Mathematics. Inverse Trigonometric Functions Session.

Mathematics

Inverse Trigonometric Functions

Session

Session Objectives

Session Objectives

2. Properties of Inverse Trigonmetric Function

1. Basic Concepts of inverse trigonometric functions

• Definition• Domain and Range

3. Conversion of one form of Inverse Trig. Fn. to other Forms

4. Identities containing inverse trigonometric functions

Basic Concepts - Definition

Inverse of a Function :

Function must be bijective

Trigonometric functions are periodic.

sin x is periodic with period equal to 2π

Not Bijective

Hence , inverse of sin x should not be valid ????


However, trigonometric functions are bijective for particular value sets in the domain.

sin-1 x is valid in these value sets

Inverse trigonometric function - principal value set

Smallest Numerical Angle

sin x is bijective in [-π/2 , π/2 ] and in [π/2 ,3π/2 ] ……. for x R

[-π/2 , π/2 ]


Inverse trigonometric functions inverse circular function.

arc sin x sin-1x ( principal value )

Inverse Trigonometric function - sin –1 x

Domain of sin –1x

( value which x can take ) is [-1,1]

Range of sin –1 x

(values which sin–1x can take ) is [- /2, /2 ]

-1 1/2

-/2

3/2

X

Y

Inverse Trigonometric function - cos–1 x

Domain of cos –1x

( value which x can take ) is [-1,1]

Range of cos –1 x

(values which cos–1x can take ) is [0, ]

X-1 1

/2

-

Y

Inverse Trigonometric function - tan–1 x

Domain of tan –1x

( value which x can take ) is (- , )

Range of tan –1 x

(values which sec–1x can take ) is (-/2, /2)

X

/2

-/2

-

Inverse Trigonometric function - sec–1 x

Domain of sec –1x ( value which x can take ) is (- -1] U [1, )

Range of sec –1 x

(values which sec–1x can take ) is [0, ] excl. x = /2

Y

X

/2

-/2

-

1-1

Inverse Trigonometric function - cot–1 x

Domain of cot –1x

( value which x can take ) is (- , )

Range of cot –1 x

(values which cot–1x can take ) is (0, )

Y

X

/2

-/2

-

Inverse Trigonometric function - Domain and Range

Function Domain Range

sin-1x [-1,1] [-/2, /2]

cos-1x [-1,1] [0, ]

tan-1x (- , ) (-/2, /2)

sec-1x (-, -1] U [1, ) [0, ] excl. /2

cosec-1x (-, -1] U [1, ) [-/2, /2] excl. 0

cot-1x (- , ) (0, )

Function Domain Range

Inverse Trigonometric function – Properties

Always remember to keep the constraint of domain and range , while solving inverse trigonometric functions.

sin( sin-1 x) = x and cos (cos-1 x) = x if x is in [-1,1]

tan( tan-1 x) = x x if x is in ( -, )

sin-1 ( sin x) = x if x is in [-/2, /2]

cos-1 ( cos x) = x if x is in [0, ]

sec-1 ( sec x) = x if x is in [0, ] excl. x = /2


cos-1 (-x) = - cos-1x if x is in [-1,1]

sin-1 (-x) = -sin-1 (x) if x is in [-1,1]

tan-1(-x) = - tan-1x if x is in ( -, )

cot-1(-x) = - cot-1x if x is in (-,)

cosec-1(-x) = - cosec-1x if x is in (-,-1] U [1,)

sec-1(-x) = - sec-1x if x is in (-,-1] U [1,)


Always remember to keep the constraint of domain and range , while solving inverse trigonometric functions.

sin-1 (-x) = -sin-1 (x) if x is in [-1,1]

Let y = sin-1(-x) ; constraint : y is in [-/2, /2]

sin y = - x x = - sin y = sin ( -y )

sin-1(-x) = sin-1 ( sin (-y))

sin-1(-x) = -y

sin-1(-x) = -sin-1 x

Class Exercise - 1

Find the principal value of

1 2sin sin

3

Solution : 1 2

Let sin sin3

2sin sin where ,

3 2 2

3sin

2

3

Class Exercise - 2

Find the principal value of sin –1 ( sin 5 )

Let y = sin-1(sin 5).Hence y is in [-/2,/2]

5 Wrong

Now , sin 5 = sin [(5/). ] = sin ( 1.59)

= - sin (2 - 1.59)

= sin ( 1.59 -2)

in [-/2, /2]

sin 5 = sin ( 5 - 2 )

sin-1(sin 5) = sin-1 ( sin ( 5 - 2 ))

= 5 - 2

Solution :

Other important properties

1 1 1 x ytan x tan y tan

1 xy

If x > 0 , y > 0 and xy < 1


1 xy

If x > 0 , y > 0 and xy > 1


1 xy

If x<0,y<0 and xy < 1

sin-1 x+ cos-1 x = /2 ;

if x is in [-1,1]

Class Exercise - 5

Find the value of

1 11 1tan tan

2 3

Solution :

1 11 1Let tan and tan

2 3

1 1

tan and tan and , ,2 3 2 2

1 12 3Now tan1 1

1 .2 3

tan 1

Class Exercise - 5

Find the value of

1 11 1tan tan

2 3

tan 1 Solution :

4

5why

4

Class Exercise - 9

In triangle ABC if A = tan-12 and B = tan-1 3 , prove that C = 450

Solution : For triangle ABC , A+B+C =

tan A B tan C

tanA tanBtanC

1 tanA.tanB

tanA 2; tanB 3

3 2tanC

1 3.2

1 tanC tanC 1

Inverse Trigonometric function – Conversion

To convert one inverse function to other inverse function :

1. Assume given inverse function as some angle ( say )

2. Draw a right angled triangle satisfying the angle. Find the third un known side

3. Find the trigonometric function from the triangle in step 2. Take its inverse and we will get = desired inverse function

Conversion - Illustrative Problem

The value of cot-1 3 + cosec-1 5 is

(a) /3 (b) /2 ( c) /4 (d) none

Step 1

Assume given inverse function as some angle ( say )

Let cot-1 3 + cosec-1 5 = x + y,

Where x = cot-13 ; cot x = 3 and

y = cosec-1 5 ; cosec y = 5



(a) /3 (b) /2 ( c) /4 (d) none

Step 2

Draw a right angled triangle satisfying the angle. Find the third unknown side

x1

3

10

y 5 1

2

cot x = 3 , tan x = 1/3

cosec y = 5 , tan y = 1/2



(a) /3 (b) /2 ( c) /4 (d) none

Step 3

Find the trigonometric function from the triangle in step 2. Take its inverse and we will get = desired inverse function

ytanxtan1ytanxtan)yxtan(

21.

31121

31

)yxtan(

tan ( x+ y) = 1

x + y = /4

tan x = 1/3 ,tan y = 1/2


Prove that

sin cot-1 tan cos-1 x = x

Step 1

Assume given inverse function as some angle ( say )

Let y = sin cot-1 tan cos-1 x

And cos-1 x = , cos = x

Hence , y = sin cot-1 tan

Solution :


Prove that sin cot-1 tan cos-1 x = x

Step 2


1

x

2x1 cos = x, tan =xx1 2

Hence , y = sin cot-1

xx1 2

y = sin cot-1 tan


Prove that sin cot-1 tan cos-1 x = x

Step 2


1 x

2x1 and y = sin

From the adjoining triangle , sin = x

Let cot-1 = , cot =xx1 2

xx1 2

Hence y = x = R.H.S.

y = sin cot-1

xx1 2

Class Exercise - 3

Find the value of

1 1tan 2tan

3

Solution : 1 1Let 2 tan

3

1

tan2 3

2

12.2 tan 32As tan , tan

11 tan 1

2 9

3tan

4

Class Exercise - 4

Find the value of

11 5tan cos

2 3

Solution : 11 5 5

cos cos22 3 3

2

511 cos2 3tan

1 cos2 51

3

2 3 5tan

3 5

223 5

tan4

3 5tan

2

Class Exercise - 6

Prove that

1 1 1 2tan x cot x 1 tan x x 1

Class Exercise - 6

Prove that


Solution :

1 1Let tan x and cot x 1

1tan x and cot x 1 tan

x 1

And L.H.S. of the given identity is +

1

x tan tanx 1tan as tan1 1 tan .tan1 x.

x 1

Class Exercise - 6

Prove that


Solution : given identity is +

1

xx 1and tan

11 x.

x 1

2x x 1

tanx 1 1

2tan x x 1

1 2tan x x 1

Class Exercise - 7

Solve the equation

1 15 12sin sin

x x 2

Solution : 1 15 12

Let sin and sinx x

5 12sin and sin

x x

given equation is2

cos 0 cos .cos sin .sin

Class Exercise - 7

Solve the equation 1 15 12

sin sinx x 2

Solution :

cos 0 cos .cos sin .sin

2

2

5 25 xAs sin cos

x x

12 144 xand sin cos

x x

2 225 x 144 x 5 12. .

x x x x

Class Exercise - 7

Solve the equation 1 15 12

sin sinx x 2

Solution : 2 225 x 144 x 5 12

. .x x x x

2 225 x . 144 x 60 22 225 x . 144 x 60

4 2x 169x 0 x 0, 13 , x 0

Now If x 13 , , 0, hence x 13

Class Exercise - 8

If sin-1 x + sin-1 (1- x) = cos-1x,

the value of x could be

(a) 1, 0 (b) 1,1/2 (c) 0,1/2 (d) 1, -1/2

Class Exercise - 8If sin-1 x + sin-1 (1- x) = cos-1x,


(a) 1, 0 (b) 1,1/2 (c) 0,1/2 (d) 1, -1/2

Solution :

1 1Let sin x and sin 1 x

sin x and sin 1 x

and given equation is + = cos-1x cos (+) = x

2

2 2

As sin x cos 1 x

and sin 1 x cos 1 1 x 2x x



(a) 1, 0 (b) 1,1/2 (c) 0,1/2 (d) 1, -1/2

Solution : cos (+) = x

2 2sin x , cos 1 x , sin 1 x , cos 2x x

2 21 x . 2x x x. 1 x x

2 2 21 x . 2x x 2x x

22 2 21 x . 2x x 2x x

2 2 22x x 1 x 2x x 0



(a) 1, 0 (b) 1,1/2 (c) 0,1/2 (d) 1, -1/2

Solution : 2 2 22x x 1 x 2x x 0

x. 2 x 1 2x 0

1x 0, ,2

2

x 2 as x sin 1

x 0,2

Class Exercise - 10

If cos-1 x + cos-1 y + cos-1z = ,

Then prove that x2+y2+z2 = 1 - 2xyz

1 1 1Let cos x A , cos y B and cos z C

Solution :

cos A x, cosB y and cosC z and given : A+B+C =

Now, L.H.S. = cos2A + cos2B +cos2C

= cos2A + 1- sin2B +cos2C

= 1+(cos2A - sin2B) +cos2C

Class Exercise - 10



Solution : Given : A+B+C =

L.H.S. = 1+(cos2A - sin2B) +cos2C

= 1+ cos(A+B).cos(A–B) +cos2C

= 1+ cos( – C).cos(A–B) +cos2C

= 1+ cosC [– cos(A–B) +cosC ]

= 1+ cosC [– cos(A–B) – cos(A+B)]

Class Exercise - 10



Solution : Given : A+B+C =

L.H.S. = 1+ cosC [– cos(A–B) – cos(A+B)]

= 1– cosC [ cos(A–B) + cos(A+B)]

= 1– cosC [2cos A. cos B]

= 1– 2xyz = R.H.S.

Thank you

Mathematics. Inverse Trigonometric Functions Session.

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