Chapter 35

04/19/23 By Chtan FYHS-Kulai 1

Prepared by : Tan Chor How (B.Sc)


04/19/23 3By Chtan FYHS-Kulai

Let y = sinx

then we have ysin -1x

2

130sin or

2

1

6sin

i.e.62

1sin 1


ysin -1xis the inverse function of

xy sin

Iff y is the 1-1 function!


ysin -1x

doesn’t mean sin

yx

Also doesn’t mean

ysin

1x


x

y1

-1

0

xy sin

2

2-

-

In this region, ,y is 1-1 function.

22

x


Now, if you flip the previous graph,

x

y

02

2-

xy -1sin

1

-1Principal valuesThe principal values of y is defined as that value lying between .

2


Similarly, check the cosine graph

x

y1

-1

0 2

2-

-

xy cos In this region, ,y is 1-1 function.

x0


Now, if you flip the previous graph,

x

y

2

2- 1

-1 0

Principal valuesThe principal values of y is defined as that value lying between 0 and ∏ .

xy -1cos


x

y

0 2

2-

-

xy tan


Graph of xy 1tan

x

y

2

2-

0

xy -1tan

Principal valuesThe principal values of y is defined as that value lying between .

2


Some books write as .

Domain of y is

Range of y is

xy -1sinxy arcsin

1,1

2,2


62

1sin 1

2

1

6sin)

2

1sin(sin 1


In general, we have

xx )sin(sin 1

1,1x


xx 11 sinsin

We also have :

1,1x


e.g. 1

Evaluate . 2

2sin 1

Soln :

2

2

4sin

42

2sin 1


e.g. 2

Evaluate .

2

3-sin 1

Soln :

2

3-

3-sin

3-

2

3-sin 1


e.g. 3 Evaluate .

3

2sinsin 1

Soln :

1,13

2x

3

2

3

2sinsin 1


Soln :

e.g. 4Evaluate .

4sinsin 1

Let 4

sin 1 x

4sin

x

44sinsin 1


Now, let see Same as .

Domain of y is

Range of y is

xy -1cosxy arccos

1,1

,0


In general, we have

xx )cos(cos 1

1,1x


xx 11 coscos

We also have :

1,1x


e.g. 5

Evaluate .

1800 andSoln :Between

gives .

2

3-sin 1

6

5cos150cos

or

2

3



Domain of y is

Range of y is

xy -1tanxy arctan

,

22-

，



Domain of y is

Range of y is

xy -1cotxarcy cot

,

，0


xx )tan(tan 1

xx )cot(cot 1

,x


e.g. 6Evaluate .

2

3sintan 1

Soln :

602

3sin 1

360tan


e.g. 7Evaluate .

5

4sincos 1

Soln :Let

5

4sin 1

5

4sin

3

54

5

3cos

5

4sincos 1


e.g. 8Find the value of the following Expression :

2

1cos

5

3sinsin 11


Soln : Let and 5

3sin 1a

2

1cos 1b

2

1cos,

5

3sin ba

ba3

5

4 1

23


2

1cos

5

3sinsin 11

ba sinbaba sincoscossin

2

3

5

4

2

1

5

3

10

343


e.g. 9Find the value of the following Expression :

2tan3tan 11


Soln : Let 2tan,3tan 11 ba

2tan3tan banda

ba

baba

tantan1

tantantan

161

5


There are 2 possible answers.

135451tan 1 or

4

3135

orba

[because a and b are both positive values, a+b must be positive value.]



2cossin 11

xx

1,1x

Identity (1)


2cottan 11

xx

x

Identity (2)


Let prove the identity #1

To prove :2

cossin 11 xx

Same as to prove :

xx 11 cos2

sin

A


LHS of A : xx 1sinsinCheck slide #14

RHS of A : xxx

11 coscoscos2

sin

xx 11 cos2

sinsinsin

We have,

and

2sin

2- 1-

x

x-1cos0 [ x(-1) ]

B


0-cos- -1 x

02

cos-2

-2

1-

x

2cos-

22- 1-

x C

B and C state that both and are .

x1sin

x1cos-2

2,2


xx 11 cos2

sin

i.e.

2cossin 11

xx


Let prove the identity #2

To prove :2

cottan 11 xx

Same as to prove :

xx 11 cot2

tan

A


xx 1tantan

xxxc

11 cotcotot-2

tan

xxc 11 tantanot-2

tan

But 2

tan2

1 x

and x1cot0 [ x(-1) ]


0cot 1 x

2cot

221

x

Both and x1tan x1cot2

2,2


e.g. 10Prove that

11

7tan

4

1tan

3

1tan 111


Soln :BA

4

1tan,

3

1tan 11

Let

then 4

1tan,

3

1tan BA

BA

BABA

tantan1

tantantan

11

7

41

31

1

41

31


4,0,

BA

2,0

BA

i.e.11

7tan 1BA

11

7tan

4

1tan

3

1tan 111


e.g. 11Prove that

3

2tan2

13

5cos 11

Soln :BA

3

2tan,

13

5cos 11

Let

LHS:13

5

13

5coscos 1


RHS: B2cos3

2tan2cos 1

1cos2 2 B

B2

3

13

13

51

13

92

2,02,

BA

3

2tan2

13

5cos 11



Do keep in mind :

Equation Range of solution

The only solution

x1sin

x1cos

x1tan

x1cot

2,2

,0

2,2

,0

sinx

cosx

tanx

cotx


e.g. 12Solve the equation .

61sin 1 x

Soln :

2,26

6sin1

x

2

1

2

3 x


e.g. 13Solve the equation .

4tan 21

x

Soln :

4tan2

x

12 x

1 x


e.g. 14Solve the equation , assuming that all the inverse tangents are positive acute angles.

2tan2tantan 111 xx


Soln : 2tantan2tantantan 111 xx

Let xBxA 2tan,tan 11

2tantan1

tantan

BA

BA

xBxA 2tan,tan


221

22

x

xx

02322 2 xx

02122 xx

2,22

1 xx

[reject]



Differentiation of an inverse function :

If 53)( xxfy

then 3dx

dy

The inverse function is : xy 35

3

5y

x


3

5)(1

yxyf

3

1

dy

dx

i.e.

dxdydy

dx 1


So, in general, )(xfy

0)(' xf)(1 yfx is the inverse

function of )(xfy

dxdydy

dx 1


Differentiation of an inverse circular function :

xy 1sin 1,1x

Its inverse function :

yx sin

2,2

y


0cos ydy

dx

22

y

dydxdx

dy 1

yy 2sin1

1

cos

1

21

1

x


So,

2

1

1

1sin

xx

dx

d


Another way to derive this formula :

xy 1sin yx sin

ydx

dx

dx

dsin

dx

dyy

dy

dsin1


dx

dyycos1

ydx

dy

cos

1

y2sin1

1

21

1

x


Similarly,

2

1

1

1cos

xx

dx

d

21

1

1tan

xx

dx

d

21

1

1cot

xx

dx

d


e.g. 153

sin 1 xy

Find the differentiation of y.Soln :

3

31

12

x

dx

d

xdx

dy

29

1

x


e.g. 16Find dy/dx if .

x

xy

1

1tan 1

Soln :

x

x

dx

d

xxdx

dy

1

1

11

1

12


222

2

1

)1()1(1

11

1

x

xx

xx

x

22 1

1

22

2

xx



(1)cx

x

dx

1

2sin

1

(2)cx

x

dx

1

2tan

1


In a general format :(1)

ca

x

xa

dx

1

22sin

(2)

ca

x

axa

dx

1

22tan1


e.g. 17Evaluate . 241 x

dx

Soln :

22

414

1

41 x

dx

x

dx

2

2

214

1

x

dx


cx

21

tan

211

4

1 1

cx 2tan2

1 1


Home works :

Mathematics 3 (Further Mathematics)

Ex 15a, Ex 15d, Misc Ex.

Documents

Chapter 35