Inverse circular function

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 equation 1

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 is 1

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

111

1

Inverse Function Definition

1.01

111

1

Inverse Trigonometric Function

Important Points


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

111

1

Inverse Trigonometric Function

1.03 Domain, Range And Graphs of Inverse Functions


–1 1

cos –1

cos 1 0

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

–1cosy 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 one–one

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: x

Range : (0, )

i e y x

R

y


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. – 0

2y .

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

Illustration 1 : Find the value of –1 –11 1tan cos tan –

2 3.

Solution :

Illustration 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 – tan 3 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


–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

v –1 –1ec ) , –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)

Illustration 3 : Fin the values of –1 3cos cot cot

4ec .

Solution :

–1

–1 –1

Property – 2( )

( ) sin(sin ) ; – ( ) cos (cos ) ; 02 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 2

v 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

111

1

Properties of Inverse Trigonometric Functions

–1

–1 –1

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

4

3 3 cot(cot ) , cot cot

4 4

from equation (i), we get

3 cos

4

2

y ec i

x x x R

y ec

y

.Ans


Illustration 4 : Find the value of –1 3tan tan

4.

Solution :

–1

–1

3 from the graph we can see that if ,

2 2

then tan (tan ) can be written as –

3 3 tan tan – – 123

4 4 4

x

y x y x

y y

Illustration 5 : Find the value of –1sin (sin 7) .

Solution :

–1

–1

–1

–1

3Let tan tan

4

Note tan (tan ) if – ,2 2

3 – ,

4 2 2

3 3 3 3 tan tan ,

4 4 4 2 2

graphs of y = tan (tan ) is

y

x x x

x

as :

–1

–1

–1

Let sin (sin 7)

: sin (sin 7) 7 7 – ,2 2

5 2 7

2

graph of sin (sin ) is as :

y

Note as

y x


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

Illustration 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 :

– 2

sin (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 (– sin 5) cos (– ) – cos , 1

– cos cos – 5 .........( )2

– 2 – 5 –2

graph of cos (cos ) is as :

y

x x x

i

x


–1

–1

from the graph we can see that if –2 x –

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

5 5 from 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

Illustration 7 : Find the value of

–1 –2tan cot

3.

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

–2 tan – 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


Illustration 8 : Find the value of –1 –1 1sin(2cos sin ) when

5x x x .

Solution :

–1 –1

–1 –1

–1 –1

–1

–1

–1

Let sin[2cos sin ]

sin cos , 12

sin 2cos – cos2

sin cos2

1 cos(cos )

5

1 cos 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

1 from 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 ) , 1–1

F

i x x x x ii x x x R xx

xiii ec x ec x x

x


Illustration 9 : Find the value of –1 3sin tan

4.

Solutions :

Illustration 10 : Find the value of –11 5tan cos

2 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

3 sin

5

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

5

0, sin (sin )2

equation (ii) can be written as :

3 sin

5

ii

–1 –1 –1

–1

3 3 3 tan tan sin

4 4 5

3 from 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)3 tan2 1 cos 45 3 5

13

3 – 5 tan .........( )

2 2

0, 0,2 2 4

tan 02

iii


Illustration 11 : Find the value of –1 –1 1cos(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 ........( )

5

sin(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 24 sin cos 1–

5 25 5

24 from equation (i), we get y –

5

1 1: cos cos and 0,

5 5 2

24 sin

5

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

5

x

Aliter Let

ii

–1

–1 –1

–1 –1

0, sin (sin )2

equation (ii) can be written as

24 1 sin cos

5 5

1 24 cos 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 – 5 from equation (iii), we get tan

2 2

3 – 5 from equation (ii), we get

2y


–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 x –1 –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 xy –1 –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

Illustration 12 : Show that –1 –1 –13 15 84sin sin – sin

5 17 85.

Solution :

1.06 Identities of Addition and Subtraction

2 2

–1 –1 –1

–1

–1

3 15 3 15 8226 0, 0 and 1

5 17 5 17 7225

3 15 3 225 15 9 sin sin – sin 1– 1–

5 17 5 289 17 25

3 8 15 4 – sin . .

5 17 17 5

84 – sin

85


Illustration 13 : Evaluate –1 –1 –112 4 63cos sin – tan

13 5 16.

Solution :

Illustration 14 : Evaluate –1 –1 5tan 9 tan

4.

Solution :

–1 –1 –1

–1 –1

–1 –1 –1

–1 –1 –1

12 4 63Let cos sin – tan

13 5 16

4 4 sin – cos

5 2 5

12 4 63 cos – 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 – cos

5 13 5 13 25 169 65

equation (i) can be written as

63 63 – cos – tan

2 65 16

63 63 sin – tan

65 16

z

z

–1 –1

–1 –1

.........( )

63 63 sin tan

65 16

from equation (ii), we get

63 63 tan – tan 0 Ans.

65 65

ii

z z

–1 –1 –1

–1

–1 –1

5 5 9 0, 0 and 9 1

4 4

59

5 4 tan 9 tan tan54

1 – 9.4

tan (–1) –4

5 3 tan 9 tan .

4 4


Illustration 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

2–1

2

os if –1 0

2 tan if 12

(iii) tan = 2 tan if 1 1 –

–( 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


–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

–1 cos (cos3 ) 3


3 if 3y

–1

–1

3 if 03

1 3cos if 1

2

( ) if 3 2 cos (cos3 ) 2 – 3


2 – 3

y

y x x

ii

y

–1

–1

if 3 2

2 2 – 3 if

3 3

1 1 2 – 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

2 –2 3 if

3

1 –2 3cos if –1 –

2

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

13cos ;

2

cos (4 – 3 )

y

y x x

x

y x x –1

–1

1

1 12 – 3cos ; –

2 2

1–2 3cos ; –1 –

2

x

x x

x x


–1 3

–1

2 –1/2

2

Graph :

For cos (4 – 3 )

domain :[–1,1]

range :[0, ]

1( ) if 1, 3cos

2

–3 –3(1– ) ..........( )

1–

1 0 if ,1

2

1 decreasing 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 1 0 if ,1 concavity downwards if ,1

2 2

1 1( ) if – , 2 – 3cos .

2 2

3 1 1 0 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

1 concavity downwards if – ,0

2

1( ) if 0, then 0

2

1 concavity 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 yiii x

dx dx

y x x


–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


Inverse Trigonometric Functions

Some Useful Graphs






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 –1 3sin 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

111

1

General Definitions

EXERCISE-3

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

4 2

3

4

2–1 2 2 – –11

cos 1– . 1– cos – cos holds for2 4 2

x xx x x


(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 inequality –1 2 –1[cot ] – 6[cot ] 9 0x x

, where [.] denotes

greatest integer function, is

(a) – ,cot 3

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

(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

ab

–1tan –1–

a b

ab

–1tan1–

a b

ab

–1– tan1–

a b

ab

–1 –1 22sin sin 2 1–x 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


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

2

x (d) 2

x

15. –11 3sin 2

If sin , then tan is equal to2 5 4cos 2 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 of –1 1– sin 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 –1sin cos (1– ) sin (– ), is/arex x x

.

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

19. The number of solutions of the equation –1 –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 1

If 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,

( )9–1

( )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 of –11 1

cos cos2 8 is equal to


(a) 3/4 (b) –3/4 (c) 1/16 (d) ¼

Part : (B)

24. , and are three angles given by –1 –1 –11 1

2tan ( 2 –1), 3sin sin – and22

–1 1

cos . 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 )

2x

(d)

–1 5 –1tan(cos )

2x

26. For the equation

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

(a)

2 2a 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) a – b = 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


(a)

2f

3 3 (b)

–12 2f 2cos –

3 3 3

(c)

1f

3 3 (d)

–11 1f 2cos –

3 3 3m

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

a b c

sin A sin B sinC

Illustration 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 2yr

y.

Q4: In a ABC prove that, 2

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

Q5: In a ABC prove that 2

1 2 3 –rr rr rr ab bc ca s .

1.01

111

1

SINE RULE

EXERCISE-3

Education

Inverse circular function