082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

144 Chapter 3 The trigonometric functions
TechnicalComputingExercises3.7
1 Ploty=sin2tfor0t 2π.
2 Ploty=cos3tfor0t 3π.
( )
t
3 Ploty = sin for0t4π.
2
( )
2t
4 Ploty = cos for0t6π.
3
5 Ploty = sint +3cost for0 t 3π.Byreading
from yourgraph,statethe maximumvalue of
sint+3cost.
6 (a) Ploty=2sin3t−cos3tfor0t 2π.
Useyourgraph to findthe amplitudeof
2sin3t −cos3t.
(b) On the same axes ploty = sin3t.Estimate the
time displacement of2sin3t −cos3t.
3.8 TRIGONOMETRICEQUATIONS
Weexaminetrigonometricequationswhichcanbewritteninoneoftheformssinz =k,
cosz = k or tanz = k, wherezis the independent variable andkis a constant. These
equationsallhaveaninfinitenumberofsolutions.Thisisaconsequenceofthetrigonometric
functions being periodic. For example, sinz = 1 has solutions z = ..., −7π
2 ,
−3π
2 , π 2 , 5π 2 ,....Thesesolutionscouldbeexpressedasz = π 2 ±2nπ,n=0,1,2,....
Sometimesitisuseful,indeednecessary,tostateallthesolutions.Atothertimesweare
interestedonlyinsolutionsinsomespecifiedinterval,forexamplesolvingsinz = 1for
0 z2π. The following examples illustratethe method of solution.
Example3.15 Solve
(a) sint = 0.6105 for0 t 2π (b) sint = −0.6105 for0 t 2π.
Solution Figure 3.20 shows a graph ofy = sint, with horizontal lines drawn aty = 0.6105 and
y = −0.6105.
(a) From Figure 3.20 we see that there are two solutions in the interval 0 t 2π.
These aregiven bypoints Aand B. We have
sint = 0.6105
and so, usingascientific calculator, wehave
t = sin −1 (0.6105) = 0.6567
This is the solution at point A. From the symmetry of the graph, the second solution
is
t = π −0.6567 = 2.4849
This isthe solution atpointB. Therequired solutions aret = 0.6567,2.4849.
(b) Again from Figure 3.20 we see that the equation has two solutions in the interval
0 t 2π.ThesearegivenbythepointsCandD.Onesolutionliesintheinterval
π to 3π 2 ; the other solution liesinthe interval 3π 2
sint = −0.6105
to2π. We have