A Level Further Mathematics for AQA Student Book 1

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

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

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

m

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

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

e

1 Complex numbers

6 Matrices

pl

Introduction.............................................................. iv How to use this book................................................ v

3 Graphs and inequalities

ft

sa

1: Cubic and quartic inequalities...........................55 2: Functions of the form y = ax + b .......................58 cx + d 2 ax 3: Functions of the form y = 2 + bx + c ...............62 dx + ex + f 4: Oblique asymptotes...........................................65 5: Reciprocal transformations of functions...........66 6: Modulus transformation.....................................71

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

4 Hyperbolic functions

ra

1: Defining hyperbolic functions and hyperbolic identities...........................................80 2: Hyperbolic identities..........................................85 3: Solving harder hyperbolic equations................87

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

1: Sigma notation....................................................92 2: Using standard formulae....................................94 3: Method of differences........................................99 4: Maclaurin series.................................................103

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

D

5 Series

Focus on … Proof 1............................................. 111 Focus on … Problem solving 1...........................113 Focus on … Modelling 1.....................................115 Cross-topic review exercise 1.............................117

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

4 Hyperbolic functions

Before you start…

pl

•• 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.

e

In this chapter you will learn how to:

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

You should be able to work with natural logarithms.

AS/A Level Mathematics Student Book 1, Chapter 3

You should be able to solve quadratic equations.

2 Solve x 2 + 3 x = 1.

AS/A Level Mathematics Student Book 1, Chapter 10

You should be able to work with the symmetries of trigonometric functions.

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

AS/A Level Mathematics Student Book 1, Chapter 10

You should be able to work with trigonometric identities.

4 Simplify 3 sin 2 2 x + 3 cos2 2 x.

AS/A Level Mathematics Student Book 1, Chapter 5

You should be able to interpret transformations of graphs graphically.

5 What transformation changes f( x ) to 3 f ( x + 1) ?

AS/A Level Mathematics Student Book 1, Chapter 9

You should be able to complete binomial expansions of brackets with positive integer powers.

6 Expand (2 + x )3 .

D

ra

ft

sa

m

S/A Level Mathematics A Student Book 1, Chapter 7

y 1 (cos , sin )

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 equation x 2 + y 2 = 1) at an angle θ to the positive x-axis, has coordinates (cos θ , sin θ ).

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.

2

–1

1

O

x

–1 y

2

(cosh , sinh ) O

x

© 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.

79

A Level Further Mathematics for AQA Student Book 1

Section 1: Defining hyperbolic functions and hyperbolic identities Fast forward

Although the geometric definition of hyperbolic functions gives some helpful insight, a more useful definition is related to the number e.

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.

Key point 4.1 x −x • cosh x = e + e 2

e

x −x • sinh x = e − e 2

Tip

You can define the tanh x function by analogy with the trigonometric definition (of tan x ).

pl

Cosh is pronounced as it reads, sinh is pronounced ‘sinsh’ or sometimes ‘shine’ and tanh is pronounced ‘tansh’.

tanh x ≡

sinh x e x − e− x ≡ cosh x e x + e− x

m

Key point 4.2

sa

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) = 0 and tanh (0) = 0.

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

Did you know?

The tanh function is frequently used in physics, particularly in the context of special relativity and the study of entropy. y

y

y

1

x

ra

O

ft

1

y = sinh x

O

y = tanh x

Did you know?

Key point 4.3

Function

Domain

Range

cosh x

All real numbers

y

sinh x

All real numbers

All real numbers

tanh x

All real numbers

−1 < y < 1

80

x

–1

y = cosh x

D

O

1

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.

© 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.

x

4 Hyperbolic functions WORKED EXAMPLE 4.1 Find the range and domain of tanh 2 x + 1.

Tip Just as in trigonometry, tanh 2 x ≡ (tanh x )2.

The domain of tanh2 x is any real number.

e

y <1

So the range of tanh2 x + 1 is 1

The range of tanh x is from −1 to 1, but its square will always be positive.

pl

The range of tanh2 x is 0

Since the domain of tanh x is any real number, the expression tanh 2 x will likewise not be restricted in any way.

y <2

You just add one to the previous result.

m

The inverse functions of the hyperbolic functions are called arsinh x , arcosh x and artanh x .

sa

The graphs of these functions look like this: y

y

y

x

O

x

O

x

ra

ft

O

D

y = arsinh x

y = artanh x

y = arcosh x

Tip 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.

© 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.

81

A Level Further Mathematics for AQA Student Book 1

Key point 4.4 Function

Domain

Range

arcosh x

x

y

0

arsinh x

All real numbers

All real numbers

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....

pl

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

( arsinh x = ln ( x +

) +1 )

(

1 1+ x • artanh x = 2 ln 1 − x

)

PROOF 4

(

sa

x2

m

2 • arcosh x = ln x + x − 1

•

e

1

)

Let y = arcosh x

ra

Then cosh y = x

ft

Prove that arcosh x = ln x + x 2 − 1 .

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

Take cosh of both sides.

e y + e− y = x 2

Use the definition of cosh y .

e y + e− y = 2x

Rearrange into a disguised quadratic in e y.

D

e y + 1y = 2x e

(e y )2 + 1 = 2x e y

(e y )2 − 2x e y + 1 = 0

Continues on next page

82

© 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.

4 Hyperbolic functions

So

Use the quadratic formula.

2x ± (2x )2 − 4 e = 2 y

2 = 2x ± 4 x − 4 2

2x ± 4( x 2 − 1) 2

2 = 2x ± 4 x − 1 2 = x ± x 2 −1

But arcosh x is a function so it can only take one value.

e

Use the algebra of surds to simplify the expression.

Conventionally, you take the positive root, so this makes e y > 1 and y > 0.

pl

=

∴ ey = x + x2 −1 But y = arcosh x

So arcosh x = ln ( x + x 2 −1 )

m

y = ln ( x + x 2 −1 )

sa

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 .

y

y=k 1 –arcosh k

O

arcosh k

x

ft

WORKED EXAMPLE 4.2

ra

Given that cosh 2 x − cosh x = 2, express x in the form ln ( a + b ) or ln ( a − b ). cosh2 x − cosh x − 2 = 0

(cosh x − 2)(cosh x + 1) = 0

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.

D

cosh x = 2 or cosh x = −1

But cosh x 1so the only relevant solution is cosh x = 2. x = ± cosh−1 2

= ±ln ( 2 + 22 − 1 ) = ± ln ( 2 + 3 )

Use the inverse cosh function to find x. Remember that you need a plus or minus ( ±).

Then use the identity from Key point 4.4 to convert it to logarithmic form.

Continues on next page

© 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.

83

A Level Further Mathematics for AQA Student Book 1

( )

So

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

x = ln ( 2 + 3 ) or ln  1   2+ 3      2− 3 ln  1  = ln    2+ 3   (2 + 3 )(2 − 3 ) 

Simplify the second solution by rationalising the denominator to produce the required form.

  = ln  2 − 3   1 

e

So x = ln (2 + 3 ) or x = ln (2 − 3 )

1

pl

EXERCISE 4A

Use your calculator to evaluate each expression where possible.

.

2

5 tanh 1 + 8 c i 3 sinh 0.2 + 1 ii 2 cosh −1 0 e i tanh −1 2 ii

b

3

a i 3 sinh x − 2

cosh −1 2 d i sinh −1 0.5 ii

sa

Find the domain and range of each function. b

cosh 2 3 b i tanh 2 1 ii

ii sinh 3

m

a i cosh (−1)

ii 4 cosh x + 1

b i 3 tanh x + 2 ii 2 tanh x − 1

c i 2 cosh 2 x − 3 ii sinh 2 x − 1

d i 3 tanh 2 x + 1 ii tanh 2 x − 1

e i 2 cosh −1 x ii 3 sinh −1 x + 2

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

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

b

a i sinh x = −2

ii sinh x = 0.1

ft

b

c i 4 tanh x = 3 ii tanh x = 0.4

b i 2 cosh x = 5 ii cosh x = 4 3 cosh x = 1 d i 3 tanh x = 4 ii

4

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

Find and simplify the exact value of cosh (ln 2).

D

5

ra

e i sinh −1 x = 5 ii cosh −1 x = 4

6

Find and simplify a rational expression for tanh (ln 3).

7

Solve the equation sinh 2 3 x = 5.

8

Solve the equation 2 tanh 2 x + 2 = 5 tanh x.

9

Find and simplify an expression for cosh (sinh −1 x ).

10 Prove that sinh −1 x = ln ( x + x 2 + 1).

84

© 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.

4 Hyperbolic functions

(

)

11 Prove that tanh −1 x = 1 ln 1 + x .

2 1− x 12 In the derivation of cosh −1 x you found that two possible expressions were ln ( x + x 2 − 1) and ln ( x − x 2 − 1).

Show that their sum is zero and hence explain why the chosen expression is non-negative.

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.

pl

WORKED EXAMPLE 4.3 Prove that cosh 2 x − sinh 2 x ≡ 1.

2

 x −x  cosh2 x =  e + e   2 

m

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.

2

 x −x  sinh2 x =  e − e   2 

x −x −2 x 2x = e − 2e e + e 4

Since e x e− x = 1.

sa

x −x −2 x 2x = e + 2e e + e 4

2x −2 x = e +2+ e 4

e

Section 2: Hyperbolic identities

ft

2x −2 x = e −2+ e 4

e x e− x = 1 again.

Combine the two terms and simplify.

D

ra

2x −2 x 2x −2 x cosh2 x − sinh2 x ≡ e + 2 + e − e − 2 + e 4 4 e2 x + 2 + e−2 x − (e2 x − 2 + e−2 x ) ≡ 4 2x −2 x + + − 2 e e e2 x + 2 − e−2 x ≡ 4 4 ≡ 4 ≡1

Repeat with the sinh 2 x term.

Did you know?

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.

Rewind The result in Worked example 4.3 proves that (cosh x , sinh x ) lies on the hyperbola x 2 − y 2 = 1, as described in the introduction to this chapter.

© 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.

85

A Level Further Mathematics for AQA Student Book 1 WORKED EXAMPLE 4.4 Prove that sinh 2 x = 2 sinh x cosh x .

2x −2 x LHS ≡ e − e 2

On the LHS use the definition of sinh x and replace each x with 2 x .

x −x x −x RHS ≡ 2 × e − e × e + e 2 2

(e x − e − x ) × (e x + e − x ) 2

≡

(e x )2 − (e− x )2 2

e

≡

Then work from the RHS. Substitute the definitions of sinh x and cosh x .

pl

Multiply out the brackets, using the difference of two squares.

2x −2 x ≡ e −e 2 ≡ LHS

m

Using the rules of indices.

EXERCISE 4B Prove that cosh x − sinh x ≡ e− x.

2

Simplify 1 + sinh 2 x .

3

Prove that cosh 2 x ≡ cosh 2 x + sinh 2 x.

4

Prove that 1 − tanh 2 x ≡

5

Prove that tanh 2 x ≡

6

Prove that cosh x − 1 ≡ 1 ( e0.5 x − e−0.5 x ) . Hence prove that cosh x 2

1 . cosh 2 x

ft

2 tanh x .

1 + tanh 2 x

ra

7

sa

1

(

2

) (

1.

)

Prove that cosh A + cosh B ≡ 2 cosh A + B cosh A − B .

2 2 Use the binomial theorem to show that sinh 3 x ≡ 1 sinh 3 x − 3 sinh x.

4 4

9

a Explain why (cosh x + sinh x )3 ≡ cosh 3 x + sinh 3 x and (cosh x − sinh x )3 ≡ cosh 3 x − sinh 3 x.

Hence show that cosh 3 x ≡ cosh 3 x + 3 cosh x sinh 2 x.

D 8

b Write cosh 3 x in terms of cosh x .

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

86

© 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.

4 Hyperbolic functions

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.

WORKED EXAMPLE 4.5

m

sinh x + cosh x = 4

Use the definitions of sinh x and cosh x .

x = ln 4

When you are dealing with the sum or difference of two hyperbolic functions, it is often useful to use the exponential form.

sa

e x − e− x + e x + e− x = 4 2 2

ex = 4

e Tip

Solve sinh x + cosh x = 4.

2e x = 4 2

You will find useful hyperbolic identities in the formula book provided in the examination.

pl

It is only with experience that you will develop an instinct about which method will be most efficient.

Tip

ft

WORKED EXAMPLE 4.6

ra

Solve cosh 2 x + 1 = 3sinh x, giving your answer in logarithmic form.

The equation involves two types of function. You can use the identity cosh 2 x − sinh 2 x ≡ 1 to replace the cosh 2 term.

sinh2 x − 3 sinh x + 2 = 0

This is a quadratic.

D

cosh2 x + 1 = 3 sinh x (1+ sinh2 x ) + 1 = 3 sinh x

(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 x = ln (1+ 2) or x = ln (2 + 5 )

Use the logarithm form of arsinh.

© 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.

87

A Level Further Mathematics for AQA Student Book 1 WORK IT OUT Solve sinh 2 x cosh 2 x = 6 sinh 2 x .

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 = 3 sinh x sinh x cosh x − 3 sinh x = 0

Dividing by sinh 2 x : cosh 2 x = 6 2 x = cosh −1 6

sinh 2 x cosh 2 x = 6 sinh 2 x sinh 2 x cosh 2 x − 6 sinh 2 x = 0

e

x = 1 ln (6 + 35) 2

x = sinh 0 or x = cosh 3 −1

2 x = sinh −1 0 or 2 x = ± cosh −1 6 x = 0 or x = ±1.24

pl

−1

sinh 2 x (cosh 2 x − 6) = 0 sinh 2 x = 0 or cosh 2 x = 6

= ln (6 + 35)

sinh x (cosh x − 3) = 0 sinh x = 0 or cosh x = 3

EXERCISE 4C

m

x = 0 or x = 1.76

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.

3

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

4

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

5

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

6

Solve sinh 2 x = cosh x + 1, giving your answer in logarithmic form.

7

Solve tanh x =

9

ft

1 , giving your answer in logarithmic form. cosh x

ra

8

sa

1

Solve sinh x =

1 , giving your answer in logarithmic form. cosh x

sinh x + sinh y = 21 8

cosh x + cosh y = 27 8

D

a Show that e x = 6 − e y and e− x = 0.75 − e− y .

b 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

© 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.

4 Hyperbolic functions

Checklist of learning and understanding • Definitions of hyperbolic functions: • cosh x =

e x + e− x 2

• sinh x =

e x − e− x 2

• tanh x ≡

sinh x e x − e− x ≡ cosh x e x + e− x

Domain

cosh x

All real numbers

sinh x

All real numbers

tanh x

All real numbers

arcosh x

x

arsinh x

All real numbers

artanh x

−1 < x < 1

1

Range y

1

pl

Function

e

• Domain and range of hyperbolic and inverse hyperbolic functions:

All real numbers −1 < y < 1 y

0

m

All real numbers All real numbers

• Logarithmic form of inverse hyperbolic functions:

( • arsinh x = ln ( x +

) +1 )

(

1 ln 1 + x 2 1− x

)

D

ra

ft

• artanh x =

x2

sa

2 • arcosh x = ln x + x − 1

© 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.

89

A Level Further Mathematics for AQA Student Book 1

Mixed practice 4 1

F ind the range of y = 3 tanh 2 x + 2.

Choose from these options.

A 2

2

y <5

B −1

y

5

C −1 < y < 5

D All real numbers

S olve sinh x + cosh x = k , giving your answer in terms of k.

Choose from these options.

( )

A arsinh k 2

C e k

D ln k

e

B arcosh (2 k )

S implify tanh (1 + ln p).

4

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

5

a Express 5 sinh x + cosh x in the form Ae x + B e− x, where A and B are integers.

pl

3

[©AQA 2008]

a Use the definitions sinh θ = 1 (eθ − e−θ ) and cosh θ = 1 (eθ + e−θ ) to show that 1 + 2 sinh 2 θ = cosh 2θ .

2 2 b Solve the equation 3 cosh 2θ = 2 sinh θ + 11, giving each of your answers in the form ln p.

sa

6

m

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

Solve sinh 2 x = 2 cosh 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 2 cosh x + sinh x = 2.

ft

7

[©AQA 2009]

ra

10 Prove that 2 sinh 2 A ≡ cosh 2 A − 1.

11 Prove that cosh x > sinh x .

D

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

13 Use the binomial theorem to show that cosh 4 x ≡ 1 cosh 4 x + 1 cosh 2 x + 3 .

2 8 8 14 a Sketch the graph of y = tanh x . 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 ln 1 + u .

2 1−u 3 c i Show that the equation + 7 tanh x = 5 can be written as 3 tanh 2 x − 7 tanh x + 2 = 0. cosh 2 x

( )

ii Show that the equation 3 tanh 2 x − 7 tanh x + 2 = 0 has only one solution for x .

Find this solution in the form 1 ln a where a is an integer.

2 90

© 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.

4 Hyperbolic functions 15 a Using the definition sinh θ = 1 (eθ − e−θ ), prove the identity 4 sinh 3 θ + 3 sinh θ = sinh 3θ .

2 b Given that x = sinh θ and 16 x 3 + 12 x − 3 = 0, find the value of θ in terms of a natural logarithm. c Hence find the real root of the equation 16 x 3 + 12 x − 3 = 0, giving your answer in the form 2 p − 2 q , where p and q are rational numbers. [©AQA 2014] 16 a Prove the identity cosh 3 x = 4 cosh 3 x − 3 cosh x.

e

b If 48u 3 − 36u − 13 = 0 and u = cosh x find the value of x .

pl

c Hence find the exact real solution to 48u 3 − 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

m

cosh x + cosh y = 3.85

D

ra

ft

sa

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

© 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.

91