2 cos z)"",
when X < -=-.
Ji
246. The same supposition of n = 0, gives sin 2 mre A- = 0, and (475) gives
therefore,
„, , m (m — 1) . ,
0 = P'l = sin OTX 4- m sm (m — 2)x-\ ^-^ '- sin (m- 4) z -f &c. (477)
R remarkable property of this series of sines of multiple arcs, which holds for all
TT
values of m, provided x < -^.
247. To develop any power of the cosine of the simple angle in a aeries of ainea or coainet
df the multiple angles, the cosine of the simple angle being negative.

MULTIPLE ANGLES 143
If the denominator of re is even, there is no real value of (2 cos x)*" when cos x
IS negative ; but we may put
(2 cos x) = ( -2 cosz)"' (— l)m
= (— 2 cos x)-" [cos m (2 re' -f 1) fl- 4- .^ — l"sin m (2 re' -j- 1) r]
which, substituted in equation (a) of Art. 244, gives
(— 2 cos x)" cos m (2 re' -f 1) fl- = P,„,+^
(— 2 cos x)"- sin m (2 re' -f 1) fl- = P',„,+,
Making z = 3-, cos z = — 1, (— 2 cos z)"" = 2"', and the series become, by the
process shown in Art. 244,
• and we have
-Pd n+o , = 2"- cos m (2 re 4- 1) fl-
-P'(o n+i)i = 2"' sin m (2 re 4- 1) fl-
2"' cos m (2 re' 4- 1) fl- = 2-" cos m (2 re 4- 1) A-
2-" sin m (2re' -f- 1)fl- = 2'" sin m{2n+l)ir
whence, as before, n = re', and our formulse are
(— 2 cos z)"- = ^'" -^Y ,. (478) -
^ ' cos m (2 re-f- 1) A- ^ '
{r- 2 cos z)"- = -. ^^"'^j-' . (479)
sin m (2 re -|- 1) fl- ^ '
by which it appears that the real value of (— 2 cos z)" is also expressed either by
a series of cosines or of sines of multiple arcs, which series have the constant relation
248. If re = 0, (478) and (479) give
/"•in^r+x sin m (2n + l)ir
F^yi^^x cos m (2 re 4-1) A-
P 1
(— 2 cos z)"" = — = (cos mx -I- OTcos (m— 2) z -J- &c.) (4801
COSOTfl- cos OTA- ^ / ' / \ I
P' 1
(— 2 cos z) = -;—^ = -; • (sin OTX-4- m sin (m — i.) z 4- &c.) (481)
sm m fl- sm OT A-
In this case sin OTA- is not zero, unless m is an integer, so that the series i", does
ir
not become zero when x > -^, and both (480) and (481) maybe employed as the true
developments of (— 2 cos z)"".
249. When m is an integer, the series (476) and (480) always terminate at the
(m + l)th term.; and, since in (480) cos rejA- = ± 1, according as OTIS even or odd,
and (— 2 cos x)"* = ± (2 cos z)"" in the same cases, both (476) and (480) becoruo
(2 cos z)" = cosmx4- mcos (m — 2) X4 ^—s "os (m —4)z4- &c. (482)
But the series (481) becomes zero, so that (482) is the only series by which
(2cosz)"* can be developed in functions of the multiple arcs, when m is integral.

144 PLANE TRIGONOMETRY.
250. To develop any power of the sine of the simple angle, in a series of sines or cosines
of the multiple angles.
If y = cos z 4- y/ — 1 sin z, we have, by (435) and the Binomial Theorem,
(.^ _ 1) (2 sin z)"- = [y —
y~')'"
, m (m — 1) „_. „
= ym _ my^-^+ —^—^ .V"" — &o.
in which y", y"-^, &c. have the same values as in Art. 244, but the signs of the
coefEcients are alternately 4- and —, so that if we put
9,„„^.^ = cosm(2refl-4-z) —mcos(m —2) (2re A-4-z) 4- &c.
§',„„^.x = sinm (2n fl- 4- z) — OT sin (OT — 2) (2 re A- 4- z) 4- &c.
we have
(V* — l)" (2 s ^)'" = 9»»' + '+V
— '^ Q'-iu^+x
Substituting the value of (v* — 1)"" by (446), and comparing the real and imaginary
terms, we find
-~ . . m (in' + 1) TT ^
(2 sm z)»> COB —^—Y— = V!...

MULTIPLE ANGLES. 145
253. The series (485) and (486) become zero when m is an integer, as follows:
If m = 2m', 0 = sin mx — m sin (m — 2) x -f- &c. (491)
m = 2 m' 4- 1, 0 = cos mz — m cos (m — 2) z -)- &c. (492)
The reason why these series are zero is obvious, since they terminate ai, the
(m 4-l)th term, the terms equally distant from the first and last are equal with
opposite signs, and the middle term of (491)_is zero.
254. Given the equation
tan z = p tan y (493)
to express z ^ly ina series of multiples of y.
Substituting the values of tan x and tan y given by (431)
whence
or rutting
gixv~i — I g-ji/y-. 1
•->+ 1 ^- e'»>'-' +i
{p+l)e-yV-^—{p—l)
p+1 — [p — 1) e'!'>'-'
? = 7+1 (494)
fixy-t -. ^ _ _ ,jy_, ( i_ tl \
(..)
gi (I-!,) y -. = ^ 7 ^
1 —ye'!"'-'
Taking the Naperian logarithms of both members,
2 (2; —y)-^/—1 =log (1 —g-e-'si'-') —log (1_ je^yV-.)
and developing the second member by the formula
we have
log (1 — re) = — re — J re' — J re' — &c.
2 (z —y) v"-1 = — qe-'yy-' — ^ q^ e-'yV-' — ^ q' e-'vy— — &c.
Substituting in the second member by (430),
+ qfeyV-i + 1 g''g'yV-^ _|_ 1. jSgOjy-. + &
(? (^*y) y—• = ?
1 e—'y/—>
g
from which, by taking the logarithms and substituting as before,
(J)
3; ,„ sin2y sin4y sin6y
z + y - ^ 2 ^ 3j' '*"• ^"^
In this investigation, we have, in effect, used Moivre's formula, in its limited or
less general form; but the requisite generality may be given to our results, by observing,
that (498) would hold if we were to substitute tan z = tan (re'B--j-z), tany
-T^ tan (n"ir+y), and therefore we may substitute for the first member of (A),
19 N

146 PLANE TRIGONOMETRY.
n'lT +x — (re"fl- 4- y) = 2; — y — («" — re') fl- = z — y — mr, re being (like re and n") at
arbitrary integer or zero. Hence, the required general development of x — y in
series is
X — y = nw+ q sin 2y-f- ^ j' sin 4y-|-J j' sin 6 y-j- &c. (495)
In like manner, since tan x = tan (x — re'fl-), tan y = tan (y — re"fl-),we may substitute
in the first member of (c), x — n'lr + y —ra"fl- = z -f- y — nir, and the general development
otx + yi-a series is
sin2« sin4y sin6y , „ ,,„„,
x + y^nir - J ^ - - ^ - ^ 4 - & C . (496)
In these formulse z and y are supposed to be expressed in arc, and to obtain
z qr y in seconds, the terms of the series must be divided by sin 1".
255. The preceding problem is particularly useful in finding z when p and y are
given, and z is nearly equsjl to y; in which case p is nearly equal to unity, either
y or — is a small fraction, and one of the series (495), (496) converges rapidly.
EXAMPLKS.
1. Given y = 50° and/> = 1-00065, to find z from (493).
Taking only the first term of the series (495), and assuming re ^ 0,
2y = 100° log sin 2y 9-99335
-00065
^ = -2^00065 ^"S^ ^-""^^
ar 00 log sin 1" 5-31443
x — y = 65"-995 log {x — y) 1 -81952
z = 50° 1' 5"-995
2. Given y = 50° and^ = — 1-00065, to find z from (493). In this case
_ 2-00065
^ ~ -00065
and the computation by (496), if we assume re = 0, is
2 y = 100° log sin 2 y 9-99335
1 -00065
q 2-00065 ,,(-i)- ^ \ q
6-51174
arCO log sin 1" 5-31443
z 4- y = — 65"-995 log (z -4- y) — 1-81952
z = —y — 65"-995 = — 50° 1' 5"-995
or, ifre = l,z=]SO° —.50° 1'5".995 = 129° 58'54".005.
In general, (493) is to be solved by (495) when p is positive, and by (496) when/)
is negative.
256. Given the equation
sin (a 4- z) = m sin z (497)

which is reduced to (493) by putting
MULTIPLE ANGLES. 147
1 1 1 m4-l
z = z4-Ja y = } * ^ = ^ ^
whence q = ?—,— = —
^-f-1 m
and (495) becomes
sin* sin2a sin 3a
z=inirA. ^ -=,—, 4- - ^ + kc. (498)
' m ' 2 m' ' 3 m^ ' ' •'
which is to be employed when m > 1; and (496) becomes
z4- « = rea- — m sina — J m'sin 2 a — Jm'sin 3 a—&c. (499)
which is to be employed when m

148 PLANE TRIGONOMETRY.
259 In a plane triangle A B G, given a, b and G, to find A or B by a series of multiplet
efC.
By (260)
a
T ^'^
tan A =
a
1 r- cos C
Which, compared with (503), gives, by (505),
b
a . ^ , a' sin 2 (7 , ffi3 sin 3 C , . ,^„„^
^ = -j-smC'4-y,.—^—4--p.—^-4-&c. (500)
n being necessarily = 0 in this case. Bis found by the same series, interchanging
a and b.
260. In a plane triangle, A B G, given a, b and G, to find cby a series cf multiples of C.
We have
•
c' == a' 4- 6' — 2 ab cos 0 (507)
c' i' 2S •,
— = -^ — cos C 4- 1
a a' a
by(451) = f-^ —(cosC4-^—IsinC)"]
X ri._(oosC'—v' —lsinC)1
f-= [l-.l(cos04-^-lsinC)]x[l-y
(oosC-v/-lsin(7)'|
Taking the common logarithms, employing in the second member the formula
log (1 — re) = — Jf (re 4- J re' 4- J re' 4- &c.)
and applying Moivre's Formula (440) in expressing the powers of cos C ± v^ — 1 sin 0,
we have
2 log c — 2 log b =
— JWr-| (C03C4-V' —IsinC) 4- -|^ (

PAET II.
SPHERICAL TEIGONOMETllY.
CHAPTEE I.
GENERAL FORMULAE.
1. SPHERICAL TRIGONOMETRY treats of the methods of computing
the unknown from the known parts of a spherical triangle.
It is sho-wn in geometry,* that a spherical triangle may, in general,
be constructed when any three of its six parts are given, (not
excepting the case where the three angles are given). We are now
to investigate the methods by which, in the same cases, the unknown
parts may be computed.
We shall at first confine our attention to such triangles only as
are treated of in geometry, namely, those whose sides are each less
than a semicircumference, and whose angles are each less than two
right angles; that is, those in which every part is less than 180°.
2. It is shown in geometry, that if a solid angle is formed at the
center of a sphere by three planes, the three arcs in which these
planes intersect the surface of the sphere form a spherical triangle.
Now the real objects of investigation in spherical trigonometry are
the mutual relations of the angles of inclination of the faces and
edges of a solid angle ; but, for convenience, the spherical triangle
which forms the base of the solid angle is substituted for it. The
sides of the triangle being proportional to the angles of inclination
of the edges of the solid angle, are taken to represent those angles ;
And the angles which those sides form with each other are regarded
* The student is here supposed to be acquainted with Spherical Geometry, at
least so much of it as is to be found in Legendre's treatise, or in that of Prof. Peirce,
of H arvard University.
n2 149

150 SPHERICAL TRIGONOMETRY.
as identical vfith the angles of inclination of the faces of the solid
angle. But, since varying the radius of the sphere would not, in
any respect, change the solid angle, or the values of the angles which
enter into it, the mutual relations in question ought to be deduced
without any reference to the magnitude of the radius of the sphere.
In fact, we shall deduce our fundamental formulse from a direct consideration
of the solid angle itself.
3. In a spherical triangle, the sines of the sides are proportional
to the sines of the opposite angles.
Let AB 0, Fig. 1, be a spherical
Fig. 1.
triangle, 0 the center of the sphere.
The angles of the triangle are the
inclinations of the planes AOB,
AO Q and BOG, to each other, and
> a will be designated by A, B and Q;
their opposite sides respectively will
be designated by a, b and c, as in
plane triangles. The trigonometric
functions of these sides will be the same as those of the angles
BOG, AOG, AOB, which they subtend at the center of the sphere.
(PI. Trig. Art. 20.)
From any point B' in 0 B, let fall B'F perpendicular to the plane
AO G; and through B'P let the planes B'PA', B'PQ' be drawn
perpendicular to 0^ and 0 G, intersecting the plane OAQ in the
lines PAI, PG', and the planes .405, BOQ in the lines A!B', B'O'.
The plane triangles A'P B', B'P G' are right angled at P ; aiid
OA'B', 0 C'B' are right angled at A' and G' The angle B'A'P,
being formed by two lines perpendicular to OA, is the measure of
the inclination of the planes AOB, AO G, or .of the angle A; and
B G'P is the measure of the angle G.
We ha^e therefore, by PI. Trig. Art. 15,
sm A = sin B'A'P ••
sin G= sin B'O'P-
B'P
WA'
B'P
B'G'
whence
sin A
sin G
B'P
B'A'
B'G' _ B'G'
B'P ~ B'A'
(m)

GENERAL FORMULA.
l&l
Again,
B'O'
sin a = sin B'OG' = -r,,^
B (J
sine ^sinB'OA'
B'A'
— B'O
Comparing (m) and [n).
sin a B'G' B'O B'G'
B'O ^ B'A' B'A'
sma sin 4
sin 0 sin G
which in the form of a proportion is
sin a : sin c = sin 4 : sin G
which is the theorem that was to be proved.
4. In Fig. 1, A, a, G and c, are each less than 90°, but the construction
would not vary if any of these parts were greater than 90°,
except that the points A' and G' might be found in the lines AO, GO,
produced through(9; and one or more of the right triangles A'B'P,
kc, would contain the supplements of A, a, G, or a instead of these
quantities themselves. But the sine of an angle and of its supplement
being the same, the preceding demonstration would still be
valid, so that the theorem is applicable to any spherical triangle.
Indeed, according to PI. Trig. Art. 49, this result follows from
the nature of the trigonometric functions themselves, and the demonstration
of the preceding theorem might therefore be considered as
general, without requiring a special examination of the various positions
of the lilies of the diagram.
6. In a 'spherical triangle, the cosine of any side is equal to the
product of the cosines of the other two sides, plus tJie continued product
of the sines of those sides and the cosine of the included angle.
Let the plane B'A'G', Fig. 2, be ^ig- 2.
drawn perp. to OA, intersecting the
planes J.O.B, BO CmdAOG, in the
lines A'B', B'G' and A'G'. Then the
angle B'A'G' = A, and B'OG' = a,
and by PI. Trig. Art. 119, in the triangles
J.'.S'C", OB'Q', we have
B'G" = A'B'^ + A'G" - 2 A'B'.
A'G' cos A
B'G'^ ==0B'^+ 0G'^-20B' 0 Q' cos a
(1.1

152 SPHERICAL TRIGONOMETRY.
Subtracting the first of these equations from the second, and observing
that in the right triangles OA'B', OA'G',
we have
OB'^-A'B"=0A'% OG'^-A' C"^ = 0 A""
0 = 2 OA'^ 4- 2 A'B'. A'G' coa A - 2 0 B' 0 G' cos a
OA'.OA' , A'B'.A'G'
whence cos a = jy-gr-Qgr + Q j^, Qg, cos J.
Substituting the trigonometric functions derived from the right triangles
OA!B', OA'G',
cos a = cos 5 cos e -f sin 5 sin c cos A (2)
which is the theorem to be proved. It may be regarded as the fundamental
theorem, for the preceding (1) can be deduced from it, but
as the process is somewhat circuitous, we have preferred deducing
the two theorems from independent constructions.
6. In the construction of Fig. 2, both b and o are supposed less
than 90°, while no restriction is placed upon A and a; but the equation
(2) is no less applicable to all the other cases if the principle of
PI. Trig. Art. 49 be granted. As that principle may not be suifi-
Fig. 3. ciently evident to the student unacquainted with analytical
geometry, we shall verify it in this case, as follows.*
1st. In the triangle ABG, (Fig. 3), let b < 90° and
c > 90°. Produce BA, BG to meet in B', forming the
lune BB'; then A B'= 180° — c, a.ni b are both < 90°,
and the preceding demonstration would apply to the
triangle A B'G. Therefore, applying (2) to A B'G, we
have
cos(180°-a) = cos5co8(180°-c)4-sin5sin(180°—c)cos(180°-A)
or by PI. Trig. (64),
— cos a = — cos b cos c — sin b sin c cos A
and changing all the signs
' cos a = cos 5 cos c -f sin b sin c cos A
the same result that would have been found by applying (2) directly
to ABG.
* Hymer's Spherical Trigonometry. Cambridge, 1841.

GENERAL FORMULA. 153
2d. In the triangle ABO, Fig. 4, let b > 90°, c > 90° ;
produce AB and AG to meet in A'; then A'B andA'Q
being both less than 90°, the formula (2) is applicable to
A'BG. Therefore
cos a = cos (180° - b) cos (180° - c)
4- sin (180° - b) sin (180° - c) cos A
= (— cos b) (— cos c) 4- sin 6 sin c cos A
= cos b cos c 4- sin 5 sin c cos A
the same result as before.
7. The theorems expressed by (1) and (2) being applied successively
to the several parts of the triangle, give the two following
groups:
sin a ainB = sin 5 sin A
sin b ain G = sin c sin 5 I (3)
sin c sin 4 = sin a sin G
cos a = cos b cos e 4- sin 5 sin c cos A
cos b = cos c cos a 4- sin e sin a cos B^ I (4)
cos c = cos a cos 5 -f sin a sin b cos G
8. Let A'B'G', Fig. 5, be the polar triangle
of ABG, and designate its angles and sides by
A', B', G', a', b' and ci Then, by geometry,
A' = 180° -a, a' = 180° - A
B' = 180° - b, b' = 180° - B
O' = 180° -c,
c' = 180° - C
and applying the first equation of (4) to A'B'G',
or by PL Trig. (64),
cos a' = cos b' cos c' -f sin 5' sin c' cos Jl
— cos J. = (— cos B) (— cos C) 4- sin B sin C (— cos a)
— cosA= cos B cos (7 — sin 5 sin G cos a
Changing the signs of this, we have the first of the following group:
cos A — — cos B cos 0 -{- sin B sin 0 cos a
cos B = — cos C cos JL 4- sin (7 sin JL cos b > (5)
cos (7 = — cos .4 cos .S 4- sin A sin .5 cos c
20
J
')

154 SPHERICAL TRIGONOMETRY.
It is thus that, by means of the polar triangle, any formula of a
spherical triangle may be immediately transformed into another, in
which angles take the place of sides, and sides of angles.
9. Several other important fundamental groups of formulse are
obtained from the preceding with the greatest ease.
The first of (4) multiplied by cos c is
cosa cose = cos b cos^ c -f sin 5 sine cos c cos A
and the second of (4) is the same as
the difierence of which is
cos a cos G 4- sin a sin o cos B — cos b
sin a sin e cos .B = (1 — cos^ o) cosb — sin 5 sin e cos c cos A
Since 1 — cos^ c = sin^ c, this may be divided by sin o, and gives
sin a cosB = sin e cos 5 — cos c sin 5 cos .4 "j
whence sin b cos G = sina cos c — cos a sin c cos B V (6)
sin G cos 4 = sin 5 cos a — cos b sin a cos G
If we interchange B and G, and therefore also b and c, the group
becomes
sin a cos G = sinb cos c — cos b sin c cos A
sin b cos A = sin c cos a — cos c sin a cos B
sin c cQsB = sin a cos 5 — cos a sin b cos Q
10. If (6) and (7) are applied to the polar triangle, they give,
after changing the signs of all the terms,
and
sin Acoab — sin G cos B 4- cos Csin.B cos a
sin B cos G =ainA cos (74- cos J. sin G cos b \ (8)
sin (7 cos a = sin 5 cos J. 4-cos-B sin J. cos e
sin.4.cosc = sinB cos C-f cosjBsin 0 cosa )
sin Bcosa== sin (7 cos J. 4- cos Gain A cos 5 V (9)
sin (7 cos 5 = sin J. cos ji5 4-cos J. dn.B cose
11. Dividing the first of (6) by the following derived from (3),
sin a sin 5 . ,
-.—J— = sm 0
smA
^
J
J
J
(7)

we find the first of the following group
GENERAL FORMULA. 155
sin AcotB = sin c cot 6 — cos c cos J.
sin-B cot (7 = sin a cot e — cosacosi? V (10)
sin G cot J. = sin 5 cot a — cos b cos 0
and in the same way from (7), or by interchanging the letters B and
0, b and e in (10), we find
sin J. cot C = sin 5 cot c — cos b cos A
sin B cot 4 = sin e cot a — cos c cos B
sin G cot B = sin a cot b — cos a cos G
(11)
If (10) are applied to the polar triangle, we find (11), so that no
new relations are elicited.
12. The preceding formulse are sufiicient to furnish a theoretical
solution for every case of spherical triangles, but some transformations
are required to facilitate their application in practice.
In the first of (4) substitute, by PI. Trig. (139),
we find, by PI. Trig. (39),
cos J. = 1 — 2 sin^ 1A
cos a = cos [b — e)—2 sin b sin c sin^ J A (12)
and we have similar expressions for cos b and cos c.
If we substitute in (4), by PI. Trig. (138),
we find, by PI. Trig. (38),
cosJ. = -l
4-2cos2iJ.
cos a — cos (6 4- e) 4- 2 sin b sin e cos^ J A (13)
and, of course, similar expressions for cos & and cos e.
13. Substituting in (5)
cos a = 1 — 2 sin^ J a = — 1 4- 2 cos^ J a
we find by the same process
cos J. = — cos(.B -h G) — 2ainB sin Gsin^ J a (14)
cos J. = — cos {B — G) + 2 sin B sin G cos^ J a (15)
which might have been obtained by applying (12) and (13) to the
polar triangle.

156 SPHERICAL TRIGONOMETRY.
14. If ii) (12) we substitute cosa = 1 — 2 sin'J a, cos (b — o) = 1 —-2 sin'^(6 — c),
we obtain the first of the following equations; and the others are obtained by a
similar process from (12), (13), (14) and (15).
sin' ^ a = sin' J (6 — c) -)- sin i sin c sin' ^ A (16)
sin' i a = sin' i (b + c] — sin i sin c cos' J A (17)
cos'i a = cos' J (6 — c) — sin 6 sin c sin' ^ A (18)
cos'^ a = cos'J (J -j- c) -(- sin 6 sin 0 cos'J A (19)
sin' J ^ = cos' J (.B 4- C) 4- sin B sin G sin' J a (20)
sin' J ^ = cos' i (B—G) — sinB sin C cos' J a (21)
cos' } ^ = sin' J (5 -f C) — sin .S sin C sin' J a (22)
oos'^^ = sin'J (i?— G)+ sin .S sin Ccos'J a (23)
15. By PI. Trig, we have
1 = cos' iA + sin'iA
cos A = cos' ^ A — sin' J A
whence
COB 6 COB c = COS b cos c cos' J ^ 4- (ios 5 cos o sin' J jl
sin J sin c cos ^ ^ sin 6 sin c cos' ^ A — sin 5 sin c sin' J ^
the sum of which is, by (4),
cos a = cos (6 — c) cos' J .4 -f- cos (J -f- c) sin' J ^ (24)
and substituting 1 — 2 sin' J a, kc, for cos a, &c.
sin' J a = sin' J (i — c) cos' iA + sin' J (i -f c) sin' j ^ (25)
cos' I a = cos' J (6 — c) cos' J ^ 4- cos' J (i -f- c) sin' J ^ (26)
In the same manner we deduce from (5)
cos J4 = — cos {B — G) sin' J a — cos (JS -f C) cos' J a (27)
sin'J.4= cos'J(5—C)sin»Ja4-cos'J(.B-f- C)cos'Ja (28)
COB'J^= Bin'J(J3—C)sin'Ja4-sin'i(.B4-C)cos'Ja (29)
It is hardly necessary to add that each of the equations (12 to 29) gives a group
of three, by applying it successively to the three sides or three angles of the triangle.
16. From (12) we find
. ,, . cos (6 — c) — cos a
am^A = ^ . '
2 sm 6 sm e
If, in PI. Trig. (108), we put
X =a, y = b — 0
whence
i{^ + y) = i{^+b-c), ^{x-y) = ^{a-b-i-c).
we find
cos (5 — c) — COS a = 2 sin J (a — 5 -f e) sin J (a 4- S — c)
which, substituted in the above equation, gives
si„a X A = -i^H^-i + o)sir.i{a-hb~^
sin 0 sin e
^

then
GENERAL FORMULAE.
Let s denote the half sum of the sides, that is, let
a4-54-c = 2s,
J(a4-54-c) = s
a—b-\-c = a-hb-}-c-2b = 2s-2b=2{s-b)
a4-5 — (j = a4-J4-e — 2c=2s — 2c=2(s — e)
which substituted in (30) give
whence also
. sin (s — b) sin (s — c)
sm- •^ sm 0 sm c
sin^ J B
17. From (13) we find
sin (s — e) sin (s — a)
sm e sm a
21 n — sin (s — a) sin (s — b)
. •9- 0 — ; : J
sm- sm a sm b
157
> f^l)
cos^J J- =
cos a — cos (b 4- c)
2 sin b sin c
and from PI. Trig. (108), by making
a;=54-e, y ^ d
J (a; 4- 2/) = i (a 4- 5 4- c), J (a; — «/) = J (S 4- c — a)
we find
cos a — cos (5 -f c) = 2 sin J (a -f 5 -f e) sin J (6 4- c — a)
which, substituted above, gives
, , , 2 ^ ^ = '^^^''^^-^^^^^^^'-'''^ (32)
^ sm 0 sm e ^ ''
Introducing, as in the preceding article, s = J (a -f J 4- c).
cos^ iA =
003=" iB =
cos^ i (7 =
.{s-a)
sin 6 sin c
sin s sin (a — b)
sin e sin a
sin s sin (s — c)
sin a sin J
(33)

158 SPHERICAL TRIGONOMETRY.
18. The quotient of (31) divided by (33) gives
sin (s — b) sin [s — c)
tan^ IA =
sin s sin (s — a)
tanH5 =
tan= J G ••
sin (s — c) sin (s — a)
sin 8 sin (s — 5)
sin (s — a) sin (s — 6)
sin s sin (s — e)
V (34)
19. From (14) we find
sin^ J a
cos A 4- cos (.g 4- G)
2 sin B sin G
from which, by PI. Trig. (107), we deduce
and if we put
^ - cos i{A-i-B+ G) cos i (.5 4- G- A)
sin" * a
sin^BsinC
2xh—
sm-
i{A-i-B-i-G)
= S
(85)
sm-
j^ _ — cos«ycos {S — A)
^ sin B sin G
sm-'f e •
~COSAS'COS(*S'—5)
sm G'sm.A
— cos S cos {S— G)
sin A sin B
(36)
The first member of each of these equations being a square, the
second member must be essentially positive, although its algebraic
sign is negative ; in fact, since by geometry 2*S^> 180°, >S^> 90°,
cos S is negative, and — cos S is positive.
20. From (15) we find
cos^ J a
cos A 4- cos {B — 0)
2 sin .B sin G
from which we deduce, by a process similar to the preceding,
cos^ J a
cos I (J.-^4-(7) cos J
sin B sin G
(A-^B-G)
(37)

GENERAL FORMULA. 159
.._oo^{S-B)coa{S-G)
^
sm B sm G
^_cos(,S'- G)cos{S-A)
cos 2 ^
1 sin G sin A
_ cos {S-A) cos (S-B)
,2 1
cos ^ " sin J. sin B
21. Frcm (36) and (38)
(38)
21 _ — COS S COS (S — A)
^^'^ ^ "^ ~ cos (^-5) cos (,S-(7)
, ,, , — cos S cos (iS' — B)
tan-" * 0 = — . „—vv,—5^^-:=—'—r.
" cos (^S* — G) cos (^ - A)
(39)
2 ^ _ — COS *? cos (iS* — C)
^^"^ ^ "^ ~ cos(^-^)cos(;S'-.B)
We might have deduced (36), (38), (39), by applying (31), (33),
(34) to the polar triangle.
22. Napier's Analogies. Dividing the 1st of (34) by the 2d, we
find
tan J A sin (s — I)
tan J B sin (s — a)
Regarding this as a proportion, we have, by composition and division.
tan ^ A -\- tan ^ B _ sin (s — 5) 4- sin (s — a)
tan ^ A — tan J B sin (s — b) — sin (s — a)
[m)
In PI. Trig. (109), if we put x=s
we have
x-hy
and by PI. Trig. (126),
— 2s — a — b = c
X —y =' a — b
— b, y = s — a, whence
sin (s — 5) -f sin {s — a) _ tan J e
sin (s — 5) — sin (s-a) ~ tan J (a — 6)
tan 1 A 4- tan|^ _ sin |( J. + ^)
tan 1^ — tin | B ~ sin J (^ — B)

160 SPHERICAL TRIGONOMETRY.
Therefore (m) becomes
sin |(J. 4- -B) _, tan|c
sin ^{A — B) ~ tan J {^b) \- (40)
or sin J (J. -f i?): sin J (J. — i?) = tan J e : tan i (a — J)
which is the first of Napier's Analogies.
23. Again, the product of the 1st and 2d of (34) gives
, . , T, sin (s — e)
tan iAt&nlB = ^^ -^
sm s
or 1: tan J A tan ^ B — sin s : sin (s — c)
whence, by composition and division,
1—tan J J. tan J .B sin s — sin (s — e)
1 -f tan J J. tan J 5 ~~ sin s 4- sin (s — c)
{n)
By PI. Trig. (109), if a; = s, «/ = s — e, we have
and by PI. Trig. (127),
Therefore (n) becomes
sin s — sin (s — c) tan ^ c
sin s 4- sin (s — e) tan J (a -f J)
1 —tanJJ.tan|5 _ cos J (J. 4--8)
1-F tanJJ.tanJjB ~ cos J (J. — i?)
cos ^ (^ 4- ^) _ tan I e ]
cosi(.4—^) ~ tan i (a 4- b) (41)
or cosi(J.4--B):cosJ(J. —^)=tanie:tanj(a 4-J) j
which is the second of Napier's Analogies.
24. If (40) and (41) are applied to the polar triangle, we sha'l find
or
or
sin 1 (a 4- S) _ cot | C
sin I {a- b) ~ tan J (J.-.B) } (42)
sin J (a 4-5): sin J {a —b) = cot J (7: tan J (J.—.B)
cos I (a 4- 5) cot J G
cosi{a-b) ^ Unl{A-\-B) } (43)
cos J (a -f 6): cos J (a — 5) = cot J G: tan J (J. -f JB)
which are the third and fourth of Napier's Analogies.

25. Gauss's Theorem. If
GENERAL FORMULA.
IGl
p = coaicsini(A + B) P = cos J CcosJ (a — b)
J = cos J c cos J (.4 + B) § = sin J C cos J (a 4- b)
r = ain ic ain i (A — B) R = coa ^ G sin J (a — S)
« = sin J c cos J (.4 — £) S = ain i G sin ^ (a + b)
then the products P X g, P X r, p X ^, g X r, q X s, r X s, are respectively equal to the
vroducta P X Q, J'X Ji, J'X S, Q X M, Q X S, R x S.
First. From (3) we have
sin c (sin A ± sin B) = sin C (sin a =fc sin b)
which, by PI. Trig. (105), (106) and (135), are reduced to
Bin J c cos J c sin J (^ -f B) cos J (^ — .B) = sin J C cos J C sin J (a 4- 6) cos J (a — i)
sin J c cos J c cos J (.4 -f .S) sin J (^ —.B) = sin J C cos J C cos J (a 4- i) sin J (a — A^
or
Second. From (6) and (7)
pa = PS and qr = §5
sin c (cos 5 rt cos A) = (Izp. cos C) sia (a dt J)
which, by PI. Trig, are reduced to
sin J c cos J c COB ^(A + B) cos ^(A —.B) = sin J C sin J C sin J (a-}- b) cos § (a |. i)
sin J c cos J c sin J (4 4- .8) sin J (^ — -S) = cos J C cos J (7 sin J (a — i) cos § (a — 4)
or
Third. From (8) and (9)
ys = QS and pr = PR
(1 ± cos c) sin (.4 ± 5) = sin G (cos 5 =t: cos a)
which, by PI. Trig, are reduced to
cos J ccos J c sin J (.44-.B) cos J (.4-1- B) = sin J Ccos J G cos ^ (a + b) cos J (a — i)
sin J c sin J c sin J (A — B) cos i (A—B) =:sin^ Ccos J (7 sin J (a 4- S) sin J (a — A)
or
pq = P§ and rs = RS
26. The notation of the preceding article being still employed, the quantitiea p'', q', r*, *•,
are respectively equal to i", §', i?", (S"-
We have
and
the quotient of which ia
and in the same way
pg X pr = PQ X PR
j7-=
s' = 6"
21 o2
QR
pt _ pa -pfhence ^ = ± P
j' = §' ? = ± «
r'= R' r = ±R
« = ± S

162 SPHERICAL TRIGONOMETRY.
27. In these last equations, the positive sign must be used in all the second members, or
the negative sign in all of them. For if we take
^ = 4-P
the equations
pq = PQ, pr = PR, pa = PS
being divided by this, give
g = +Q, T = + R, s = + S
and if we talce
i> = - P
the same equations, divided by this, give
g=—Q, r = —R, a = — S
We have therefore the following, which are generally cited as Gauss's Equations.
cos ieaini (A + B) = cos J C cos J (a — b)
cos ^ecosi (A + B) = sin J C cos J (a + b)
sin ^c ain i (A — B) = cos i G aini (a — b)
or
sin i c coa i (A — B) = sin i C sin i [a + b)
- "
cos icsin^ (A + B) = —cosi G cos i (a — b)
cos iccos^ (A + B) = — sin J C cos ^ (a + b)
Bin i c sin i (A — B) = — cos J C sin J (a — b)
sin J c cos J (4 — B) = — sin J (7 sin i {a+ b)
If, however, we consider only those triangles whose parts are all less than 180°, tho
first of these groups, (44), is alone applicable, for we must then have p = +P; since
cos J c, sin i (A + B), cos ^ G, cos J (a — J) are then all positive quantities. The use
of (45) will be seen in the chapter on the solution of the general spherical triangle.
Napier's Analogies, (40), (41), (42) and (43) can be deduced directly from (44).
ADMTIONAL FonwvLJE.
28. We shall here add some formulas which, though not so frequently used as the
preceding, are either remarkable for their elegance and symmetry, or of importance
in certain inquiries of astronomy and geodesy.
29. The product of (30) and (32) gives
4 sin s sin (« — a) sin (s —
Bin'.4 —
*) sin
sin' b sin' c
Put
n' = sin s sin (s — a) sin (s — *) sin
then
. 2re
sin b sin c
and in the same manner
the quotient of which is
- D 2re
sin a sin c
din A
ein B
sin a
sin b

30 We have also from (35) and (37)
and if
ADDITIONAL FORMULA. 16-3
sin' a = — ^ g°s g cos (S — A) cos (S — B) cos (S — 0)
sin' B sin' C
From (48) and (51),
(49)
iV' = _ cos 5 cos (S — A) cos (5" — B) cos (-S—C) (50)
2Ar
sin B sin C
re sin a sin 5
N sin ^ sin B sin C
81. If we develop (47) and (50) by PI. Trig. (173) and (174)
(51)
(52)
4 re' =1 — cos' a — cos' b — cos' '^ 4- 2 cos a cos 5 cos c (53)
4iV = 1 — cos'^ — cos'5 —cos'C—2cos^ cos J? cos G (54)
32. The following simple results are easily deduced from the equations (31 to 38;
cos J A cos J B
sin J G
sin s
sin c
cos J ^ sin J If sin (s — a)
cos J G
sin e
sin J A cos J P sin (s — 6)
cos J C
sin c
sin J J4 sin J P sin (s — c)
sin J C
sin e
sin J a sin J 5 — cos 5"
cos J c sin C
sin J a cos J S cos (S — A)
sin J c
sin G
cos J a sin J 5 cos (S — B)
sin J c
sin G
cos J a cos J i cos (,S— G)
cos J c
sin G
• (55)
83. By means of (55) and (56) we can deduce expressions for the functions of
t, a — a, &c., in terms of the angles, or of S, S — A, kc, in terms of the sides
We have, from (51),
. 2N N_^
sin ^ sin P 2 sin J ^ cos ^A sin J P cos J P
which, substituted in (55), gives
N ,.-,
2Bin J.4smJ5siii J(7
^
sin (. - 0) = ^.^^--^-^--J-^__^-^ (581
whence, by interchanging the letters, we have also sin (s — a) and sin (s — b).

16-i SPHERICAL TRIGONOMETRY.
Again, we have
whence
sin (s — c) = sin s cos c — COB « sin e
sin s cos c — sin (a — c)
COB a = :
Bin c
which, by (55), is reduced to
and from the equation
cos i J1 COS i P COS c — sin J .4 sin J P
cos a =; ? ? -.—=--=
sm J 0
•we find, by substituting (55) and (59),
COS (a — c) = cos s COB c + sin a sin e
, , — sin i .4 sin i P cos c 4- cos i A cos J P .^^
cos (s — c) = ^ i-T—r-?;—' —- I w)
^ ' sm J C '
To eliminate c from the second members of (59) and (60), we have, by (5),
cos G + cos A cos P
sin A sin P
(69)
whence
, . , _ cos C + cos A cos P
cos * .4 cos * P cos c = —-.—;—^—,—-.—-, _ •
-* •* 4 sm J ^ sin J P
. , , . , _ cos C? -f- cos .4 cos P
sm * ^ sin * P cos c = —^ -^j-^ , „
•* -* 4 cos J .4 cos J P
which, substituted in (59) and (60), give
cos A + cos B + cos G — 1 1 — sin' ^ A — sin' J P — sin' J ^ rn-i\
°°^ ' ~ 4 sin ^ ^ sin J P sin J (T" ~ ' 2 sin J ^ sin } P sin J G *• '
, cos ^ 4- cos P — COS G + 1 cos' ^ .4 -j- cos' ^ P — cos' ^ G ,„„.
cos (a — c) _ 4 cos J ^ cos J P sin J (7 ~ 2 cos J ^ cos } P sin J (7 *• ^
From the preceding we easily deduce
, 2 N sin e ,„g,
COB ^-|- cos P 4" cos C — 1 cos c — tan J ^ tan J P ^
. 2 JV" sin c ..,.
tan (s — c) -- ^^^ A+OOSB — COBG+1~ cot^AcosiB- cos e '' '
84. The equations (57 to 64) applied to the polar triangle, giv3,
2 cos i a cos i b cos } c ' -'
(oos^-C7) = 2iS^^lI^jr53Fj7 (66)
. „ sin J a sin J i cos G + cos J a cos J 5
•'"^=—^^ cosiT ^^ ^ (67)

ADDITIONAL rORMUL.ai. 165
sin /S = 1 -4- cos a -f cos a -f cos c _ cos' ^ a -f- cos' ib + cos' j e — 1 ,„„,
4 cos J a cos J 6 cos J c 2 cos J o cos J 6 cos J c ^ -'
sin (S—G) = "QS ^ a cos ^ & cos C+ sin ^ a sin ^ & .
cos J c
* '
. ,^ fn 1 — cos a — cos i -t- cos c sin' ^ a -f- sin' J 6 — sin' J c .
4 sin J a sin J 6 cos J c 2 sin ^ a sin J 6 cos J c *• '
_ , ^ 2re sin C
1 4- cos a -j- cos i 4- '"'s c cos £7 -{- cot J a cot J 6 ^ '
cot (S-G)= -, ^" , , = „ , f°f , , . (72)
"• ' 1 — cos a — cos 0 -j- cos c cos G + tan j a tan Jo ^ '
85. From (73) we find
, . „ 2 cos i a cos i 6 cos J c — cos' A a — cos' i i — cos' J c 4- 1 ,nn^
1 — sm 5" = ^ ? ^ , VT -,—? 2_a_ (78)
2 cos J a cos J 0 cos J c ^ '
, , . „ 2 cos i a cos J 6 cos i c-4-cos'i a 4-cos'i J 4-cos'ic — 1 ,_,,
1 -J- sm 5 = ^ ^ =^—^ VTT-—r-^—-^ (74)
' 2 cos J a cos J 0 cos J c ^ ^
the numerators of which may be reduced by PI. Trig. (173) and (174), by making
z = Ja, y = J6, z=Jc, whence u = i(a+b
&c.: therefore,
+ c) = is, v — i = J (s — a),
1 — Bin S = 2 sin J 8 sin J (s-a) sin ^ (s — b) ain j (s — c)
cos J a cos J 6 cos J c
1-1- M .5 — ^ °°^ iscoai(s — a) cos } (a — i) cos J (a— c) •
""" cos J a cos J i cos J c ^ ^
The product of these equations reproduces (65); their quotient is, by PI. Trig. (154),
tan' (45° — J S) = tan J s tan J (s — a) tan J (s — b) tan J (» — c) (77)
86. Gagnoli'a Equation.—Multiplying the first equation of (4) by cos A, we fii d
*• ^
COB a cos A =
COB 6 cos c cos ^4 -J- sin b sin c — sin b sin c sin' A
and from (5) in a similar manner,
cos a cos J. ^ — cos P cos C cos a -J- sin P sin 0 — sin P sin C sin' .i
Observing that by (3) we have sin 5 sin c sin' ^ = sin P sin C sin' a,
these two equations give,
sin i sin c 4- cos b cos c cos 4 = sin P sin G — cos P cos G cos a (78)
a relation between the six parts of the triangle, first given by CAGNOLI. It is a
property of this equation that either member is a function which has the same value in a
given spherical triangle and its polar triangle. Thus, if we distinguish the sides and
angles of the polar triangle by accents,.we have*
sin b sin c + cos b cos c cos A = sin b^ sin c' + cos i' cos c' cos A' (79)
* See Matbematioal Monthly, (Cambridge, Mass.,) vol. i. p. 282.

166 SPHERICAL TRIGONOMETRY.
S7. To dettvJ". the formulae of plane triangles from those of spherical triangles.
The analogy of many of the preceding formulas with those of plane triangles is
Bufficieutly obvious. We can, in fact, deduce the plane formulfe from those of this
chapter, by regarding the plane triangle as described upon a sphere whose radius is infinite,
the triangle being an infinitely small portion of the sp/iere. The quantities a, b and
c, must, in this case, express the absolute lengths of the sides ; and the angles which
a b c
they subtend at the center of the sphere, expressed in arc, will be —, —, —, r being
the radius of the sphere. When r is very large, —, —, —, are very small; md we
r r r
may express the values of sin —, cos —, &c. approximately, by one or two terms of
r r
their expansions in series, PI. Trig. (405) and (406), and if their values be substituted
in our spherical formulae, we shall obtain approximate relations between the
sides and angles of the triangle. If we then make r infinite we shall obtain exact
relations between the sides and angles of a plane triangle.
Thus we have
. a a a' . „ a' , „
Bin^ ^"'7 ---211^ +^°- "-TITf+^osin
P b b b'
sin — 2.3 r, .. 7-' ., 4- ^ &c. b — 2.3 r'
and making r infinite, we find the formula of PI. Trig.
In the same manner
sin^
sin P
a b e
COB cos — cos —
, r r r
cos A = . i . c / b b
sin — sm —
a
b
^l-~ + kc.-{l-^-^ + ~-kc.)
{~-2^+-'^-\~~-^+'^-)
'
Ve
ibc-^-kc.
2r'
and making r infinite, we have the formula of PI. Trig.
S« 4- c' — a'
cos A =
2 be
Formulse that involve only the sines or tangents of the sides may be reduced im.
mediately to the plane formulae by substituting a, b, kc, for sin a, tan a &o. Thus.
(31 to 34) give the corresponding formulae of PI. Trig, by omitting the symbol sin.
and (40), (41), by omitting the symbol tan. when these symbols are prefixed to sides

SOLUTION OF SPHERICAL RIGHT TRIANGLES. 167
CHAPTER II.
SOLUTION OF SPHERICAL RIGHT TRIANGLES.
38. WHEN one of the angles of a spherical triangle is a right
angle, the general formulae of the preceding chapter assume forms
that are remarkably analogous to the relations established for tho
solution of plane right triangles, and equally simple in their application.
39. Let (7=90°, Fig. 6. From (3) we
have
. . sina .
sin J1 = -. sm G
sine
but since Q= 90°, sin (7= 1; therefore,
sin a
and, in the same manner, sm A = —.—•
(801
. sm sin e 5
smB = sine J
that is, the sine of either oblique angle of a spherical right triangle
is equal to the quotient of the sine of the opposite side divided by the
sine of the hypotenuse. Compare PI. Trig. (1).
40. From (11), we find
sin5 cote—sinJLcotC
cos A — TTTT"—•
cos 0
but if C= 90°, cot C= 0; therefore,
or
. ,sin 6 cot e ^ , ^
cos A = i— = tan 6 cot e
tan J cos b tanrt
cos A = 7 cos B = 1—^ (ol)
tan e tan c ^ '

16&-
SPHERICAL TRIGONOMETRY.
that is, the cosine of either angle is equal to the tangent of the adjacent,
side, divided by the tangent of the hypotenuse. Compare
PI. Trig. (1).
41. From (10), we have,
sin 6 cot a — cos b cos Q
cot A =
: 7y
sm O
which, when O = 90°, becomes
1. A - I ^ sin 6
cot A = sm 6 cot a =
tana
or, taking the reciprocals,
tana .„ tan 6 ,_„,
tanJ.=-=—5; tz.nB = -. (82)
sm 0 sm a ^ '
that is, the tangent of either angle is equal to the tangent of the opposite
side, divided by the sine of the adjacent side. Compare
PI. Trig. (1).
42. From (5), we find,
. . cos 54-cos (7 cos J.
sin.A = J—;—7=
cos b sin G
and if (7= 90°,
cos 5 . _ cos J. ,„„,
sm A = r sm B = (83)
cos b cosa ^ '
that is, the cosine of either angle, divided by the cosine of its opposite
side, is equal to the sine of the other angle. In PI. Trig, we have
sin A — cos B.
43. From (4), we have,
cos e = cos a cos 6 4- sin a sin b cos G
or, when (7=90°,
cos e = COS a cos b (84)
that is, the cosine of the hypotenuse is equal to the product of the cosines
of the two sides. In Vl. Trig, e^ = a^ 4- b^.
44. From (5),
cos G-{- cosA cos B
cos c = -.—1—-.—^
am. A smB
or, when (7 =90°,
cos A cos B
cos c = -.—1—;—„- = cot J. cot 5 (8.5)
sm A sm B
^ '

SOLUTION OF SPHERICAL RIGHT TRIANGLES. 169
hat is, the cosine of the hypotenuse is equal to the product of the cotangents
of the two angles. In PL Trig., 1 = cot JL cot B.
45. No difficulty will be found in remembering the preceding formula
for spherical right triangles, if they are associated with the
corresponding ones for plane triangles: thus.
In plane right triangles.
sin J.=
cos J. =
tan^ =
a
G
b
G
a
T
smA = coa B,
sin-B
cos-B
tan 5
c^ = a^ + §2
e
a
c
b_
a
sin.B = cos^
1 = cot J.cot.B
In spherical righ^ triangles.
, sm a
sin^ = -^—
sin e
cos JL = tan b
tan c
tana
tan A = -:—r
sm 6
sin.A = cos-B
cos h
. -_ sm b
sm B = ^—
sm e
cos-B =
tana
tan e
tan.B =
tan 6
sin a
cos J.
sin 5 =
cos a
cos e = cos a cos b
cos e = cot J. cot-B
.46. Napier's Rules. By putting these ten equations under a difi'erent form, Napier
uontrived to express them all in two rules, which, though artificial, are very generally
employed as aids to the memory.
In these rules, the complements of the hypotenuse and of the two oblique angles
are employed instead of the hypotenuse and the angles themselves. The right angle
not entering into the formulae, they express the relations of five parts, but in the
rules the five parts considered are a, b, co. c, co. A and co. P. Any one of these
parts being called a middle part, the two immediately adjacent may be called adjacent
parts and the remaining two, opposite parts. The right angle not being considered,
the two sides including it are regarded as adjacent parts. The rules are :
I. The sine of the middle part is equal to the product of the tangents of the adjacent
parts.
II. The sine of the middle part is equal to the product of the cosines of the opposite parts.
The correctness of these rules will be sho-wn by taking each of the five parts as
middle part, and comparing the equations thus found with those already demonstrated.
1st. Let CO. c be the middle part; then co. A and co. B are the adjacent parts,
a and b the opposite parts, and the rules give
sin (co. c) = tan (co. A) tan (co. P)
sin (co. c) = coa a cos b
cos c = cot J1 cotP
cos c = COS a cosi
which are (85) and (84).
2d. Let CO. A be the middle part; then co. c and b are the adjacent parts, co. P
»nd a the opposite parts, and the rules give
an (co. A) = tan (oo. c) tan b or cos A = cot e tan b
sin (co. A) = cos (co. P) cos a
22 P
cos .4 = sin P cos a

170 SPHERICAL TRIGONOMETRY.
In tht same maimer, if co. B is taken as the middle part,
sin (oo. P) = tan (co. c) tan a or cos P = cot c tan a
sin (co. P) = cos (co. A) cos b
cos P = sin .4 oo? b
and these four equations are the same as (81) and (83).
3d. Lei a be the middle part; then co. B and b are the adjacent parts, co. A
and CO. c the opposite parts, and the rules give,
sin a =: tan (oo. P) tan b or sin a = cot P tan b
sin a = cos (co. A) cos (co. c)
sin a = sin .4 sin e
In the same manner, if b is taken as the middle part,
sin b = tan (co. A) tan a or sin J = cot ^ tan a
sin b = cos (co. P) cos (co. c)
sin 6 = sin P sin c
and these four equations are the same as (80) and (82).
It appears, therefore, that these rules include all the ten equations previously
proved; and they include no others, since we have taken each part successively as
the middle part.
In the application of these rules, it is unnecessary to use the notation co. A, co. P,
CO. c, since we may write down at once sin A for cos (co. A), kc*
47. In order to solve a spherical right triangle, two parts must
be given, and from the equations of Art. 45, that equation must be
selected which expresses the relation between these two parts and
the required part.
When Napier's Rules are employed, it is only necessary to determine which of tho
three parts—the two given and the one required—is to be taken as the middle part.
" These three parts are either all adjacent to each other, in which case the middle
one is taken as the middle part, and the other two are adjacent parts; or one is
separated from the other two, and then the part which stands by itself is the middle
part, and the other two are opposite parts."f
48. In order to distinguish the functions of parts less than 90°
from those greater than 90°, it will be necessary carefully to observe
their algebraic signs, according to PI. Trig. Art. 40. But when a
required part is determined by its sine, since the sine of an angle
and of its supplement are the same, there will be two angles, both
of which may be regarded as solutions, except when this ambiguity
is removed by either of the following principles.
* If we employ as the five parts, the hypotenuse, the two angles, and the complements
of the two sides including the right angle, these parts will be the complements
of those used in Napier's Rules, and we shall have
MAUDOTT'S RULES.—I. The cosine of the middle part is equal to the product of the cotangents
of the adjacent parts.
II. The cosine of the middle part is equal to the product of the sines of the opposite parts.
With a little attention at the commencement, however, and by observing the analogy
exhibited in Art. 45, the student will find that he will have little use for either
of these artificial rules.
f Peirce's Spherical Trigonometry.

SOLUTION OP SPHERICAL RIGHT TRIANGLES. 171
49. In a right spherical triangle, an angle and its opposite side
are always in the same quadrant, that is, either both less or both
greater than 90°. For, by (83),
cos B
sin A = cos b
m which, since sin A is always positive, {A < 180°), cos B and
cos b must have the same sign; that is, B and b must be either both
less or both greater than 90°.
50. When the two sides including the right angle are in the same
quadrant, the hypotenuse is less than 90°, and when the two sides
are in different quadrants, the hypotenuse is greater than 90°.
For, by (84),
cos e = cos a cos b
in which, if a and b are in the same quadrant, cos a and cos b have
like signs, and cos e is positive, that is, c < 90° ; but if a and h
are in difi'erent quadrants, cos a and cos b have difi'erent signs, and
cos e is negative, that is, c > 90°
We proceed now to the solution of the several cases.
51. CASE I. G-iven the hypotenuse and one angle, or e and A,
Fig. 6.
To find a. The relation among the three
Mg. 6.
parts, e. A, and a, (as in PL Trig, with the
same data), is given by the sine of A; and
by Art. 45,
from which we find*
sm.A =
sm c
sin a = sin e sin A (86;
There will be two values of a corresponding to the same sine, but,
by Art. 49, the true value is that which is in the same quadrant
as A.
To find b. The relation among the three parts, e, A, and b, (as
in PI. Trig, with the same data), is given by the cosine of-4,,or,
from whichf
cos-4 = tan 6
tan c
tan b = tan c cos A (87)
* This equation would be found by Napier's Rules, taking a as the middle part,
f We find the same result by Napier's Rules, taking co. A as the middle part

172 SPHERICAL TRIGONOMETRY.
To find B, We have, by (85),*
cos e = cot A cot B
from which cot B = . = cos c tan A (88')
cot A
^ '
The quadrants in which b and B are to be taken, will be determined
by means of the signs of tan b and cot B, according to PI.
Trig. Art. 40.
Gheck. To guard against numerical errors, it is often expedient
to compute the same quantity by two difi'erent and independent
methods. In many cases, however, -sve may test the accuracy of
several operations by a single formula, which may be called the
check. In the present instance, when the three parts, a, b, and B,
have been found, we should have, by (82), the relation
sin a = tan b cot B
so that if the work is correct, we shall find
log sin a = log tan b 4- log cot B
EXAMPLES.
1. Given e = 110° 46' 20", A = 80° 10' 30", to solve the triangle.
By (86). By (87). By (88).
e, log sin 9-9708106 log tan — 0-4210061 log cos — 9-5498045
A, log sin 9-9935833 log cos 4- 9-2320794 log tan 4- 0-7615038
log sin a 9-9643939 log tan b — 9-6530855 log cot B — 0l3113083
log tan b — 9-6530855
Gheck. log sin a 4-9^9643938
Ans. a = 67° 6'52"-7, b = 155°46'42"-7, B = 153° 58'24"-5
2. Given c = 120°, A = 120° ; solve the triangle.
Ans. a = 131° 24' 34"-7 b = 40° 53' 36"-2 B = 49° 6' 28 '-S
52-. If A = 90°, we must also have, by (85), c = 90°, and then
0 0
tan 0 = — tan P = -jr-
BO that b and P are both indeterminate; that is, there is an indefinite number of
triangles which satisfy the given values of e and A; but since
cos P = cos b sin A = cos b
we always have B z=h; and since
sin a = sin c sin ^ = 1
we have a = 90°, and all the parts of the triangle are equal to 90°, except b and P
H only c is .given = 90°, all the parts of the triangle are equal to 90°, except A
and a ; and we have Ax= a.
* Or by Napier's Rules, taking co. c as the middle part.

SOLUTION OF SPHERICAL RIGHT TRIANGLES. 173
53. CASE II. Given the hypotenuse and a side, or e and a.
To find A. We have by (80),
To find B. By (81),
• . sin a . _^
sm A = -. = cosec e sm a (o9}
sm c
^ '
^ tana ,.„,
cos B = = cot e tan a (90)
tan c
^ '
To find b. By (84),
cos c = COS a cos b
from which cos o = = cos c sec a (91)
cos a
^ '
Check.
We have between J., B, and 6, the relation
cos -B = sin ^ cos b
EXAMPLES.
1. Given e = 140°, a = 20° ; solve the triangle.
By (89). By (90). By (91).
c,log cosec 0-1919325 log cot - 0-0761865 log cos - 9-8842540
a, log sin 9-5340517 log tan 4- 9-5610659 log sec 4- 0-0270142
log sin A 9-7259842 log cos B - 9-6372524 log cos b - 9-9112682
log sin A 4- 9-7259842
Gheck. log cos B - 9-6372524
Ans. A= 32° 8'48"-l
.B = 115°42'23"-8
5 = 144°36'28"-4
2. Given e = 101° 16' 16".7, b = 115° 42' 38"-5 ; find A.
Ans. ^ = 65° 32'56".4
64. When a = c and consequently both = 90°, sin ^ = 1, A ^ 90°, and
cos P = -rr- cos J = -jT- COS P = COS J
go that B = b, but both are indeterminate as in Art. 52.
P2

174 SPHERICAL TRIGONOMETRY.
55. CASE III. Given one angle and its opposite side, or A and a.
We shall have
sin J. • sm a whence sin e = cosec A sin a (92)
tan .4 = tan a
sin 6
sin J = cot ^ tan a (93)
sin.B =
cos J.
cosa
sin J5 = cos J. sec a (94)
Glieck. sin b = sin c sin B
In this, case, there are always two solutions, all the required parts
being determined by their sines, and the ambiguity not being removed
by either Art. 49 or Art. 50. This also appears from Fig. 7.
If AB and AG\>e produced to meet in A!, ABA' and
AG A! are semicircumferences and A = A!; the triangles
ABG and A'BG both contain the given parts A and a,
but c', b' and B' are respectively the supplements of c,
b and B. It must not be inferred that in every case all
the required parts are less than 90° in one triangle, and
greater than 90° in the other; but the proper values for
each triangle must be selected by Arts. 49 and 50.
EXAMPLES.
1. Given A = 100°, a = 112°; solve the triangle.
Ans. e= 70°18'10"-2 ) ( e = 109°41'49".8
J = 154° 7'26"-5 V or < b= 25°52'33".5
B = 152° 22,' l"-3 J I B= 27°36'58".7
2. Given A = 80°, a = 68° ;
Ans. e= 70°18'10".2
6= 25°52'33".5
B= 27°36'58"-7 .
solve the triangle.
r e = 109°41'49"-8
>• or < 5=154° 7'26"-5
I .B = 152°23' 1".3
3. Given B = 150°, b = 160° ; solve the triangle.
Ans. e = 136°50'2.3"-3 ] r c= 43° 9'36".7
a= 39° 4'50"-7 > or < a = 140° 55' 9".3
A= 67° 9'42"-7 i I .A = 112° 50'17"-3

SOLUTION OF SPHERICAL RIGHT TRIANGLES. 175
56. CASE IV. Given one angle and its adjacent side, or A and b.
We shall find the required parts by the equations
Check,
cos B = sin^ cos b (95)
tan a = tan A sin b (96)
cot e = cos A cot b (97)
cos B = tan a cot e
EXAMPLES.
1. Oiven A = 80° 10' 30", b = 155° 46' 42".7; solve the triangle.
Ans. B = 158° 58'24:"-5
a= 67° 6'52"-6
c = 110°46'20"-0
2. Given B = 152° 23' 1".3, a = 112° 0' 0" ; solve the triangle.
Ans. A = 100°
5 = 154° 7'26"-5
c= 70°18'10"-2
57. CASE Y. Given the two sides, a and b.
We find the required parts by the equations
cos c = cos a cos 5 (98)
cot A = cot a sin b (99)
cotB = sina cot b
Check, cos e = cot J. cot B
0-^^)
Given a = 116°,
EXAMPLE.
b = 16°; solve the triangle.
Ans. e= 114° 55'20'-4
A= 97°39'24"-4
B= 17°4r39"-9
58. CASE VI. Given the two angles, A and B,
The required parts are found by the formulse
cos e = cot J. cot B
(1*^1)
cos a = cos ^ cosec B (102)
cos b = cosec A cos B
Check, cos e = cos a cos h
(}-^^)

176 SPHERICAL TRIGONOMETRY.
EXAMPLE.
Given J. = 60° 47' 24".3,^ = 57° 16' 20".2.: solve the triangle.
Ans. G = 68° 56' 28"-9
a = 54° 32' 32"-l
b = 51° 43' 36".l
ADDITIONAL FO-RMVLIE FOB THE SOLUTION OF SPAEEIOAL RIOHT TRIANGLES.
59. As in plane trigonometry, cases occur in which particular solutions of greater
accuracy than the ordinary ones are required. (PI. Trig. Art. 112.)
BO. From (89) we find
1 — sin A sin c — sin a
1 -}- sin ^ sin c 4- sin a
which by PI. Trig. (154) and (109) is reduced to
tan«(45°-J^)=*-"^iii:^! (104)
^ ' tan i (c + a)
which will give a more accurate result than (89), when A is nearly 90°.
61. From (91) we find
1 — cos b cos a — cos c
1 4- cos b cos a 4- cos c
or tan' J J = tan ^ (e + a) tan J (c — a) (105)
whi

QUADRANTAL AND ISOSCELES TRIANGLES. 177
of which (110) may be used when c is small, and (111) when c is nearly 180°, instead
of (101).
64. The equations (92), (98), and (94), of CASE III. give
tan' (45° - J c) = t^°H^-°) ^^^^^
^ ^ ^ tan i (.4 4-a) ^ '
tan' (45° - * 6) = ''",^'^7"^ (113)
^ ' sm (A + a) ^ '
tan' (45° — J P) = tan J (^ — a) tan J (.4 4- a) (114)
The roots of these equations having the double sign, we may take the angles
45° —• J e, etc. either with the positive or negative sign, whence the two solutions
of the problem, as in Art. 55.
65. Some of the solutions may be adapted for computation by the table of natural
sines. Thus from (86), (95), and (98),
sin a = .J [cos (c — A) — cos (c + A)J (115)
cosP= J [sin (6 4-^) —sin (5 —.4)] (116)
cos c = J [cos (a 4- *) 4- cos (a — 6)] (117)
66. The following relations are occasionally useful:
From (83) we have
cos a sin .4 cos ^ sin 2 A
cos b sin P cos P sin 2 B
(118)
From (80) and (83),
sinP sin A cos A sin 2 ^
sm c sin a cos a sin 2 a
(119)
From (80) and (84),
sin A sin a cos a sin 2 a
cos b sin c cos c sin 2 c
(120)
67. Yarious relations may be deduced from the general formulas of the preceding
chapter by making G = 90°. The following are easily obtained :
sin (e — a) = cos c tan b tan J P = cos a sin b tan J P
sin (c 4- a) = cos c tan b cot J P = cos a sin b cot J P
cos (c — a) = cos b + sin a sin b tan J B
cos (c+ a) = cos b •— sin a sin b cot J P
sin (a — b) = 2 sin c sin ^ (A + B) sin f (A —B)
sin (a 4" *) = 2 sin c cos J (^ -f- -^) '^"^ i (-^ — -^)
. „ cos i a cos i i „ sin i a sin i i
sm S = . — cos S" = —-,—^—
cos ^ C
COS ^ c
tan S = — cot J a cot J 5 (121)
QUADBANTAL AND ISOSCELES TBIANOLES.
68. The polar triangle of the right triangle is a quadrantal triangle, one side (the
side opposite the angle G) being equal to 90°. The solution of such triangles is as
simple as that of right triangles, the formuljB for the purpose being obtained from
the preceding, by the process of Art. 8. It is unnecessary to produce them here, as
quadrantal triangles are generally avoided in practice, and when unavoidable are
"-eadily solved by means of the polar triangle.
An isosceles triangle is easily solved by dividing it into two right triangles by a
oerpendicular from the angle included by the equal sides.
23

]7a SPHERICAL TRIGONOMETRY.
CHAPTER IIL
SOLUTION OF SPHERICAL OBLIQUE TRIANGLES.
69. In the solution of spherical oblique triangles, a required part
may sometimes be found by its sine, in which case there will be two
values of that part, answering to the conditions, unless the proper
value can be determined by other considerations. In certain cases,
the true value can be selected by applying one or more of the folio-wing
principles, some of which are demonstrated in geometry. We
still consider only those triangles each of whose parts is less than 180°.
I. The greater side is opposite the greater angle, and conversely.
II. Bach side is less than the sum of the other two.
III. The sum of the sides is less than 360°.
IV. The sum of the angles is greater than 180°.
V. Hach angle is greater than the difference between 180° and
the sum of the other two angles.
For, by IV., A + B+ G> 180°
whence,
A > 180° — {B-\-G)
But if -B 4- G> 180°, we have, in the polar
triangle, A'B'G', Fig. 8, by II.,
a' < 6' 4- e'
180°-J. < 180°-5 4-180°- 0
-A< 180° - (^ 4- {B+ (7)-180°
VI. A side which differs more from 90° than another side, is in
the same qvadrdtnt as its opposite angle.
For, by (4), we have
. cos a — cos b cos e
cos A = ;—,—;
sm 6 sm e
In which the denominator is always positive.
If, then, a differs

SOLUTION OF SPHERICAL OBLIQUE TRIANGLES. 179
more frcm 90° than b or than c, we have, (neglecting the signs for
a moment),
and still more
cos a > cos 5 or >• cos e
cos a > cos J cos e
Hence cos a being numerically greater than cos b cos e, the sign of
the whole numerator, and therefore the sign of cos A, is the same
as that of cos a; that is, A and a are in the same quadrant.
VII. An angle which differs more from 90° than another angle,
is in the same quadrant as its opposite side. For, by (5),
cos a-
cos A -f cos B cos G
sin B sin G
in which, if A difiers more from 90° than B, or than G, cos A determines
the sign of the -whole fraction, and therefore the sign of cos a.
VIIL In every spherical triangle there are at least two sides which
are in the same quadrants as their opposite angles respectively. This
follows from VI. and VII.
IX. The sum of two sides is greater than, equal to, or less than,
180°, according as the sum of the two opposite angles is greater than,
equal to, or less than, 180° In other words, the half sum of two
sides is in the same quadrant as the half sum of the opposite angles.
For, by (41),
tan J (a 4- b) cos J (J. 4- B) = tz.nl c coal {A — B)
the second member of which is always positive, so that tan J (a -"- b)
and cos \[A -\- B) must have the same sign.
70. CASE I. Given tivo sides and the ineluded
angle, or b, c and A. (Fig. 9.)
First Solution; when the third side and
one of the remaining angles are required.
To find a. The relation between the given ^^^^ ^-^ B
parts b, 0, A and the required part a is expressed
by the first equation of (4),
cos a = cos c cos S 4- sin e sin b cos A
by which a may be found by computing separately the two terms of
the second member and adding |;heir values to form the natural cosine
of a ; but we should thus be required to use, besides the table
of log. sines, also the table of logarithms of numbers, and the table
of natural sines and cosines. To adapt it for logarithmic computation
by the table of log. sines exclusively, we employ the prir^ess of
(M)

IgQ
SPHERICAL TRIGONOMETRY.
PL Trig., Arts. 174, 175. Thus, let ^ be a number and $ an auxiliary
angle such that
A sin (B = sin & cos .A )
then (M) becomes
J , r ^'"^
^ cos

SOLUTION OF SPHERICAL OBLIQUE TRIANGLES. 181
In the use of these formulse, as indeed of all that follow, tho
sighs of all the functions must be carefully observed, according to
PI. Trig. Arts. 37 and 40.
We may take (p between 0 and 180°, less or greater than 90°,
according as the sign of its tangent is positive or negative; or we
may take it numerically less than 90° in all cases, but positive or
negative according to the sign of its tangent, (PL Trig. Arts. 37
and 174).
Gheck. The quotient of (ra) divided by (m*) is
cot B tan (c — $)
cos a sin b sin A
which multiplied by the following, from (3),
sin a sin ^ = sin b sin A
gives tan a cos B = tan (c —

182 SPHERICAL TRIGONOMETRY.
EXAMPLES.
1. Given 5 = 120° 30'30", c = 70°20'20", .4 = 60° 10'10";
find a and B.
By (122).
b = 120° 30' 30" log tan 5 - 0-2297071
A = .50° 10' 10" log cos A 4- 9-8065322
(p = 132° 36' 44"-2* .log tan $ - 0-0362393
c= 70°20'20"-0
c_cp=_ 62°16'24".2
By (122). By (123). By (124).
logcos(c—?)-}- 9-6676893 log sin (c — j) —9-9470304 logtan (c — ?) —0-2793410
w 00 log COST-? — 04693898 ar co log sin p 4-0-1331505 log tan a 4-0-4291648
log cos 6 —9-7055761 log cot ^-f 9-9212038 log cosP—9-8501762
log cos a 4:9-5426552 log cot P—0-0013847 Gheck. —0-2793410
a = 69°34'55"-9
P = 135°5'28"-8
2. Given 5 = 120° 30'30", e = 70° 20'20", ^ = 50° 10'10";
find a and G.
Ans. a = 69° 34' 55"-9
(7= 50° 30' 8"4
3. Given 5 = 99° 40'48", c = 100° 49'30", ^ = 65° 33'10";
find a and B,
Ans. a = 64° 23' 15".0
5 = 95° 38' 4".0
4. Given b = 99° 40'48", e = 100° 49' 30", A = 65° 33' 10";
find a and G.
Ans. a = 64° 23' 15".0
(7=97°26'29".l
5. Given b = 98° 2' 20", e = 80° 35' 40", A = 10° 16' 30";
find a and G.
Ans. a = 20° 13' 30"-l
(7=30°35'56"-7
72. If B, (7 and a were all required, we might find a and (7 by
(125), and then B by Art. 3, which gives
sin a: sin t = sin A : sin B
. „ sin b sin A
or sm xj = :
sm a
*We may also take? = —47° 2.3' ]5"-8, whence c —j = 117° 43' 35- 8, -which
will give the same values of a and B as found in the text.

SOLUTION OF SPHERICAL OBLIQUE TRIANGLES. 183
Of the-two values of B less than 180° given by this formula, the
proper one may generally be selected by the principles of Art. 69.
There are cases, however, in which all the conditions there given are
satisfied by both values of -B,* and on this account it is preferable,
in general, to combine (123) and (125), or to employ the following
solution, when the three unknown parts are all to be found.
73. CASE I. Given b, e and A. Second Solution ; when the two
remaining angles are required, or when the three unknown parts are
all required.
We have, by Napier's Analogies, (42) and (43),
whence
sin J (6 4- e) : sin |(6 — e) = cot J J.: tan ^{B— G)
cos J (5 4- e): cos J (5 — c) = cot J J.: tan J (-B 4- (7)
*^^H^-^)=:-i^ff^]cotjA (126)
*^^H^+^) = ^^^f^jcotJ^ (127)
which determine ^{B— G) and l{B-\- G); then the half difi'erence
added to the half sum gives the greater angle, and the half difi'erence
subtracted from the half sum gives the less angle.
If c > &, we may write c — b, G— B,in the place oi b — c, B — G.
We may now find a by either of Napier's Analogies, (40), (41),
which givef
sini(5 4-C)
*^^ * "^ = sinl {B-G) *^° * (5 - c^) (128)
cos IIB+G)
^^'^ * "^ = cosl{B-G) *'''' ii^ + ") (129)
* By Art. 69, "VL, if b difi-ers more from 90° than c, B is in the same quadrant
aa b, and all ambiguity is removed. If e differs more from 90° than b, we may fina
a and B by (122) and (123), and then G by the formula
. _ sin c sin A
sm G = : •
sm a
G being taken in the same quadrant as c:
f We may also find a from any one of Gauss's Equations (44), which become, vr,.
the present ease,
cos J a sin'J [B + G) = cos J A cos J (i — c)
cos J a cos J (P 4- G) = sin ^ A cos i (b + c)
sin J a sin J (P —• G) = cos ^ A sin ^ (b — c)
sin J a cos J (P — d) = sin ^ A ain ^ (b + c)

184 SPHERICAL TRIGONOMETRY.
1. Given b =
find B, 0 and a
By (126).
irCO log sin 1(6 4- c) 4- 0-0019487
logsini(6-c)4-9-6273228
log cot i-^ 4-0-3296529
EXAMPLES.
120° 30' 30", c = 70° 20' 20", A
We have
1 (J 4- c) = 95° 25' 25"
1 (6 - c) = 25° 5' 5"
ij. = 25° 5' 5"
log tan \{B-G) 4- 9-9589244
\(B-G)= 42° 17' 40"-2
^=135° 5'28"-8
50° 10' 10";
By (127).
ar cologcosi(5 + e) - 1-0244829
logcosJ(6-e) 4-9-9569757
log cot J J. 4-0-3296529
log tan \{B^C) -173"llill5
1 (^ 4- (7) = 92° 47' 48"-6
(7= 50° 30' 8".4
By (128).
By (129).
arCO log sini(.B- (7)4-0-1720227 ar colog cos \{B- (7)4-0-1309469
log sin i(5-hC)4-9-9994824 log C0SK.B4- Q)- 8-6883709
logtani(5-c)-F9-6703471 log tan i(J 4-c)-1-0225342
log tan J a 4-9-8418522
log tan la 4-9-8418520
1 a = 84° 47' 28"-0
2. Given b = 99° 40' 48", e
find B, 0 and a.
Ans. B = 135° 5' 28"-8
0 = 60° 30' 8".4
a = 69° 34' 56"-0
100° 49' 30", A = 65° 33' 10";
Ans. B = 95° 38' 4"-0
G = 97° 26' 29"-l
a = 64° 23' 15"-1
74. It may be remarked with regard to (128) and (129) that, when b and e (and
consequently P and G) are nearly equal, a small error in the previous determination
of the small angle J (P — G) may produce alarge one in log sin \(B — 0), and
consequently in log tan J a found by (128). In that case, therefore, (129) must be preferred.
In like manner, if J(J -|- c), and consequently ^(B + G), are nearly equal to
90°, (129) -will become inaccurate, and then (128) is to be preferred.
Formula (128) would fail entirely if P = C, and formula (129) would fail if
J (P 4- C) = 90°, since the second members in these cases would assume the indeterminate
form —.
75. CASE I. Given b, c and A. Third Solution. Wlien the
third side is alone required, the computation by(122)is inmost cases
as convenient as any other; but theie are various other methods

SOLUTION OF SPHERICAL OBLIQUE TRIANGLES. 185
derived from the formulse of the preceding chapter, which hav» been
employed with advantage in particular applications. Among the
most convenient are the following, from (12) and (13):
cos a = cos (6 — e) — 2 sin b sin e sin^ J A (130)
cos a = cos (5 -f e) 4- 2 sin b sin c cos^ J A (131)
The computation of these requires the use of natural cosines and
numbers, the signs of which must be carefully observed.
EXAMPLE.
Given b = 99° 40' 48", c = 100° 49' 30", A = 65° 33' 10";
find a.
By (130).*
IA= 32° 46' 35" log sin'-1 J. = 2 log sin J A 9-4669752
6 _ c = - 1° 8'42" log si'n c 9-9922023
log sin b 9-9937722
log 2 0-3010300
- 2 sin J sin e sin^ i J. = - 0-5675181 log 9-7539797
nat cos (6 — c) = 4- 0-9998003
nat cos a = 4- 0-4322822 a = 64° 23' 15"
By (131).
1 ^ = 32° 46' 35" log cos^ J ^ = 2 log cos i J. 9-8493748
6 4- e = 200° 30' 18" log sin c 9-9922023
log sin 6 9-9937722
log 2 0-3010300
•f 2 sin b sin c cos^ J JL = 4- 1-3689240 log 0-1363793
nat cos (5 4- c) = — 0-9366416
nat cos a = 4- 0-4322824 a = 64° 23' 15"
76. In Art. 14, we have deduced several formulse by which J a may be computed
VVe may adapt (17) and (18) for logarithmic computation, as follows :
sin' 0 = sin S sin c cos' ^ A 1
sin' J a = sin' J (i -f- c) — sin' 0 I (132)
= sin [J (b+c) + f\ sin [J (J + c) _ 0] J
sin' (6 = sin i sin c sin' ^A ]
COB' J a = cos' J (4 — c) — sin' «i I (133)
= cos [J (b — c)+ 0] cos [J (b — c) — 0] J
of wHch (132) is to be preferred when J a < 45°, and (133) when J a > 45°.
* The computation of (130) is facilitated by the use of a special table (given in
many treatises on navigation), from which, with the arguments is taken the logarithm
of 2 sin' J -4 = versin A. [PI. Trig. (4) and (139)].
•2i P 2

186 SPHERICAL TRIGONOMETRY.
Fig. 9.
^B
77. CASK II. Given two angles and the
included side, or A, 0 and b. (Fig. 9).
Pirst Solution ; when the third angle and
one of the remaining sides are required.
To find B. The relation between A, G, b
and B, is, by (5),
cos B = — cos (7 cos A -f sin Gain A cos b
(M)
which is adapted for logarithms by the method employed in the preceding
case. Thus, let
then (M) becomes
A sin 9- = cos A )
AcosS- = sin^ cos 5
coaB = h (sin Ccos9- — cos CsinB)
= h sin ((7-9-)
,. . . , cos A
or, eliminating h = -:—^, the formulse for finding B are
(m)
(m')
cot 9" = tan A cos b
cos
B= sin {G — B-) coa A
sinS-
To find a. From the third equation of (10), we find,
sin OcotA + cos (7 cos 5
cot a =
sin 6
sin GcosA -{- cos Gain A cos 6
sin A sin b
which, by (m), becomes
cot a
_ ACQS ((7—9-)
sin A sin b
(134)
(«)
«^ T • J.- I sin J. COS 5
or, ehmmatmg h = ^^^ ^ , we have, for finding a,
cot 9- = tan A cos J
cot a
cos ((7— 9-) cot 5
cos 9-
(135)
As in the preceding case, we may either take 9- always between 0
and 180°, less or greater than 90° according as its tangent is posi-

SOLUTION OF SPHERICAL OBLIQUE TRIANGLES. 187
tive or negative; or we may take 9- numerically less than 90° in ail
cases, positive or negative, according to the sign of its tangent.
(PL Trig. Art. 174.)
Gheck. The quotient of (w) by (w') is
which, multiplied by
gives
cot a _ cot ((7—9-)
cos B sin A sin b
sin JB sin a = sin J. sin b
tan -B cos a = cot ((7 — 9-) (136)
by which the values of B and a, found by (134) and (135), may be
verified.
78. If B and e were required, the solution would be similar, only
interchanging a and e, A and G. By the fundamental formulae, we
should have,
cos B = — cos A cos G -\- ainA sin G cos b
sin A cos G -{- cos ^4 sin (7 cos b
cot c = •
sin G sin b
and denoting the auxiliary angle by ^, the logarithmic solution
would be
cot ^= tan Ccos b
sin (A — ^) cos C
cos B =
sin^
cos {A — ^) cot b
cot e =
cos ^
Gheck. tan JB COS C = cot (J. — ^)
EXAMPLES.
(0)
(137^
1. Given A = 135° 5'28"-8, C= 50° 30'8"-4, 6 = 69° 34'55".9,
find .B and a.
By (134).
J.= 135° 5'28"-8 log tanJ.-9-9986154
b = 69° 34' 55"-9 log cos b 4- 9-5426553
9- = 109° 10' 31"-0* log cot 9- -'9-5412707
0 = 50° 30' 8"-4
G-B- = — 58° 40'22'-6
» -We may also take3- = — 70° 49' 29"-0, whence C—9- = 121° 19' 37"-l, which
will evidently give the same results as those obtained in the text.

188 SPHERICAL TRIGONOMETRY.
By (134). By (135). By (136).
logsin(C—3-) —9 9315664 logcos(C—3-)4-9-7159386 logcot(C'—S-) —9-7843722
arcologsin9--f-0-0247897 ar co log cosa—0-4835187 log tan P4-0-0787962
logcosJ:—9-8501762 log cot J 4-9-5708352 log cos a—9-7055757
log cos P4-9-8065323 log cot a—9-7702925* Glieck. —9-7843719
P= 50°10'10"-0 a=120°30'29"-9
Ans.B= 50°10'10"-0
a=120°30'29"-9
2. Given J. = 135° 5'28"-8, C=50°30'8"-4, 5 = 69°34'55"-9;
find B and c.
Ans.B= 50°10'10"-0
c = 70°20'20"-0
3. Given A = 65° 33' 10", C = 95° 38' 4", b = 100° 49' 30";
find 5 and a.
Ans.B= 97° 26'29"
a = 64° 23'15"
4. Given J. = 97° 26'29", C = 95°38'4", 5 = 64° 23'15";
find B and a.
Ans.B= 65° 33'10"
a = 100° 49' 30'
79. If a, e and B were all required, we might find B and c by
(137), and then a by Art. 3, which gives,
Bin B: sin. J. = sin 6 : sin a
sin ^ sin 5
^„„,
sm a = ^^— (138)
sm i? ^ '
Of the two values of a given by this equation, the proper one is to
be selected, if possible, by the principles of Art. 69.* But as cases
occur in which all the conditions there given are satisfied by both
values of a, it is preferable, in general, to combine (135) and (137),
or to employ the following solution when the three unknown parts
are all to be found.
-*• By Art. 69, -VII., when A differs more from 90° than G, a must be taken in the
same quadrant with A, and all ambiguity is removed. If, then, by A we always
denote that angle which differs more from 90° than the other given angle, we may
always solve this case by means of (187) and (138), without meeting with any difficulty
in determining the quadrant in which c is to be taken.

SOLUTION OF SPHERICAL OBLIQUE TRIANGLES. 189
80. CASE II. Given J,, C and 5. Seco7id Solution ; when the two
remaining sides, or when the three unknown parts are all required.
We have, by Napier's Analogies, (40) and (41),
sin 1 (J. 4- C): sin i (.4 - C) = tan i 5 : tan i {a - c)
.cos 1 (J. 4- C): cos J (J. - C) = tan J 5 : tan J (a 4- c)
whence
tani(a-e) = ^!^iC^^)
sin 1{A+ G) tan I b
1 / , cos i (J. — C)
tani(a4-e)= cos 1 (J. 4- C) tani5
>(139)
which determine J (a — e) and J (a -f c); then the half difi'erence
added to the half sum gives the greater side, and the half difi'erence
subtracted from the half sum gives the less side. If C> -4, we may
-write C — A, e — a in the place of ^ — G, a — e.
We may now find B by either of Napier's Analogies, (42) and
(43), which give*
sin i (a 4- c) , , .
coti-B= • . ^ ^ ^ . tan 1 (^ - C) (140)
sin J (a — e)
cot J ^ = — ^('*'^'')tani(-^+C) -
(141)
cos J (a — c)
EXAMPLES.
1. Given A = : 135° 5'28"-6, C= 50° 30' 8"-6, b •• 69° 34' 56"-2;
find a, G and B.
We have 1{A+ C) = 92°17'48"-6
i(^-C) = 42°17'40"-0
Then, by (139),
i5 = 34°47'28"-l
ar colog sinl (J.+ C)4- 0-0005176 ar co log cosi(J.4- C)-l-3116286
log sinl (J.- (7)4- 9-8279768 log cosi(^- C)4- 9-8690535
log tan I b 4-9-8418527 log tan i b -f 9-8418527
iogtanJ(a — e) 4-9-6703471
i(a-c)= 25° 5' 5"-0
a = 120°30'30"-0
logtani(a 4-e)-1-0225348
J(a4-e) = 95°25'25"-0
e = 70°20'20"-0
* We may also find P by any one of Gauss's Equations, (44), interchanging B
iTiiI C. // and c.

190 SPHERICAL TRIGONOMETRY.
By (140). By (141).
arCO logsini(a -e) 4- 0-3726772 arcologcosi( a - c) 4- 0-0430243
logsini(a 4-c) 4-9-9980523 log cosi( a 4-c)-8-9755171
logtani(J.-C)4- 9-9589234 logtani(^4-C) -1-3111110
logcotJ5 4-0-32C6529 *log cot i .B 4-0-3296524
i.B = 25°,V5"-0
Ans. a = 120° 30' 30"
e= 70° 20'20"
B = 50° 10' 10"
2. Given A = 95° 38'4", C = 97° 26' 29", b = 64° 23' 15";
find a, G and B.
Ans. a= 99° 40'48"
e = 100° 49' 30"
B= 65° 33'10"
81. CASE II. Given A, G and b. Third Solution. When the
third angle B is zlone reqwi'^ed, the computation by (134) is'in most
cases as convenient as «ny -"ther, but there are other methods (corresponding
to those given ifi Art. 75 for finding a) which may occasionally
be serviceable. By jl4) and (15) we have
coaB = — cos (A 4- C) - 2 sin A sin Csin^ J b (142)
cos .B = - cos (J. - ,C) 4-' 2 sin A sin C cos^ i b (143)
the computation of which is similar to that of (130) and (131).
EXAMPLE.
Given A = 95° 38' 4", C = 97° 26' 29", b = 64° 23' 15";
find-B.
By (142).
ij= 32°ir37"-5 log sin^ 1 5 = 2 log sin J 5 9-4531022
A4-C = 193° 4'33" log sin J. 9-9978967
log sin C 9-9963268
log 2 0-3010300
- 2 sin J. sin Csin^ J 5 = - 0-5602162 log 9-7483557
— nat cos (J. 4- C) = 4- 0-9740715
nat cos B = 4- 0-4138553 B = 65° 33' 9"-9
-* For the reasons given in Art. 74, (141) is, in this example, not so accurate
fls (140).

SOLUTION OF SPHERICAL OBLIQUE TRIANGLES. 191
82 In Art. 14, several formulse are given, by which J P may be computed By
(21) and (22) we have
sin' J P = cos' i (A — G) — sin A sin G cos' J b
cos' J P = sin' J {A + G) — sin A sin G sin' J b
which may be adapted for logarithms, thus:
sm"
sin' i^ P : cos' i (A— C) — sin'

192 SPHERICAL TRIGONOMETRY.
The auxiliary cp will be fully determined by (m), being taken between
0 and 180°, and always positive (PI. Trig. Art. 174); but, as
the cosine of an angle is also the cosine of the negative of that
angle [PL Trig. (56)], we may take 0' in (m') either with the positive
or the negative sign, so that c = cp ± cp'- There will thus be
two values of e answering to the same data, both of which will be admissible,
except when

SOLUTION OF SPHERICAL OBLIQUE TRIANGLES. 193
cot 9- = tan A cos b \
cos 9-' = cos 9- tan 6 cot a |- (14T)
C = 9- ± 9-'
J
To find B. We have several methods : Ist, directly by (3),
_ sin A sin b
sm B = -. (118)
sin a
^
which gives two values of B, supplements of each other, corresponding
respectively to the two values of c and C. We shall presently
see how to determine which are the corresponding values of e,
C and B.
2d. In (123), cp has the same value as in (146), and therefore putting
in (123), e — cp = cp', we have
.,^ sin Cp' cot^
cot B = r— (149)
sm (p ^ '
which gives two values of B by the positive and negative values
of cp'.
3d. By (124),
cos B = tan

194 SPHERICAL TRIGONOMETRY.
nse of (149), (150), (151) and (152), it is only necessary carefully to
observe the signs of the several terms.
Checks. Of the various formulae above given for finding B, one or
more may be employed for the purpose of verification. When c and 0
have been found, the most simple check is the following, from (3),
^ = ^ (153)
sm c sm-a ^ -'
which, indeed, might have been employed to find C, after e was
found, and reciprocally, but for the ambiguity attaching to the sines.
85. According to Art. 69, VI., if b differs more from 90° than a,
B must be in the same quadrant as b, and, since but one of the twc
values of B can satisfy this condition, there will be but one solution.
In that case c and C will each be found to have but one admissible
value.
86. The problem will be altogether impossible, when a differs more
from 90° than b, and is yet not in the same quadrant with A. In
such case, we should find that cp 4-cp'> 180°, and $ —cp' 1.
EXAMPLES.
1. Given a =40° 16', J = 47° 44', J. = 52° 30'; find .B.
By (148).
a= 40° 16' arCO log sin a 0-1895350
b = 47° 44' log sin b 9-8692449
4= 52° 30' log sin A 9-8994667
B= 65° 16'35" log sin .B 9-9582466
or B = 114° 43' 25"
2. With the same data, find c and B.
By (146.)
a= 40° 16' logcosa4-9.8825499
}== 470 44' ]og tan b 4-0-0414996 ar co log cos 64-0-1722647
4== ,52° 30' log cos J.4-9-7844471
(p= 33°48'51"-4 logtancp4-9-8259467 log cos cp4-9-9195201
(p'=±19° 30' 29"-0
e,= 53°19'20"-4
".= 14°18'22"-4
log cos (p'4-9'9743250

SOLUTION OF SPHERICAL OBLIQUE TRIANGLES., 195
By (149). Gheck. (150).

19« SPHERICAL TRIGONOMETRY.
5. Given a = 99° 40' 48", b = 64° 23' 15", A = 95^ 38' i"; find
e, C and B.
Ans. c = 100° 49' 30"
C= 97° 26'29"
B= 65° 33'10"
6. Given a = 40° 5' 25"-6, b = 118° 22' 7"-3, A = 29° 42' 33"-8 ;
find e, C and B.
Ans, e=153°38'42"-4 ) T c = 90° 5'41"-0-
C = 160° r24"-4 > a C= 50°18'55"-2
B= 42°37'17"-5 ) .B = 137°22'42"-5
7. Given a = 69° 34' 56", b = 120° 30' 30", A = 50° 10' 10";
find c and C.
Ans. c = 70° 20' 20'
(7= 50° 30' 8"-4
8. Given a = 120° 30' 30", b = 69° 34' 56", A = 50° 10' 10";
find c and C.
^ws. Impossible.
9. Given a = 40°, b = 60°, ^ = 50° ; solve the triangle.
Ans. Impossible.
87. CASE III. Given a, b and A. Second Solution. We find
B by the formula
sin A sin b
ainB =. sm a
and then by Napier's Analogies, (41) and (43),
or by (40) and (42),
^ cos i (^ 4- -B) ^ 1/ , .A
^'^^^^ =cosH^--g) ^"^^''"^^^
cot : *''-Sll^'""i(^ cos J (a — J) + ^)
. J sin J (J. 4- -B)
tant e = -:;—f4--j j^; tan^ (a — b)
^ sml{A — B) ^ ^ ''
(154]
• (155)
m which we employ successively the two values of B, and obtain
two solutions, except when for one of these values the second members
become negative, for J e and J C being less than 90°, their
tangents must be positive.

SOLUTION OF SPHERICAL OBLIQUE TRIANGLES. 197
We leave it to the student to a,pply these formulse to the preceding
examples.
88. To determine by inspection of the data a, b and A, whether there are two solutions,
or but one.
1st. It has already been seen. Art. 85, that when b differs more from 90° than a,
B must be in the same quadrant as b, and there can be but one solution. It remains
to show,
2d. That when a differs more from 90° than b, there will necessarily be two solu
tions. We have, by the first of (4),
cos a — cos b cos c
Bin c = :—
sin b cos A
Two solutions exist so long as both values of c are positive, and less than 180°, that
is, 80 long as sin c is positive. Now when a differs more from 90° than b, we have,
(neglecting the signs for a moment),
cos a > cos b > cos b cos c
therefore the numerator of the above value of sin c has the sign of cos a. But by
Art. 69, \1., a and A are in the same quadrant, and cos a and cos A have the same
sign; consequently also, the numerator and denominator have the same sign, and
the value of the fraction, or of sin c, is positive, as was to be proved.-^
Hence, there is but one solution when the side opposite the given angle differs less from 90°
than the other given side, and two solutions when the side opposite the given angle differs
more from 90° than the other given side.
89. CASE IV. Given two angles and a side opposite one of them,
or A, B and b. (Fig. 9).
First Solution, in which each required Hg.g. a
part is deduced directly from the fundamental
formulse.
To find c. We have, by (10),
sin e cot b — cos e cos ^ = sin ^ cot B
or multiplying by sin b,
sin e cos 6 — cos e sin 5 cos J. = sin A cot B sin b (M)
to solve which we take
fc sin cp = sin b cos A I , s
r ^^
k cos cp = cos 0
I
* The same proposition may be otherwise proved thus. By the equations (m)
»nd (m') Art. 84, we have
cos b COS a , sin b .
Cos^ = - ^ COS0'=-^ A = -^cos^
from the third of which we see that k has the sign of cos ^ ; if then a differs more
from 90° than b, that is, if cos a and cos A have the same sign, cos (p' is positive,
and 0' < 90°. Also since, (neglecting signs), cos a > cos b, we have cos '/ > cos f,
or 'p' differs more from 90° than and 0
and ip+ Ip' < 180°, or both values oi c are between 0 and 180°.
ii2

198 SPHERICAL TRIGONOMETRY,
then, putting e —

SOLUTION OF SPHERICAL OBLIQUE TRIANGLES. 199
To find a. We have several methods : 1st, .directly by (3),
which gives
sin b sin A
2d. By (146), where cp and ip' have the same values as in this case,
3d. By (150),
cos

200 SPHERICAL TRIGONOMETRY,
sin C
sin c
sin B
sin b
(163)
which might have been employed for finding C after c was found, or
reciprocally, but for the ambiguity attaching to the sines.
90. According to Art. 69, VII., if A differs more from 90° than
B, a must be in the same quadrant with A. But since the two
values of a are supplements of each other, only one of them can
satisfy this condition, and there will then be but one solution. In
such case c and C will each be found to have but one admissiblo
value.
91. The problem will be impossible when B differs more from 90°
than A, and yet is not in the same quadrant with b. In such case
we should find both values of c (and both values of C) to be greater
than 180°, or both negative.
The problem will also be impossible when sin 6 sin ^ > sin B,
since by (158) we shall then have sin a > 1.
EXAMPLES.
1. Given A = 132° 16', B = 139° 44', b = 127° 30'; find a.
• By (158).
B = 139° 44' 0" ar co log sin B 0-1895350
A = 132° 16' 0" log sin A 9-8692449
b = 127° 30' 0" log sin b 9-8994667
a = 65° 16' 35"-l log sin a 9-9582466
or a = 114° 43' 24"-9
2. With the same data, find C and a.
By (157).
J5= 139° 44' 0" log cos £-9.8825499
A= 132° 16' 0"logtanJ.-0-0414996arcologcosJ.-0-1722.547
J= 127°30• 0" logcos 5-9-7844471
9-= 56° 11' 8"-6 log cot9-4-9-8259467 log sin 9-4-9-9195201
9-'^=4- 70°29'31"0 1
'
&''=4-109°30'29"-0 I ' • log sin 9-'4-9.9743250
C^= 126°40'39"-6
t7= 165°41'37"-6

SOLUTION OF SPHERICAL OBLIQUE TRIANGLES. 201
By (161). Cheek. (162).
9- = 56° 11' 8"-6 arCO log cos 9-4-0-2545328 log cot 5-0-0720848
tCim° f o m ] ^°S cos&'±9-5236676 log cot y±9-5493427
6 ^=127° 30' 0" log cot 5-9-8849805 rp9.6214275
a' =^65° 16'34"!9 }
^°^ °°* «4=9-6631809 log cos «.F9-6214275
^ws. C=126°40'39"-6 "I T C=165°41'-37"-6
a = 114°43'25"-l j "'^ I a= 65°16'34"-9
3. Given A = 110°, B = 60°, 6 = 50° ; find c and a.
By (156).
B =
A =
b =
60°
110°
50°
logcosJ--9.5340517
logtan6 4-0-0761865
log cot5 4-9-9614394
logtan.4-0-4389341
cp = 157°49'26".4 logtancp-9-6102382 log sin cp 4-9-5768627

202 SPHERICAL TRIGONOMETRY.
J. Given A = 70°, B = 120°, 5 = 80° ; solve the triangle.
Ans. Impossible.
8. Given A = 60°, B = 40°, b=50°; solve the triangle.
Ans. Impossible.
92. CASE IV. Given A, B and b. Second Solution. We find a
by the formula
sin b sin A
sm a ••
sinB
and then by Napier's Analogies we find e and C, precisely as in
Case III., Art. 87, employing successively, in (154) or (155), the
two values of a given by the preceding equation. There will be but
one solution, if one of these values renders the second members of
(154) or (155) negative.
The student should apply this method to the preceding examples.
93. To determine by inspection of the data A, B and b, whether there are two solutiont
or but one.
1st. It has already been seen. Art. 90, that when A differs more from 90° than P,
a must be in the same quadrant with A, and there can be but one solution. It
remains to show that,
2d. When P differs more from 90° than A, there will necessarily be two solutions
We have, by (5),
. „ cos P 4-cos J4 COS C
sm C = +-: J
sm A COS 0
Two solutions exist so long as both values of G are less than 180°, and both positive,
that is,BO long as sin G is positive. Now when P differs more from 90° than A, we
have, (neglecting signs for a moment),
COB P > COB A > COB A COB O
therefore the numerator of the value of sin G has the sign of cos P. But by
Art. 69, -YIL, P and b are in the same quadrant, consequently the numerator and
denominator have the same sign, and the value of the fraction, or of sin C is always
positive, as was to be proved.*
Hence, there is but one solution when the angle opposite the given aide differs less from
Q0° than the other given angle; and two solutions when the angle opposite the given aide
differs more from 90° than the other given angle.
94. CASE IV. might have been reduced to Case III. by means of the polar triangle
of Art. 8. For there will be known in the polar triangle two sides and an angle
opposite one of them, being the supplements of the given angles and side of the
proposed triangle. The polar triangle being solved, therefore, by Case III., and its
twc remaining angles and third side found, the supplements of these parts will be
the required sides and third angle of the proposed triangle.
-*• It may be shown that both values of C will be admissible, by a process of reasoning
similar to that employed in the note on page 197, applied to the equations
of Art 8'.>.

SOLUTION OF SPHE.IICAL OBLIQUE TRIANGLES.
95. CASE V. Given the three sides, or a, b
and e. (Fig. 9.) We have three methods
for computing the half angles :
ist. By the sines, from (31), remembering
that
s = 1 (a 4- 6 4- e)
sin IA=\ ('^^^i^3^{LriA\
^ \ sin 6 sm e /
sin J 5 •
sin i C
2d. By the cosines, from (33),
sin (s — c) sin (s — a)^
sin e sin a /
sin (s - a) sin (s — 5)^
sin a sin b
Mg. 9.
203
(164)
cosi^= //Z^^^^^MfJ^N
^ >/ \ sin 6 sin e /
sin (s — 5)
cos J 5 >J V SI
sm e sm a
cos J'^-JC
sin s sin (s — e)
sin a sin 6
3d. By the tangents, from (34),
tan IA _
/ , sin (s — b) sin {s — c)\
*>* V sin s sin {s-a) '
tan»5=/(^-^f-^)f^(\-^))
b)
tan|C=/C(f~^)^;°(^7Jl)
^ V sin sm s .s sin am (« ix — c) el
'
(165)
(166)
When only one of the angles is required, the simplest method will
be by (165), but if the required angle is less than 90°, it will be
found more accurately by (164), for then ^ A

204 SPHERICAL TRIGONOMETRY.
No ambiguity can arise in these solutions, since the half angles
must be less than 90°; they require therefore no attention to tho
algebraic signs.
EXAMPLES.
1. Given a = 100°, b = 50°, e = 60° ; find A.
a =100°
b= 50°
e= 60°
2 8 = 210°
s = 105°
8 — a= 5°
IA= 69°
A = 138°
7'52' -7
15'45" •4
2. With the same data, find all the angles.
By (166).
log cosec 0-1157460
log cosec 0-0624694
log sin 9-9849438
log sin 8-9402960
2)9-1034552
logcos 9-5517276
8=105° 1. cosec 0-0150562 L cosec 0-0150562 h cosec 0-0150562
s-a= 5° h cosec 1-0597040 1. sin 8-9402960 h sin 8-9402960
8-5=55° 1. sin 9-9133645 1. cosec 0-0866355 h sin 9-9133645
s-c^ 45° 1. sin 9-8494850 1. sin 9-8494850 L cosec 0-15051-50
IA=
2)0-8376097 2)8-8914727 2)9-0192317
1. tan 0-4188049 1. tan 9-4457364 \. tan 9-5096159
69° 7'52"-7 i5=15°35'37"-0 JC= 17°54'59"-1
Ans. J. = 138°15'45"-4 .B=31°ll'14".0
C= 35° 49'58"-2
8. Given a = 10°, J = 7°, e = 4°; find the angles.
Ans.A = 12ii°4:4:'i5"-l
B = 33° 11' 12"-0
C= 18°15'31"-1
96. The method by (166), may be put under the following convenient form. Let
I/Bin (s — a) sin (s — J) sin (s — c)\
N \ sin« /
then . (167)
P P P
ia.niA = -^— r, tanJP= -. — y-, tanJC = -;—-^ ^
sm (a — a) •* sin (a — b) ^ sin (s — c)
which are similar to the formulae of PI. Trig. Art. 146, and are computed in the same
manner.

SOLUTION OF SPHERICAL OBLIQUE TRIANGLES. 205
97. CASK V. Given a, b and c. Second Solution. If the whole angle is re.juired
directly,* we have
cos a '— cos b cos c
coa A =
sin b sin c
which may be adapted for logarithms by an auxiliary thus:
Or thus,
cos

2Q6
SPHERICAL TRIGONOMETRY.
SOLUTION OF OBLIQUE SPHEKIOAL TEIANQLES BY MEANS OF A PERPENDICCHB.
101. All the cases of oblique spherical triangles may be solved by dividing tho
triangle into two right triangles by a perpendicular from one of the vertices to the
"riposite side, and solving these partial triangles by the methods of the preceding
chapter. Bowditch has given two rules, based upon Napier's Rules, (Art. 40), by
which the application of this method is facilitated.
102. Bowditch's Rules for Oblique Triangles. "If in a
Hg. 10. Q spherical triangle, (Fig- 10), two right triangles aro
formed by a perpendicular let fall from one of its vertices
upon the opposite side ; and if, in the two right
triangles, the middle parts are so taken that the perpen-
\j^ dicular is an adjacent part in both of them ; then
The sines of the middle parts in the two triangles are proportional
to the tangents of the adjacent parts.
But if the perpendicular is an opposite part in both the triangles, then
The sines' of the middle parts are proportional to thecosines of the opposite parts.
To prove which rules, let M denote the middle part in one of the right triangles,
A an adjacent part, and 0 an opposite part. Also, let m denote the middle part in
the other triangle, a an adjacent part, and o an opposite part; and \ot p denote the
perpendicular.
First. If the perpendicular is an adjacent part in both triangles, we have, by
Napier's Rules, (Art. 46,)
whence
or
sin M = tan yt tan ^
sin m = tan a tan p
sin if tan A tan p tan A
sin m tan a tan p tan a
sin Hf: sin m = tan A : tan a
Secondly. If the perpendicular is an opposite part in both triangles, we have, bj
Napier's Rules
sin M = cos 0 cos p
whence
sm m = cos o coap
Bin M cos 0 cosjo cos 0
sin m cos o coap cos o
or sin M: sin m = cos 0 : cos o" *
We proceed to solve the six cases of spherical triangles with the aid of a perpen-
•iicular. It will be seen, however, that Bowditch's Rules are applicable but in the
first four cases.
•* Peirce's Spherical Trigonometry, Art. 44.

SOLUTION OF SPHERICAL OBLIQUE TRIANGLES.
'•ZOI
i03. CASE I. Given b, c and A. Let the perpendicular G P, Fig. 10, be drawn
from C, (that is, in such a manner as to put two given parts in one of the right
triangles). Then the right triangle AGP gives, by Napier's Rules, if we put
AP = Ip,
tan *= tan 6 cos .4 (112)
then talking co. b and co. a as middle parts in the two triangles, A P =i p and
B P= c —-1^ * are the opposite parts, whence, by Bowditch's Rules,
whence
cos Ip : cos (c —ip) =
cos b : cos a
cos (c — *) cos 5
cos a = !^ Li (173)
cos 0 ^ '
Again, taking A P and PB as middle parts, co. A and co. P are adjacent parts,
whence, by Bowditch's Rules,
whence
.sin 0 : sin (c — ip) = cot ^ : cot P
, „ sin (c — (P) cot A
cot P = !^ r^ (174)
sm 0 * '
and the formulae (172), (173), (174), agree entirely with (122) and (123).
The triangle P GP gives as a check
tan a cos P = tan (c —

208 SPHERICAL TRIGONOMETRY.
Again, taking co.^CPandco.PCPas middle parts, and therefore co. 6 and co. •
us adjacent parts, Bowditch's Rules give
whence
cos S : cos (G — S) = cot b : cot a
cos (C — S) cot t
cot a ^ (178)
and (176), (177), (178), agree entirely with (134) .and (185).
The triangle B G P gives
tan P cos a = cot (C— 3)
(179)
which agrees with (136).
By drawing the perpendicalar from A, we may in the same manner obtain the
formuloB (137).
The side c may be found from the proportion
sin A : sin G = sin a : sin o
and Art. 69 ; or c being found by means of a perpendicular from A, we may find a
by a similar proportion.
105. CASE III. Given a, b and A. Let the perpendicular
be drawn from G, Fig 10, as in the preced­
Fig.10.
ing cases, and let A P = ip, B P=

SOLUTION OF SPHERICAL OBLIQUE TRIANGLES. 2011
and by Bowditch's Rules,
whence
and since in Fig. 11,
cot b : cot a = cos ^ ; cos 3'
cos 3' 7= COS 3 tan 4 cot a (184)
C = ^CP = ^CP 4- BCP = 5 + y, oT G = ACB' = AGP — B'CP = 3 — 3',
we have
C = 3 ± .y (185)
and the formulse (180), (181), (182), (183), (184), (185), agree entirely with (146)
and (147).
After c was found, we might have found Cfrom the proportion
and B is found from the proportion
sin a : sin c = sin 4 : sin G (186;
sin a : sin J = sin .A : sin P (187)
The two values of P determined by (187), are both admissible when c has two values
as above. It is also evident, from Fig. 11, that the two values of P are supplemental.
To determine the corresponding values of c and P, we observe that, by Art. 49, the
perpendicular CP is in the same quadrant with A and with CBP and OB'P, and
therefore GB'A is in a different quadrant from A. Hence, that value of P lohich is in
the same quadrant as A corresponds to the value of c = ip + ip', and that value of B
which is in a different quadrant from A corresponds to the value of c = ip —^ 0'; which
agrees with what is shown in Art. 84.
In computing (186), the two values of c must be employed successively, and the
formula computed twice. At each computation we shall have two values of G found
from the sine, one of which must be selected by Art. 69. But as the application of
the principles of Art. 69 is tedious and embarrassing, it is better to find C by (184)
and (185).
The formulije (149), (150), (151), (152), for finding P, may easily be deluced by
Napier's and Bowditch's Rules.
106. CASE IV. Given A, B and b. Let the perpendicular be dra-wn as before,
Fig. 10, and let AP = ip, BP = ip', then as before,
and by Bowditch's Rules,
whence
tan 0 = tan b cos A (188)
cot ^ : cot P = sin 'P : sin /
sin 0' = sin 0 tan .4 cot P 1 ,.„„,
c = 0 -t- 0' / ^ ^
which agree with (156). But 0' having two supplemental values determined by the
sine, e has two values, as already explained in Art.
To show the same geometrically, let BP,
Fig. 12, be the acute value of ip', and about
(7 as a pole, let a small circle be described
passing through P, and intersecting the ^"^
great circle AB again in B". Let P"Cbe
drawn, and produced to meet AB again in
B', foi:ming the lune B"B'. Then we have
P' = P" =
27 s 2
OBA

210 SPHERICAL TRIGONOMETRY.
Kg. 12.
In the triangle AGB'vc have
C =
and in the triangle A CB' we have
so that in both triangles, -4e'P and ACB,
the value of the angle opposite the side i
.jji is the same, that is, both triiingles contain
the same data. A, B and b. Now
180° = B"B' = B"P + B'P =zBP\- B'P,
BO that BP and B'P are supplements of
each other.
lp+i
= ^P 4- PP
c = 0 4- 0' = ^P 4- B'P
and hence the two values of c are found by giving 0' its acute and obtuse values
successively, as already shown analytically.
By Art. 49, CP must be in the same quadrant with A; hence, if P is in the same
quadrant with A, P falls between A and B, as in the figure, and for the same reason,
between A and P'. But if A and P were in different quadrants, both points,
P and P', might fall between A and P. The two values of c would then be found
by the formula
c =

SOLUTION OF SPHERICAL OBLIQUE TRIANGLES. 211
107. CASE-v. Given a, i and c. The perpendicular Kg. 13.
cannot be drawn, in this case, so that two of the given
parts shallbe in one triangle; nevertheless the case can
be solved by means of a perpendicular. Let the perp.
be drawn from any angle, as G, Fig. 13, and as before,
put^P = 0, BP = 0'; then by Bowditch's Rules,
cos 0 : cos 0' = cos b : cos a
, cos 0' — cos I
Whence cos 0' -j- cos 0 cos a + cos b
or, by PI. Trig. (110),
whence, since 0 -|- 0' = c,
tan \(

212 SPHERICAL TRIGONOMETRY.
Downey's Table XXII., with the argument log 2:, the difference of tiis given logarithms,
gives log (1 4" 2;), wliich being added to log q, the less logarithm, gives the
required log. sum, or log (p + q). Table XXIII., with the argument log x, gives
log M -|
) which, being added to log^, the greater logarithm, gives the required
log. sum. Either table may, in general, be employed, but one or the other may be
found more convenient in a particular application, and therefore both are given
Again, we have
p-q=p{l-l)
p
BO that, putting, as before, z = —, we have
log z = log jci — log q
log (p — q) = logp + log {l ——)
Downes's Table XXIV., with the argument, log z, gives log (1
j which, being
added to the greater logarithm, gives the required log. difference, or log (p — q).
With these tables, then, we may readily compute any of the preceding formulso
which contain two terms in the second member, without the aid of auxiliary angles
EXAMPLES.
1. Given b = 120° 30' 30", c = 70° 20' 20", A = 50° 10' 10" ; find a. (Sams
as Ex. 1. p. 182).
The formula is
cos a = cos b cos c 4- sin 5 sin e cos A
which will be thus computed :
log COB b — 9-70557
log cos c + 9-52693
log q — 9-28250
log sin b + 9-93529
log sin c + 9-97891
logcos^ 4- 9-80054
log^ 4- 9-71574
log.? — log 2 = log X — 0-48324
The terms p and q have opposite signs, and although, by the formula, they are to be
added (algebraically), an arithmetical difference is required. By marking the signs
of all the quantities, as above, we shall always know whether a sum or difference
is required by the sign before log z. In this case this sign being negative, we are
to find a difference, and therefore, by Table XXIV., we take
log^l ^ 9-82694
log;; 4- 9-71574
log cos a + 9-54268
a = 69° 34' 52"

SOLUTION OP SPHERICAL OBLIQUE TRIANGLES. 213
2. With the same data, find P. The formula is
which mast here be put under tlie form
and is thus computed:
or, by Table XXIII., we have
. „ sin c cot b — cos c cos A
•cot P =: :—;
sm A
^ „ sin c cot S
cot B = —T—3
sva. A
(which is to be added to the greater log.)
cos c cot A
log sin c + 9-97391
log cot b — 9-77029
arCO log sin^ -j- 0-11467
log p — 9-85887
log (— cos c) — 9-52693
log cot ^4- 9-92121
log q — 9-44814
logp — log g- = log z + 0-41073
where the sign before log z being positive, the tables for log. sum must be used. By
Table XXII., we have -
log (1 4- z) 0-55325
(wliich is to be added to the less log.)
logq — 9-44814
log cot P — 0-00139
0-14252
logp -- 9-85887
log cot P — 0-00139
P = 135° 5'81"
In these isolated examples, the labor of computation is very little less than with
the use of an auxiliary angle, as on p. 182 ; but the Gaussian Table has greatly the
advantage when the same formula is to be repeatedly computed with successive
values of one of the data while the others remain constant. Thus, in the first of
the preceding examples, if successive values of a are to be found corresponding to
successive values of A, while b and c are constant, log q will be constant, and log x
•will take successive values, corresponding to those of log cos A, so that .after the
first value of a is found the succeeding ones are rapidly obtained. On the other hand,
as the auxiliary 0 in the formulse (122), depends upon A, the whole process would
have to be repeated in finding each value of a.
For other forms of the Gaussian Table, see the original table, (to five places of
decimals), by Gauss, published in Zaoh's Mona.tUche Gorrespondenz, Nov. 1812; Matthiessen's,
(to seven places), Altona, 1817 ; in Ye'ga,'a Sammlung maihematischer Tajkln,
(five places), Leipzig, 1840; Zech, (seven places), Leipzig, 1849; Shortrede's
C.'llection of Tables, (seven places), Edinburgh, 1849 ; Gray's Tables for the Computation
of Life Contingencies, (six places), London, 1849; Schumacher's Hiilfstafeln,
new ed. (four places).

214 SPHERICAL TRIGONOMETRY.
CHAPTER IV.
SOLUTION OF THE GENERAL SPHERICAL TRIANGLE.
109. WE have thus far, following the usual coui-se, considered those spherical
tiiangles only whose sides and angles are less than 180°. In the applications of this
subject in astronomy, however, it is often necessary to consider triangles whose
sides or angles exceed 180°. (For example, the right ascension of a heavenly body,
admitting of all values from 0° to 360°, may be one part of such c triangle). Vfc
may, it is true, in such cases, always substitute another triangle whose parts are the
supplements to 180° or 360° of those of the proposed triangle; but this modcr
although very generally regarded as the simplest, is not really so in the cases
alluded to. The construction of figures for discovering the supplemental triangles
is often embarrassing and liable to mistake, while the solutions, when obtained, are
mostly deficient in generality, and can only be regarded as solutions of the particular
cases of a general problem. But if we proceed by a method that is as applicable
when the parts of the triangle exceed as when they are less than 180°, we may investigate
a problem under the simplest supposition of the values of these parts, and
rely upon the generality of the method to cover all the particular cases.
110. We shall first endeavor, in an elementary manner, to give the student a conception
of the nature of the general spherical triangle.
ng. 14.
Let^PC7, Fig. 14, be any spherical triangle whose partt
are all less than 180°; then the remainder or complemen*
of the sphere is also a spherical triangle whose sides are
II, b und c, iind whose angles are 860° — A, 860° — B
and 860° — C. We shall distinguish these triangles from
each other by means of accents, writing the letters wiilm
the triangle to which they respectively belong, as ip
Fig. 14. The sides are common, but when referred to ne
sides of A'B'G', they will be denoted by a', b' and c'.
Again, one of the sides may exceed 180°, as the side a of the triangle ABG,
Fig. 15. In this triangle, it is evident that we must have A > 180°, so long as P,
C, b and c are each < 180°. In the triangle A'B'G' we have A' < 180°, whil(
B' > 180°, C" > 180°.
Kg. 16
If we next suppose two of the sides to exceed 180°, as a and b. Fig 16, these sides
intersecting in two points whose distance is 15^0°, the figure ceases to present the

SOLUTION OF THE GENERAL SPHERICAL TRIANGLE. 215
triangle as an enclosed surface, but it will presently appear that such triangles are
solved by tho same general methods that apply in other cases. To form a just conception
of the triangle in this case, we may conceive Fig. 16 to be obtained from
Fig. 15 by carrying the point A along the arc CA produced until it crosses the side a
the points ^1 and P may then be joined either by an arc less than 180°, as in Fig. 16,
or by its supplement to 360°, as in Fig. 17, in Fig. 17
which last case every side exceeds 180°. In •^'•'fc -•^'
these figures, to avoid confusion, the point A p, /^^\ IA\
is not placed in its true position according to ,d / \ /
perspective. I \ \ /
In each figure we have two triangles, whose N^ »'-'
sides are common, and whose angles are sup- \ ^ —-w^
plements to 360°. It will be easy to trace the \ \ /
two triangles signified by ^PC and ^'P'C, by \^ ^^^y y '
remarking that the letters in each case are all ^ ^^--^.^^^-^
on the same side of the perimeter of the triangle.
We may go farther, ana suppose the arc joining A and P to be a circumference
+ the arc ^4P, or any number of circumferences 4- -^B; and similarly the angles
may be supposed to be altogether unlimited; but since the relative positions of any
three points of the sphere must be fully determined by arcs and angles less than
300°, nothing is gained by passing beyond this limit.
111. All the formulce of Chapter I. are applicable to the general spherical triangle.
This proposition might be considered as established by the principle of PI. Trig.
Art. 49, but it is also very easily established by a continuation of the process of
Spher. Trig. Art. 6, where tho fundamental equation was shown to apply to all
triangles whose parts are less than 180°.
It was proved in Art. 29, that all the equations of Chap. I. may be deduced from
the fundamental one,
cos a = cos b cos c + sm b sin c cos A
We have then only to prove the generality of this single equation.
Ist. Let all the sides be < 180, but A' >180°, Fig. 14. The formula being true for
the triangle ABC, we have
cos a = cos 5 cos c + sinb sin c cos (860° — A')
or in the triangle A'B'G', by PI. Trig. (76),
cos a' ^
cos V COB c' + sin b' sin c' cos A'
2d. Let «> 180°, Fig. 15, and produce a to complete the great circle. The triangles
ABC and A'B'G' are respectively the difference and sum of " hemisphere and the
triangle A'ik, all of whose parts are < 180°. In the triangle A'ik we have, in terms
of the parts of ABG,
cos (360° — a) = cos i cos c 4" sin b sin c cos (360° — A)
and in terms of the parts of A'B'G',
cos (360° — a') = cos V cos c' 4- sin b' sin d cos A'
both of which reduce to the form (M). But it is here necessary to show that the
formula may also be applied to each of the other angles: thus the triangle A'ik
COB b = cos (360° — a) cos c + sin (360" — a) sin c cos (180° — P)
cos b'= cos (860° —a') cos c'+ sin (860° — a') sin c'cos (P' —180°)
both of which reduce to the form (M).
(M)

216 SPHERICAL TRIGONOMETRY.
3d Let -I > 180°, i>180°. Fig. 18: those arcs intersect at (, and the triangle
A' B' I gives
Fig. 18.
cos (a — 180°) = cos (b — 180°) cos c + sin (5 — 180°) sin c cos (300° — A)
cos (a'— 180°) = cos (V — 180°) cos c' + sin (b' — 180°) sin c' cos A'
which reduce to the form (M) ; and in the same way the formula applies to the angle
P. We have also
cos c = cos (a -^ 180°) cos (b —180°) 4- sin (a —180°) sin (5 — 180°) cos i
and since cos i = cos C = cos (860° — C") = cos G', this also reduces to the form
(M) for both ABC and A' B' C.
4th. Let a > 180°, b > 180°, c > 180°, Fig. 19; the side e being produced to
complete the circle, the triangle ikl gives
cos (a —180°) = cos (b —180°) cos (860°—c) + sin (b — 180°) sin (360° — c) cos I
and since cos I = cos (180° — A) = — cos A = cos (A' —180°) = — cos A', this
reduces to the form (M) for both ABG and A'B' G'; and in the same way the formula
applies to the angle P. We have also
cos (360° — c) = cos (a — 180°) cos (b —180°) 4- sin (a — 180°) sin (b —180°) cos i,
and since cos i = cos G = cos C", this reduces to the form (M) for both ABO
and A' B' C.
The cases in which the angles or sides exceed 360° are included in the preceding,
in consequence of PI. Trig. Art. 45.
112. The preceding demonstration, though tedious, has the advantage of giving a
definite conception of the figures which our formulse represent. But perhaps the
most satisfactory (as it is the most elegant) method, is to rest the demonstration
of our fundamental equations themselves upon the principles- of analytical
geometry, and, for the sake of those who are acquainted with that subject, we add
the following investigation:
Any point of the sphere may be referred by rectangular co-ordinates to three
planes passing through the centre of the sphere at right angles to each other. Let
O be the centre of the sphere. Fig. 20, and ^ P C a spherical triangle upon its
surface. Let one of the co-ordinate planes, as JT Y, coincide with the gr«at cir-
•i\e A P. and let the axis of JTpass through P. If CP be drawn perpendicular

SOLUTION OP THE GENERAL SPHERICAL TRIANGLE.
to the plane XT, and GP' and PP' to the axis OX, the oo-ordinat-s of the
pomt G are
X = OP', y = PP', z=GP
rig. 20. I'ig. 21.
the values of which (0 C being taken = 1) are
X = cos a
y = sin a cos P
2 = sin a sin P
If now the axis of X be made to pass through A, Fig. 21, without changing
the position of the plane X F we shall have for z', y', z', the co-ordinates of Creferred
to the new axes,
x' = cos b
y' = — sia b cos A
z' = sin b sin A
The axis of z being unchanged, the relations between z', y', and x, y, are expressed
simply by the formulae for the transformation of co-ordinates in a plane; the inclination
of the new axes to the first is here expressed by c, and the formulse of
transformation are therefore
X = z' cos c — y' sin c
y = x' sin c + y' cos c
substituting the values of the co-ordinates, we have at once the three following
' indamental equations :
cos a = cos c cos b + a\n c sin b cos A
sin a cos P = sin c cos b — cos c sin b cos A
(N)
sin a sin P ^
sin b sin A
which are identical with (4), (6), and (3).
113. Having established the complete generality of our fundamental equations,
we may now employ for the solution of the general triangle any of those deduced
from them in Chap. I.
As a single trigonometric function is not sufEcient to determine an unlimited
angle or arc, (PI. Trig. Art. 53), it becomes necessary in most oases to deduce expressions
for both the sine and cosine of the required part.
It will be found that all the six cases of the general triangle admit of two solutions,
but that they all become determinate, when, in addition to the other data, the sign of
the sine cr msine of one df the required parts is given. In the practical applications in
Bstronomj', 't mostly happens that the conditions of the problem supply this sign.
28 T

218 SPHERICAL TRIGONOMETRY.
114. CASE I. Given b, c and A. First Solution; when one of the remaining angles,
as B, and the third side a are required. The relations between the given and
required parts are
cos a = cos c cos b + sin c sin b cos A 1
sin a cos P = sin c cos b — cos c sin b cos A
Y (l^*')
sin a sin B =i sin i sin jl J
The signs of the second members will be known from their computed numerical
values; the sign of cos a is therefore known. If the sign of sin a ia also
given, the quadrant in which a must be taken will be known ; the second and third
equations will determine the sign of the sine and cosine of B, and therefore the
quadrant in which P is to be taken.
In like manner, if the sign of either cos P or sin P is given, that of sin a becomes
known, and the problem is determinate. If no conditions are atttached to the required
parts, there must be two solutions.
The numerical solution will be conducted as follows: The values of the second
members (or simply their logarithms) are to te separately computed, and their
signs carefully noted; then the quotient of the 3d by the 2d (or the difference
of their logs.) will give tan P, and hence P, which will be taken in the quadrant
indicated by the signs of the sine and cosine. Then the 3d divided by sin ,P,
or the 2d by cos P, will give sin a, which, agreeing with tho value from the
1st equation, will serve to verify the correctness of the whole process.
This solution may be adapted for logarithms by the methods employed in the preceding
chapter.
1st. Let k and 0 be determined by the equations
k sin 0 :^ sin 6 cos A
k cos 0 = cos b
k being a positive number (PI. Trig. Art. 174); then
cos a ^ k cos (c — 0)
sin a cos B = k sin (c — 0)
sin a sin P = sin S sin A
[ (197)
2d. Eliminating k, and taking 0 < 180°, (PI. Trig. Art. 174),
tan 0 = tan S cos A (ip < 180°)
cos b
cos a = cos (c — 0)
„ cosb . ,
Bin a cos P = sm (c — 0)
OOS0 ^ '
sin a sin P = sin b sin A
\ (198)
3d If tli quadrant in which a, is to be taken is given, we may give the preceding
equations the following form:
tan 0 = tan b cos A (0 < 180°)
tan a cos P = tan (c — 0)
sin 0 tan A
tan a sin P =: cos ( C 0)
(19i>;

SOLUTION OF THE GENERAL SPHERICAL TRIANGLE. 219
4th. If both a and b are less than 180°, as not unfrequentlj happens in tho applications
of this problem, let
k
sin b
sm a
tht-u jrj and n are both positive numbers [k being positive) and (197) gives
Check. We find
m sin 0 = cos A
m. cos 0 = cot b
n sin P = sin 0 tan A
n cos P = sin (e — 0)
cot a = cot (c — 0) cos P
(200)
Bin (c — 0)
ain 0
cos (c — 0)
cos 0
sin a cos P
sin b cos >4
cos a
cos b
tan ^
tan B
(201)
besides which we may employ, in connection with (200), the equation sin a sin P
= sin 4 sin ^ ; or in connection with (197) or (198) the equation tan a cos P =
tan (c — ip). Or when (197) and (198) are employed, we may find a both by its
sine and its cosine.
115. The angle C may be found in the same manner as P, interchanging P and G,
b and c, in the preceding formulie. But when P and 0 are both required, the Second
Solution to be given presently is preferable.
EXAMPLE.
Given A = 261° 16' b = 45° 54', c = 138° 32', and a < 180°; to find a and P.
We shall first employ (197). The first column of the following computation, con
taining the symbols expressing the operations to be performed, should be prepared
before opening the tables:
A 261° 16'
b 45° 54'
c 138° 32'
log sin .4 —9-9949352
logcos^ —9-181.3744
• log sin b + 9-8562008
log cos J = log/c cos 0 + 9-8425548
log sin i cos ^ = log/c sin ip — 9-0375752
log tan 0 — 9-1950204
log cos 0 4- 9-9947336
log* 4- 9-8478212
0 351° 5'42"-6
*c —0 147°26'17"-4
' As 0 > c, we take c == 138° 32' 4- 360°, so that c — 0 may be a.positive angle;
but it would be equally convenient to take c — 0 = — 212° 33' 42"-6

220 SPHERICAL TRIGONOMETRY.
log sin (c — 0) 4- 9-7309514
log cos (c — 0) — 9-9257303
log A cos (c-r-

SOLUTION OF THE GENERAL SPHERICAL TRIANGLE.
22 i
Same as in Art. 115. a < 180°.
EXAMPLE.
A
b
e
J (* -

222 SPHERICAL TRIGONOMETRY.
Adapting (208) for logarithms, we find
Ist.
*k sin 9 z= cos A
k cos 9 = sin A cos b
cosP= ksin (C —9)
sin P cos a z= A; cos (0 — 9)
sin P sin a = sin A sin b
2d.
cot 9 = tan A cos b
cosP =
sin P cos a -.
cos A
sin (C —3)
sin 3
cos A
cos (C —3)
sin 3
sin P sin a = sin A sin d
3d. When th quadrant in which B is to be taken is given
cot 3 = tan A cos b
tan P cos a = cot (C — 3)
tan 6 COB 9
tan P sm a = -. ~ -= -
sm (C — 3)
(k positive)
(3 < 180°)
(3 < 180°)
(204)
y (205)
(206)
4th. When A and B are both less than 180°, let
k
P = sin A
sin P
I
thenj7 and q are positive numbers, and we have from (204),
Checks. We have
cos
/) sin 3 =: cot A
p cos 9 =: cos b
q sin a = tan b cos 3
q cos a = cos (G — 3)
cotP= tan (0—3) cos a
(Ccos
3
sin (C-
s)
sin 3
s)
sin P COB a
sin .4 cos b
COS P
COS A
tan b
tan a
^•(207)
L (208)
'* The same factor k is used here and in (197), although the auxiliaries 0 and 9 are
different. To show that k has the same value in (197) and (204), let the squares of
the equations
A sin 0 = sin b cos A
k cos 0 = cos b
be added ; we find
F (sin" 0 -f- cos' 0) = F = cos' b + sin' b cos' 4 = 1 — sin' 6 sin' .4
and in the same manner, from the equations
k sin 9 = cos A
k cos 9 = sin 4 cos b
we find
^ = 1 — sin' b sin' 4
and therefore, in both cases, J = v' (1 — sin' b sin' 4)

SOLUTION OF THE GENERAL SPHERICAL TRIANGLE. 223
besides which we may employ, with (207), the equation sin P sin a = sin A sin b;
or with (204) and (205), the equation tan P cos a = cot (C-3)! Also, wlieu
(204) or (205) is employed, we m.ay find P both by its sine and its cosine.
These formulse are computed in the same manner as thosfe of preceding case.
119. CASE II. Given 4, C and b. Second Solution; when the two sides, a and c,
or all the remaining parts, are required. We employ Gauss's Equations in the following
form:
sin i P sin J (a 4- c) = cos ^ (A — C) sin J b
sin J P cos J (a -j- c) z= cos ^ (A + G) cos J b
cos J P sin J (a — c) = sin J (4 — C) sin J b
cos J P cos J (a — c) = sin i (A + G) cos J 6
(209)
whicii are solved in the same manner as (202).
ExAItlPLB.
G./^ 4 = 121° 30' 19"-8, G = 42° 15' 13"-7, 5 = 40° 0' 10", and P > 180°.
By (209).
b
A
G
iiA-c)
i{^+0)
40° 0'10"-0
121°36'19"-8
42° 15' 13"-7
39° 40' 33"-0
81° -55' 46"-7
20° 0' 5"-0
d = log cos i (A — C) + 9-8863038
log e = log sin J (4 — (7) 4- 9-8051224
log/ = log cos I (4 -f c] 4- 9-1473326
logg = log sin J (4 -f C) -j- 9-9956775
log sin J 4 4- 9-5340806
log cos J 6 4- 9-9729820
log dsin^b = log sin J P sin i (a+ c) + 9-4203844
log/ cos I 6 = log sin J P cos J (a -[- c) + 9-1203146
log tan I (ffl 4- c) -i- 0-3000698
i (a+ c) 63° 23' 8"-8
log sin ^ (a 4- c) -f 9-9518526
log sin J P 4- 9-4690318
JP
162°52'28"-6
'log (— e sin J 6) = log (— cos J P) sin ^ (a — c) — 9-3392080
*log (—ff cos J 5) = log (— cos I P) cos i(a — c) — 9-9686595
log tan J (a — c) + 9-3705435
i(a — e) 193°12'32"-9
r a = 256° 35'36"-2
fAns. J (; = 280°10'30"-4
[ P= 325°44'57"-2
* The sign of eacn of these factors is changed because P > 180°, and cos J P is
negative.
f It was necessary to increase J (a 4- o) by 360°, to obtain c. The corresponding
value of b would be 016° 3-5' 30 "-2. See note at the end of this chapter, p..227

224 SPHERICAL TRIGONOMETRY.
120. When P only is required, we may employ the methods of Arts. 81 and 8i,,
which are determinate when the sign of sin P is given; or when that of either sin a
or sin c is given, since we may then find that of sin P by inspecting the equations
sin 4 sin b sin C sin 4
sin P ;
sm a sm c
121. CASE III. Given a, b and 4. First Solution; when the three remaining parts
P, C, and c are all required.
We find P by the equation
sin P = sin 4 sin b sm a
(210)
which is determinate when the sign of cos P is given.
Then, to find C, we have
— cos 0 cos 4 + sin C sin 4 cos b = cos P
sin Coos4 -j- cos C sin A cos 6 = sin P cos a
which have already been employed and adapted for logarithms in Art. 118. If we
denote the auxiliary by 9, and put C — 3 = 3', we find, from (204),
A sin 9 = cos A (k positive)
k cos 9 ^ sin A cos b
k sin 3' = cosP \ ;211)
k cos 9' = sin P cos a
C = 9 -f 9'
To find c, we have
cos c cos b + sin c sin b cos 4 = cos a
sin c cos b — cos c sin b cos A = sin a cos P
which have already been employed and adapted for logarithms in Art. 113.
denote the auxiliary by 0, and put c — 0 = 0', we find, from (197),
Checks. We have
sin a
sin ¥'
sin Ip
k sin 0 = sin b cos 4
k cos 0 = cos b
k sin 0' = sin a cos P
A OOS0' = cos a
c =0 4-*'
cos 4
cosP
tanP
tan 4
COB 9
cos 9'
cos 0
COS0'
(k positive)
tan a
tan b
cos i
If we
(212
\ (213)
One of which maybe used as a check when either C or c has been alone computed.^
When both C and c have been found, the obvious check is
sin_^ _ sin4.
sin c sm a ^ '
* The following relations deserve a passing notice:
sin 0 cos 3'
cos 0 sm 9
sin b sin P
tan 0 tan 0'
tan 9 tan 3'
sin 0' cos 3
= sin a sin .4
cos 0 sin 3'
sin' a sin' P = sin' 4 sin'4
sin 2 0 sin 2 3'
sin 2 9 sin 2 0'
sin' b
sin'a

SOLUTION OF THE GENERAL SPHERICAL TRIANGLE. 225
EXAMPLE.
Given a = 126° 25' 6"-6. b = 138° 82' 0', 4 = 26r 16' 0", and cos P negative
a 126° 25' 6"-6
b 138° 32' 0"-0
4 261° 16' 0"-0
^y (210)> log sin a + 9-9056351
log sin b -f 9-8209788
log sin 4 —9-9949352
log sinP — 9-9102789
P 234°2.5'29"-3
^y (^ll)> log cos 6 — 9-8746795
log sin Aoosb = log A cos 3 + 9-8696147
log cos4 = log 4 sin 3 —9-1813744
log tan 3 — 9-3117597
9 848° 24' 58"-0
log cos a — 9-7735515
log sin P COB a = log k cos 3' + 9-6838304
log cos P = log i sin 3' — 9-7647520
log tan 9' — 0-0809216
9' 809° 41' 83"-7
9-f. 9' = C 298° 6'26"-7
By (212), log sin 4 cos 4 = log A sin 0 — 9-0028532
log cos 4 =: log A cos 0 — 9-8746795
log tan 0 4- 9-1276737
0 187°38'81"-3
log sin a cos P = log A sin 0' — 9-6703871
log cos a = log k cos 0' — 9-7785515
log tan 0' 4- 9-8968356
0' 218°15'28"-6
0 4-0' =

226 SPHERICAL TRIGONOMETRY.
i sin 3 = cos 4
(k positive)
k cos 3 = sin 4 cos 4
cos 3' = cos 3 cot a tan 4
(3' < 180° with the sign of cos P)
(7 = 9 -f 3'
(215)
To find c, we observe that sin 0' has the sign of sin a cos P, so that we have the
foUo-wing formulse:
/c sin 0 = sin 4 cos A (k positive)
k cos 0 = cos 4
cos 0 cos a ,„,„,
(0' < 180° with the sign of sin a cos P)
c = 0 4- 0'
Gheck. The equation (214).
128. CASE IV. Given 4, P and 4. First Solution,- when the three remaining parts
a, c and 0 are all required.
We find a by the equation
Bin A sin 4
sin P
which is determinate when the sign of cos a is given.
is by (211) and (212).
(2171
The remainder of the solution
124. CASE IV. Given 4, P and 4. Second Solution; when 0 and G are required,
without finding a.
We easily find, from (211),
A sin 3 = cos A
(k positive)
k cos 3 = sin A cos 4
And from (212),
CWeA The equation (214)
sin 9' , sin 9 cos P
cos 4 - (218)
(COB 3' and sin P cos a to have the same sign)
C =
9-}-y
A sin 0 = sin 4 cos 4
k cos 0 == cos 4
sin 0' = sin 0 tan 4 cot P
(cos 0' and cos a to have the same sign)
c = 0 4- $'
125. CASE V. Given a, 4 and c. The formula
cos a — cos 4 cos c
cos 4 =
sin 4 sin c
(A positive)
(219)
(220)
determines A when the sign of sin 4 is known. If the sign of sin P or of sin 0 if
gi fsn, that of sin A becomes known by the equation
sin 4
sin a
sin P
sin 4
sin 0
sin c

SOLUTION OF THE GENERAL SPHERICAL TRIANGLE. 227
The formulse (31), (33), (34), may be used, each of which will become determinate
when the sign of either sin 4, sin P, or sin 0 is kno-wn.
126. CASE VI. Given 4, P and C. The formula
cos A + coa B cos Q
cos ffl = . ' .—^= (221)
sm P sm C '• ^
determines a when the sign of sin a is given. If the sign of sin 4 or of sin c i?
given, that of sin a becomes known by the equation
sin 4 sm c
sin 4 sin P sin C
The formulse (36), (38), (39), may be used, each of which will be determinate
when the sign of either sin a, sin b, or sin c is known.
NOTE UPON GAUSS'S EQUATIONS.
In the unlimited spherical triangle, we may consider any part, as a, to have an
infinite number of values, viz. a, a 4- 360°, a + 720°, &c., expressed generally by
the formula a + 2 n jr, n being any whole number or zero ; and since
sin a = sin (a -f 2 re rr)
cos a = coa (a + 2 nn-)
all those equations of Chap. I. that involve only sin a and cos a will not be changed
by the substitution of a -|- 2 n sr for a. A similar substitution may be made for
each of the parts, or for all of them, at the same time, so that there is an infinite
series of triangles to which these equations are applicable.
But the substitution of a -f- 360° for a, in Gauss's Equations, (202), will change
the sign of all of them, since
sin J (a 4- 360°) = — sin J a
cos ^ (a + 860°) = — cos J a
while the substitution oi a + 720° for a will not change their sign, since
sin J (a 4- 720°) = sin J a
cos J (a'4- 720°) = cos J a
In general, their sign is changed by the substitution of a + (A n + 2) TT for a, and
it is not changed by the substitution ot a + i n rr. The same results follow like
substitutions for each of the parts. It follows that these equations taken only with
the positive sign, do not include all the triangles of the infinite series above spoken
of, and that they are complete only when taken with the double sign, and expressed
in two distinct groups, as (44) and (45) of Art. 27.
In practice, however, we may take them with tlie positive sign only; for they will then
give at least one of the triangles of the series, from which all the others, (and particularly
that whose parts are less than 360°), may be directly deduced by the appli
cation of 360°.-*
This will be illustrated by the example of Art. 119, p. 223 ; we there find
J (ffl -f c) = 63° 28' 3"-3
J (a — c) = 193° 12' 32"-9
or rather, since J (a -j- c) should be greater than J (a — c),
J (a 4- c) = 428° 23' 3"-3
J (a — c) = 193° 12' 32"-9
•* Gauss (Tiieoria Motus Corp. Cml. Art. 54) recommends the use of the positive
sign only, observing that any side or angle may be diminished or increased by 360°,
as the case may require, but confines himself to the statement of this practical presept,
without explaini-ig the grounds upon which it rests.

228 SPHERICAL TRIGONOMETRY.
whence
a = bl6° 35' 36"-2
c = 280° 10' 30"-4
which is the proper solution of the equations taken with the positive sign.
we deduct 360° from a, and take, as on p. 223,
a = 256° 35' 86"-6
e = 230° 10' 30"-4
If now
we have the solution that would have been obtained by taking the negative sign in
all the equations ; for we now have
J (a 4- c) = 248° 23' 3"-3
J (a — c) = 13° 12' 32"-9
which, differing from the former values by 180°, must change the sign of all the
equations.
I have given some further particulars respecting unlimited spherical triangles,
and a fuller discussion of Gauss's Equations, in an essay which the reader will find
in the Astronomical Journal, Vol. I., published at Cambridge, Mass.

AREA OP A SPHERICAL TRIANGLE. 229
CHAPTER V.
AREA OF A SPHERICAL TRIANGLE.
127. Given the three angles of a spherical triangle, to compute
the area.
This problem is solved in geometry, -where it is proved that the
surface of a spherical triangle is measured by the excess of the sum of
its three angles over t-wo right angles, by which is meant, that the area
is as many times the area of the tri-rectangular triangle as there are
right angles in the excess of the sum of the angles over two right angles.
To express this analytically, let
r = radius of the sphere
T = surface of the tri-rectangular triangle
= \ surface of a sphere = J TT /"^
2S= A-\-B-¥ G
K = area of the triangle ABG.
Also, let the angles A, B and C be expressed in the unit of Art. ] 1,
that is, let A, B, C denote the arcs -which measure the angles in a
circle -whose radius is unity. The right angle expressed in the same
unit is -^, therefore the number of right angles in2 S is
2S '^ ^'^
•2 TT
and we have, according to the above theorem of geometry,
,4 ^ ^x 2T
K = 2'xf-^-2) = --(2^-.)
or ir = r'(2/S'-3-) (222)
and if the radius of the sphere is taken = 1
K=2S-% :223)
128. In a plane triangle the sum of the angles is equal to %,
and in a spherical triangle the sum exceeds -K by K; hence this
quantity, K, is commonly called the spherical excess.

23(? SPHERICAL TRIGONOMETRY
129 Given the three sides, to find the area.
By (223), we have
in ^ K =s sin (S — -^) = — cos
cos 1.2"= cos (s' —y) = sinS
tan i K = — cot S
> (2245
in which we havo only to substitute the values of cos S, sin S, and cot S, given in
Art. 34, to obtain the required solution. We find, [s = J (a -f- 4 -f- c)].
. , „ v" [sin s sin (s — a) sin (s — 4) sin (s — c)]
2 cos J a COS J 4 cos J o
(225)
cos J .£•=
cos a 4- cos 4 -^ cos c 4- 1
4 cos J a cos J 4 cos Jc
cos' J ffl-f- cos' J 4 -f- cos' J c — 1
2 cos J a cos J b cos Jc
(226)
The numerator of (225) being denoted by n, we find,
^ , „ 1 4- cos a -i- cos 6 -I- cos c
cot J .2^ = —'• '-^ C
which is known as De Gua's formula. 2 n
Again, from the formulse of Art. 35, since 1 — sin /S = 2 sin 'Jj^, 1 4- sin iS:
2 cos "iK, we find
[
sin I s sin J (s — a) sin J (s — 4) sin ^ (s — c)~\
cos J a cos J 4 cos j c
J
(227)
cos i.K'
tan ^ K^-^
rcosJ«cosJ(s — a) cos|(s — 4) cosJ(s — c]
^ L cos J a cos J 6 cos J c ]
[tan Js tan } (s — a) tan J (a — b) tan J (s —

AREA OF A SPHERICAL TRLANGLE. 281
131. it' we admit more than three parts of the triangle into the expression of K,
we have, by (56) and (67),
the quotient of which gives (229).
sin J a Bin J 4 .
. * A = —;—=— sm G
cos J 0
COB i K = cos J ffl COS J 6-f-siu Ja sin J 4 cos 0 cos J 0
> (230)
132. Since there are always two triangles upon the surface of the sphere which
have the same three sides, (Art. 110), the angles not being limited to values less than
180°, the formulse (225), (226)_, (227) should give the areas of both of them, and
their sum should be equal to the surface of the sphere = 4 jr. In fact, by (225),
Bin J .ff- may be either positive or negative, while by (226) the cosine is fully determined,
so that these formulse give two values ot^K whose sum is 2 a-, and therefore
two values of K, whose sum is 4^.
It lollows that (225) alone is not sufSciently determinate when the triangle is unlimited,
since it gives four solutions. The most convenient formula is therefore
(228), for we must always have J .ff < a-, and the double sign of the radical give«
the two values of J K, one lesB and the other greater than -3-.

232 SPHERICAL TRIGONOMETRY.
CHAPTER VI.
LIFFERENCES AND DIFFERENTIALS OF SPHERICAL TRLVNGLES.
133. Two parts of a spherical triangle being constant, and a third
receiving an increment, it is required to deduce the corresponding
increments of the remaining three parts. As in plane triangles,
(PL Trig. Chap. XII.), this will be effected by a comparison of two
triangles having two parts in common. The triangle formed from
the given one by applying the increments to the variable parts will
be distinguished as the derived triangle.
We shall first consider the increments as finite differences, and
give them the positive sign, (PI. Trig. Art. 187).
134. CASE I. A and c constant. The parts of
ABG, Fig. 22, being A, c, B, 0, a, b, those of the
derived triangle ABG' are J., c, i? 4- h.B, C -\- AC,
a -f A«, 5 -f A5; and the parts of the differential
triangle BCQ' are a,a-\- [^a,^b, 180° — G, G + LG
and A5. We have, then, in BOG', by (3),
sin A6 sin {a + i^d) sin a
sin Ai? ~ sin C sin(C4-AC^
Also, in BCG', by (40), we have
>vhence
sin i (180° - C 4- C 4- AC) _
tan|A6
sinJ(180°-C-C-AC) ~ tan J (« 4- Aa - a)
By (41) we
By (42),
tanjAa
taniA5
cos(C-fiAC)
COSJAC
find in a similar manner,
tan J A5
sin J AC
tan (a 4- J Aa)
sin (C 4-J AC)
(231)
(232)
(233)
sin J Aa sin (a -f |- Aa)
(234)
taniA.B tan(C4-iAC) ^ '

By (43),
DIFFERENCES OP SPHERICAL TRIANGLES, 233
tan|AC_ __ cos(a4-|Aa)
tanjA.B cos ^ La ^
By combining (232) and (233),
tan I Aa ^ _ tan {a + j Aa)
.^„„.
tan 1 AC tan (C 4-i AC) • ^
As these formute involve the increments in the second members,
they are to be computed by successive approximations. (See
PI. Trig. Art. 201).
135. CASE II. A and a constant. The given
triangle being ABC, Fig. 23, the parts of the
derived triangle A'BG are ^, a, ^ 4- A -S, 5 4- Ab,
C-f AC, C -f Ac. Although .the figure appears to'^'V
show that the angle B is diminished, it is still proper -^
to represent the angle A'B G hj B -h AS, to preserve uniformity
in the algebraic signs of the increments; the essential signs being
given by the equations of diiferences themselves. Hence we put
the angle AB A'= AB G - A'B G= B - {B-{• AB) = - A B.
Joining A A' we have in ^ J. J.' and CA A', by (43),
cos (c 4- i Ac) : cos 1 Ac = — cot i AB : tan i {A'AB 4- A A'B)
cos (5 4- i Ab) : cos J A6 = cot J AC : tan J {A'AC-hA A'G)
but since A is constant, or B A G = B A' G, Vfc find that the fourth
terms of these proportions are equal; whence
tan ^ AB _ cos {b -\- ^ A6) cos | Ac ^o-a"\
tan J AC COS (c 4- i Ac) cos J A6 \ ' )
In the polar triangle oi AB G, the constants are still an angle
and its opposite side, and the preceding equation applied to this
polar triangle (by Art. 8) gives
tan J Ab
tan J Ac
cos (S 4- i AB) COS I AC , ,
cos(C4-jAC)co3iA^ ^ '
In J. .B C and A'B C we have
sin tl sin S = sin A sin b
sin a sin {B f AB) = sin A sin {b -i- A?^
30 u2

£34 SPHERICAL TRIGONOMETRY,
the difference and sum of which give
sin a cos (S 4- J AB) sin J AB = sin A cos (5 -4- i AJ) sin J Ai
sin a sm (-B 4- J AB) cos J Ai? = sin A sin (S 4- J A5) cos J A5
from which, by division, we find
tan J A5 _ tan (S 4- J A5)
tan i AB ~ tan (54-1 AB)
(2391
and in the same manner
tan J Ac _ tan (c 4- ^ Ae)
tan J AC ~ tan (C4- i AC)
(240)
The product of (237) and (239) gives
whence also*
sin I Ab ^ _
tan J AC
sin {b -\- ^ Ab) cos | Ac
cos (e 4- J Ac) tan (-B 4- J AB)
(241)
sin J Ac
tan ^AB
_ sin (c -f I Ac) COS | .A5
cos(6 4-J A6)tan(C4-JAC)
(242)
Fig. 24.
a 136. CASE III. b and c constant. The given
c-,triangle being ABG, Fig. 24, the parts of the
derived triangle ABC are J, c, a -f Aa, S -f AB,
C4- AC, A 4- AJ.. Joining C G' we have in B C C,
ty (42),
sin (a 4- J Aa): sin J Aa = cot | AB : tan ^{BCG'—B G'G)
But observing that J. C = J. C, J. CC = J. C'C, we have
BCG' = AGG' - G
BC'C = AC'C + O-if AC
i {B OC -BG'0) = -{Q+l AG)
and the above proportion gives, therefore.
sin I Aa _
tan J AB
sin (a -f J Aa)
cot (C 4-i AC)
(243)
* The equations (239), (240), (241), and (242), contain each two factors less thin
the corresponding equations given by Cagnoli.

DIFFERENCES OF SPHERICAL TRIANGLES. 235
In the same manner we should find
sin I Aa ^ _ sin (a 4- j Aa)
tan J AC cot(B4-jAB) ^''*'"
The quotient of (243) and (244) gives
tan I AB _ tan (S 4- | AB)
tan J A"C ~ tan (C 4" J AC)
(245)
In J -B C and J. S C, by (4), we have
cos a = cos 6 cos e -f sin 5 sin e cos A
cos (a -f Aa) = cos 6 cos e 4- sin b sin c cos (-4. -f AA)
the difference of which gives
sin J Aa _ sin b sin o sin (A -j- ^
sin -J A J. sin (a -f J Aa)
The quotient of .(243) divided by (246) gives
AA)
(246)
sin i AA _ _ sin^ (a 4- | Aa) tan (C 4- j AC)
tan J AB sin b sin c sin (J -f J AA)
(247)
and from (244) and (246), in the same manner,
sin I AA sin^ {a + j Aa) tan (B + | AB)
tan J AC ~ sin b sin e sin (A -{- ^ AA)
(248)
137. CASE IV. B and C constant. The equations of the preceding
case (243 to 248), applied to the polar triangle, give
sin ^ AA _ sin (J 4- j- A.4)
tan § Ab cot (c 4- J Ac)
sin I- A J _ sin (A 4- i A^)
tan ^ Ac cot (5 4- i A6)
tan J A5 _ tan (5 4- | A5)
tan J Ac tan (c -f |- Ac)
sin J A J. _ sin B sin C sin (a 4- | Aa)
sin J Aa sin (J 4- i A^)
(249)
(250)
(251)
(252)

286 SPHERICAL TRIGONOMETRY.
sin i Aa _ sin' (-A 4- | AA) tan (c 4- | Ac)
tan J Ab ~ sin B sin C sin (a 4- J Aa)
sin i AA ^ sin^ (-A 4- | A.A) tan {b + j Ab)
tan J Ac sin B sin C sin (a -f- J Aa)
(253)
(254)
FINITE DIFFERENCES OF SPHERICAL RIGHT TRIANGLES.
138. All the preceding equations are, of course, applicable to
right triangles, or to quadrantal triangles, and in some cases they
assume simpler forms. Thus in Case I., if the variable C = 90°
(231) and (232) become
sin Ab = sin (a 4- Aa) sin
AB
tan J Aa = — tan J Ab tan J AC
and similar modifications take place in other cases.
139. When one of the constants is 90°, the preceding equations
do not generally assume any simpler forms, but they may be transformed
so as to involve the same variables in both members, which is
generally desirable in their practical applications.*
The method that we shall follow is so simple that it will be unnecessary
to repeat it in every case. A single example will suffice
to explain it.
Let C (= 90°) and b be the constants; to find the relation of As
and A B,^ we have between the two variables and the constant b, the
equations
sin B = sin 5 cosec c
sin (B -f AB) = sin b cosec (e 4- Ac)
the difference and sum of which, by PI. Trig. (105), (106), (131),
and (132), are
o rr, , -, •r.N . , r, 2 sin J cos (e -f i Ac) sin i Ac
2 cos (B 4- i AB) sm A AB = — -.—\ / ' ^—
sm c sm (c 4- A c)
n • rr, , -, -r-v , ' -r. 2 sin 5 sin (c 4-J Ac) cos i Ac
2sin(B4-*AB)cosiA.B=
-.—K—r^—^—r-^—
sm c sin (c -f Ac)
» Cagnoli gives these equations reduced so as to involve the same variables in
Doth members ; but in almost every instance his formulse involve two factors more
than are necessary, and are far less simple and convenient than those here given.

DIFFERENCES OF SPHERICAL TRIANGLES.
and the quotient of these is
23;
tan i AB
tan {B 4-1 AB)
tan J Ac
tan (c 4- J Ac)
which gives the first equation of the following article. This process
always eliminates the constant, and is applicable in every case.
When the equation to be differenced involves cosines, we employ
PI. Trig. (107) and (108); if tangents, (115) and (116); if cotangents,
(122); if secants, (129) and (130). The results are as
follows:
140. CASE I. C = 90° and b constant.
tanjAe
tanjA^
tan(e-|-jAc
tan(B4-iAB
tanjA c
taniAa
cot( e 4-JA c
cot(a -f jAa
(255)
sin A a
sin A A ~
sin(2 a-f A a
sin(2J.4-AJ.
tanJ^A a
sinAB^
tan(a -\-^Aa
'ain{2B+AB
(256)
sin Ac
tanjAJ ^
sin(2 e -f A c
cotp~+|AZ
tan^A^
tan|AB
ta,n{A 4-J A J.
• cot (B4-iAB;
(257)
141. CASE II. C = 90
sin(2JL-fAJ-
'sin{2B+AB
smA J _
sinAi?
tan JA a
tanjAj4
sm A a
tan|AB""
tan(a -f jAa
tan(J.4-iAJ
sin (2 a 4-A a
cot(B4-iA5
142. CASE III. C = 901° and A constant,
tan|A e
tan|Aa ~
tanjA e
sinAB
tan( c -FJA c
tan(a-i-2-Aa
cot(c 4-jAc
sin(2B4-AB
sinAe sin (2 c -t-Ac
sin A 5 sin (2 5-f A 5
and c constant.
tan JA a
tan JA b
tanJA b
tanjAB
sin A 5
taniA4
sin A a
tanjA b
tanJA b
tanjAB
tanJA a •
tanjAB
cot (a 4-JAa
"cot(64-jA5
tan( b+^Ab
tan(B4-iAB
sin(2 5 4-A5
'cot(J.4-iAJ.
sin (2 a-f A a
tan(5 4-jA6
cot{b-hiAb
cot(B4-iAB
cot(a4-jAa
tan(B4-jAB
(258)
(259)
(260)
(261).
(262)
(263)
143. If a constant side is 90°, the equations of finite differences
for the triangle may be obtained by applying the preceding eq-iations
to the jiolar triangle.

238 SPHERICAL TRIGONOMETRY.
DIFFERENTIAL YARIATIONS OF SPHERICAL OBLIQUE TRIANGLES.
144. To obtain the differential variations, we have only to maka
the increments infinitely small in the equations of finite differences,
observing the principles of PL Trig. Art. 192. Or we may differentiate
the equations of spherical triangles directly, employing the
differentials of the trigonometric functions given in PI. Trig. Art. 192
For example, A and c being constant, to find the relation of c^ a
and dB, we have
the differential of which is
da
5C
sin Aaxnc = sin a sin C
0 = sin a

DIFFERENTIAL VARIATH-lNS OF SPHERICAL TRIANGLES. 23!)
146. CASE II. A and a constant.
dB
d
d b
dB~
d b
d G
G~'
'
cos b
cos e
tan b
tanB
sin b
cos c tan B
147. CASE III. b and c constant.
d b
d G ~
d c
d c
da . ^ da
dB~' — sin a tan C
dG~
dB
d G~
dA
dB~
tanB
tan C
sin A
sin B cos C
d a
d~A~
dA
d C~
cosB
cos C
tan G
tan C
sine
cos b tan C
• sin a tan B
sin b sin C
sin J.
sin C cos B
(267)
(268)
(269)
(270)
(271)
(272)
1. CASE IV.
dA
d b~
d b
d G
d a
d b~
B and C constant
sin A tan c
tan 6
tan c
sina
sin 6 cos e
dA
d c
dA_
d a
d a
d G~
sin ^ tan b
ain B sin 'e
sin a
sin c cos 6
(273)
(274)
(275)
DIFFERENTIAL VARIATIONS OF SPHERICAL RIGHT TRIANGLES.
The preceding may also be used for right triangles; but it
may be desirable to have the same variables in both members, as in
the following formulae derived from those of Arts. 140, 141, and
142:
149. CASE I. C = 90° and b constant.
d 0 tan c d c
d B ~ ~ tan B d a ~
d a sin 2 a da
JA ^ sin 2 A CTB "
cot c
cot a
2 tan a
sin 2 5
do sin 2 c d^ _ tan A
d^ '^ TcotA ~d~B "" ~ coT^
{2m
(277)
^278^

240 SPHERICAL TRIGONOMETRY,
150. CASE II. C = 90° and c constant.
d A sin 2 A da
d~B^~ sin 2B ~dT
d a tan a d b
(TA. ~ tan A d B
da sin 2 a d b
d B 2 cot 5 (^ ^
cot a
cot b
tan b
tan B
sin 2 b
2 cot J.
151. CASE III. C = 90° and A constant.
d c tun c djz_ _ sin 2 a
Ta ^ tana a

APPROXIMATE SOLUTION OP SPHERICAL TRIANGLES. 241
CHAPTER VII.
APPROXIMATE SOLUTION OF SPHERICAL TRIANGLES IN CERTAIN
CASES.
154. WHEN some of the parts of the triangle are small, or nearly 90°, or nearly
180°, approximate solutions may be employed with advantage. These are generally
found by means of series.
155. In a spherical right triangle (the right angle being C), given A and c, to find b
We have
tan J = cos A tan c (288)
which is of the form in PL Trig. (493), and may therefore be developed by (495)
and (496) by putting x = b, y = c, p = cos A, whence
p — 1 1 — cos A , ,
g = I T = — rn i = — tan' J A
p + \ 1 -4- cos ^1 2
and (495) and (496) become, [taking n = 0 in (495), and n = 1 in (496)],
i = c — tan' \Aa\n2c+\ tan' J ^ sin 4 c — &c. (289)
b = TT—c 4- cot' J 4 sin 2 0 — J oof .} A sin 4 o 4- &c. (290)
If .4 is small, cos A is nearly equal to unity, and b exceeds c by a small quantity
which is approximately found by one or more terms of the series (289).
If A is nearly 180°, or cos^ nearly = — 1, S exceeds^-—c by a small quantity,
which is found by (290).
For examples of the mode of computation, see PI. Trig. Art. 255.
156. Althougli these solutions are termed approximate, it must not be inferred that
they are less accurate in practice tlian the direct solution of (288) by the tables ; for
the logarithmic tables are themselves only approximate, and the neglect of the
higher powers in such series as (289) and (290) may involve a less theoretical error
than the similar neglect of the higher powers in the series by which the tables are
computed. In the examples of PI. Trig. Art. 255, the thousandths of a second were
found with accuracy, which could not have been effected by a direct solution with
less than eight decimal places in the logarithms.
These considerations lead to the frequent employment of approximate solutions
in astronomy.
157. If A and b are given, to find c, we have
tan c = sec J. tan 5
whicli is reduced to PI. Trig. (493), by putting z •== c, y = b, p = sec A,
and the series -will be
sec.4 — 1 1 — 003^ , „, ,
secA + \ l-|-cos4. ^
c = b + tan' J 4 sin 2 i 4- J tan' J A sin 4 6 -f &c. (291)
c =!r—i — cot' J 4 sin 2 5 — J cot* I A ainib — &o. (292).
31 V

242 SPHERICAL TRIGONOMETRY.
158. Similar solutions apply to the equations of right triangles,
the last being solved under the form
tan o = sin 6 tan A
cotB = cos c tan A
tan (90° — B) = cos c tan A
We may also compute, in the same manner, the auxiliaries ip and S in (122) and
(134), so frequently employed in the solutions of oblique triangles.
159. In a right spherical triangle, given c and A, to find a, when A is nearly 90°.
We have
sin a = sin A sin c (293)
from which we deduce
tan J (c — a) = tan' (45° — ^ A) tan ^ (c + a) (294)
From this we may find c —a, which is supposed very small, by successive approximations.
For a first approximation, let a = ein the second member, and find thence
the value of c — a and of a; for a second approximation substitute in the second
member the value of a just found; and so on until two successive values agree as
nearly as may be desired.
EXAMPLE.
Given A = 89°, c = 87° ; find a.
Here 45° — J ^ = 0° 30', and for the first approximation i (o + a) = 87°.
log tan J (c 4-ffl) 1-28060
log tan' (45° — ^ A)
arCO log sin 1"
J (e _ ffl) == 299"-74 log i(c — a)
a = 87° — 9' 59"-48 = 86° 50' 0"-52
5-88172
5-31443
2-47675
2D APPKOX.
3D APPEOX.
4TH APPEOX.
i(c+a)
log tan ^ (c + a)
, tan' (45° — i A)
log ^ . .„ —-
Bin 1
log J (c — ffl)
i(c — a)
c
ffl
a
86° 55' 0"
1-26868
1-19615
2-46488
29r'-63
9' 43"-26
86° 50' 16"-74
86° 55' 8"
1-26899
1-19615
2-46514
291"-83
9' 43"-66
86° 50' 16"-34
86° 55' 8"-17
1-26900
1-19815
2-46515
291"-84
9' 43"-68
86° 50' 16"-32
The direct solution of (293) gives a = 86° 50' 16", but cannot give the fractions
of a second without tables of more than seven figure logs. We have given this problem,
however, not so much on account of its particular utility, as for the purpose
of introducing the method of approximation to which it leads, and which is often
employed.
The process here explained may obviously be applied to any equation if the form
sm X = m sm y
when m is nearly equal to unity.

APPROXIMATE SOLUTION OF SPHERICAL TRIANGLES. 243
160. In ffl spherical oblique triangle, given two sides and the included angle, to find the
other angles and side by series.
If a, b and C are the data, to find c, we have
Substituting half arcs,
cos c = cos fflcos i 4- sin a sin b cos G
sin' J c = sin' J a cos' J S 4- cos' J a sin' J b
— 2 sin J fflcos J 6 cos J ffl sin J S cos C
which is of the form Pi; Trig. (507), and may be developed by (508) by substituting
Bin I c for c, sin J fflcos J 6 for a, and cos J a sin J 6 for b ; so that (508) becomes
lo,Binic = logcosiaainib-M[^^cosC + { ^ J ^ ^ + .c] (295)
To find A and B, we have,
t^ni(A + B) = ''°'i(;~''KotiG
' cos J (ffl 4- 4) '
whence
t^nl(A-B) = '2^^^^lcotlG
^ ' sm J (a 4" o)
tan lln-l(A + B)-\ = ^-"^ f " "^^^"f f) tan J C
^ ^ Vcot Jo 4-tan Ji/ '•
. ri 1 , A T,^^ /tan J ffl4-tan iJ\ , „
*^°^^"-H^-^)] = Unjffl.-tanjJ^^°*^
Comparing these equations with PL Trig. (493), and developing by (495), 7» = 0,
irr-UA^B) = ,C-^^J^C+ (SH-^y ^ ^ - . c . (290)
i.-H.-.)=i. + £^.n.4-(Sf^y^%.c. (297)
If we develop by (496), we find
T 1 , , . ^ . ^ . /cot i ffl \ . „ /cot i a\ sin 2 C , „ ,„„„,
J._J(^ + ^ ) = _ j a 4 - ( ^ ) s m ( 7 - ( ^ - ^ J _ _ + &c. (298)
1 1 / .1 T,N 1^ /tanlfflX . ^ /tan Jffl \' sin2 C „ ,„„„^
from which a selection will be made in any particular case, according to the convergenoy
of the series. The terms of the series are in arc, and must be reduced to
seconds, by dividing by sin 1"-
This solution may be applied to the case where two angles and the included side
are the data, by means of the polar triangle.
161. To express the area of a spherical triangle in series.
Companug (229) with PI. Trig. (500), and developing by (502), we find
J £"= tan J ffl tan J 6 sin C — J tan' J a tan' J 6 sin 2 C 4- &c.
(3i)Ci)

244 SPHERICAL TRIGONOMETR":^.
162 LEGENDRE'S THEOBEM. If the sides of a spherical triangle are very smuil co'ipur'd
with the radius of the sphere, and a plane triangle be formed wliose sides are eqi'-al
to those of tlie spherical triangle, then each angle of the plane triangle is equal to the eorrisponduig
angle of the spherical triangle minus one-third of the spherical excess.
Let (.-, b and c be the sides of the spherical triangle expressed in .arc, the radius
of the sphere being unity; and let A', B' and C" be the angles of the plane triangle
whose sides are a, b and c. Then we have, in the spherical triangle,
. cos a — cos b cos c
cos ..4 =
sin b sin c
Substitute in the second member of this, the values of cos a, &c., in series, by
PI. Trig. (405) and (406), neglecting only powers above the fourth, viz.
we find
cos a = 1 — ^ a''+ 2j ffl'
cos 6 = 1 — J S' 4- jL i* sin 5 = * — -J 6'
cos c = 1 — J c' 4- j)^ c' sin c = c — J c'
cos A
io[l-i(6'4-c')]
• 6 J' c')
Multiplying the numerator and denominator hj 1 + i (b'' + c'), and neglecting
terms of a higher order than the fourth, as before, we have
cos 4 _ i'4- c' —"'.I ffl'4-i'4-(;' —2ffl'&' —2ffl'c' —2i'c'
2 o b ;, c , "T" ' 24 6 c
which, by PI. Trig. (225) and (239), becomes
cos A = cos A' — i be sin' A'
Let A = A' + z, then since x is small, we may put cos x = 1, so that, by
PI. Trig. (38),
cos A = cos A' — X sin A'
whence
a: = J J c sin A
But i be sin A — area of the plane triangle = very nearly area of the spherical
triangle = K, whence
=' = iS: A' = A — iK
The same reasoning applies to each of the other angles, so that
S'^B-iK
G'=G-iK
which proves the theorem.
^ 103. This theorem is applied in geodetical surveying, and is found to be sufficiently
accurate for triangles whose sides are considerably greater than 1°. It is to
be remembered that the sides are to be expressed in arc; and if they are given in
feet (for example), they must be reduced to arc by dividing by the radius in feet,
or, which is equivalent, the area must be divided by the square of this radius. If
then 7-= radius of the earth in units of any kind, a, b and e the sides of the triangle
in units of the same kind, and k the area of the plane triangle, we shall have
K in seconds, by the equation
K=-
k
r' sin 1"
EXAMPLE.
In a triangle upon the earth's surface, given b = 183496-2 feet, c — 1561221 feet
and A = 48° 4' 32"-35 ; to find the remaining parts.

APPROXIMATE SOLUTION OF SPHERICAL TRIANGLES. 245
We have k = }, b c sin ..l.and the mean value of ?- = 20888780 feet. Hence
log b 5-26363
log c 5-19346
log sin ^ 9-87159
arCO log 2 r' sin 1" 0-37356
K = 5"-04 log K 0-70224
It is evident that great accuracy in the value of r and of the other data is not
required in computing K. We now have ^ K = l"-68. A' = 48° 4' 30'--67, and by
solving the plane triangle with the data A', b and c, we iind
ffl = 140580-0 feet B' = 76° 12' 22"-19 C" = 55° 43' 7"-13
Adding ^ K to each of these angles, the angles of the spherical triangle are
B = 76° 12' 23"-87 C = 55° 43' 8"-81.
For further details respecting geodetical triangles, and for the methods of solving
spheroidal triangles, special works upon geodesy must be consulted, such as
Legendre's Analyse des Triangles traces sur la surface d'une sphSroide; Puissant's Traile
de Geodesic; Vnisssmt's Nouvel essai de trigonometric spheroidique; Wisohex's Lehrbuch
der hoheren Geoddsie; various papers by Gauss, Bessel, &c.
164. To solve a spherical triangle when two of its sides are nearly 90°.
If a and b are nearly 90°, c and C are nearly equal, and it will be expedient to
compute the small quantity O— c by an approximate method. We have, by (25),
sin' J c = sin' ^ (a + b) sin' J C -f sin' J (ffl — 5) cos' J C
and by PI. Trig.
sin' J C= [sin' ^ (a + b) + cos' J (ffl 4- *)] sin' } G
the difference of which equations is
sin J (C 4- c) sin ^(0 — c) = cos' i (a + b) sin' J C — sin' } (a — 5) cos'* 0
Let
ffl' = 90° — ffl i' = 90° — b
a' and b' being very small: also, since 0 and c are nearly equal, put
i(C+c)=.C
then the above equation becomes
sin C sin J (C—c) = sin' J (ffl' 4- b') sin' J C — sin' | (a' — b') cos' J C
Dividing by sin C = 2 sin J C cos J C, and substituting the arcs J {C — c),
J (ffl' -|- b'), J (ffl' — b'), for their sines, we find
C-c = Bin 1" [ ( ^ y tan J C _ ( ^ 7 cot J c] (301)
which is the required approximate formula for the case when ffl', b' and G are given
to find e.
If a', 4' and c are given, to find C, we may exchange G for c in the second member,
whence
G-c = sin 1" [ ( ^ ' ) ' tan J c - ( ^ ) ' cot J .] (802)
V2

246 SPHERICAL TRIGONOMETRY.
CHAPTER VIII.
MISCELLANEOUS PROBLEMS OF SPHERICAL
TRIGONOMETRY.
165. In a given spherical triangle, to find the perpendicular from one of the angles upon
the opposite side.
Kg. 25.
Denoting the perpendicular upon the side e (Fig. 25)
by p, we have
sin^ = sin b sin A (303;
If the three sides or the three angles are given, we
find by (48), or (51), and
2n 2N
sm p 5= sin sm -: c =
sin (7 n ^ '
in which n and iVare given by (47) and (50).
If we admit more than three parts of the triangle into the expression of p, we
have, by (55), (56), and (303),
„•„.„ 2 sin J ^ sin J B . 2 cos} ffl cos ii
smp = cos -,„ J 1 C n sm s = r? , _ ^ cos S (805)
sin J c
166. To find the radius of the circle described about a given spherical triangle.
Kg. 26. The radius here understood is the aro £)^ = 0 B r=
0 C, Fig. 26, drawn from the pole of the small circle
AB G to either of the angles. Let
then
0,4.8= OBA = x
G=OGA+OCB=OAG+OBG
= A — X + B — z
putting S=l(A + B+0).
z = i(A + B—G) = S—G
The triangle AOB being isosceles, the perpeu-licular
0 P bisects the side c, therefore if 0.4 =: iS, we have
or, by (70),
, _ tan 1 e tan i c
tan R = i— =
cosz cos (^S — 0)
_ 2 sin J ffl sin J b sin J c
tan li ^= ~—
(806)
(3071
By applying the principles of Art. 37, this will give the corresponding formulse of
PI. Trig. (285).
167. From (69) and (70) we find
cos (S— 0) = — cos 5cot J ffl cot J b
by which (306) is reduced to
. „ tan J a tan J b tan J c
•— cos JS
'?Off.

MISCELLANEOUS PROBLEMS.
Also, by the last equation of (56), (306) becomes
or, by (50),
tan R =
J. r> sin J c
tan .S = ±
cos J fflcos ,} b sin G
168. Substituting in (306) for tan J c by (39),
— cos S
tan R — ^\oo6(S (S—A) — cos (-S— B) cos (Scos
S
If
169. Let the sides of the triangle ABC, Fig. 27, be produced
to meet in A', B', and C" ; and denote the radii of
the circles circumscribed about 4'7? C, B'A 0, O'AB by
R', R", R'" respectively. Then if 2 S' denote the sum of
the angles of A' B 0, (A, B and C being the angles of
ABC),
j
2S'= 2^ — B— C+A
S' — A' = w — i (A + B + C) z= TT — S
so that (306) applied to A' B G gives
and in like maimer
tan R' =
tan M" =
tan J ffl
cos (-S" — A')
tan ^ 5
— cos S
tan i c
tan R'" =
— cos S
• cos S
Substituting for tan § ffl, &c., by (39), or for cos S by (69),
ro)
Fig. 27
247
(809,
(310)
(311)
cos (S — A) 2 sin J a cos J b cos J c
tan B':
iV
COS (S — B) 2 cos I ffl sin | & cos } e
tan iJ" =
I (312)
N
cos (5 — G) 2 cos ^ fflcos } 5 sin } c
tan R'" =
N
n
170. Combining (310) with (312), we find the relation
cot R cot R' cot R" cot B'" = iV"'
(313)
If this be multiplied successively by the squares of (310) and (312), we obtain
tan R cot .8' cot B" cot R'" = cos' 5
cot B tan B' cot R" cot J!'" = cos' (S-A)
(314)
cot iZ cot R' tan iJ" cot iS'" = cos' (S — B)
cot R cot ii' cot R" tan i?'" = cos' (S — C)

248 SPHERICAL TRIGONOMETRY.
171. Again, from (310) and (312) we find
— tan iJ 4- tan R' + tan R" + tan R'"
_ cos S + cos (S — A) + cos (S — B)+ cos (S •
= ~ N
_ 2 cos ^ A cos i (B + C) + 2 cos ^ A cos i (B
~ N
whence
4 cos J A cos J B cos } G
— tan iJ 4- tan R' + tan R" + tan R'" =
(315)
We shall find in a similar manner
4 cos } ^ sin } 5 sin J C
tan R — tan R' + tan R" + tan R'"
If
4 sin J J4 COS ^ B sin^ G
tan B + tan i2' — tan R" + tan R'" =
(316i
If
4 sin J ^ sin J B cos J 0
tan .K 4- tan R' + tan i2" — tan R'" =
N
It is also easily shown that
tan' R + tan' R' + tan' R" + tan' i2"' = 2 4- 2 cos J1 COS B COS (7 (317)
172. To find the radius of the circle inscribed in a given spherical triangle.
Kg. 28. ^ In Pig, 28, 0 being the pole of the required circle,
draw OP', OP" and OP'" to the points of contact, and join
OA, OB. We have OP" = OP'" and the triangles
ip, A OP" and A OP'" right-angled at P" and P'"; hence
Bin OAP" =
sin OP" sin OP'"
sin ^ 0 sin .4 0 = sin0.4P"'
therefore 0 A P" = 0 A P'", (for we cannot have
0A F" := TT — OA P'"), and the pole of the inscribed circle is consequently found
by the same construction as inplano, namely, by bisecting the angles of the triangle.
If then we put s z=^ ^ (a + b + c), and r = radius of the inscribed circle, we
have
AP'" + BP' + GP' = AP'"+ a = a, AP'" = s — a
and the right triangle A 0 P'" gives
tan r = sin (a — ffl) tan } A (818)
corresponding with the formula of PI. Trig. (288).
Substituting, in (318), the value of tan } A,
^sin (s — a) sin (s — 6) sin (s — c)\
sin s /
tan r -.
'J(c
tan r = •
Substituting, in (318), the value of sin (s — a) given by (58),
^^^ ^
^
2 cos J A cos J B cos J G
Also, by (51), we have iV = } sin B sin G sin a, which reduces (320) to
Bin i B sin i C .
tan r = =—=~-r^— sm a
'.OS ^ A
(319)
(320)
(321)

MISCELLANEOUS PROBLEMS. 249
173- Let the radii of the circles inscribed in the three triangles A'BG, B'AG,
CAB of Fig. 27, be r', r" and r'". Then if V denote the half sum of the sides of
A'BG, we have
2s' = 27r — b — c+a
s' — a =z 7r — ^ (a + b + c) =z TT — a
BO that (318) applied to the throe triangles, gives
tan r" = sin a tan J B (322)
Substitutiug in these the values of tan \ A, &c., or of sin s,
Also, by (321),
tan /
tan r"
tan r''
n
sin (s — a)
n
sin (s — b)
n
sin. (s — c)
N
2 cos J .4 sin f iS sin J G
N
2 sin \ A cos \ B s-i.n\ G
iv;
2 sin J J. sin J iJ cos J C
cos If B cos if G .
tan r = = -,—i-^— sm a
cos J A
cos J C cos \ A . ,
tan r" = -— T—n — ^ *
cos f B
cos J -4 cos i B .
tan r'" = ^ r-PT— ^m c
cos J 0
174. The product of (319) and (323) gives
tan r tan / tan r" tan /" = -^-^ = «*
whence, as in Art. 170,
cot r tan r' tan /' tan r'" := sin' a
tan ?- cot / tan r" tan /" = sin' (s — a)
tan r tan r' cot r" tan /" = sin' (s — 6)
tan r tan r' tan r" cot »-'" = sin' (s — c)
- (323)
. (324)
(325)
(326)
175. We find from (319) and (323), as in Art. 171,
4 sin J a sm J J sin J c
— cotr + cot r' -f cot r" + cot /" =
cot r — cot / + cot J-'-f cot r"' =
4 sin J fflcos J 6 cos } c
4 cos J a sin J S cos J e
cot r -4- cot r' — cot r" 4- cot r'" =
(32V)
4 cos } a cos } S sin J e
cot r 4- cot r' 4- cot r" — cot r"' =
oof r 4- cot V 4- cot' r" + cot' j'" = 2 ' — 2 cos fflcos J cos c
82

250 SPHERICAL TRIGONOMETRY.
176. From (309) and (321), we find
tan r
tan R
From (307) and the first of (327),
From (315) and (320),
4 Bin J J1 sin J .S sin J C cos J a cos J h cos J o
— cot r + cot r' 4- cot r" + cot ?-'" :
— tan iB -J- tan R' + tan R" + tan .B"
: 2 tan iJ
: 2 cot »-
(328)
(329)
(330)
and other similar relations are found by comparing (312) with (327), and (31G)
with (323).
177. The following relations are also worth remarldng.
If f is the perpendicular from G upon c.
tan R a\np =
cot r sin^ =:
2 sin J a sin \ b
cos J c
2 cos ^A cosJJS
sin J 0
y (331)
178. TAe ^o?c of the circle inscribed in a spherical triangle is also the pole of the circle
circumscribed about the polar triangle ; and the radii of these circles are complements of each
other.
The arcs bisecting the angles of a given triangle will evidently bisect the sides
of the polar triangle, and wUl be perpendicular to those sides respectively; the
common intersection of these arcs is therefore at once the pole of the circle inscribed
in the first and circumscribed about the second.
Again, if we join the angular points of the polar triangle with this common pole,
the arcs thus drawn, being produced to meet the sides of the first triangle, are
perpendicular to those sides, and therefore pass through the points of contact of
the inscribed circle. Each of these arcs = 90°, and is at the same time the sum
of the two radii of the circles in question.
This latter property is also obvious from the analytical expressions of the two
radii. By means of it, we might have deduced all the formulse for the inscribed
from those for the circumscribed circle, or vice versa.
179. To find the are joining the poles of the circles inscribed in, and cirmmseribed about
a qiven spherical triangle.*
Let O be the pole of the circumscribed circle,
Fig. 29, and 0' that of the inscribed circle. Put
00' = D; then
coaD = cos J. 0 cos .40'4" sin ^0 sin ^0'cos OAO'
By Art. 166, we have 0.4.B = S — G, whence
OAO' = S— G—iA = i(B—G)
We have also
cos .40' = cos O'P cos AP = cos r cos (s — a)
sin O'P
Bin AC : sin O'AP
sin r
sin ^ A
* Hymer's Spherical Trigonometry

Ther'fore,
cos D
cos R sin T
Substituting by (819), (307), and (44),
MISCELLANEOUS PROBLEMS. 251
1. , % I J. r. COS i (i? — C,
:= cot r COS (s — a) + tan .K -.—,—^—'
^ ' sm^ A
COS D sin s cos (-s — a) -|- 2 sin J i sin J c sin \(b + c)
I as R sin r
whence, jy (53),
sin a 4- sin b + sin c
2 n
' cosi? V - 1 4-sin a sin 6 4-sin a sin c 4-! sin b sin c — cos a cos 6 cos e
\cos R sin r)
2 n'
by PI. Trig. (17
sin s 4- 2 sin J ffl sin J 5 sin J c
(
)•
= (ootr 4-tani?)'
cos' D = cos' {R — r) + cos' iJ sin' r
sin' 7) = sin' (ie — ?•) — cos' iJ sin' r (832)
If the inscribed circle is inscribed in A'BG, Fig. 27, and its radius = /, w«
have, by a similar process,
sin' D' = sin' (R + r') — cos' R sin' /
(0-U)
180. To find the equilateral spherical triangle inscribed in a given circle.
U R = radius of the given circle, and A = one of the angles of the equilateral
triangle, we have, by (310), and PI. Trig. Art. 76,
• cos f ^
tan' R = •
cos" J A
3 COS J -4 — 4 cos' J A
cos' J A
whence
0Si-4=J(; 4 4- tan' B
(334)
181. To find the equilateral spherical triangle circumscribed about a given circle.
If 7- = radius of the given circle, and a = one of the sides of the triangle, we find
sm J • = J( 4 4- cot' r
(335)
182. Given the base and area of a -tphirical triangle, to find the locus of the vertex.
Kg. ao. Let a s= the given base, and K = area of ABC, Fig. 30.
Produce AB and ^O to meet in A'. Let 0 be the pole
of the circle described about A'BG. The radius of this
circle is given by the first equation of (311), which, by
(224) becomes
tan R'
tan J ffl
sin J K
(336)
The second member of this equation, being constant for
all the triangles of the same base a, and the same area A",
shows that R'ia also constant, and consequently, that the

252 SPHERICAL TRIGONOMETRY.
point A is always found upon the circumference of the same small circle A'BC.
But ^4 and A' being the extremities of the same diameter of the sphere, A is als»
found upon a small circle, equal and parallel to the circle A'BC.
The perpendicular distance (p') of 0 from the base BO, is found by the equation
cos R'
coap = r—
cos J ffl
and the pole of the locus of ^ is in the same perpendicular, at a distance from
BG = B- —p' = p, whence
cosig'
COS P = 1— (°°'J
•^ COS J ffl ^ '
The equations (336) and (337) determine the radius and position of the pole of the
required locus, wliich may therefore be constructed.
This elegant proposition is due to Lexell.
183. To find the angle between the chords of two sides of a spherical triangle.
Kg. 31.
In Fig. 31, 0 being the center of the circumscribed circle, the angle between tho
chords of the sides AG and BG ia half the spherical angle AOB. If, then,
we have
0, == angle between the chords of a and b
cos 0, = cos A OP = sin OAF cos AP
or, by Art. 166, cos 0, = sin (S — C) cos J c (338)
By (72) this becomes
cos 0, = sin J ffl sin J 6 -|- cos J ffl cos J 4 cos 0 (339)
184. The preceding problem is employed for geodetical triangles, in which C,
differs very little from C, in which case it is expedient to compute the small difference
G — 01 = 2;. We easily reduce (339) to the following:
.SOS 0, := COS J (ffl — b) cos' JO— COS J (ffl 4- i) sin' J C
= COB' J O —2 sin' i(a — b) cos'J 0—-sin' J 04-2 sin' i(a+b) sin' J 0
Subtracting cos 0 = cos' JO — sin' J 0, we have,
Bin J (04- 0.) sinj (0—0.) = sin'i(ffl4-J) sin'J 0—sin'^ (ffl—J)cc8'JO
or approximately, taking
sin J (O4- 0,) = sin 0 = 2 sin J 0 cos J O
and sin J(0—0.) = J3;sinl"
X being expressed in seconds,
1 1
" ^ alTF ^'"'^ (" -t- i) tan J O— ^j^-p sin' J (a — i) cot } O '840)

MISCELLANEOUS PROBLEMS. 253
18!^ TJ a great circle (DE, Pig. 32) bisect the ba.^e of - spherical triangle at rigln
angles, any great circle (FG), perpendicular to it, dividc.-< the sides [AC, BC) into segments
whose sines are proportional; that is,
sin FA : sin FC = am GB : sin GC
Jjet P be the pole of ED, (DP = 90°), and
PGF any great circle drawn through P, and
therefore perpendicular to DE. Then, since
we have, by (3),
PB + PA = 2 PD = 180°
(341)
sin F sin FA = sin P sin PA = sin P sin PB
= sin G sin GB.
sin .f sin FC = sin O sin GO
whence, by division, the theorem (341). The arc FG is analogous to the parallel to
the base in plane triangles.
186. If two arcs of great circles, (AB, CD, Fig. 33), terminated by any circle, intersect,
the products of the tangents of the semi-segments are equal to one another; that is,
tan J AE tan ^ EB = tan J CE tan J ED
(342)
Let P be the pole of the ctrcle DACB. Join PE ^'S- ^•
and draw the perpendiculars PF, PG, bisecting the
arcs AB and CD. Then we have
cos FE cos PE cos PE cos GE
cosi^i? COS PB cos PD coa GD
cos FE — cos FB
cos FE + cos FB
cos GE — cos GD
cos GE + cos GD
which, by PI. Trig. (110), gives (342).
187. If three arcs be drawn from the angles of ffl spherical triangle through the same
point, to meet the opposite sides, the products of the ainea of the alternate segments of the sides
will be equal.
Thus, in Fig. 34, we shall have
sin AB' sin 0.4' sin .BO' = sin GB' sin BA' sin A G' (343)
For we easily find
sin AB' _ sin AP _ sin APB'
sin GB' ~ sin GP ' sin CPB'
sin 0.4' sin GP sin 0P.4'
sin BA! sin BP sin BPA'
ain BC
'a\n AG'
sin BP
sin BPG'
smAPO'
Multiplying these equations together, the product of the second members is unity,
whence (343).
Tho same property is easily extended to the segments of the angles.
188. It follows, that when three arcs .are drawn from the three angles, so as to
satisfy the condition (343), they must intersect in the same point. This occurs in
the same cases as in plane triangles, that is, when the angles are bisected - when the
sides are bisected ; when the three arcs are drawn from the angles to the points of
W

254
SPHERICAL TRIGONOMETRY.
contact of the inscribed circle ; and when the three arcs are the three perpendiculars
upon the sides.
The first three of these cases are obvious. To prove the last, if A', B' and C,
Fig. 34, are right angles, we have
cos .45' cos 0.4' cos BO' cos AB cos CA cos BC = 1
cos CB' cos BA! " sin P"
Bin a = .—p.— sin P = "--.
sm y • sin J.

whence
and similarly
.MISCELLANEOUS PROBLEMS.
255
sin it =
sin ^
sin y
sin /S'
sin a.' =
sin y'
, sin y" sin P"
sin Q
sin )." sin P"
sin g
sin /3" _ sin y" sin P"
sin at!' =
sin y" sm Q
which, substituted above, give
^-^ • , , m , sin^' • I , » , sin R" .
-: sm (a' — a") -i !_ sm (a" — a) 4- jv^-^ sin (a _ a') = 0 (345
smy ^ ' smy
Again, if we express (344) in the notation of this article, it becomes
cos (/S4- y) sin (a'—a")4-cos (/S'-l- y) sin (a"—a) 4- cos (fS'+ y") sin (a—a')=0 (346)
which, added to (
45), gives
-^)sin(a'-a")4-"'^^^'+>'^sin(a"-a)-F^^f "+„>")
t&ny ' tany ^ ' ' tany
191. If P is the pole of ^ Q, we have
and (345) and (347) both give
^ 4- J. = /S' 4- y = ^"4- y = 90°
sin(a—a')=0 (347)
tan /3 sin (a! — a") + tan /S' sin (a" — a) -f- tan /S" sin (a — a') =5 0 (348)
192. To find the inclination of two adjacent faces of a regular polyhedron, and the radii
of the inscribed and circumscribed spheres.
Let 0 and E, Fig. 36, be the centres of two adjacent faces whose common edge is
AB; 0 the centre of the inscribed and circumscribed spheres. Draw 0 D bisecting
AB at right angles; draw CD, ED, which wiU also evidently be perpendicular
to J1 P ; and put
/ = inclination of the faces = GD E
R = radius of the circumscribed sphere = 0A = OB
r •= radius of the inscribed sphere =00=: 0 E
a = one of the edges = .4 P
m = number of faces that form a solid angle
n = number of sides of a face
oi
Suppose a sphere to be described about the centre 0
with any radius, and cad the triangle formed upon its surface
by the planes GOD, COA, AO D; this triangle is
right-angled at d and gives
But cos c d ^ cos GOD, and
cos cad
cos ed =s sin acd
O 01? = 90° — GDO =
-^-\I
cad •= \ angle of the planes 0 AG and 0 A!E
2 TT
m
tcd=
^AOB-

256
SPHERICAL TRIGONOMETRY.
therefore
sin J i' ^
(3-J'B)
Then from the triangles 0 0 D, AG D, kc,
we find
!- = — tan \ I cot •
2 ^
(350)
J- = P cos ac = R cot ac(icotca(Z=: P cot — cot —
n m
P = — tan i / tan —
2 ^ m
(-351)
193. Jlj .^n(Z the surface and volume of ffl regular polyhedron.
Let / = number of faces of the polyhedron ; 5 = the surface, and V= the
volume; then the area of each face (the notation of the preceding article being
continued) is equal to
whence
A AB X OD xn = a'- — cot —
i n
S = ffl' - -f cot —
4 n
(352)
and since V=:: S i