Download study material on Inverse Circular Function for IIT-JEE

7/29/2019 Download study material on Inverse Circular Function for IIT-JEE

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Progression / APEX INSTITUTE FOR IIT-JEE / AIEEE / PMT, 0120-4901457, +919990495952, +919910817866 Page 1

INVERSE CORCULAR FUNCTIONS

If a function is one to one and onto from A to B, then function g which associates each element y B to

one and only one element x A, such that y = f(x), then g is called the inverse function of f, denoted by x

= g(y). Usually we denote 1fg {Read as f inverse}

1f ( ).x y

We have seen that the trigonometric functions, sin, cos etc. are all periodic and thus, each of them

achieves the same numerical value at an infinite number of points. Thus, the equation1

sin2

x has an

infinite number of solutions, viz., , 6 6

x etc. If one is to answer the question : What is the angle

whose sine is1

2?, there is no unique answer. The difficulty arises as the function f : defined byR R

f ( ) sinx x is not one to one and thus, does not admit of an inverse. To achieve a unique answer to the

aforesaid question we restrict the domain of sin x so that the resulting function is invertible. Thus, the

function : , [1, 1] defined by ( ) sin2 2

g g x x is one to one and onto and admits of an inverse

(denoted by 1sinh and read as sin inverse or arc sin) defined as :[1, 1] , where2 2

h

( ) if sinh y x y x . The function 1sin is the inverse of the sin function when the sin function is viewed in

a restricted sense.

We similarly define the other inverse trigonometric functions

1: 1sin x is an angle and denotes the smallest numerical angle, whose sine is x.

2: If there are two angles one positive and the other negative having same numerical value. Then we

shall take the positive value.

1.00

Inverse Function Definition

1.01

Inverse Trigonometric Function

Important Points


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Here, 1 1 1sin ,cos , tanx ec x x belongs to I and IV quadrant.

Here, 1 1 1cos , sec , cotx x x belongs to I and II quadrant.

1. I quadrant is common to all the inverse functions.2. III quadrant is not used inverse function.3. IV quadrant is used in the clockwise direction i.e., / 2 0y .

1. If sin y = x, then 1siny x under certain condition.1 sin 1; but siny y x

1 1x

Again, sin y = 1 y = /2 and sin y = 1 y = /2

Keeping in mind numerically smallest angles or real numbers.

/2 y /2

These restrictions on the values of x and y provide us with the domain and range for the function

1siny x .

i.e., Domain : [1, 1]

Range : [ / 2, / 2]

x

y

2. Let cos y = x then 1cosy x under certain condition 1 cos 1y .

1.02

Inverse Trigonometric Function

1.03 Domain, Range And Graphs of Inverse Functions


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1 1

cos 1

cos 1 0

0 {as cos is a decreasing function in [0, ]; hence cos cos cos 0}

x

y y

y y

y x y

These restrictions on the values of x and y provide us the domain and range for the function1cosy x .

i.e., Domain : [1, 1]

Range : [0, ]

x

y

3. If tan y = x then 1tany x , under certain conditions.Here, tan

tan / 2 / 2

Thus, domain

Range ( / 2, / 2)

y R x R

y y

x R

y

4. If cot y = x, then 1coty x (under certain conditions)cot ;

cot 0

y R x R

y y

These conditions on x and y make the function, cot y = x oneone

and onto so that the inverse function exists.

5. If sec y = x, then 1sec , where 1 and 0 , / 2x x y y Here, Domain : (1, 1)

Range: [0, ]{ / 2}

x R

y

6. If cosec y = x then 1cos ,y ec x where 1 and / 2 / 2, 0

Here, Domain : (1, 1)

Range: [ / 2, / 2]{0}

x y y

R

1. ., cot is meaningful.

i.e., Domain: xRange : (0, )

i e y x

Ry


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Function Domain Range

1

1

( ) sin where 1 1 2 2

( ) cos where 1 1 0

i y x x y

ii y x x y

1

1

( ) tan where 2 2

( ) cos where 1 or 1

iii y x x R y

iv y ec x x x

1

1

, 02 2

( ) sec where 1 or 1 0 ;2

( ) cot where

y y

v y x x x y y

vi y x x R 0 y

Note :

(a) 1st quadrant is common to the range of all the inverse functions.

(b) 2nd quadrant is not used in inverse functions.

(c) 4th quadrant is used in the clockwise direction i.e. 02

y .

(d) No inverse function is periodic. (See the graphs on page 17)

I llustration 1 : Find the value of 1 11 1tan cos tan 2 3

.

Solution :

I llustration 2 : Find the domain of 1 2sin (2 1)x .

Solution :

1.04 Principal values & Domains of Inverse Trigonometric / Circular Functions

1 11 1Let tan cos tan

2 3

tan tan3 6 6

1.

3

y

y Ans

1 2

2

2 2

Let sin (2 1)

For y to be defined

1 (2 1) 1

0 2 2 0 1

[1, 1]

y x

x

x x

x


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1 1

1 1

Property 2( )

( ) sin(sin ) , 1 1 ( ) cos(cos ) , 1 1

( ) tan(tan ) , ( ) cot(cot ) ,

( ) sec(s

A

i x x x ii x x x

iii x x x R iv x x x R

v1 1

ec ) , 1, 1 ( ) cos (cos ) , 1, 1x x x x vi ec ec x x x x

These functions are equal to identity function in their whole domain which may or may not be R.

(See the graphs on page 18)

I llustration 3 : Fin the values of 13

cos cot cot4

ec .

Solution :

1

1 1

Property 2( )

( ) sin(sin ) ; ( ) cos (cos ) ; 0

2 2

( ) tan (tan ) ; ( ) cot (cot ) ; 02 2

B

i x x x ii x x x

iii x x x iv x x x

1 1( ) sec (sec ) ; 0 , ( ) cos (cos ) ; 0,

2 2 2v x x x x vi ec ecx x x x

These are equal to identity function for a short interval of x only.(See the graphs on page)

1.05

Properties of Inverse Trigonometric Functions

1

1 1

3Let cos cot cot .........( )

4

3 3cot(cot ) , cot cot

4 4

from equation (i), we get

3cos

4

2

y ec i

x x x R

y ec

y

.Ans


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I llustration 4 : Find the value of 13

tan tan4

.

Solution :

1

1

3from the graph we can see that if ,

2 2

then tan (tan ) can be written as

3 3tan tan 123

4 4 4

x

y x y x

y y

I llustration 5 : Find the value of 1sin (sin7) .

Solution :

1

1

1

1

3Let tan tan

4

Note tan (tan ) if ,2 2

3 ,

4 2 2

3 3 3 3tan tan ,

4 4 4 2 2

graphs of y = tan (tan ) is

y

x x x

x

as :

1

1

1

Let sin (sin7)

: sin (sin7) 7 7 ,2 2

52 7

2

graph of sin (sin ) is as :

y

Note as

y x


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The function 1 1 1sin , tan and cosx x ec x are odd functions and rest are neither even nor odd.

I llustration 6 : Find the value of 1cos sin(5) .

Solution :

1

1

1

5From the graph we can see that if 2 then

2

sin (sin ) can be written as :

2sin (sin 7) 7 2

Similarly if we have to find sin (sin(5)) then

3 2 5

2

from the graph

x

y x

y x

1

1

of sin (sin ), we can say that

sin (sin(5)) 2 (5)

2 5

x

1 1 1 1

1 1 1 1

Property 2( )

( ) sin ( ) sin ; 1 1 ( ) tan ( ) tan ,

( ) cos ( ) cos , 1 1 ( ) cot ( ) cot ,

C

i x x x ii x x x R

iii x x x iv x x x R

1

1 1 1

1

1

Let cos sin(5)

cos (sin5) cos ( ) cos , 1

cos cos 5 .........( )

2

2 5 2

graph of cos (cos ) is as :

y

x x x

i

x


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1

1

from the graph we can see that if 2 x

then y=cos (cos ) can be wriiten as 2

5 5from the graph cos cos 5 5 2 5

2 2 2

from equation (i), we ge

x y x

t

5 3 5 5 Ans.

2 2y y

1 1 1 1

1

1

1

Property 2( )

1 1( ) cos sin ; 1, 1 ( ) sec cos ; 1, 1

1tan ; 0

( ) cot1

tan ; 0

D

i ec x x x ii x x xx x

xx

iii x

xx

I llustration 7 : Find the value of

1 2tan cot3

.

Solution :

1 1 1 1

1 1

Property 2( )

( ) sin cos , 1 1 ( ) tan cot ,2 2

( ) cos sec , 12

E

i x x x ii x x x R

iii ec x x

1

1 1

1

2Let y = tan cot .........( )

3

cot ( ) cot ,

equation (i) can be written as

2tan cot

3

i

x x x R

y

1 1 1

1

2 1 tan cot cot tan 0

3

3 3 tan tan

2 2

y x ifxx

y y


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I llustration 8 : Find the value of 1 11

sin(2cos sin ) when5

x x x .

Solution :

1 1

1 1

1 1

1

1

1

Let sin[2cos sin ]

sin cos , 12

sin 2cos cos2

sin cos2

1cos(cos )5

1cos cos

5

y x x

x x x

y x x

x

x x

y

1

1

.........( )

cos(cos ) if [1, 1]

1 1 1[1, 1] cos cos

5 5 5

1from equation (i), we get

5

i

x x x

y

1 1 2 1 1

1 1

2

Property 2( )

1( ) sin(cos ) cos(sin ) 1 , 1 1 ( ) tan(cot ) cot(tan ) , , 0

( ) cos (sec ) sec(cos ) , 11

F

i x x x x ii x x x R xx

xiii ec x ec x x

x


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I llustration 9 : Find the value of 13

sin tan4

.

Solutions :

I llustration 10 : Find the value of 11 5

tan cos2 3

.

Solution :

1

1

1 1

1

3Let sin tan .......( )

4

: To find y we use sin(sin ) , 1 1

For this we convert tan in sin

3 3Let tan tan and 0,

4 4 2

y i

Note x x x

x x

1 1

1

1

3sin

5

3sin (sin ) sin ........( )

5

0, sin (sin )2

equation (ii) can be written as :

3sin

5

ii

1 1 1

1

3 3 3tan tan sin

4 4 5

3from equation (i), we get sin sin

5

3

5

y

y

1

1

1 5Let tan cos ...........( )2 3

5 5Let cos 0 0, and cos

3 2 3

equation (i) becomes

tan .........(2

y i

y ii

22

)

51

1 cos 3 5 (3 5)3tan2 1 cos 45 3 5

1

33 5

tan .........( )2 2

0, 0,2 2 4

tan 02

iii


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I llustration 11 : Find the value of1 1 1

cos(2cos sin ) when 5x x x .

Solution :1 1

1 1

1 1 1

1

1

1

Let cos[2cos sin ]

sin cos , 12

cos 2cos cos cos cos2 2

1 sin(cos )

5

1 sin cos ........( )

5sin(cos ) 1

y x x

x x x

y x x x

x x

y i

x x

2

1

1

1 1

, 1

1 1 24sin cos 1

5 25 5

24from equation (i), we get y

5

1 1: cos cos and 0,

5 5 2

24sin

5

24sin (sin ) sin .........( )

5

x

Aliter Let

ii

1

1 1

1 1

0, sin (sin )2

equation (ii) can be written as

24 1sin cos

5 5

1 24cos sin

5 5

Now equation (i) ca

1

1

n be written as

24 sin sin ........( )

5

24 24 24[1, 1] sin sin

5 5 5

from equation (iii), we get

24

5

y iii

y

3 5from equation (iii), we get tan

2 2

3 5from equation (ii), we get

2y


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1 1 1 2 2 2 2

1 2 2 2 2

2 2 1 1

2 2 1 1

1

.

( )sin sin sin 1 1 , 0, 0 & ( ) 1

sin 1 1 , 0, 0 & 1

Note that : 1 0 sin sin2

1 sin sin2

( ) cos c

A

i x y x y y x x y x y

x y y x x y x y

x y x y

x y x y

ii x1 1 2 2

1 1 1

1

1 1

os cos 1 1 , 0, 0

( ) tan tan tan , 0, 0 & 11

tan , 0, 0 & 11

, 0, 0 & 12

Note that : 1 0 tan tan ;2

y xy x y x y

x yiii x y x y xy

xy

x yx y xy

xy

x y xy

xy x y xy1 1

1 1 1 2 2

1 1 1 2 2

1 1 1

1 tan tan2

.

( ) sin sin sin 1 1 , 0, 0

( ) cos cos cos 1 1 , 0, 0,

( ) tan tan tan , 0, 0

1

: For 0 and

x y

B

i x y x y y x x y

ii x y xy x y x y x y

x yiii x y x y

xy

Note x 0 those identities can be used with the help of preperties 2( )

. . change and to and which are positive.

y C

i e x y x y

I llustration 12 : Show that 1 1 13 15 84

sin sin sin5 17 85

.

Solution :

1.06 Identities of Addition and Subtraction

2 2

1 1 1

1

1

3 15 3 15 82260, 0 and 1

5 17 5 17 7225

3 15 3 225 15 9sin sin sin 1 1

5 17 5 289 17 253 8 15 4

sin . .5 17 17 5

84 sin

85


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I llustration 13 : Evaluate 1 1 112 4 63

cos sin tan13 5 16

.

Solution :

I llustration 14 : Evaluate 1 15

tan 9 tan4

.

Solution :

1 1 1

1 1

1 1 1

1 1 1

12 4 63Let cos sin tan

13 5 16

4 4sin cos

5 2 5

12 4 63cos cos tan

13 2 5 16

4 12 63 cos cos tan .......( )

2 5 13 16

4 12 0, 0 and

5 13

z

z

z i

1 1 1 1

1 1

1 1

4 12

5 13

4 12 4 12 16 144 63

cos cos cos 1 1 cos5 13 5 13 25 169 65

equation (i) can be written as

63 63 cos tan

2 65 16

63 63sin tan

65 16

z

z

1 1

1 1

.........( )

63 63sin tan

65 16

from equation (ii), we get

63 63tan tan 0 Ans.

65 65

ii

z z

1 1 1

1

1 1

5 59 0, 0 and 9 1

4 4

59

5 4tan 9 tan tan

54 19.4

tan (1) 4

5 3tan 9 tan .

4 4


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I llustration 15 : Define 1 3cos (4 3 )y x x in terms of 1cos x and also draw its graph.

Solution :

1

1 2 1

1

1

1 2

.

12sin if

2

1( ) sin 2 1 2sin if

2

1( 2sin ) if

2

2cos if 0 1(ii) cos (2 1) =

2 2c

C

x x

i x x x x

x x

x xx

1

1

1 1

2

1

21

2

os if 1 0

2 tan if 12

(iii) tan = 2 tan if 11

( 2 tan ) if 1

1 (iv) cos

1

x x

x xx

x xx

x x

x

x

1

1

2 tan if 0`

2 tan if 0

x x

x x

1 3

1

1 3

1

Let cos (4 3 )

Note Domain : [1,1] and range : [0, ]

Let cos [0, ] and cos

cos (4 3cos )

cos (cos3 ) .........( )

y x x

x x

y cos

y i


15/27


1 1

[0, ] 3 [0,3 ]

to define cos (cos3 ), we consider the graph of cos (cos ) in the interval [0,3 ].

Now from the above graphs we can see that

( ) if 0 3

y x

i

1cos (cos3 ) 3


3 if 3y

1

1

3 if 03

13cos if 1

2

( ) if 3 2 cos (cos3 ) 2 3


2 3

y

y x x

ii

y

1

1

if 3 2

22 3 if 3 3

1 12 3cos if

2 2

( ) 2 3 3 cos (cos3 ) 2 3


2

y

y x x

iii

y

1

1

1 3

3 if 2 3 3

22 3 if

3

12 3cos if 1

2from (i), (ii) & (iii), we get

13cos ;

2

cos (4 3 )

y

y x x

x

y x x1

1

1

1 12 3cos ;

2 2

12 3cos ; 1

2

x

x x

x x


16/27


1 3

1

2 1/ 2

2

Graph:

For cos (4 3 )

domain :[1,1]

range:[0, ]

1( ) if 1, 3cos2

33(1 ) ..........( )

1

10 if ,1

2

1decreasing if ,1

2

y x x

i x y x

dyx i

dx x

dyx

dx

x

2

2 2 3/2

again if we differentiate equation (i) w.r.t. ' ', we get

3

(1 )

x

d y x

dx x

2

2

1

2

1 10 if ,1 concavity downwards if ,1

2 2

1 1( ) if , 2 3cos .

2 2

3 1 10 if ,

2 21

increasing if

d yx x

dx

ii x y x

dy dyx

dx dxx

x2

2 2 3/2

2

2

2

2

1 1 3, and

2 2 (1 )

1( ) if ,0 then 0

2

1concavity downwards if ,0

2

1( ) if 0, then 0

2

1concavity downwards if 0,

2

d y x

dx x

d ya x

dx

x

d yb x

dx

x

2

2

1 3

1

( ) Similarly if 1 then 0 and 0.2

the graph of cos (4 3 ) is as

dy d y

iii x dx dx

y x x


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1 1 1 1

1 1 1

1 1 1

.

If tan tan tan tan if , 0, 0, 0 & ( ) 1

1

NOTE:

( ) If tan tan tan then

(ii) If tan tan tan then 12

(iii) I

D

x y z xyzx y z x y z xy yz zx

xy yz zx

i x y z x y z xyz

x y z xy yz zx

1 1 1

1 1 1

f tan 1 tan 2 tan 3

1 1( ) tan 1 tan tan

2 3 2iv


18/27


Inverse Trigonometric Functions

Some Useful Graphs


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21/27



22/27



23/27


1. 1 1 1sin , cos , tanx x x etc. denote angles or real numbers whose sine is x , whose cosine is x and

whose tangent is x, provided that the answers given are numerically smallest available. These are

also written as arc sin , cosx arc x etc.

If there are two angles one positive & the other negative having same numerical value, then

positive angle should be taken.

Part : (A)

1. 1 1 1If cos cos cos 3 then is equal tov v v .

(a) 3 (b) 0 (c) 3 (d) 1

2. 1 1 1Range of f ( ) sin tan sec is.x x x x

(a)3

,4 4

(b)3

,4 4

(c)3

,4 4

(d) none of these

3. The solution of the equation 1 13

sin tan sin 0 is.

4 6x

(a) x = 2 (b) x =4 (c) x = 4 (d) none of these

4. The value of 1 1 1sin [cos{cos (cos ) sin (sin )}], where , is2

x x x

(a)2

(b)4

(c) 4

(d) 2

5. The set of values of k for which 2 1x kx + sin (sin 4) > 0 for all real x is

(a) {0} (b) (2, 2) (c) R (d) none of these

6.

(a) 0 (b) (c) (d)

7.

1.07

General Definitions

EXERCISE-3

1 1 1 1sin (cos(sin )) cos (sin(cos )) is equal tox x

4 2

3

4

21 2 2 11

cos 1 . 1 cos cos holds for 2 4 2

x xx x x


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(a) (b)

(c) (d)

8.

(a) (b)

(c) (d)

9. The set of values of x for which the formula is true, is.

(a) (1, 0) (b) [0, 1]

(c) (d)

10. The set of values of a for which has at

least one solution is

(a) (b)

(c) R (d) none of these

11. All possible values of p and q for which

(a) (b) (c) (d) none of these

12. If , where [.] denotes the greatest integer function, then complete set of

values of x is

(a)cos1, 1

(b) (cot 1, cos 1) (c)cot1, 1

(d) none of these

13. The complete solution set of the inequality1 2 1[cot ] 6[cot ] 9 0x x

, where [.] denotes

greatest integer function, is

(a) ,cot3

(b) [cot 3, cot 2] (c)cot3,

(d) none of

these

14.1 11 1

tan cos tan cos , 0 is equal to4 2 4 2

x x x

1x x R

0 1x 1 0x

1 1tan tan , where 0, 0, 1, is equal toa b a b ab

1tan1

a b

ab1tan

1

a b

ab

1tan1

a b

ab

1 tan1

a b

ab

1 1 22sin sin 2 1x x x

3 3 ,

2 2

1 1 ,

2 2

2 1 2 1 2sin ( 4 5) cos ( 4 5) 0x ax x x x x

, 2 2 , , 2 2 ,

1 1 1 3cos cos 1 cos 1 holds, is

4p p q

11,

2p q

11,

2q p

10 1,

2p q

1 1[cot ] [cos ] 0x x


25/27


(a) x (b) 2x (c)

2

x (d) 2

x

15.11 3sin 2If sin , then tan is equal to

2 5 4cos2 4 .

(a) 1/3 (b) 3 (c) 1 (d) 1

16.1 1If cot tan tan tan , tan is equal to

4 2

uu then .

(a) tan (b) cot (c) tan (d) cot

17. The value of1 1sin 1 sin

cot , , is :21 sin 1 sin

x xx

x x

(a)

2

x

(b) 2 2

x(c) 2

x

(d)2

2

x

18. The number of solutions of the equation,1 1 1

sin cos (1 ) sin ( ), is/arex x x.

(a) 0 (b) 1 (c) 2 (d) more than 2

19. The number of solutions of the equation1 1 1

2

1 1 2tan tan tan is

2 1 4 1x x x .

(a) 0 (b) 1 (c) 2 (d) 3

20.1 1 1 1 11 1 1 1If tan tan tan ........ tan tan , then is equal to.

1 2 1 2.3 1 3.4 1 ( 1)n n

(a) 2

n

n (b) 1

n

n (c)1n

n (d)1

n

21.1If cot , , then the maximum value of 'n' is :

6

nn N

(a) 1 (b) 5 (c) 9 (d) none of these

22. The number of real solutions of (x, y) where,

( )91

( )3

sin , cos (cos ), 2 2 , is :C

C

y x y x x

(a) 2 (b) 1 (c) 3 (d) 4

23. The value of11 1cos cos

2 8 is equal to


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(a) 3/4 (b) 3/4 (c) 1/16 (d)

Part : (B)

24. , and are three angles given by

1 1 11 12tan ( 2 1), 3sin sin and22

1 1cos . Then3

(a) (b) (c) (d)

25.

1 1cos tan thenx x

(a)2 5 1

2x

(b)2 5 1

2x

(c)

1 5 1sin(cos )2

x(d)

1 5 1tan(cos )

2x

26. For the equation

1 1 1 32 tan(2 tan ) 2tan(tan tan )x a a a, which of the following is invalid?

(a)

22a x a x

(b)

2 2 1 0a ax(c)

0a(d)

1, 1a

27.

1

4 21

4The sum of tan is equal to

2 2n

n

n n

(a)

1 1tan 2 tan 3(b)

14tan 1

(c)/ 2

(d)

1sec 2

28. If the numerical value of

1tan(cos (4 / 5)) is a/b then.

(a) a + b = 23 (b) ab = 11 (c) 3b = a + 1 (d) 2a = 3b

29. If satisfies the in equation

2

2 0,x x then a value exists for

(a)

1sin(b)

1cos

(c)

1sec(d)

1cosec

30.

1 1 21If f ( ) cos cos 3 3 then:

2 2

xx x x


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(a)

2f

3 3 (b)

12 2f 2cos 3 3 3

(c)

1f

3 3 (d)

11 1f 2cos 3 3 3

m

In a triangle ABC, the sides are proporticnal to the sines of the angles opposite to them i.e.

a b c

sinA sinB sinC

I llustration 1 : In any ABC, prove that

Q1: In a ABC, prove that cot cot cot 2( )a A b B c C R r .

Q2: In a ABC, prove that 4 1 1 1s s s r

a b c R.

Q3: If , , are the distances of the vertices of a triangle from the corresponding points of contact with

the incircle, then prove that 2y

ry

.

Q4: In a ABC prove that,2

1 2 2 3 3 1r r r r r r s .

Q5: In a ABC prove that2

1 2 3 rr rr rr ab bc ca s .

1.01

SINE RULE

EXERCISE-3

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