CH07 - Inverse Functions

Domain: Set of all possible x values (the independent variable) of a function

Horizontal line test: Used to test for the existence of inverse functions. A function’s inverse relation is a function if any horizontal line cuts the original function in one place only

Inverse function: An inverse function is the opposite of (or undoes) the original function. The domain becomes the range and the range becomes the domain. It is a result of reflection of the original function in the line y = x

Mutually inverse functions: Two functions are mutually inverse if f -1[f(x)] = f [f -1(x)] = x

Range: Set of all possible y values (the dependent variable) of a function

Restricted domain: Domain restricted to the x values that will make the inverse relation a function

TERMINOLOGY

Inverse Functions

7

335Chapter 7 Inverse Functions

DID YOU KNOW?

INTRODUCTION

In thIs chapter you will study inverse functions. In particular, you will learn about inverse trigonometric functions, including differentiation and integration of these functions.

From left to right: Kepler, Herschel and Newton on the Astronomers Monument at the Griffith Observatory in Los Angeles

When solving or changing the subject of an equation, we use inverse operations. For example, to solve ,x 3 7+ = we subtract 3 from both sides.

Inverse functions are formed by taking the inverse operation or ‘undoing’ the operation of the function. however, the inverse is not always a function.

Inverse Functions

The notation for inverse functions was first used by the astronomer Sir William Herschel (1738–1822). Born in Germany, Herschel discovered Uranus, two of its satellites, and two satellites of Saturn. He also discovered infrared radiation. His sister and his son were also astronomers.

336 Maths In Focus Mathematics Extension 1 HSC Course

The inverse of multiplying by 2 is dividing by 2.

EXAMPLES

1. the inverse relation of y x2= is .y x2

=

2. the inverse relation of y x= is .y x2=

3. the inverse relation of y ex= is .logy xe=

Write down the inverse relation of each of the following functions.

7.1 exercises

1. y x3=

2. y x= −

3. ( )f x x5

=

4. y x= 3

5. y x7=

6. ( )f x x 1= +

7. y x 5= −

8. ( )f x x 3= +

9. y x3=

10. ( )f x 2x=

11. logy x4=

12. y x5=

13. ( )f x x 9= −

14. ( )f x x5= −

15. y x3= −

16. y x2=

17. y x= 7

18. logy xe=

19. y x9

=

20. y x8=

It is harder to find the inverse relation when more than one operation is involved.

EXAMPLES

Find the inverse relation of each of the following:

1. y x2 1= +

Solution

changing the subject of the equation to x by using inverse operations willshow what these operations are.

y x

y x

yx

2 1

1 2

2

1

= +− =−

=

Both y x2 1= + and xy

2

1=

− represent the same relation.


the inverse operations of ‘multiplying by 2 then adding 1’ are ‘subtracting 1 then dividing by 2’.this means that the inverse relation of y x2 1= + is

.y x2

1= −

changing the subject of this inverse relation gives .x y2 1= +

2. y x 23= −

Solution

changing the subject of the equation to x by using inverse operations will show what these operations are.

y x

y x

y x

2

2

2

3

3

= −+ =+ =3

Both y x 23= − and x y 2= +3 represent the same relation.the inverse operations of ‘cubing then subtracting 2’ are ‘adding 2 then finding the cube root’.this means that the inverse relation of y x 23= − is .y x 2= +3

changing the subject of this inverse relation gives .x y 23= − Can you find an easy way to find the inverse relation?

EXAMPLES

1. Find the inverse relation of .y x3 8= −

Solution

x y

x y

x y

3 8

8 3

38

= −+ =+ =

2. Find the inverse relation of ( ) .f x x2 75= +

Solution

x y

x yx y

x y

2 7

7 2

27

27

5

5

5

= +− =− =

− =5

the inverse relation of ( )y f x= can be found by interchanging the x and y values of the function.


sometimes the inverse relation is harder to find.

EXAMPLES

Find each inverse relation

1. y x 32= +

Solution

–

±

x y

x y

x y

3

3

3

2

2

= +=

− =

2. y x x4 72= + −

Solution

–

( )

±

±

x y y

x y y

x y y

x y

x y

x y

4 7

7 4

7 4

11 2

11 2

11 2

4 4

2

2

2

2

= ++ = +

+ = ++ = ++ = +

+ − =

+ +

Notice that in these examples, the inverse is not a function.

1. y x5=

2. y x2 3= −

3. y x 53= +

4. ( )f x x 17= −

5. y x 2= −3

6. y x2=

7. yx 5

3=+

8. y x2

1= +

9. ( )f x x 2= +

10. y x 7= −3

11. yx

3=

12. ( )f x x5 13= +

13. y x3 25= −

14. ( ) 2 5f x x= +

15. 3y x3 2 1= +

16. y ex=

17. y e x2=

18. lny x=

19. ( )lny x 1= +

20. ( )f x e 1x3= +

21. y x2=

22. y x2 4=

23. y x 52= +

24. – y x 36=

7.2 exercises

Find the inverse relation of each of the following functions

Complete the square on 4y y2 +


Class Investigation

sketch pairs of functions with their inverse relations on the same number plane.

What do you notice about their graphs?

on the number plane, the inverse relation can be represented by a reflection of the original function in the line .y x=

For example, y x2= gives x y2= when reflected in the line .y x= notice that, x y2= is not a function.

yy 5 x2

x 5 y2

y 5 x

x

25. y x x82= +

26. y x x4 2= −

27. y x x2 32= − +

28. y x x10 12= + −

29. y x x6 32= − −

30. y x x12 112= + −

Class Investigation

Find the sketch of the inverse relation of a function by reflecting it in the line .y x=

EXAMPLE

y x3=use thin (e.g. tracing) paper and fold along the line y x= to see the inverse relation.

Graph of Inverse Functions


Graph the inverse relation of each function below. Function Inverse

y xy x

y e

y x x

y x

13

21

x

2

2

= += +

== + −

=

x yx y

x e

x y y

x y

13

21

y

2

2

= += +

== + −

=

• Whichoftheseinverserelationsarefunctions?• Howcouldyoutesttheoriginalfunctiontoseeifitsinverseisa

function?

Horizontal line test

a function has a unique y value for every x value. since the inverse relation is an exchange of the x and y values, the inverse is a function if there is a unique value of x for every y value in the original function.

this means that the original function must be a one-to-one relation—that is, there is a unique x value for every y value, and a unique y value for every x value. this occurs when a graph is monotonic increasing or decreasing.

In the preliminary course, you used a vertical line test to check a function. to check if the inverse relation is a function, we use a horizontal line test.

EXAMPLES

are the inverse relations of the following curves functions?

1. y x3=

a horizontal line cuts the curve in only one place. thus the inverse will be a function.


2. y x2=

a horizontal line can cut the curve more than once, so the inverse is not a function.

1.

2.

3.

4.

7.3 exercises

Do these functions have an inverse function?


5.

6.

x

y

7.

x

y

8. y

x

9. y

x

10. y

x

11. y

x

12. y

x


notice the functions that have inverse functions do not turn around, but either always increase or decrease. We call this monotonic increasing or decreasing.

When an inverse function exists, the domain of the inverse function is the range of the original function, and the range of the inverse function is the domain of the original function.

Notation

Given a function ( ),y f x= the inverse function is written as ( ) .y f x= 1− sometimes we can just write f 1− .

You studied monotonic increasing and decreasing

curves in Chapter 2.

If ( )y f x= is a one-to-one (monotonic increasing or decreasing) function with domain a x b and range ( ) ( ),f a f by the inverse function ( )y f x= 1− has domain ( ) ( )f a x f b and range .a by

EXAMPLES

1.

Write down the domain and range of the function (a) .yx 2

1=−

Find the inverse function.(b) Find the domain and range of the inverse function.(c)

Solution

Domain: all real (a) ≠x 2 range: all real 0≠y

(b) :f xy

y x

y x

21

2 1

1 2

=−

− =

= +

1−

Domain: all real (c) 0≠x

: ≠

≠

x

x

1 0

1 2 2

Range

` +

` range: all real 2≠y

The domain and range of the inverse function are the range

and domain, respectively, of the function.

CONTINUED

This is different from

( )

1[ ( )] .

f xf x

1 =−

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2.

Find the domain and range of (a) ( ) 3 .f x x=What is its inverse function?(b) Write down the domain and range of the inverse function.(c)

Solution

Domain: all real (a) x range: 0y >

(b) :

( )

log loglog

log

f x

xy

y

f x x

3

33

y

y3 3

3

3`

====

=

1

1

−

−

Domain: (c) 0x > range: all real y

If we could restrict the domain of a function to a monotonic increasing or decreasing curve, then it will have an inverse function over that domain.

EXAMPLES

1. By restricting the domain of the function ,y x2= find the inverse function defined over that domain.

Solution

y x2= is monotonic increasing for 0.≥xthe inverse relation is

±

x y

x y

2==

For f to have a domain of ,≥x f0 1− must have a range of 0.≥yy x` = is the inverse function


similarly, if the domain of y x2= is restricted to 0≤x (monotonic decreasing) then the range of the inverse function is 0.≤y` the inverse function is y x= −

2. By restricting the domain of ( ) ,f x x x42= − find the inverse function defined over that domain.

Solution

Inverse relation:

`

( )

±±

x y y

x y y

y

x y

y x

4

4 4 4

2

4 2

4 2

2

2

2

= −+ = − +

= −+ = −

= + +

the graph of ( )f x x x42= − is monotonic increasing for 2.≥x If the function is restricted to this domain, the range of its inverse function will be 2.≥y` ( )f x x 4 2= + +1−

The axis of symmetry is x = 2. You studied the axis of symmetry of the parabola in the Preliminary Course.

CONTINUED


similarly, if the domain of f is restricted to ,≤x 2 the inverse function will be ( ) .f x x 4 2= − + +1−

sometimes you may need to use calculus or other methods to find where a curve is monotonic increasing or decreasing. you then need to look at the domain and range closely to find the inverse function.

EXAMPLE

1. Find where the function – y x x6 12= + is monotonic increasing and state the inverse function over that domain.

Solution

For increasing curve:

dx

dy

x

x

x

0

2 6 0

2 6

3

>

>>>

−

so the function is monotonic increasing when 3≥x .When :

( )

x

y

3

3 6 3 18

2

=

= − += −

When the domain is all real ,x 3> the range is all real ≥y 8−f 1:

–

–

–

( )

±

±

x y y

x y y

x y y

x y

x y

x y

6 1

1 6

1 6

8 3

8 3

8 3

9 9

2

2

2

2

= +− =

− =+ = −+ = −

+ + =

+ +

so the inverse function could be 8 3y x= + + or 8 3y x= − + + .

You could use the axis of symmetry to find where the curve is increasing.

When x ,3= the curve is stationary so can be included.

-


Looking at the domain and range:the inverse function will have domain all real 8≥x − and range all real 3≥yFor 3, 3≥y y x 8= + +so the inverse function is 8 3y x= + + .

1. Find the inverse function of each of the following functions, and state the domain and range of the inverse function.(a) y x3=(b) 3 2y x= −(c) y ex=

(d) ( )f x x2=

(e) yx 1

1=+

2. If the domain of each of the following functions is restricted to a monotonic increasing curve, find the inverse function, and state the domain and range of the inverse function.(a) y x2 2=(b) y x 22= +(c) ( )y x 3 2= −(d) y x x22= −(e) y x6=(f) y x1 2= −

(g) y x 14= −

(h) yx1

2=

3. (a) Find the domain over which the function y x x62= + is monotonic increasing.

Find the inverse function over (b) this restricted domain, and state its domain and range.

Find the domain over (c) which y x x62= + is monotonic decreasing.

Find the inverse function over (d) this restricted domain, and state its domain and range.

4. restrict the domain of each function to a monotonic decreasing curve and find the inverse function over this domain.(a) y x2=(b) y x3 12= −(c) ( ) ( )f x x 2 4= −

(d) yx3

2=

(e) ( )f xx2

4=

5. By restricting the following functions to (i) monotonic increasing and (ii) monotonic decreasing curves, write down the inverse function.(a) ( )f x x x22= −(b) y x 24= −(c) y x2 14= +(d) ( )f x x x6 12= − +(e) y x x4 32= + − .

7.4 exercises


Properties of inverse functions

Inverse functions have different properties. some of these you have seen already, such as the domain and range being the reverse of the original function. here are some others.

Inverse operations ‘undo’ each other. For example,

( ) ( )x x x2 2= =

a function and its inverse are mutually inverse functions. that is,

[ ( )] [ ( )]f f x f f x x= =1 1− −

EXAMPLES

1. show ( )f x x3= and its inverse function are mutually inverse functions.

Solution3( )

[ ( )] ( )

( )

[ ( )] ( )

( )

f x xf f x f x

xx

f f x f x

xx

3

3

3

====

===

3

3

3

1−

1

1

1

−

−

−

` and f 1− are mutually inverse functions.

2. If 2 5,y x= − find the inverse function and show that [ ( )] [ ( )] .f f x f f x x= =1 1− −

Solution

:

:

f y x

x yx y

x y

2 5

2 55 2

25

= −

= −+ =+ =

f 1−

f


[ ( )] ( )( )

[ ( )]

[ ( )] [ ( )]

f f x f xx

x

x

f f x f x

x

x

x

f f x f f x x

2 5

22 5 5

22

25

22

5 5

5 5

`

= −

=− +

=

=

= +

= + −

= + −=

= =

1 1

1

1 1

− −

−

− −

c

c

m

m

a useful property of differentiation is

dx

dy

dydx 1=

given that ( )y f x= is a differentiable function (i.e. it is able to be differentiated).

You will use this property to prove the derivatives of the inverse trigonometric

functions.

there are different ways of showing that this result is true. here is an informal method.

Proof

Let ( , )P x y be a point on the curve ( ) .y f x= the tangent at P has x-intercept Q and y-intercept R.

Gradient of : αtanPRdx

dy=

changing the subject of the equation to x, the curve can also be represented by ( ).x g y=

( )

( )

β

αβ α

tandydx y

OQR

90

the curve makes with the axis

Now

sum of

-

`

+

+

+ ∆

=

== −

( )vertically opposite s+

´

°


`

` ( )β αα

α

tan tan

cot

tan

dydx

dx

dy

dx

dy

dydx

90

1

1

1

= −=

=

=

=

the formal proof of this result uses calculus.

Proof

Let ( )y f x=Let xδ and yδ be small changes in x and y respectively.

`

,δδ

δδ

δ δ

δδ

δδlim lim lim

x

y

yx

x y

x

y

yx

dx

dy

dydx

1

0 0

1

1

Taking limits of both sides as and

δ δ δx y x0 0 0

" "

=

=

=

" " "

EXAMPLES

1.

Find (a) .3

dx

dyy xif =

1

express (b) x in terms of y.

Find (c) dydx in terms of x.

show that (d) .dx

dy

dydx 1=

Solution

(a) 3

dx

dyx

31=

− 2

(b) 3y xy x3

==

1

(c)

3

3

( )x

x

dydx y3

3

3

2

2

=

=

=

1

2

(d) −

3 3

dx

dy

dydx x x

31 3

1

=

=

2 2

You studied differentiation from first principles in the Preliminary Course.

α and c90 α− are complementary angles. You studied these in the Preliminary Course.

°

´

´

´

´

´

´ ´


2.

Find (a) .dx

dyy eif x2=

change the subject of the equation to (b) x.

Find (c) dydx and write your answer in terms of x.

show that (d) .dx

dy

dydx 1=

Solution

(a) dx

dye2 x2=

(b)

ln lnln

ln

y e

y ex e

x

x y

2

2

21

x

x

2

2

`

====

=

(c) dydx

y

y

e

21 1

21

21

x2

=

=

=

(d) dx

dy

dydx e

e2

21

1

xx

22

=

=

1. For each function and its inverse, show that

[ ( )] [ ( )]f f x f f x x= =1 1− − .(a) ( ) 7f x x= +(b) 3y x=(c) y x=(d) ( )f x ex=(e) 3 1y x= +

2. (a) Find the domain and range of

x

y1

2−

= .

Find the inverse function.(b) state the domain and range of (c)

the inverse function.

3. (a) sketch lny x= on the number plane.

Find the inverse function and (b) sketch this on the same number plane.

Write down the domain and (c) range for both functions.

7.5 exercises

´

´

´ ´


4. For each of the following functions,

find (i) dx

dy

change the subject to (ii) x

find (iii) dydx in terms of x

show (iv) dx

dy

dydx 1=

(a) ( )f x x5=

(b) y x1=

(c) ( )lny x 1= +

(d) ( )f x e x= −

(e) 3y x= −

5. For each of the following functions, (i) write the inverse function f 1−

with x in terms of y

(ii) write f 1− with y in terms of x

(iii) find dx

dy of the inverse

function

(iv) find dydx of the inverse

function in terms of x

(v) show dx

dy

dydx 1=

(a) 3 1y x= +(b) ( )f x x3 5=

(c) ( )f xx 3

2=+

(d) y x 73= −

(e) y e x5 1= +

the trigonometric functions only have inverse functions when they have a restricted domain.

Inverse sine function

Class Investigation

consider siny x= and its reflection in the line .y x=

Does the reflection represent a function?1. can you restrict the domain of 2. siny x= so that the inverse function exists?Is this domain the only one that gives an inverse function for 3.

?siny x=

Inverse Trigonometric Functions

´

´


the inverse sine function is written as .siny x1= − If we restrict the curve

siny x= to a monotonic increasing curve, it will have domain 2 2

xπ π− and range 1 1.y−

x1−sin can also be written as arc sin. Some calculators

use this notation.

this means that siny x1= − has domain 1 1£ £x− and range .π πy

2 2−

Inverse cosine function

Class Investigation

consider cosy x= and its reflection in the line .y x=

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Does the reflection represent a function?1. can you restrict the domain of 2. cosy x= so that the inverse function exists?Is this domain the only one that gives an inverse function for 3.

?cosy x=

1−cos x can also be written as arc cos. the inverse cosine function is written as .cosy x1= − If we restrict the curve

cosy x= to a monotonic decreasing curve, it will have domain 0 x π and range .y1 1−

this means that cosy x1= − has domain 1 1x− and range .πy0

y

x−1 1

π

π2

Inverse tangent function

Class Investigation

consider tany x= and its reflection in the line .y x=

y = tan x

x = tan y

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Does the reflection represent a function?1. can you restrict the domain of 2. tany x= so that the inverse function exists?Is this domain the only one that gives an inverse function for 3.

?tany x=

the inverse tangent function is written as .tany x–1= If we restrict the curve

tany x= to a monotonic increasing curve, it will have domain π πx2 2

− and range all real y values.

tan 1 x can also be written as arc tan.

this means that tany x–1= has domain all real x values and range .π πx

2 2−

EXAMPLES

1. Find ( ) .tan 1–1

Solution

( )

π

tantan

yy

y

11

4

IfThen

–1

`

==

=

CONTINUED

< <

< <x


2. Find .sin2

11 −−e o

Solution

.

( )π π π

sin

sin

y

y

y y

21

21

4 2 2

If

Then

the range is restricted to

1

`

= −

= −

= − −

−e o

3. evaluate .cos231 −− e o

Solution

( )π π π

π

cos

cos

y

y

y y

23

23

60

65

Let

Then

in the range

1

`

= −

= −

= −

=

− e o

4. evaluate [ ( )] .sin cos 11 −−

Solution

[ ( )] πsin cos sin10

1 − ==

−

5. show .πsin cos21

21

21 1+ =− −c cm m

Solution

`

π π

π

π

π

sin cos

sin cos

21

21

6 3

63

2

21

21

2

LHS

RHS

1 1

1 1

= +

= +

=

=

=

+ =

− −

− −

c c

c c

m m

m m

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6. sketch .siny x3 21= −

Solution

`

`

`

sin

x

x

x

1 1

1 2 1

−−

−

:

:

:

:

π π

π π

π π

sin

sin

sin

sin

y x

y x

x

y x

y x

y

3 2

21

21

2 2

3 223

23

23

23

Domain of

Domain of

Range of

Range of

1

1

1 1

1 1

==

−

=

=

−

−

−

− −

− −sin x3 2−

y

x

3π2

3π2

−

12

− 12

Class Investigation

can you find the domain and range of, and sketch1. (a) ?siny a bx1= −

(b) ?cosy a bx1= −

(c) ?tany a bx1= −

What is the domain and range of 2. ( )?sin siny x1= − can you sketch its graph? Does ( )sin siny x1= − have the same domain and range?

Properties of inverse trigonometric functions

( )sin sinx x1 1− = −− −

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Proof

`

`

`

( )

( )

( )

π πsin

sin

sin

sin

sin

sin

sin sin

y x

y x y

x y

y

y x

y x

x x

2 2

Let

Then for

1

1

1

1 1

= −

= − −

= −= −

− == −

− = −

−

−

−

− −

b l

You can see this property from the graph of y sin x.1= −

( ) πcos cosx x1 1− = −− −

Proof

`

`

`

( )( )

( )

( )

π

ππ

ππ

coscos

cos

cos

cos

cos

cos cos

y x

y x y

x y

y

y x

y x

x x

0LetThen for

1

1

1

1 1

= −= −= −= −

− == −

− = −

−

−

−

− −

( )tan tanx x1 1− = −− −

Proof

( )

( )

( )

π πtan

tan

tan

tan

tan

tan

tan tan

y x

y x y

x y

y

y x

y x

x x

2 2

Let

Then for < <

1

1

1

1 1

`

`

`

= −

= − −

= −= −

− == −

− = −

−

−

−

− −

b l

Class Investigation

sketch siny x1= − and cosy x1= − on the same number plane. By adding ordinates, or other methods, sketch .sin cosy x x1 1= +− −

can you write down a property of sin cosx x1 1+− − from this?

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πsin cosx x2

1 1+ =− −

Proof

( )

( )

π π

π

π

π

π

sin

sin

cos

cos

sin cos

y x

x y y

y

y x

x x y y

1

2 2

2

22

2

2

Let

Then for

From(1)and(2):

1

1

1 1

`

=

= −

= −

− =

+ = + −

=

−

−

− −

b

b

l

l

EXAMPLES

1 show ( ) ( ) .πcos cos1 11 1− = −− −

Solution

( )

( )

( ) ( )

π

πππ

π

cos

cos

cos cos

1

10

1 1

LHS

RHS

LHS RHS

1

1

1 1`

= −=

= −= −==

− = −

−

−

− −

2 prove that .πsin cos54

54

2–1 1+ =−c cm m

Solution

Let αsin541 =−c m and βcos

541 =− c m

then αsin54= and .βcos

54=

b ( )

.

α β π

πsin cos

2

54

54

2

ut sum of

1 1`

+ ∆+ =

+ =− −c cm m

£ £


General solution of trigonometric equations

you studied general solutions of trigonometric functions in chapter 5.We can write these general solutions using inverse function notation as

follows:

the solution for θsin b= is given by ( )θ π sinn b1 n 1= + − − where n is an integer.

the solution for θcos b= is given by ±θ π cosn b2 1= − where n is an integer.

the solution for θtan b= is given by θ π tann b1= + − where n is an integer.

EXAMPLES

Find the general solution of these trigonometric equations. 1. 1θtan =

Solution( )

( )

θ ππ

θ π π

tan

tan

n

n

1

14

4

1

1

`

= +

=

= +

−

−

2. θcos23=

Solution

±

±

θ π

π

θ π π

cos

cos

n

n

223

23

6

26

1

1

`

=

=

=

−

−

e

e

o

o


1. evaluate, giving exact answers.(a) ( )sin 11−

(b) ( )tan 01−

(c) ( )cos 11−

(d) sin211−c m

(e) ( )tan 11 −−

(f) ( )sin 11 −−

(g) ( )cos 01−

(h) cos2

11−e o

(i) tan3

11−e o

(j) ( )tan 31 −−

(k) sin2

11−e o

(l) cos231 −− e o

(m) tan3

11 −−e o

2. evaluate(a) [ ( )]tan cos 11−

(b) [ ( )]cos cos 11 −−

(c) ( )πcos sin1−

(d) cos cos2

11−e o> H

(e) [ ( )]sin tan 11−

(f) tan cos211−c m< F

(g) ( )sin tan01−

(h) tan sin231 −− e o> H

(i) cos tan3

11 −−e o> H

(j) [ ( )]tan tan 11 −−

3. evaluate, in radians, correct to 2 decimal places.(a) .sin 0 41−

(b) .tan 1 721−

(c) .cos 0 5691−

(d) ( . )sin 0 61 −−

(e) ( . )tan 3 71 −−

4. evaluate, correct to 2 decimal places.(a) ( . )sin sin0 671−

(b) ( . )tan tan 0 141 −−

(c) ( . )cos cos1 641−

(d) ( . )sin cos 0 261−

(e) ( . )tan sin 0 671 −−

5. sketch the graph of(a) cosy x1= −

(b) tany x1= −

(c) siny x1= −

(d) cosy x21= −

(e) siny x31= −

(f) cosy x2 1= −

(g) cosy x5 31= −

(h) siny x32

1= −b l

(i) siny x2 41= −

(j) cosy x2 71= −

(k) tany x4 51= −

(l) ( )sin siny x1= −

(m) ( )cos cosy x1= −

(n) ( )tan tany x1= −

6. state the domain of(a) ( )siny x1 2= −

(b) cosy x x1= −

(c) [ ]sin tany x1= −

7. Find exact values for(a) ( ) ( )cos cos1 11 1− +− −

(b) ( ) ( )sin sin1 11 1+ −− −

(c) ( ) ( )tan tan1 11 1+ −− −

(d) sin cos21

211 1+− −

c cm m

(e) cos sin23

231 1+− −e eo o

(f) sin cos2

12

11 1+− −e eo o

7.6 exercises


8. Find exact values for

(a) sin sin541−c m< F

(b) cos sin541−c m< F

(c) tan cos13121−c m< F

(d) sin tan731−c m< F

(e) πcos sin4

1−b l< F

(f) ( )πtan cos1−

9. (a) Write θsin 2 in terms of θ.hence find the exact value of(b)

.sin cos532 1−c m< F

10. Find whether each of the following functions is even, odd or neither(a) siny x=(b) siny x1= −

(c) cosy x=(d) cosy x1= −

(e) tany x=(f) tany x1= −

(g) ( )sin siny x1= −

11. show that(a) ( )tan tan1 11 1− = −− −

(b) ( )sin sin1 11 1− = −− −

(c) ( )tan tan3 31 1− = −− −

(d) πcos cos21

211 1− −=− −

c m

(e) sin sin2

12

11 1− = −− −e o

12. Write down the general solution for the following equations

(a) θsin23=

(b) 1θcos =(c) θtan 3=(d) θsin

21=

(e) 1θtan = −

(f) θcos23=

(g) 0θcos =

(h) θsin21=

(i) θsin2

1= −

(j) θtan3

1= −

13. (a) Find the general solution for 1.θsin = −

Find the solutions when (b) , .n 0 1=

14. (a) Find the general solution for

.θcos21=

Find the solutions when (b) ±1.n =

15. (a) Find the general solution for .θtan 3= −

Find the solutions when (b) ,n 1 2= and 3.

16. Find the general solution to the following equations correct to 2 decimal places(a) .cos x 0 045=(b) .sin x 0 378=(c) .tan x 1 3=(d) .sin x 0 86= −(e) .tan x 0 93= −(f) .cos x 0 227= −


17. (a) Draw siny x1= − and cosy x1= − on the same set of axeson the same set of axes, sketch (b)

the graph of sin cosy x x1 1= +− − to show graphically that

πsin cosx x2

1 1+ =− − .

18. prove that

(a) πsin cos73

73

21 1+ =− −c cm m

(b) sin sin95

951 1− = −− −

c cm m

(c) πcos cos52

521 1− = −− −

c cm m

(d) tan tan107

1071 1− = −− −

c cm m

Differentiation of Inverse Trigonometric Functions

( )sindxd x

x1

11

2=

−−

Proof

sinsin

cos

cossin cos

cos sin

cos sin

y xx y

dydx y

dx

dy

dydx

yy y

y y

y y

x

dx

dy

x

1

1

1

1

1

1

1

1

LetThen

1

2 2

2 2

2

2

2

`

`

==

=

=

=

+ == −= −= −

=−

−

sindxd

ax

a x

11

2 2=

−−a k


Proof

`

`

`

sin

sin

sin

cos

cossin cos

cos sin

cos

y ax

ax y

x a y

dydx a y

dx

dy

dydx

a yy y

y y

ax

aa x

ya

a x

aa x

dx

dy

aa a x

a x

1

1

1

1

1

1

1

Let

Then

1

2 2

2 2

2

2

2 2

2

2 2

2 2

2 2

2 2

=

=

=

=

=

=

+ == −

= −

= −

= −

= −

=−

=−

−

a k

( )cosdxd x

x1

11

2= −

−−

Proof

`

coscos

sin

sinsin cos

sin cos

sin cos

y xx y

dydx y

dx

dy

dydx

yy y

y y

y y

x

dx

dy

x

1

1

1

1

1

1

1

1

LetThen

1

2 2

2 2

2

2

2

`

==

= −

=

= −

+ == −= −= −

= −−

−


cosdxd

ax

a x

11

2 2= −

−−a k

Proof

cos

cos

cos

sin

sinsin cos

sin cos

sin

y ax

ax y

x a y

dydx a y

dx

dy

dydx

a yy y

y y

ax

aa x

ya

a x

aa x

dx

dy

aa a x

a x

1

1

1

1

1

1

1

Let

Then

1

2 2

2 2

2

2

2 2

2

2 2

2 2

2 2

2 2

`

`

`

=

=

=

= −

=

= −

+ == −

= −

= −

= −

= −

= −−

= −−

−

a k

( )tandxd x

x111

2=

+−

Proof

tantan

sec

sec

y xx y

dydx y

dx

dy

dydx

y

1

1

LetThen

1

1

2

==

=

=

=

−

−


`

sec tany y

x

dx

dy

x

1

1

11

2 2

2

2

= += +

=+

tandxd

ax

a xa1

2 2=

+−a k

Proof

`

tan

tan

tan

sec

secsec tan

y ax

ax y

x a y

dydx a y

dx

dy

dydx

a yy y

ax

aa x

dx

dy

aa

a x

a xa

1

1

1

1

1

Let

Then

1

2

2

2 2

2

2

2

2 2

2

2 2

2 2

=

=

=

=

=

=

= +

= +

= +

=+

=+

−

e o

EXAMPLES

1. Differentiate .cosy x21= −

Solution 1

( )( ) ( )

dx

dy

x dxd x

x

1 2

1 2

1 4

2

function of a function rule2

2

= −−

= −−

´


Solution 2

( )

cos

cos

dxd

ax

a x

dxd x

x

x

x

x

1

2

21

1

411

41 4

1

1 4

2

1

2 2

1

22

2

2

2

`

= −−

= −−

= −−

= −−

= −−

−

−

a

c

k

m

2. Find the gradient of the tangent to the curve tany x1= − at the point where 3.x =

Solution

`

,dx

dy

xx

dx

dy

m

11

3

1 31

101

When

2

2

=+

=

=+

=

3. Find the equation of the normal to the curve tany x x1= − at the point , π1

4b l

Solution

..1

tan

tan

tan

y x x

dx

dyu v v u

xx

x

xx

x1

1

1

1

12

12

=

= +

= ++

= ++

−

−

−

l l

CONTINUED


at , π14b l:

π

π

π

π

tandx

dy

m

1 11

21

4

42

4

42

42

12

1

1`

=+

+

= +

= +

= +

= +

−

normal is perpendicular to tangent.

π

π

m m

m

m

1

42 1

24

So 1 2

2

2`

= −+ = −

= −+

c m

equation:( )

( )

( ) ( ) ( )

( )

( )

( )

ππ

π π π

π π π

π π ππ π π

y y m x x

y x

y x

y x

y x

x y

4 24 1

2 24

4 1

22 4

4 4

4 2 2 16 16

16 4 2 2 16 0

1 1

2

2

2

− = −

− = −+

−

+ − + = − −

+ − − = − +

+ − − = − ++ + − − − =

1. Differentiate(a) cos x–1

(b) sin x2 –1

(c) tan x–1

(d) ( )cos x3–1

(e) ( )sin x4 2–1

(f) ( )sin x–1 2

(g) ( )tan x2 1–1 −(h) ( )cos x5 8–1

(i) cos x3

–1

(j) tan x2

1−

(k) sin x36

1−

(l) cos x3 1−

(m) cos x7

1−

(n) ( )sin x5 3 2–1 +(o) cosx x–1

(p) ( )tan x 1–1 5+

2. Find the derivative of(a) ( )sin cosx–1

(b) ( )cos cosx–1

(c) ( )sin log x–e

1

(d) ( )tan ex1−

(e) ( )ln sin x1−

(f) tan x

11−

7.7 exercises


(g) ( )tan cos x 11 1 +− −

(h) tan x11−

(i) sin x2

11 +−b l

(j) ecos x1−

3. Find the gradient of the (i) tangent (ii) normal to the curve(a) cosy x1= − at the point 0, π

2b l

(b) tany x21= − at the point

where x41=

(c) ( ) ( )sinf x x1 3= − at the point

where x21=

(d) cosy x3

1= − at the point , π02b l

(e) tany x5

1= − at the point where

.x 0=

4. Find the equation of the tangent to the curve siny x21= − at the point where .x 0=

5. Find the equation of the normal

to the curve 5 ., πtany x51

4at1= − c m

6. (a) show that

( ) .sin cosdxd x x 01 1+ =− −

(b) Why do we get this result?

7. (a) show that siny x x1= − has a stationary point at (0, 0).

Find the domain and range of (b) the curve.

sketch (c) siny x x1= − on the number plane.

8. Find the second derivative of

(a) cos x3

1−b l

(b) ( )log tan xe1−

9. Find the first derivative of ( )cos siny x1= − in the domain .π πx−

10. (a) Find .tan tanxdxd

x11 1+− − b l; E

Draw the graph of(b)

.tan tany x x11 1= +− − b l

11. Differentiate(a) cos e x1 2−

(b) ( )log tan xe1−

(c) ( )tan ln x1−

(d) sin x11 2−−

(e) etan x1−

12. a 6 metre long ladder is leaning up against a wall at a height of h and angle θ as shown.

6 m

h

θ

show that (a) θ sin h6

1= −c m .

the ladder slips down the wall (b) at a constant rate of 0.05 ms 1. Find the rate at which the angle is changing in degrees and minutes when the height is 2.5 m.

£ £

-


sin

sin

cos sin

x

dx x C

a x

dxax C

x

dx x C x C

1

1

2

1

2 2

1

2

1 1

−= +

−= +

−− = + = − +

−

−

− −

#

#

#

cos sin

tan

tan

a x

dxax C a

x C

xdx x C

a xdx

a ax C

11

2 2

1 1

21

2 21

−− = + = − +

+= +

+= +

− −

−

−

#

#

#

Integration of Inverse Trigonometric Functions

Integration is the inverse of differentiation.

We make the coefficient of x2 equal to 1 to put it into the form of the standard integral.

EXAMPLES

1. evaluate x

dx

4 20

2

−#

Solution

π

π

sin

sin sin

x

dx

x4

21 0

20

2

20

2

1

0

2

1 1

−

=

= −

= −

=

−

− −

; E

#

2. Find x

dx4 9 2+#

Solution

tan

tan

xdx

xdx

x

dx

x C

x C

4 9

994

99

91

94

91

321

32

61

23

2

2

2

1

1

+

=+

=+

= +

= +

−

−

c

f

m

p

#

#

#


3. Find the volume, correct to 2 decimal places, of the solid of revolution

formed if the curve x

y1

12+

= is rotated about the x-axis from 0x =

to .x 3=

Solution

( )

.

π

π

ππ

tantan tan

yx

yx

V y dx

xdx

x

1

1

11

1

3 0

3 92 units

a

b

2

22

2

20

3

103

1 1

3

`

=+

=+

=

=+

== −=

−

− −

6 @

#

#

1. Find the integral (primitive function) of

(a) x1

12−

(b) x1

22−

−

(c) x1

12+

(d) x9

12+

(e) x4

12−

(f) x4

52+

(g) x2 1

32−

(h) x5 16

12−

(i) x 3

12 +

(j) x5

12−

2. Find

(a) x

dx36 2+#

(b) x

dx

1 4 2−#

(c) t

dt

9 2−#

(d) x

dx1 9 2+#

(e) x

dx

4 25 2−#

(f) x

dx9 16

32+

#

(g) t

dt25 4

22−

#

(h) x

dx1 5 2+#

(i) x

dx4 3 2+#

(j) x

dx5 25 9

22−

−#

7.8 exercises


3. Find the exact value of

(a) x

dx

1 20

1

−#

(b) x

dx1 20

1

+#

(c) x

dx

4 21

1

−−#

(d) t

dt923 +−

0#

(e) x

dx81 20

42

1

−#

(f) x

dx4

221

3

−#

(g) x

dx2 1

32

2

3

2

3

−−#

(h) x

dx4 9 2

3

1

3

1

−−#

(i) x

dx1 49

320

7 2

1

−#

(j) x

dx5 2

0

15

+#4. Find the area bounded by the

curve ,x

y1

12+

= the x-axis and the

lines 0x = and 2,x = to 1 decimal

place.

5. (a) Find the area enclosed

between the curve ,x

y1

12−

=

the x-axis and the lines

0x = and .x21=

this area is rotated about the (b) y-axis. Find the exact volume of the solid formed.

6. (a) Differentiate ( ) .log sin xe1−

hence find (b) sinx x

dx1 2 1

4

1

3

1

− −#

correct to 1 decimal place.

7. use simpson’s rule with 3 function values to find an

approximation to ,sin x dx1

0

−.0 4# correct to 1 decimal place.

8. Find the area enclosed between the curve ,cosy x1= − the y-axis

and the lines 0y = and .πy4

=

9. Find x

x dx1 6

2

−# by using the

substitution .u x3=

10. Find the equation of the curve

with dx

dy

x9

12−

= and passing

through , .π37b l

11. (a) show that

( ) ( )

.x xx

x x1 42 5

11

41

2 2

2

2 2+ ++

++

+=

hence evaluate(b)

( ) ( )x xx dx

1 42 5

2 2

2

1

2

+ ++# correct to

2 decimal places.

12. For the curve ,x

y1

12+

= find the

volume of the solid formed by rotating the curve about the

x-axis from .x x3

1 1to= =

13. (a) Differentiate .cosx x x11 2− −−

Find the area bounded by the (b) curve ,cosy x1= − the x-axis and

the lines 0x = and .x21=

14. use sinx x +dxd

x11 2−−> H to help

find the area enclosed between the curve ,siny x1= − the x-axis and the lines 0x = and 1.x =

15. the acceleration of a particle

is given by ,x

x1

12+

=p where x is

displacement in metres over time t seconds. If the particle starts at rest at the origin and always has positive velocity, find the equation of its velocity in terms of x.


1. evaluate(a) tan 11−

(b) sin231− e o

(c) cos2

11 −−c m

(d) ( )cos sin 11−

(e) πtan tan651−

c m

2. Find the indefinite integral of

(a) x1

32+

(b) x16

12−

3. Differentiate(a) sin x1−

(b) tan x31−

(c) cos x2 51−

4. Find the exact area bounded by the curve

,x

y1

12−

= the x-axis and the lines x21−=

and .x2

1=

5. Find the inverse function of ( ) .f x x3 2= −

6. Find the general solutions of .θcos2

1=

7. Find the inverse function of ( ) .f x e 1x3= +

8. (a) evaluate ( ) .sin cosdxd x x1 1+− −

explain this result.(b)

9. Differentiate .tanx x1−

10. (a) state the domain and range of .y x 1= −

Find the inverse of this function and (b) state its domain and range.

11. sketch the graph of .tany x2

1= −

12. show that .πcos cos23

231 1− = − −− −e eo o

13. (a) Find the domain and range of

.yx 2

1=+

Find the inverse function.(b) Find the domain and range of the (c)

inverse function.

14. (a) Find the area bounded by the curve

,x

y1

12−

= the x-axis and the lines 0x =

and .x21=

Find the volume of the solid formed (b) if this area is rotated about the x-axis.use simpson’s rule and give your answer correct to 2 decimal places.


(a) sin cos23

231 1+− −e eo o

(b) tan cos531−

c m

16. (a) Find the domain and range of .siny x2 1= −

sketch the graph of (b) .siny x2 1= −

17. Find the equation of the tangent to the

curve siny x3

1= − at the point 1 , .π21

6c m

18. (a) Find the domain over which the curve y x x42= − is monotonic increasing.

Find the inverse function over this (b) domain.

Test Yourself 7


1. By restricting the domain of each of the following functions to a monotonic increasing curve, find the inverse function, and state the domain and range of the inverse function.(a) ( )f x x1 2= −

(b) yx 1

12

=−

2. Differentiate .sinx

xy1

=−

are there any stationary points on the curve?

3. By using the substitution ,u x2= find

.x

x dx1 4+#

4. (a) show that .πtan tan54

45

21 1+ =− −

Find (b) .tan tandxd x x

11 1+− −b l

show that (c) πtan tanx x1

21 1+ =− − for

all x.

5. the acceleration of a particle is given by

.dtd x

x11

2

2

2=

− If initially the velocity

is 0.5 ms–1 when the particle is 0.1 m to the right of the origin, find its velocity, correct to 1 decimal place, when it is 0.8 m to the right of the origin.

6. By considering the difference between the area of a rectangle and the area enclosed between the curve siny x1= − and the y-axis, find the exact area bounded by the curve ,siny x1= − the

x-axis and the lines 0x = and .x23=

7. prove that sin cosx x11 1 2= −− − for 0 1.x

8. Find ,x

x dx1 9 4−

# using the substitution

.u x3 2=

9. sketch ( )sin cosy x1= − for 2 2 .π πx−

10. Find the volume of the solid formed if the curve secy x1= − is rotated about

the (a) x-axis from 0x = to 0.5,x = using the trapezoidal rule with 5 subintervals (give your answer correct to 1 decimal place)

the (b) y-axis from 0y = to πy4

= .

Challenge Exercise 7

19. If ( )p x x 13= − and ( ) ,q x x2 5= + evaluate(a) ( )p 71−

(b) ( ( ))q p 31−


(a) x

dx1 2

2

1

2

1

−−#

(b) x

dx1 7

32

0

21

1

+#

£ £

£ £

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CH07 - Inverse Functions