A Level Fur ther Mathematics for AQA

Brighter Thinking 

A Level Fur ther 

Mathematics for AQA 

Student Book 1 (AS/ Year 1) 

Stephen Ward and Paul Fannon 

This resource has been entered into the AQA approval process.

Contents 

Contents 

Introduction.............................................................. iv 

How to use this book................................................v 

1 Complex numbers 

1: Definition and basic arithmetic of i......................1 

2: Division and complex conjugates.......................6 

3: Geometric representation....................................9 

4: Locus in the complex plane...............................20 

5: Operations in modulus–argument form...........24 

2 Roots of polynomials 

1: Factorising polynomials......................................31 

2: Complex solutions to polynomial 

equations.............................................................34 

3: Roots and coefficients........................................38 

4: Finding an equation with given roots...............43 

5: Transforming equations......................................48 

3 Graphs and inequalities 

1: Cubic and quartic inequalities...........................55 

2: Functions of the form y = 

ax + b 

cx + d .......................58 

2 

3: Functions of the form y = 

ax + bx + c 

2 

...............62 

dx + ex + f 

4: Oblique asymptotes...........................................65 

5: Reciprocal transformations of functions...........66 

6: Modulus transformation.....................................71 

4 Hyperbolic functions 

1: Defining hyperbolic functions and 

hyperbolic identities...........................................80 

2: Hyperbolic identities..........................................85 

3: Solving harder hyperbolic equations................87 

5 Series 

1: Sigma notation....................................................92 

2: Using standard formulae....................................94 

3: Method of differences........................................99 

4: Maclaurin series.................................................103 

Focus on … Proof 1.............................................111 

Focus on … Problem solving 1...........................113 

Focus on … Modelling 1.....................................115 

Cross-topic review exercise 1.............................117 

6 Matrices 

1: Addition, subtraction and 

scalar multiplication..........................................120 

2: Matrix multiplication.........................................126 

3: Determinants and inverses of 

2× 

2 matrices..................................................... 131 

4: Linear simultaneous equations........................140 

7 Matrix transformations 

1: Matrices as linear transformations................... 147 

2: Further transformations in 2D..........................155 

3: Invariant points and invariant lines..................162 

4: Transformations in 3D....................................... 167 

8 Further vectors 

1: Vector equation of a line..................................177 

2: Cartesian equation of the line.........................184 

3: Intersections of lines.........................................190 

4: Angles and the scalar product.........................193 

5: The vector product...........................................202 

9 Polar coordinates 

1: Curves in polar coordinates.............................212 

2: Some features of polar curves.........................215 

3: Changing between polar and Cartesian 

coordinates........................................................219 

10 Further calculus 

1: Volumes of revolution.......................................225 

2: Average value of a function.............................230 

Focus on … Proof 2............................................ 236 

Focus on … Problem solving 2.......................... 238 

Draft sample 

Focus on … Modelling 2.....................................241 

Cross topic review exercise 2............................244 

Practice paper 1 ...................................................248 

Formulae................................................................250 

Answers to exercises ............................................253 

Glossary .................................................................296 

Index ..................................................................... 297 

Acknowledgements............................................. 000 

© Cambridge University Press 2017 

The third party copyright material that appears in this sample may still be pending clearance and may be subject to change. 

iii


In this chapter you will learn how to: 

•• 

define the hyperbolic functions sinh x, cosh x and tanh x 

•• 

recognise and use the graphs, range and domain of the hyperbolic 

functions 

•• 

work with identities involving hyperbolic functions 

•• 

write the inverse hyperbolic functions in terms of logarithms 

•• 

solve equations involving hyperbolic functions. 

Before you start… 

AS/A Level Mathematics 

Student Book 1, 

Chapter 7 



Chapter 3 



Chapter 10 



Chapter 10 



Chapter 5 



Chapter 9 

You should be able to work with 

natural logarithms. 

You should be able to solve 

quadratic equations. 


the symmetries of trigonometric 

functions. 


trigonometric identities. 

You should be able to interpret 

transformations of graphs 

graphically. 

You should be able to complete 

binomial expansions of brackets 

with positive integer powers. 

What are hyperbolic functions? 

Trigonometric functions are sometimes called circular functions. This 

is because of the definition that states: a point on the unit circle (with 

2 2 

equation x + y =1) at an angle θ to the positive x-axis, has coordinates 

(cos θ,sin θ ). 

2 2 

A related curve, with equation x − y =1, is called a hyperbola. Points on 

this hyperbola have coordinates (cosh θ,sinh θ) although θ can no longer 

be interpreted as an angle. 

x 

1 Given that 2 = 5, write x in the form ln a 

ln b 

. 

2 

2 Solve x + 3x = 1. 

3 Given that sin 40°= a find sin 220°. 

2 2 

4 Simplify 3sin 2x+ 

3cos 2x. 

5 What transformation changes f( x) to 

3f( x + 1) ? 

6 Expand (2 + x ) 3 . 

Draft sample 

–1 

y 

1 

O 1 

–1 

O 

y 

(cos , sin ) 

x 

(cosh , sinh ) 

x 



79

A Level Further Mathematics for AQA Student Book 1 

Section 1: Defining hyperbolic functions and 

hyperbolic identities 

Although the geometric definition of hyperbolic functions gives some 

helpful insight, a more useful definition is related to the number e. 

Key point 4.1 

• cosh x = 

e 

• sinh x = 

e 

e 

2 

x x 

+ − 

x x 

− e 

− 

2 

You can define the tanh x function by analogy with the trigonometric 

definition (of tan x). 

Key point 4.2 

O 

1 

y 

y = cosh x 

sinh x x 

tanh x ≡ ≡ 

e − e 

x 

cosh x e + e 

There are not many special values of these functions that you need to 

know, but from Key points 4.1 and 4.2 you should be able to see that 

cosh (0) = 1, sinh (0) = 0and tanh (0) = 0. 

To investigate the domain and range you need to consider the graphs. 

Key point 4.3 

x 

Function Domain Range 

cosh x 

All real numbers y 1 

−x 

−x 

y 

O 

y = sinh x 

sinh x 

All real numbers All real numbers 

tanh x 

All real numbers − 1< y < 1 

x 

Fast forward 

In Further Mathematics Student 

Book 2, you will see that the 

trigonometric functions can 

also be defined in a similar way 

in terms of complex numbers. 

Tip 

Cosh is pronounced as it reads, 

sinh is pronounced ‘sinsh’ or 

sometimes ‘shine’ and tanh is 

pronounced ‘tansh’. 

Did you know? 

The tanh function is frequently 

used in physics, particularly in 

the context of special relativity 

and the study of entropy. 

1 

–1 

y 

O 

y = tanh x 

Draft sample 

Did you know? 

You may think that the graph 

of cosh x looks like a parabola, 

but it is slightly flatter. It is 

called a catenary, 

which is the shape 

formed by a 

hanging chain. 

x 

80 


The third party copyright material that appears in this sample may still be pending clearance and may be subject to change.


WORKED EXAMPLE 4.1 

Find the range and domain of tanh 2 

x + 1. 

Tip 

The domain of tanh 2 x is any real number. Since the domain of tanh x is any real number, the 

expression tanh 2 x will likewise not be restricted 

in any way. 

The range of tanh 2 x is 0 y < 1 

The range of tanh x is from − 1 to 1, but its square 

will always be positive. 

So the range of tanh 2 

x + 1 is 1 y < 2 

The inverse functions of the hyperbolic functions are called arsinh x, 

arcosh x and artanh x. 

The graphs of these functions look like this: 

y 

O 

y = arsinh x 

x 

y 

O 

y = arcosh x 

Just as in trigonometry, 

2 2 

tanh x ≡ (tanh x ) . 

You just add one to the previous result. 

x 

Tip 

y 

O 

y = artanh x 

Draft sample 

On your calculator they might 

− 

be called sinh 1 − 

x, cosh 1 x and 

− 

tanh 1 x. 

Rewind 

You saw in the AS/A Level 

Mathematics Student Book 2, 

Chapter 2, how to form the 

graphs of inverse functions 

from the original function. 

x 



81


Key point 4.4 


arcosh x 

x 1 y 0 

arsinh x 


artanh x 

− 1< x < 1 

All real numbers 

You can use the inverse hyperbolic functions to solve simple equations 

involving hyperbolic functions. For example, if sinh x = 2 then 

x = arsinh 2, which you can evaluate, on a calculator, as 1.4436.... 

However, you can use the definition of sinh x to derive a logarithmic form 

of this result. You can do this for all three inverse hyperbolic functions. 

Key point 4.5 

( ) 

( 

2 

) 

2 

• arcosh x = ln x+ x −1 

• arsinh x = ln x+ x + 1 

( ) 

• artanh x = 

1 + x 

2 ln 1 1− 

x 

PROOF 4 

( ) 

Prove that arcosh x = ln x+ x −1 

2 . 

Let y = arcosh x 

Let y = arcosh x and then look to find an expression for y. 

Then cosh y = x 

Take cosh of both sides. 

y 

e + e − 

2 

y 

= x 

y − y 

e + e = 2x 

y 

e + 

1 

y 

= 2x 

e 

y 2 

( e ) + 1= 

2xe 

y 2 y 

( e ) − 2xe 

+ 1= 

0 

y 

Use the definition of cosh y. 

Rearrange into a disguised quadratic in e y . 

Draft sample 

Continues on next page 

82 




So 

e 

y 

2 

2 x± (2 x) −4 

= 

2 

= 

2x± 4x 2 −4 

2 

2x± 4( x 2 −1) 

= 

2 

Use the quadratic formula. 

2 

= 

2x± 4 x −1 

Use the algebra of surds to simplify the expression. 

2 

= x± x 2 −1 

But arcosh x is a function so it can 

only take one value. 

y 

∴ e = x+ x 2 −1 

( ) 

2 

y = ln x+ x − 1 

But y = arcosh x 

( ) 

2 

So arcosh x = ln x+ x − 1 

When you are trying to solve equations involving hyperbolic cosines, 

using the inverse function – unfortunately – does not give all the 

solutions: it just gives the positive one. Just as when taking the 

square root of both sides, you need to use a ‘plus or minus’ sign to 

get both possible solutions. This may be clear if you consider the 

graph of cosh x. 


Conventionally, you take the positive root, so this makes 

y 

e > 1 and y > 0. 

( ) 

( ) 

Given that cosh 2 x − cosh x = 2 , express x in the form ln a + b or ln a− 

b . 

cosh 2 x −cosh x − 2= 

0 

(coshx 

− 2)(cosh x + 1) = 0 

cosh x = 2 or cosh x =−1 

But cosh x 1so the only relevant solution is 

cosh x = 2. 

− 

x =± cosh 1 2 

–arcosh k 

1 

O 

y 

arcosh k 

The expression is a disguised quadratic, so 

rearrange it to make one side zero and then factorise 

it. You could also have used the quadratic formula. 

Draft sample 

Use the inverse cosh function to find x. Remember 

that you need a plus or minus ( ± ). 

x 

y = k 

2 

( 2 2 1) 

=± ln + − 

( 2 3) 

=± ln + 

Then use the identity from Key point 4.4 to convert it 

to logarithmic form. 

Continues on next page 



83

. 

b 

b 

b 

b 


So 

ln 

⎛ 

or ln 

1 

⎜ 

⎝ 2+ 

3 

x = ( 2+ 

3) 

⎞ 

⎟ 

⎠ 

−1 Use the fact that −ln x ≡ ln ( x ) ≡ ln ( 

1 

x ). 

⎛ ⎞ ⎛ 

⎜ ⎟ = 

− ⎞ 

ln 

1 

ln 

2 3 

⎜ 

⎟ 

⎝ 2+ 

3 ⎠ ⎝ (2+ 3)(2 − 3) ⎠ 

⎛ 

= 

− ⎞ 

ln 

2 3 

⎜ ⎟ 

⎝ 1 ⎠ 

So x = ln (2+ 

3) or x = ln(2 − 3) 

EXERCISE 4A 

1 Use your calculator to evaluate each expression where possible. 

2 2 

a i cosh ( − 1) 

ii sinh 3 b i tanh 1 

ii cosh 3 

c i 3sinh 0.2 + 1 ii 5tanh 1 + 8 

2 

d i 

e 

i 

−1 

tanh 2 

ii 

−1 

cosh 0 

2 Find the domain and range of each function. 

−1 

sinh 0.5 

ii 

−1 

cosh 2 

a i 3sinh x − 2 

ii 4cosh x + 1 

b i 3tanh x + 2 

ii 2tanh x − 1 

c i 2cosh 2 

x − 3 ii sinh 2 

x −1 

d i 3tanh 2 

x + 1 

ii tanh 2 

x −1 

− 

e i 2cosh 1 − 

x ii 3sinh 1 −1 −1 x + 2 

f i 2tanh ( x + 1) + 1 ii 3tanh ( x − 2) + 1 

3 Solve each equation, giving your answers to 3 significant figures. 

a i sinh x =− 2 

ii sinh x = 0.1 

b i 2cosh x = 5 

ii cosh x = 4 

c i 4tanh x = 3 

ii tanh x = 0.4 

d i 3tanh x = 4 

ii 3cosh x = 1 

− 

e i sinh 1 − 

x = 5 

ii cosh 1 

x = 4 

4 Solve the equation 3cosh( x − 1) = 5. 

5 Find and simplify the exact value of cosh (ln2). 

6 Find and simplify a rational expression for tanh (ln3). 

7 Solve the equation sinh 3x = 5 

2 

. 

8 Solve the equation 2tanh 2 

x+ 2= 

5tanh x. 

− 

9 Find and simplify an expression for cosh (sinh 1 

x ). 

−1 2 

10 Prove that sinh x = ln ( x+ x + 1) . 

Simplify the second solution by 

rationalising the denominator to 

produce the required form. 

Draft sample 

84 




11 Prove that tanh = 

+ 

( − ) 

− 1 

2 ln 1 1 

x 

x 

1 

x . 

12 

− 

In the derivation of cosh 1 x you found that two possible expressions were ln ( x+ x 

−1) 

ln ( x − x2 

−1) 

Show that their sum is zero and hence explain why the chosen expression is 

non-negative. 

Section 2: Hyperbolic identities 

You can use the definitions of hyperbolic functions given in Key point 4.1 

and Key point 4.2 to find and prove identities relating hyperbolic functions. 


2 2 

Prove that cosh x−sinh x ≡1. 

⎛ 

x 

2 

cosh x = 

e + e 

⎜ 

⎝ 2 

−x 

2 

⎞ 

⎟ 

⎠ 

− 

= 

e + 2e e + e 

4 

= 

e 

2x x x −2x 

+ 2+ e 

− 

4 

2x 

2x 

⎛ 

x 

2 

sinh x = 

e − e 

⎜ 

⎝ 2 

cosh 

−x 

2 

⎞ 

⎟ 

⎠ 

− 

= 

e − 2e e + e 

4 

= 

e 

x−sinh 

2 2 

2x x x −2x 

2x 

+ − 2x 

− 2 e 

4 

Did you know? 

2x −2x 2x −2x 

x ≡ 

e + 2+ 

e 

− 

e − 2+ 

e 

4 

4 

2x −2x 2x −2x 

e + 2 + e −( e − 2 + e ) 

≡ 

4 

2x −2x 2x −2x 

≡ 

e + 2+ e − e + 2− 

e 

4 

≡ 

4 

4 

≡ 1 

Start from the definition of one of 

the hyperbolic functions. It doesn’t 

matter which one. It is squared in the 

expression so square it and simplify. 

x −x Since e e = 1. 

Repeat with the sinh 2 x term. 

x −x e e = 1 again. 

Combine the two terms and simplify. 

Draft sample 

Rewind 

Osborn’s rule links identities for hyperbolic functions with identities for 

trigonometric functions. The rule states that you replace every occurrence 

of sine or cosine with the corresponding hyperbolic sine or cosine, except 

when there is a product of two sines. This is replaced with a negative 

product of two hyperbolic sines. This is a useful rule to help you remember 

the identities, but it will not be tested in the examination. 

The result in Worked example 

4.3 proves that (cosh x,sinh x) 

2 2 

lies on the hyperbola x − y =1, 

as described in the introduction 

to this chapter. 



85



Prove that sinh 2x = 2sinh x cosh x. 

LHS ≡ 

e 

2x 

− e 

− 2x 

2 

− 

RHS≡ 2× 

e − e 

× 

e + e 

2 2 

x x x −x 

x −x x −x 

( e − e ) × ( e + e ) 

≡ 

2 

x 

( e ) − ( e 

≡ 

− x) 

2 

2 2 

2x 

− − 2x 

≡ 

e e 

2 

≡ LHS 

EXERCISE 4B 

1 Prove that cosh x−sinh x ≡e 

−x 

. 

2 Simplify 1+ 

sinh 2 x . 

2 2 

3 Prove that cosh 2x ≡ cosh x+ 

sinh x. 

2 

4 Prove that 1−tanh 

x ≡ 

1 

cosh 

2 

x . 

2tanh x 

5 Prove that tanh 2x 

≡ 

1 + tanh 2 x . 

6 Prove that cosh −1 ≡ 

1 ( − 

2 e0.5 e− 

0.5 

) 

2 

On the LHS use the definition of sinh x and replace each x 

with 2 x. 

Then work from the RHS. Substitute the definitions of 

sinh x and cosh x. 

Multiply out the brackets, using the difference of two 

squares. 

Using the rules of indices. 

x x 

x . Hence prove that cosh x 1. 

7 Prove that cosh A+ cosh B ≡ 2cosh 

A+ B 

cosh 

A− 

B 

( 2 ) ( 2 ). 

Draft sample 

8 Use the binomial theorem to show that sinh 3 

x ≡ 

1 

− 

4 sinh 3 x 

3 

4 sinh x. 

9 a 

3 

3 

Explain why (cosh x+ sinh x) ≡ cosh 3x+ 

sinh 3x and (cosh x−sinh x) ≡cosh 3x− 

sinh 3x. 

3 2 

Hence show that cosh 3x ≡ cosh x+ 

3cosh x sinh x. 

b Write cosh 3 x in terms of cosh x. 

10 Given that tan y = sinh x show that sin y =± tanh x. 

86 




Section 3: Solving harder hyperbolic equations 

When you are solving equations involving hyperbolic functions you have 

several options: 

• Rearrange to get a hyperbolic function that is equal to a constant and 

use inverse hyperbolic functions. 

• Use the definition of hyperbolic functions to get an exponential 

function that is equal to a constant and use logarithms. 

• Use an identity for hyperbolic functions to simplify the situation to 

one of the two preceding options. 

It is only with experience that you will develop an instinct about which 

method will be most efficient. 


Solve sinh x+ cosh x = 4. 

sinh x + cosh x = 4 

− 

e − e 

+ 

e + e 

2 2 

x x x −x 

2e 

2 

= 4 

x 

= 4 

x 

e = 4 

x = ln 4 


Use the definitions of 

sinh x and cosh x. 

Solve cosh 2 

x+ 1= 

3sinh x, giving your answer in logarithmic form. 

cosh 2 x + 1= 

3sinh 

x 

(1 + sinh 2 x ) + 1= 

3 sinh x 

sinh 2 x − 3sinh 

x + 2= 

0 

This is a quadratic. 

Tip 

When you are dealing with 

the sum or difference of two 

hyperbolic functions, it is often 

useful to use the exponential 

form. 

The equation involves two types of function. You can use 

2 2 

the identity cosh x−sinh x ≡1 

to replace the cosh 2 term. 

( sinh x−2)( sinh x− 1) = 0 

Factorise (or use the quadratic formula). 

sinh x = 1or sinh x = 2 

x = arsinh 1 or x = arsinh 2 

You will find useful hyperbolic 

identities in the formula book 

provided in the examination. 

Draft sample 

Tip 

x = ln (1+ 

2) or x = ln (2+ 

5) 

Use the logarithm form of arsinh. 



87


WORK IT OUT 

Solve sinh 2x cosh2x = 6sinh2x. 

Which is the correct solution? Identify the errors made in the incorrect solutions. 

Solution 1 Solution 2 Solution 3 

Dividing by 2: 

sinh x cosh x = 3sinh x 

sinh x cosh x− 3sinh x = 0 

sinh x(cosh x− 3) = 0 

sinh x = 0orcosh x = 3 

−1 −1 

x = sinh 0or x = cosh 3 

x = 0or x = 1.76 

EXERCISE 4C 

1 Find the exact solution to cosh x = 5− sinh x. 

2 Solve cosh x− sinh x = 2, giving your answer in the form ln k. 

= 

Dividing by sinh 2 x : 

sinh 2x cosh2x 6sinh2x 

x = 

1 

+ 

2 ln (6 35) −1 −1 

2x 

= sinh 0 or 2x 

=± cosh 6 

x = 0or x =± 1.24 

cosh 2x 

= 6 

− 

2x 

= cosh 6 

= ln (6 + 35) 

sinh 2x cosh2x− 6sinh2x 

= 0 

sinh 2 x(cosh 2x− 6) = 0 

sinh 2x 

= 0 or cosh 2x 

= 6 

3 Solve 3(2sinh x −1)(cosh x − 4) = 0, giving your answers correct to 3 significant figures. 

4 Solve sinh 2x = cosh x, giving your answer in logarithmic form. 

5 Find the exact solution to 2sinh x = 1+ cosh x. 

6 Solve sinh 2 

x = cosh x + 1, giving your answer in logarithmic form. 

7 Solve tanh x = 

1 

, giving your answer in logarithmic form. 

cosh x 

8 Solve sinh x = 

1 

, giving your answer in logarithmic form. 

cosh x 

9 sinh x+ sinh y = 

21 

8 

cosh x+ cosh y = 

27 

8 

x y − 

a Show that e = 6−e 

and e = 0.75 −e 

b 

Draft sample 

x − y 

. 

Hence find the exact solutions to the simultaneous equations. 

10 Find a sufficient condition on p, q and r for p 2 cosh x+ q 2 sinh x = r 2 to have at least one solution. 

88 




Checklist of learning and understanding 

• Definitions of hyperbolic functions: 

• cosh x = 

e 

• sinh x = 

e 

e 

2 

x x 

+ − 

e 

2 

x x 

− − 

sinh x x 

• tanh x ≡ ≡ 

e − e 

x 

cosh x e + e 

−x 

−x 

• Domain and range of hyperbolic and inverse hyperbolic functions: 


cosh x 

All real numbers y 1 

sinh x 


tanh x 

All real numbers − 1< y < 1 

arcosh x 

x 1 y 0 

arsinh x 


artanh x 

• Logarithmic form of inverse hyperbolic functions: 

( ) 

( 

2 

) 

2 

• arcosh x = ln x+ x −1 

• arsinh x = ln x+ x + 1 

( ) 

• artanh x = 

1 + x 

2 ln 1 1− 

x 

− 1< x < 1 

All real numbers 

Draft sample 



89


Mixed practice 4 

2 

1 Find the range of y = 3tanh x + 2. 

Choose from these options. 

A 

2 y < 5 

B −1 y 5 

C − 1< y < 5 

D All real numbers 

2 Solve sinh x+ cosh x = k, giving your answer in terms of k. 

Choose from these options. 

A 

arsinh 

k 

2 

3 Simplify tanh (1 + ln p) 

. 

( ) B k 

arcosh (2 ) C e k D ln k 

4 Solve cosh ( x + 1) = 3, giving your answer in terms of logarithms. 

x − 

5 a Express 5sinh x+ cosh x in the form Ae 

+ Be 

x 

, where A and B are integers. 

b Solve the equation 5sinh x + cosh x + 5= 0, giving your answer in the form ln a, where a is a 

rational number. 

[©AQA 2008] 

6 a Use the definitions sinh θ = 

1 

(e θ −e 2 −θ 

) and cosh θ = 

1 

(e θ + e 

2 −θ 

) to show that 1 + 2sinh θ cosh 2θ 

b Solve the equation 3cosh2θ 

= 2sinhθ 

+ 11, giving each of your answers in the form ln p. 

7 Solve sinh 2x = 2cosh x, giving your answer in logarithmic form. 

8 Find the exact solutions to cosh 2x+ cosh x = 

27 

. 

8 

9 Find the exact solutions to 2cosh x+ sinh x = 2. 

10 Prove that 2sinh 2 

A≡cosh2A −1. 

11 Prove that cosh x > sinh x. 

12 Find and simplify an expression for tanh (arsinh x). 

13 Use the binomial theorem to show that cosh 4 

x ≡ 

1 

+ + 

8 cosh 4 x 

1 

2 cosh 2 x 

3 

. 

8 

14 a Sketch the graph of y = tanh x. 


Draft sample 

b Given that u = tanh x, use the definitions of sinh x and cosh x in terms of e x − 

and e x to show that 

x = 

1 u 

2 ln 1 + 

( 1 − u ). 

c i Show that the equation 

3 

+ 7tanh x = 5 can be written as 3tanh 2 

x− 7tanh x + 2= 

0. 

cosh 

2 

x 

ii Show that the equation 3tanh 2 

x− 7tanh x + 2= 

0has only one solution for x . 

Find this solution in the form 

1 

ln a where a is an integer. 

2 

90 




θ θ 

15 a Using the definition sinh θ = 

1 

(e −e 2 − ), prove the identity 4sinh 3 θ + 3sinh θ = sinh 3θ 

. 

3 

b Given that x = sinh θ and 16x + 12x − 3= 

0, find the value of θ in terms of a natural logarithm. 

c 

3 

Hence find the real root of the equation 16x + 12x − 3= 

0, giving your answer in the form 2 − 2 

where p and q are rational numbers. 

3 

16 a Prove the identity cosh 3x = 4cosh x−3cosh 

x. 

3 

b If 48u −36u − 13 = 0 and u = cosh x find the value of x . 

p q 

, 


3 

c Hence find the exact real solution to 48u −36u − 13 = 0, giving your answer in a form without 

logarithms. 

17 Solve these simultaneous equations, giving your answers in exact logarithmic form. 

sinh x+ sinh y = 3.15 

cosh x + cosh y = 3.85 

18 Using the logarithmic definition, prove that arsinh ( − x) = − arsinh ( x) 

. 

Draft sample 



91

A Level Fur ther Mathematics for AQA

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