A Level Fur ther Mathematics for AQA
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Brighter Thinking<br />
A <strong>Level</strong> <strong>Fur</strong> <strong>ther</strong><br />
<strong>Mathematics</strong> <strong>for</strong> <strong>AQA</strong><br />
Student Book 1 (AS/ Year 1)<br />
Stephen Ward and Paul Fannon<br />
This resource has been entered into the <strong>AQA</strong> approval process.
Contents<br />
Contents<br />
Introduction.............................................................. iv<br />
How to use this book................................................v<br />
1 Complex numbers<br />
1: Definition and basic arithmetic of i......................1<br />
2: Division and complex conjugates.......................6<br />
3: Geometric representation....................................9<br />
4: Locus in the complex plane...............................20<br />
5: Operations in modulus–argument <strong>for</strong>m...........24<br />
2 Roots of polynomials<br />
1: Factorising polynomials......................................31<br />
2: Complex solutions to polynomial<br />
equations.............................................................34<br />
3: Roots and coefficients........................................38<br />
4: Finding an equation with given roots...............43<br />
5: Trans<strong>for</strong>ming equations......................................48<br />
3 Graphs and inequalities<br />
1: Cubic and quartic inequalities...........................55<br />
2: Functions of the <strong>for</strong>m y =<br />
ax + b<br />
cx + d .......................58<br />
2<br />
3: Functions of the <strong>for</strong>m y =<br />
ax + bx + c<br />
2<br />
...............62<br />
dx + ex + f<br />
4: Oblique asymptotes...........................................65<br />
5: Reciprocal trans<strong>for</strong>mations of functions...........66<br />
6: Modulus trans<strong>for</strong>mation.....................................71<br />
4 Hyperbolic functions<br />
1: Defining hyperbolic functions and<br />
hyperbolic identities...........................................80<br />
2: Hyperbolic identities..........................................85<br />
3: Solving harder hyperbolic equations................87<br />
5 Series<br />
1: Sigma notation....................................................92<br />
2: Using standard <strong>for</strong>mulae....................................94<br />
3: Method of differences........................................99<br />
4: Maclaurin series.................................................103<br />
Focus on … Proof 1.............................................111<br />
Focus on … Problem solving 1...........................113<br />
Focus on … Modelling 1.....................................115<br />
Cross-topic review exercise 1.............................117<br />
6 Matrices<br />
1: Addition, subtraction and<br />
scalar multiplication..........................................120<br />
2: Matrix multiplication.........................................126<br />
3: Determinants and inverses of<br />
2×<br />
2 matrices..................................................... 131<br />
4: Linear simultaneous equations........................140<br />
7 Matrix trans<strong>for</strong>mations<br />
1: Matrices as linear trans<strong>for</strong>mations................... 147<br />
2: <strong>Fur</strong><strong>ther</strong> trans<strong>for</strong>mations in 2D..........................155<br />
3: Invariant points and invariant lines..................162<br />
4: Trans<strong>for</strong>mations in 3D....................................... 167<br />
8 <strong>Fur</strong><strong>ther</strong> vectors<br />
1: Vector equation of a line..................................177<br />
2: Cartesian equation of the line.........................184<br />
3: Intersections of lines.........................................190<br />
4: Angles and the scalar product.........................193<br />
5: The vector product...........................................202<br />
9 Polar coordinates<br />
1: Curves in polar coordinates.............................212<br />
2: Some features of polar curves.........................215<br />
3: Changing between polar and Cartesian<br />
coordinates........................................................219<br />
10 <strong>Fur</strong><strong>ther</strong> calculus<br />
1: Volumes of revolution.......................................225<br />
2: Average value of a function.............................230<br />
Focus on … Proof 2............................................ 236<br />
Focus on … Problem solving 2.......................... 238<br />
Draft sample<br />
Focus on … Modelling 2.....................................241<br />
Cross topic review exercise 2............................244<br />
Practice paper 1 ...................................................248<br />
Formulae................................................................250<br />
Answers to exercises ............................................253<br />
Glossary .................................................................296<br />
Index ..................................................................... 297<br />
Acknowledgements............................................. 000<br />
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iii
4 Hyperbolic functions<br />
In this chapter you will learn how to:<br />
••<br />
define the hyperbolic functions sinh x, cosh x and tanh x<br />
••<br />
recognise and use the graphs, range and domain of the hyperbolic<br />
functions<br />
••<br />
work with identities involving hyperbolic functions<br />
••<br />
write the inverse hyperbolic functions in terms of logarithms<br />
••<br />
solve equations involving hyperbolic functions.<br />
Be<strong>for</strong>e you start…<br />
AS/A <strong>Level</strong> <strong>Mathematics</strong><br />
Student Book 1,<br />
Chapter 7<br />
AS/A <strong>Level</strong> <strong>Mathematics</strong><br />
Student Book 1,<br />
Chapter 3<br />
AS/A <strong>Level</strong> <strong>Mathematics</strong><br />
Student Book 1,<br />
Chapter 10<br />
AS/A <strong>Level</strong> <strong>Mathematics</strong><br />
Student Book 1,<br />
Chapter 10<br />
AS/A <strong>Level</strong> <strong>Mathematics</strong><br />
Student Book 1,<br />
Chapter 5<br />
AS/A <strong>Level</strong> <strong>Mathematics</strong><br />
Student Book 1,<br />
Chapter 9<br />
You should be able to work with<br />
natural logarithms.<br />
You should be able to solve<br />
quadratic equations.<br />
You should be able to work with<br />
the symmetries of trigonometric<br />
functions.<br />
You should be able to work with<br />
trigonometric identities.<br />
You should be able to interpret<br />
trans<strong>for</strong>mations of graphs<br />
graphically.<br />
You should be able to complete<br />
binomial expansions of brackets<br />
with positive integer powers.<br />
What are hyperbolic functions?<br />
Trigonometric functions are sometimes called circular functions. This<br />
is because of the definition that states: a point on the unit circle (with<br />
2 2<br />
equation x + y =1) at an angle θ to the positive x-axis, has coordinates<br />
(cos θ,sin θ ).<br />
2 2<br />
A related curve, with equation x − y =1, is called a hyperbola. Points on<br />
this hyperbola have coordinates (cosh θ,sinh θ) although θ can no longer<br />
be interpreted as an angle.<br />
x<br />
1 Given that 2 = 5, write x in the <strong>for</strong>m ln a<br />
ln b<br />
.<br />
2<br />
2 Solve x + 3x = 1.<br />
3 Given that sin 40°= a find sin 220°.<br />
2 2<br />
4 Simplify 3sin 2x+<br />
3cos 2x.<br />
5 What trans<strong>for</strong>mation changes f( x) to<br />
3f( x + 1) ?<br />
6 Expand (2 + x ) 3 .<br />
Draft sample<br />
–1<br />
y<br />
1<br />
O 1<br />
–1<br />
O<br />
y<br />
(cos , sin )<br />
x<br />
(cosh , sinh )<br />
x<br />
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79
A <strong>Level</strong> <strong>Fur</strong><strong>ther</strong> <strong>Mathematics</strong> <strong>for</strong> <strong>AQA</strong> Student Book 1<br />
Section 1: Defining hyperbolic functions and<br />
hyperbolic identities<br />
Although the geometric definition of hyperbolic functions gives some<br />
helpful insight, a more useful definition is related to the number e.<br />
Key point 4.1<br />
• cosh x =<br />
e<br />
• sinh x =<br />
e<br />
e<br />
2<br />
x x<br />
+ −<br />
x x<br />
− e<br />
−<br />
2<br />
You can define the tanh x function by analogy with the trigonometric<br />
definition (of tan x).<br />
Key point 4.2<br />
O<br />
1<br />
y<br />
y = cosh x<br />
sinh x x<br />
tanh x ≡ ≡<br />
e − e<br />
x<br />
cosh x e + e<br />
There are not many special values of these functions that you need to<br />
know, but from Key points 4.1 and 4.2 you should be able to see that<br />
cosh (0) = 1, sinh (0) = 0and tanh (0) = 0.<br />
To investigate the domain and range you need to consider the graphs.<br />
Key point 4.3<br />
x<br />
Function Domain Range<br />
cosh x<br />
All real numbers y 1<br />
−x<br />
−x<br />
y<br />
O<br />
y = sinh x<br />
sinh x<br />
All real numbers All real numbers<br />
tanh x<br />
All real numbers − 1< y < 1<br />
x<br />
Fast <strong>for</strong>ward<br />
In <strong>Fur</strong><strong>ther</strong> <strong>Mathematics</strong> Student<br />
Book 2, you will see that the<br />
trigonometric functions can<br />
also be defined in a similar way<br />
in terms of complex numbers.<br />
Tip<br />
Cosh is pronounced as it reads,<br />
sinh is pronounced ‘sinsh’ or<br />
sometimes ‘shine’ and tanh is<br />
pronounced ‘tansh’.<br />
Did you know?<br />
The tanh function is frequently<br />
used in physics, particularly in<br />
the context of special relativity<br />
and the study of entropy.<br />
1<br />
–1<br />
y<br />
O<br />
y = tanh x<br />
Draft sample<br />
Did you know?<br />
You may think that the graph<br />
of cosh x looks like a parabola,<br />
but it is slightly flatter. It is<br />
called a catenary,<br />
which is the shape<br />
<strong>for</strong>med by a<br />
hanging chain.<br />
x<br />
80<br />
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4 Hyperbolic functions<br />
WORKED EXAMPLE 4.1<br />
Find the range and domain of tanh 2<br />
x + 1.<br />
Tip<br />
The domain of tanh 2 x is any real number. Since the domain of tanh x is any real number, the<br />
expression tanh 2 x will likewise not be restricted<br />
in any way.<br />
The range of tanh 2 x is 0 y < 1<br />
The range of tanh x is from − 1 to 1, but its square<br />
will always be positive.<br />
So the range of tanh 2<br />
x + 1 is 1 y < 2<br />
The inverse functions of the hyperbolic functions are called arsinh x,<br />
arcosh x and artanh x.<br />
The graphs of these functions look like this:<br />
y<br />
O<br />
y = arsinh x<br />
x<br />
y<br />
O<br />
y = arcosh x<br />
Just as in trigonometry,<br />
2 2<br />
tanh x ≡ (tanh x ) .<br />
You just add one to the previous result.<br />
x<br />
Tip<br />
y<br />
O<br />
y = artanh x<br />
Draft sample<br />
On your calculator they might<br />
−<br />
be called sinh 1 −<br />
x, cosh 1 x and<br />
−<br />
tanh 1 x.<br />
Rewind<br />
You saw in the AS/A <strong>Level</strong><br />
<strong>Mathematics</strong> Student Book 2,<br />
Chapter 2, how to <strong>for</strong>m the<br />
graphs of inverse functions<br />
from the original function.<br />
x<br />
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81
A <strong>Level</strong> <strong>Fur</strong><strong>ther</strong> <strong>Mathematics</strong> <strong>for</strong> <strong>AQA</strong> Student Book 1<br />
Key point 4.4<br />
Function Domain Range<br />
arcosh x<br />
x 1 y 0<br />
arsinh x<br />
All real numbers All real numbers<br />
artanh x<br />
− 1< x < 1<br />
All real numbers<br />
You can use the inverse hyperbolic functions to solve simple equations<br />
involving hyperbolic functions. For example, if sinh x = 2 then<br />
x = arsinh 2, which you can evaluate, on a calculator, as 1.4436....<br />
However, you can use the definition of sinh x to derive a logarithmic <strong>for</strong>m<br />
of this result. You can do this <strong>for</strong> all three inverse hyperbolic functions.<br />
Key point 4.5<br />
( )<br />
(<br />
2<br />
)<br />
2<br />
• arcosh x = ln x+ x −1<br />
• arsinh x = ln x+ x + 1<br />
( )<br />
• artanh x =<br />
1 + x<br />
2 ln 1 1−<br />
x<br />
PROOF 4<br />
( )<br />
Prove that arcosh x = ln x+ x −1<br />
2 .<br />
Let y = arcosh x<br />
Let y = arcosh x and then look to find an expression <strong>for</strong> y.<br />
Then cosh y = x<br />
Take cosh of both sides.<br />
y<br />
e + e −<br />
2<br />
y<br />
= x<br />
y − y<br />
e + e = 2x<br />
y<br />
e +<br />
1<br />
y<br />
= 2x<br />
e<br />
y 2<br />
( e ) + 1=<br />
2xe<br />
y 2 y<br />
( e ) − 2xe<br />
+ 1=<br />
0<br />
y<br />
Use the definition of cosh y.<br />
Rearrange into a disguised quadratic in e y .<br />
Draft sample<br />
Continues on next page<br />
82<br />
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4 Hyperbolic functions<br />
So<br />
e<br />
y<br />
2<br />
2 x± (2 x) −4<br />
=<br />
2<br />
=<br />
2x± 4x 2 −4<br />
2<br />
2x± 4( x 2 −1)<br />
=<br />
2<br />
Use the quadratic <strong>for</strong>mula.<br />
2<br />
=<br />
2x± 4 x −1<br />
Use the algebra of surds to simplify the expression.<br />
2<br />
= x± x 2 −1<br />
But arcosh x is a function so it can<br />
only take one value.<br />
y<br />
∴ e = x+ x 2 −1<br />
( )<br />
2<br />
y = ln x+ x − 1<br />
But y = arcosh x<br />
( )<br />
2<br />
So arcosh x = ln x+ x − 1<br />
When you are trying to solve equations involving hyperbolic cosines,<br />
using the inverse function – un<strong>for</strong>tunately – does not give all the<br />
solutions: it just gives the positive one. Just as when taking the<br />
square root of both sides, you need to use a ‘plus or minus’ sign to<br />
get both possible solutions. This may be clear if you consider the<br />
graph of cosh x.<br />
WORKED EXAMPLE 4.2<br />
Conventionally, you take the positive root, so this makes<br />
y<br />
e > 1 and y > 0.<br />
( )<br />
( )<br />
Given that cosh 2 x − cosh x = 2 , express x in the <strong>for</strong>m ln a + b or ln a−<br />
b .<br />
cosh 2 x −cosh x − 2=<br />
0<br />
(coshx<br />
− 2)(cosh x + 1) = 0<br />
cosh x = 2 or cosh x =−1<br />
But cosh x 1so the only relevant solution is<br />
cosh x = 2.<br />
−<br />
x =± cosh 1 2<br />
–arcosh k<br />
1<br />
O<br />
y<br />
arcosh k<br />
The expression is a disguised quadratic, so<br />
rearrange it to make one side zero and then factorise<br />
it. You could also have used the quadratic <strong>for</strong>mula.<br />
Draft sample<br />
Use the inverse cosh function to find x. Remember<br />
that you need a plus or minus ( ± ).<br />
x<br />
y = k<br />
2<br />
( 2 2 1)<br />
=± ln + −<br />
( 2 3)<br />
=± ln +<br />
Then use the identity from Key point 4.4 to convert it<br />
to logarithmic <strong>for</strong>m.<br />
Continues on next page<br />
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83
.<br />
b<br />
b<br />
b<br />
b<br />
A <strong>Level</strong> <strong>Fur</strong><strong>ther</strong> <strong>Mathematics</strong> <strong>for</strong> <strong>AQA</strong> Student Book 1<br />
So<br />
ln<br />
⎛<br />
or ln<br />
1<br />
⎜<br />
⎝ 2+<br />
3<br />
x = ( 2+<br />
3)<br />
⎞<br />
⎟<br />
⎠<br />
−1 Use the fact that −ln x ≡ ln ( x ) ≡ ln (<br />
1<br />
x ).<br />
⎛ ⎞ ⎛<br />
⎜ ⎟ =<br />
− ⎞<br />
ln<br />
1<br />
ln<br />
2 3<br />
⎜<br />
⎟<br />
⎝ 2+<br />
3 ⎠ ⎝ (2+ 3)(2 − 3) ⎠<br />
⎛<br />
=<br />
− ⎞<br />
ln<br />
2 3<br />
⎜ ⎟<br />
⎝ 1 ⎠<br />
So x = ln (2+<br />
3) or x = ln(2 − 3)<br />
EXERCISE 4A<br />
1 Use your calculator to evaluate each expression where possible.<br />
2 2<br />
a i cosh ( − 1)<br />
ii sinh 3 b i tanh 1<br />
ii cosh 3<br />
c i 3sinh 0.2 + 1 ii 5tanh 1 + 8<br />
2<br />
d i<br />
e<br />
i<br />
−1<br />
tanh 2<br />
ii<br />
−1<br />
cosh 0<br />
2 Find the domain and range of each function.<br />
−1<br />
sinh 0.5<br />
ii<br />
−1<br />
cosh 2<br />
a i 3sinh x − 2<br />
ii 4cosh x + 1<br />
b i 3tanh x + 2<br />
ii 2tanh x − 1<br />
c i 2cosh 2<br />
x − 3 ii sinh 2<br />
x −1<br />
d i 3tanh 2<br />
x + 1<br />
ii tanh 2<br />
x −1<br />
−<br />
e i 2cosh 1 −<br />
x ii 3sinh 1 −1 −1 x + 2<br />
f i 2tanh ( x + 1) + 1 ii 3tanh ( x − 2) + 1<br />
3 Solve each equation, giving your answers to 3 significant figures.<br />
a i sinh x =− 2<br />
ii sinh x = 0.1<br />
b i 2cosh x = 5<br />
ii cosh x = 4<br />
c i 4tanh x = 3<br />
ii tanh x = 0.4<br />
d i 3tanh x = 4<br />
ii 3cosh x = 1<br />
−<br />
e i sinh 1 −<br />
x = 5<br />
ii cosh 1<br />
x = 4<br />
4 Solve the equation 3cosh( x − 1) = 5.<br />
5 Find and simplify the exact value of cosh (ln2).<br />
6 Find and simplify a rational expression <strong>for</strong> tanh (ln3).<br />
7 Solve the equation sinh 3x = 5<br />
2<br />
.<br />
8 Solve the equation 2tanh 2<br />
x+ 2=<br />
5tanh x.<br />
−<br />
9 Find and simplify an expression <strong>for</strong> cosh (sinh 1<br />
x ).<br />
−1 2<br />
10 Prove that sinh x = ln ( x+ x + 1) .<br />
Simplify the second solution by<br />
rationalising the denominator to<br />
produce the required <strong>for</strong>m.<br />
Draft sample<br />
84<br />
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4 Hyperbolic functions<br />
11 Prove that tanh =<br />
+<br />
( − )<br />
− 1<br />
2 ln 1 1<br />
x<br />
x<br />
1<br />
x .<br />
12<br />
−<br />
In the derivation of cosh 1 x you found that two possible expressions were ln ( x+ x<br />
−1)<br />
ln ( x − x2<br />
−1)<br />
Show that their sum is zero and hence explain why the chosen expression is<br />
non-negative.<br />
Section 2: Hyperbolic identities<br />
You can use the definitions of hyperbolic functions given in Key point 4.1<br />
and Key point 4.2 to find and prove identities relating hyperbolic functions.<br />
WORKED EXAMPLE 4.3<br />
2 2<br />
Prove that cosh x−sinh x ≡1.<br />
⎛<br />
x<br />
2<br />
cosh x =<br />
e + e<br />
⎜<br />
⎝ 2<br />
−x<br />
2<br />
⎞<br />
⎟<br />
⎠<br />
−<br />
=<br />
e + 2e e + e<br />
4<br />
=<br />
e<br />
2x x x −2x<br />
+ 2+ e<br />
−<br />
4<br />
2x<br />
2x<br />
⎛<br />
x<br />
2<br />
sinh x =<br />
e − e<br />
⎜<br />
⎝ 2<br />
cosh<br />
−x<br />
2<br />
⎞<br />
⎟<br />
⎠<br />
−<br />
=<br />
e − 2e e + e<br />
4<br />
=<br />
e<br />
x−sinh<br />
2 2<br />
2x x x −2x<br />
2x<br />
+ − 2x<br />
− 2 e<br />
4<br />
Did you know?<br />
2x −2x 2x −2x<br />
x ≡<br />
e + 2+<br />
e<br />
−<br />
e − 2+<br />
e<br />
4<br />
4<br />
2x −2x 2x −2x<br />
e + 2 + e −( e − 2 + e )<br />
≡<br />
4<br />
2x −2x 2x −2x<br />
≡<br />
e + 2+ e − e + 2−<br />
e<br />
4<br />
≡<br />
4<br />
4<br />
≡ 1<br />
Start from the definition of one of<br />
the hyperbolic functions. It doesn’t<br />
matter which one. It is squared in the<br />
expression so square it and simplify.<br />
x −x Since e e = 1.<br />
Repeat with the sinh 2 x term.<br />
x −x e e = 1 again.<br />
Combine the two terms and simplify.<br />
Draft sample<br />
Rewind<br />
Osborn’s rule links identities <strong>for</strong> hyperbolic functions with identities <strong>for</strong><br />
trigonometric functions. The rule states that you replace every occurrence<br />
of sine or cosine with the corresponding hyperbolic sine or cosine, except<br />
when <strong>ther</strong>e is a product of two sines. This is replaced with a negative<br />
product of two hyperbolic sines. This is a useful rule to help you remember<br />
the identities, but it will not be tested in the examination.<br />
The result in Worked example<br />
4.3 proves that (cosh x,sinh x)<br />
2 2<br />
lies on the hyperbola x − y =1,<br />
as described in the introduction<br />
to this chapter.<br />
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85
A <strong>Level</strong> <strong>Fur</strong><strong>ther</strong> <strong>Mathematics</strong> <strong>for</strong> <strong>AQA</strong> Student Book 1<br />
WORKED EXAMPLE 4.4<br />
Prove that sinh 2x = 2sinh x cosh x.<br />
LHS ≡<br />
e<br />
2x<br />
− e<br />
− 2x<br />
2<br />
−<br />
RHS≡ 2×<br />
e − e<br />
×<br />
e + e<br />
2 2<br />
x x x −x<br />
x −x x −x<br />
( e − e ) × ( e + e )<br />
≡<br />
2<br />
x<br />
( e ) − ( e<br />
≡<br />
− x)<br />
2<br />
2 2<br />
2x<br />
− − 2x<br />
≡<br />
e e<br />
2<br />
≡ LHS<br />
EXERCISE 4B<br />
1 Prove that cosh x−sinh x ≡e<br />
−x<br />
.<br />
2 Simplify 1+<br />
sinh 2 x .<br />
2 2<br />
3 Prove that cosh 2x ≡ cosh x+<br />
sinh x.<br />
2<br />
4 Prove that 1−tanh<br />
x ≡<br />
1<br />
cosh<br />
2<br />
x .<br />
2tanh x<br />
5 Prove that tanh 2x<br />
≡<br />
1 + tanh 2 x .<br />
6 Prove that cosh −1 ≡<br />
1 ( −<br />
2 e0.5 e−<br />
0.5<br />
)<br />
2<br />
On the LHS use the definition of sinh x and replace each x<br />
with 2 x.<br />
Then work from the RHS. Substitute the definitions of<br />
sinh x and cosh x.<br />
Multiply out the brackets, using the difference of two<br />
squares.<br />
Using the rules of indices.<br />
x x<br />
x . Hence prove that cosh x 1.<br />
7 Prove that cosh A+ cosh B ≡ 2cosh<br />
A+ B<br />
cosh<br />
A−<br />
B<br />
( 2 ) ( 2 ).<br />
Draft sample<br />
8 Use the binomial theorem to show that sinh 3<br />
x ≡<br />
1<br />
−<br />
4 sinh 3 x<br />
3<br />
4 sinh x.<br />
9 a<br />
3<br />
3<br />
Explain why (cosh x+ sinh x) ≡ cosh 3x+<br />
sinh 3x and (cosh x−sinh x) ≡cosh 3x−<br />
sinh 3x.<br />
3 2<br />
Hence show that cosh 3x ≡ cosh x+<br />
3cosh x sinh x.<br />
b Write cosh 3 x in terms of cosh x.<br />
10 Given that tan y = sinh x show that sin y =± tanh x.<br />
86<br />
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4 Hyperbolic functions<br />
Section 3: Solving harder hyperbolic equations<br />
When you are solving equations involving hyperbolic functions you have<br />
several options:<br />
• Rearrange to get a hyperbolic function that is equal to a constant and<br />
use inverse hyperbolic functions.<br />
• Use the definition of hyperbolic functions to get an exponential<br />
function that is equal to a constant and use logarithms.<br />
• Use an identity <strong>for</strong> hyperbolic functions to simplify the situation to<br />
one of the two preceding options.<br />
It is only with experience that you will develop an instinct about which<br />
method will be most efficient.<br />
WORKED EXAMPLE 4.5<br />
Solve sinh x+ cosh x = 4.<br />
sinh x + cosh x = 4<br />
−<br />
e − e<br />
+<br />
e + e<br />
2 2<br />
x x x −x<br />
2e<br />
2<br />
= 4<br />
x<br />
= 4<br />
x<br />
e = 4<br />
x = ln 4<br />
WORKED EXAMPLE 4.6<br />
Use the definitions of<br />
sinh x and cosh x.<br />
Solve cosh 2<br />
x+ 1=<br />
3sinh x, giving your answer in logarithmic <strong>for</strong>m.<br />
cosh 2 x + 1=<br />
3sinh<br />
x<br />
(1 + sinh 2 x ) + 1=<br />
3 sinh x<br />
sinh 2 x − 3sinh<br />
x + 2=<br />
0<br />
This is a quadratic.<br />
Tip<br />
When you are dealing with<br />
the sum or difference of two<br />
hyperbolic functions, it is often<br />
useful to use the exponential<br />
<strong>for</strong>m.<br />
The equation involves two types of function. You can use<br />
2 2<br />
the identity cosh x−sinh x ≡1<br />
to replace the cosh 2 term.<br />
( sinh x−2)( sinh x− 1) = 0<br />
Factorise (or use the quadratic <strong>for</strong>mula).<br />
sinh x = 1or sinh x = 2<br />
x = arsinh 1 or x = arsinh 2<br />
You will find useful hyperbolic<br />
identities in the <strong>for</strong>mula book<br />
provided in the examination.<br />
Draft sample<br />
Tip<br />
x = ln (1+<br />
2) or x = ln (2+<br />
5)<br />
Use the logarithm <strong>for</strong>m of arsinh.<br />
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87
A <strong>Level</strong> <strong>Fur</strong><strong>ther</strong> <strong>Mathematics</strong> <strong>for</strong> <strong>AQA</strong> Student Book 1<br />
WORK IT OUT<br />
Solve sinh 2x cosh2x = 6sinh2x.<br />
Which is the correct solution? Identify the errors made in the incorrect solutions.<br />
Solution 1 Solution 2 Solution 3<br />
Dividing by 2:<br />
sinh x cosh x = 3sinh x<br />
sinh x cosh x− 3sinh x = 0<br />
sinh x(cosh x− 3) = 0<br />
sinh x = 0orcosh x = 3<br />
−1 −1<br />
x = sinh 0or x = cosh 3<br />
x = 0or x = 1.76<br />
EXERCISE 4C<br />
1 Find the exact solution to cosh x = 5− sinh x.<br />
2 Solve cosh x− sinh x = 2, giving your answer in the <strong>for</strong>m ln k.<br />
=<br />
Dividing by sinh 2 x :<br />
sinh 2x cosh2x 6sinh2x<br />
x =<br />
1<br />
+<br />
2 ln (6 35) −1 −1<br />
2x<br />
= sinh 0 or 2x<br />
=± cosh 6<br />
x = 0or x =± 1.24<br />
cosh 2x<br />
= 6<br />
−<br />
2x<br />
= cosh 6<br />
= ln (6 + 35)<br />
sinh 2x cosh2x− 6sinh2x<br />
= 0<br />
sinh 2 x(cosh 2x− 6) = 0<br />
sinh 2x<br />
= 0 or cosh 2x<br />
= 6<br />
3 Solve 3(2sinh x −1)(cosh x − 4) = 0, giving your answers correct to 3 significant figures.<br />
4 Solve sinh 2x = cosh x, giving your answer in logarithmic <strong>for</strong>m.<br />
5 Find the exact solution to 2sinh x = 1+ cosh x.<br />
6 Solve sinh 2<br />
x = cosh x + 1, giving your answer in logarithmic <strong>for</strong>m.<br />
7 Solve tanh x =<br />
1<br />
, giving your answer in logarithmic <strong>for</strong>m.<br />
cosh x<br />
8 Solve sinh x =<br />
1<br />
, giving your answer in logarithmic <strong>for</strong>m.<br />
cosh x<br />
9 sinh x+ sinh y =<br />
21<br />
8<br />
cosh x+ cosh y =<br />
27<br />
8<br />
x y −<br />
a Show that e = 6−e<br />
and e = 0.75 −e<br />
b<br />
Draft sample<br />
x − y<br />
.<br />
Hence find the exact solutions to the simultaneous equations.<br />
10 Find a sufficient condition on p, q and r <strong>for</strong> p 2 cosh x+ q 2 sinh x = r 2 to have at least one solution.<br />
88<br />
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4 Hyperbolic functions<br />
Checklist of learning and understanding<br />
• Definitions of hyperbolic functions:<br />
• cosh x =<br />
e<br />
• sinh x =<br />
e<br />
e<br />
2<br />
x x<br />
+ −<br />
e<br />
2<br />
x x<br />
− −<br />
sinh x x<br />
• tanh x ≡ ≡<br />
e − e<br />
x<br />
cosh x e + e<br />
−x<br />
−x<br />
• Domain and range of hyperbolic and inverse hyperbolic functions:<br />
Function Domain Range<br />
cosh x<br />
All real numbers y 1<br />
sinh x<br />
All real numbers All real numbers<br />
tanh x<br />
All real numbers − 1< y < 1<br />
arcosh x<br />
x 1 y 0<br />
arsinh x<br />
All real numbers All real numbers<br />
artanh x<br />
• Logarithmic <strong>for</strong>m of inverse hyperbolic functions:<br />
( )<br />
(<br />
2<br />
)<br />
2<br />
• arcosh x = ln x+ x −1<br />
• arsinh x = ln x+ x + 1<br />
( )<br />
• artanh x =<br />
1 + x<br />
2 ln 1 1−<br />
x<br />
− 1< x < 1<br />
All real numbers<br />
Draft sample<br />
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89
A <strong>Level</strong> <strong>Fur</strong><strong>ther</strong> <strong>Mathematics</strong> <strong>for</strong> <strong>AQA</strong> Student Book 1<br />
Mixed practice 4<br />
2<br />
1 Find the range of y = 3tanh x + 2.<br />
Choose from these options.<br />
A<br />
2 y < 5<br />
B −1 y 5<br />
C − 1< y < 5<br />
D All real numbers<br />
2 Solve sinh x+ cosh x = k, giving your answer in terms of k.<br />
Choose from these options.<br />
A<br />
arsinh<br />
k<br />
2<br />
3 Simplify tanh (1 + ln p)<br />
.<br />
( ) B k<br />
arcosh (2 ) C e k D ln k<br />
4 Solve cosh ( x + 1) = 3, giving your answer in terms of logarithms.<br />
x −<br />
5 a Express 5sinh x+ cosh x in the <strong>for</strong>m Ae<br />
+ Be<br />
x<br />
, where A and B are integers.<br />
b Solve the equation 5sinh x + cosh x + 5= 0, giving your answer in the <strong>for</strong>m ln a, where a is a<br />
rational number.<br />
[©<strong>AQA</strong> 2008]<br />
6 a Use the definitions sinh θ =<br />
1<br />
(e θ −e 2 −θ<br />
) and cosh θ =<br />
1<br />
(e θ + e<br />
2 −θ<br />
) to show that 1 + 2sinh θ cosh 2θ<br />
b Solve the equation 3cosh2θ<br />
= 2sinhθ<br />
+ 11, giving each of your answers in the <strong>for</strong>m ln p.<br />
7 Solve sinh 2x = 2cosh x, giving your answer in logarithmic <strong>for</strong>m.<br />
8 Find the exact solutions to cosh 2x+ cosh x =<br />
27<br />
.<br />
8<br />
9 Find the exact solutions to 2cosh x+ sinh x = 2.<br />
10 Prove that 2sinh 2<br />
A≡cosh2A −1.<br />
11 Prove that cosh x > sinh x.<br />
12 Find and simplify an expression <strong>for</strong> tanh (arsinh x).<br />
13 Use the binomial theorem to show that cosh 4<br />
x ≡<br />
1<br />
+ +<br />
8 cosh 4 x<br />
1<br />
2 cosh 2 x<br />
3<br />
.<br />
8<br />
14 a Sketch the graph of y = tanh x.<br />
[©<strong>AQA</strong> 2009]<br />
Draft sample<br />
b Given that u = tanh x, use the definitions of sinh x and cosh x in terms of e x −<br />
and e x to show that<br />
x =<br />
1 u<br />
2 ln 1 +<br />
( 1 − u ).<br />
c i Show that the equation<br />
3<br />
+ 7tanh x = 5 can be written as 3tanh 2<br />
x− 7tanh x + 2=<br />
0.<br />
cosh<br />
2<br />
x<br />
ii Show that the equation 3tanh 2<br />
x− 7tanh x + 2=<br />
0has only one solution <strong>for</strong> x .<br />
Find this solution in the <strong>for</strong>m<br />
1<br />
ln a where a is an integer.<br />
2<br />
90<br />
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4 Hyperbolic functions<br />
θ θ<br />
15 a Using the definition sinh θ =<br />
1<br />
(e −e 2 − ), prove the identity 4sinh 3 θ + 3sinh θ = sinh 3θ<br />
.<br />
3<br />
b Given that x = sinh θ and 16x + 12x − 3=<br />
0, find the value of θ in terms of a natural logarithm.<br />
c<br />
3<br />
Hence find the real root of the equation 16x + 12x − 3=<br />
0, giving your answer in the <strong>for</strong>m 2 − 2<br />
where p and q are rational numbers.<br />
3<br />
16 a Prove the identity cosh 3x = 4cosh x−3cosh<br />
x.<br />
3<br />
b If 48u −36u − 13 = 0 and u = cosh x find the value of x .<br />
p q<br />
,<br />
[©<strong>AQA</strong> 2014]<br />
3<br />
c Hence find the exact real solution to 48u −36u − 13 = 0, giving your answer in a <strong>for</strong>m without<br />
logarithms.<br />
17 Solve these simultaneous equations, giving your answers in exact logarithmic <strong>for</strong>m.<br />
sinh x+ sinh y = 3.15<br />
cosh x + cosh y = 3.85<br />
18 Using the logarithmic definition, prove that arsinh ( − x) = − arsinh ( x)<br />
.<br />
Draft sample<br />
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91