Chapter 4 Inverse trigonometic Functions

NOTE:

Trigonometric Table

(i) Let

⇒ [ We know which value of x when placed in sin gives us this answer ]

∴

(ii) Let

⇒ [We know which value of x when put in this expression will give us this result]

⇒

(iii) Let

⇒ [We know which value of x when put in this expression will give us this result]

∴

(iv) Let

⇒ [We know which value of x when put in this expression will give us this result]

∴

(v) Let

⇒ [We know which value of x when put in this expression will give us this result]

∴

(vi) Let

⇒ [We know which value of x when put in this expression will give us this result]

∴

(vii) Let

⇒

[We know which value of x when put in this expression will give us this result]

∴

(i) Let

⇒ [Formula: sin^-1(-x) = -sin^-1 x ]

⇒ [We know which value of x when put in this expression will give us this result]

∴

(ii) [ Formula: cos^-1(-x) = π – cos^-1 x]

Let

⇒ [We know which value of x when put in this expression will give us this result]

∴

Putting this value back in the equation

(iii) Let

⇒ [Formula: tan^-1(-x) = - tan^-1 (x)]

⇒ [We know which value of x when put in this expression will give us this result]

∴

(iv) …(i) [ Formula:sec^-1(-x) = π– sec^-1 (x) ]

Let

⇒ [We know which value of x when put in this expression will give us this result]

∴

Putting the value in (i)

(v) Let

⇒ [ Formula: cosec^-1(-x) = -cosec^-1 (x) ]

⇒

∴

(vi) … (i)

Let

⇒ [We know which value of x when put in this expression will give us this result]

⇒

Putting in (i)

=

[ Refer to question 2(ii) ]

= cos { π }

=

= -1

=

=

=

=

[Formula: sin^-1(-x) = sin^-1(x) ]

=

[ Formula: cos^-1(-x) = -cos^-1(x) ]

=

=

[ Formula: tan^-1(-x)= -tan^-1 (x) ]

[ We know that , thus ]

=

[ Formula: sec^-1(-x)= π – sec^-1(x) ]

=

=

[Formula: cosec^-1(-x) = -cosec^-1(x) ]

=

This can also be solved as

Since cosec is negative in the third quadrant, the angle we are looking for will be in the third quadrant.

=

=

[Formula: cot^-1(-x) = π – cot^-1(x) ]

=

=

[Formula: tan^-1(-x)= -tan^-1 (x) ]

=

[ Formula: sec^-1(-x)= π – sec^-1(x) ]

=

=

Putting the value directly

[ Formula: sin(π – x) = sin x )

=

[ Formula: sin^-1( sin x) = x ]

=

[Formula: tan(π – x) = -tan (x) , as tan is negative in the second quadrant. ]

=

[Formula: tan^-1(tan x) = x ]

=

[Formula: cos(2π – x) = cos (x), as cos has a positive vaule in the fourth quadrant. ]

= [Formula: cos^-1(cos x) = x

=

[ Formula: cos (2π + x) = cos x , cos is positive in the first quadrant. ]

= [Formula: cos^-1(cos x) = x]

=

[ Formula: tan( π + x) = tan x, as tan is positive in the third quadrant.]

= [Formula: tan^-1(tan x) = x ]

=

Putting the value of and using the formula

cot^-1(-x)= π-cot^-1x

=

Putting the value of

=

=

=

=

[Formula: sin^-1(-x) = -sin^-1x ]

=

=

Putting value of

=

=

=

= 1

[Formula: ]

Putting value of

= 0

[Formula: ]

Putting the value of

= 1

[Formula: ]

Putting the value of

=1

Putting the values of the inverse trigonometric terms

=

=

[Formula: cos^-1(-x)=π – cos(x) and sin^-1(-x)= -sin(x) ]

Putting the values for each of the inverse trigonometric terms

=

=

=

=

=

[Formula: sin(π – x) = sin x, as sin is positive in the second quadrant.]

= [Formula: sin^-1(sinx)=x ]

=

To Prove:

Formula Used:

Proof:

LHS … (1)

Let x = tan A … (2)

Substituting (2) in (1),

LHS

From (2), A = tan^-1 x,

= RHS

Therefore, LHS = RHS

Hence proved.

To Prove: tan^-1 x + cot^-1 (x + 1) = tan^-1 (x² + x + 1)

Formula Used:

1)

2)

Proof:

LHS = tan^-1 x + cot^-1 (x + 1) … (1)

= tan^-1 (x² + x + 1)

= RHS

Therefore, LHS = RHS

Hence proved.

To Prove:

Formula Used: sin 2A = 2 × sin A × cos A

Proof:

LHS … (1)

Let x = sin A … (2)

Substituting (2) in (1),

LHS

= sin^-1 (2 × sin A × cos A)

= sin^-1 (sin 2A)

= 2A

From (2), A = sin^-1 x,

2A = 2 sin^-1 x

= RHS

Therefore, LHS = RHS

Hence proved.

To Prove: sin^-1 (3x – 4x³) = 3 sin^-1 x

Formula Used: sin 3A = 3 sin A – 4 sin³ A

Proof:

LHS = sin^-1 (3x – 4x³) … (1)

Let x = sin A … (2)

Substituting (2) in (1),

LHS = sin^-1 (3 sin A – 4 sin³ A)

= sin^-1 (sin 3A)

= 3A

From (2), A = sin^-1 x,

3A = 3 sin^-1 x

= RHS

Therefore, LHS = RHS

Hence proved.

To Prove: cos^-1 (4x³ – 3x) = 3 cos^-1 x

Formula Used: cos 3A = 4 cos³ A – 3 cos A

Proof:

LHS = cos^-1 (4x³ – 3x) … (1)

Let x = cos A … (2)

Substituting (2) in (1),

LHS = cos^-1 (4 cos³ A – 3 cos A)

= cos^-1 (cos 3A)

= 3A

From (2), A = cos^-1 x,

3A = 3 cos^-1 x

= RHS

Therefore, LHS = RHS

Hence proved.

To Prove:

Formula Used:

Proof:

LHS … (1)

Let x = tan A … (2)

Substituting (2) in (1),

LHS

= tan^-1 (tan 3A)

= 3A

From (2), A = tan^-1 x,

3A = 3 tan^-1 x

= RHS

Therefore, LHS = RHS

Hence proved.

To Prove:

Formula Used:

Proof:

LHS … (1)

= RHS

Therefore, LHS = RHS

Hence proved.

To Prove: cos^-1 (1 – 2x²) = 2 sin^-1 x

Formula Used: cos 2A = 1 – 2 sin² A

Proof:

LHS = cos^-1 (1 – 2x²) … (1)

Let x = sin A … (2)

Substituting (2) in (1),

LHS = cos^-1 (1 – 2 sin² A)

= cos^-1 (cos 2A)

= 2A

From (2), A = sin^-1 x,

2A = 2 sin^-1 x

= RHS

Therefore, LHS = RHS

Hence proved.

To Prove: cos^-1 (2x² - 1) = 2 cos^-1 x

Formula Used: cos 2A = 2 cos² A – 1

Proof:

LHS = cos^-1 (2x² - 1) … (1)

Let x = cos A … (2)

Substituting (2) in (1),

LHS = cos^-1 (2 cos² A – 1)

= cos^-1 (cos 2A)

= 2A

From (2), A = cos^-1 x,

2A = 2 cos^-1 x

= RHS

Therefore, LHS = RHS

Hence proved.

To Prove:

Formula Used:

1) cos 2A = 2 cos² A – 1

2)

Proof:

LHS

= cos^-1 (2x² – 1)… (1)

Let x = cos A … (2)

Substituting (2) in (1),

LHS = cos^-1 (2 cos² A – 1)

= cos^-1 (cos 2A)

= 2A

From (2), A = cos^-1 x,

2A = 2 cos^-1 x

= RHS

Therefore, LHS = RHS

Hence proved.

To Prove:

Formula Used:

1)

2) cosec² A = 1 + cot² A

3)

4)

Proof:

LHS

Let x = cot A

LHS

= cot^-1(cosec A – cot A)

From (2), A = cot^-1 x,

= RHS

Therefore, LHS = RHS

Hence proved.

To Prove:

We know that,

Also,

Taking A = √x and B = √y

We get,

Hence, Proved.

_{We know that,}

Now, taking A = x and B = √x

We get,

As, x.x^1/2 = x^3/2

Hence, Proved.

To Prove:

Formula Used:

1)

2)

Proof:

LHS

= RHS

Therefore LHS = RHS

Hence proved.

To Prove:

Formula Used:

Proof:

LHS

= RHS

Therefore LHS = RHS

Hence proved.

To Prove:

Formula Used:

Proof:

LHS

= RHS

Therefore LHS = RHS

Hence proved.

To Prove:

Formula Used:

Proof:

LHS

= RHS

Therefore LHS = RHS

Hence proved.

To Prove:

Formula Used:

Proof:

LHS

= tan^-1 1

= RHS

Therefore LHS = RHS

Hence proved.

To Prove:

Formula Used:

Proof:

LHS = tan^-1 2 – tan^-1 1

= RHS

Therefore LHS = RHS

Hence proved.

To Prove:

Formula Used:

Proof:

LHS

{since 2 × 3 = 6 > 1}

= π

= RHS

Therefore LHS = RHS

Hence proved.

To Prove:

Formula Used:

Proof:

LHS

= tan^-1 1

= RHS

Therefore LHS = RHS

Hence proved.

To Prove: ⇒

Formula Used:

Proof:

LHS

= RHS

Therefore LHS = RHS

Hence proved.

To Prove:

Formula Used:

Proof:

LHS

= RHS

Therefore, LHS = RHS

Hence proved.

To Prove:

Formula Used:

Proof:

LHS

= sin^-1 1

= RHS

Therefore, LHS = RHS

Hence proved.

To Prove:

Formula Used:

Proof:

LHS … (1)

Let

Therefore … (2)

From the figure,

… (3)

From (2) and (3),

Substituting in (1), we get

LHS

= RHS

Therefore, LHS = RHS

Hence proved.

To Prove:

Formula Used:

Proof:

LHS … (1)

Let

Therefore … (2)

From the figure,

… (3)

From (2) and (3),

Substituting in (1), we get

LHS

To Prove:

Formula Used:

Proof:

LHS … (1)

Let

Therefore … (2)

From the figure,

… (3)

From (2) and (3),

Substituting in (1), we get

LHS

= tan^-1 1

= RHS

Therefore, LHS = RHS

Hence proved.

To Prove:

Formula Used:

Proof:

LHS … (1)

Let

Therefore … (2)

From the figure,

… (3)

From (2) and (3),

… (3)

Now, let

Therefore … (4)

From the figure,

… (5)

From (4) and (5),

… (6)

Substituting (3) and (6) in (1), we get

LHS

= RHS

Therefore, LHS = RHS

Hence proved.

To Prove:

Formula Used:

1)

2)

Proof:

LHS … (1)

… (2)

Substituting (2) in (1), we get

LHS … (3)

Let

Therefore … (4)

From the figure,

… (5)

From (4) and (5),

… (6)

Substituting (6) in (3), we get

LHS

= tan^-1 1

= RHS

Therefore, LHS = RHS

Hence proved.

To find: value of x

Formula Used:

Given:

LHS

Therefore,

Taking tangent on both sides, we get

⇒ 62x = 16 – 8x²

⇒ 8x² + 62x – 16 = 0

⇒ 4x² + 31x – 8 = 0

⇒ 4x² + 32x – x – 8 = 0

⇒ 4x × (x + 8) – 1 × (x + 8) = 0

⇒ (4x – 1) × (x + 8) = 0

Therefore, are the required values of x.

To find: value of x

Given:

LHS = cos(sin^-1 x) … (1)

Let sin θ = x

Therefore θ = sin^-1 x … (2)

From the figure,

… (3)

From (2) and (3),

… (4)

Substituting (4) in (1), we get

LHS

Therefore,

Squaring and simplifying,

⇒ 81 – 81x² = 1

⇒ 81x² = 80

Therefore, are the required values of x.

To find: value of x

Formula Used:

Given:

LHS = cos(2sin^-1 x)

Let θ = sin^-1 x

So, x = sin θ … (1)

LHS = cos(2θ)

= 1 – 2sin² θ

Substituting in the given equation,

Substituting in (1),

Therefore, are the required values of x.

To find: value of x

Given:

We know

Let

Therefore,

Therefore,

Squaring both sides,

⇒ x² – 64 = 225

⇒ x² = 289

⇒ x = ± 17

Therefore, x = ±17 are the required values of x.

To find: value of x

Given:

LHS

Therefore,

Squaring both sides,

Therefore, are the required values of x.

To find: value of x

Given:

We know that

Therefore,

Substituting in the given equation,

⇒ x = 1

Therefore, x = 1 is the required value of x.

Given:

We know that

So,

Substituting in the given equation,

Rearranging,

Therefore, is the required value of x.

Principal value branch of sin^-1 x is

Principal value branch of cos^-1 x is [0, π]

Principal value branch of tan^-1 x is

Principal value branch of cot^-1 x is (0, π)

Principal value branch of sec^-1 x is

Principal value branch of cosec^-1 x is

To Find:The Principle value of

Let the principle value be given by x

Now, let x =

cos x=

cos x=cos() ()

x =

To Find: The Principle value of

Let the principle value be given by x

Now, let x =

cosec x =2

cosec x=cosec() ()

x =

To Find: The Principle value of

Let the principle value be given by x

Now, let x =

cos x =

cos x= - cos() ()=)

cos x=cos() ())

x =

To Find: The Principle value of

Let the principle value be given by x

Now, let x =

sin x =

sin x= - sin() ()=)

sin x=sin() ())

x =

To Find: The Principle value of

Let the principle value be given by x

Now, let x =

cos x =

cos x= - cos() ()=)

cos x=cos() ())

x =

To Find: The Principle value of

Let the principle value be given by x

Now, let x =

tan x =

tan x= - tan() ()=)

())

x =

To Find: The Principle value of

Let the principle value be given by x

Now, let x =

cot x =-1

cot x= - cot() ()=)

cot x=cot() ())

x =

To Find: The Principle value of

Let the principle value be given by x

Now, let x =

sec x =

sec x= - sec() ()=)

sec x=sec() ())

x =

To Find: The Principle value of

Let the principle value be given by x

Now, let x =

cosec x =

cosec x= - cosec() ()=)

cosec x=cosec() ())

x =

To Find: The Principle value of

Let the principle value be given by x

Now, let x =

cot x =

cot x= - cot() ()=)

cot x=cot() ())

x =

To Find: The value of

Now, let x =

sin x =sin ()

Here range of principle value of sine is [-]

x = [-]

Hence for all values of x in range [-] ,the value of

is

sin x =sin () (sin ()= sin () )

sin x =sin () (sin ()= sin as here )

x =

To Find: The value of

Now, let x =

cos x =cos ()

Here ,range of principle value of cos is [0,]

x = [0,]

Hence for all values of x in range [0,] ,the value of

is

cos x =cos (2) (cos ()= cos () )

cos x =cos () (cos ()= cos )

x =

To Find: The value of

Now, let x =

tan x =tan ()

Here range of principle value of tan is []

x = []

Hence for all values of x in range [] ,the value of

is

tan x =tan () (tan ()= tan () )

tan x =tan () (tan ()= tan )

x =

To Find: The value of

Now, let x =

cot x =cot ()

Here range of principle value of cot is []

x = []

Hence for all values of x in range [] ,the value of

is

cot x =cot () (cot ()= cot () )

cot x =cot () (cot ()= cot )

x =

To Find: The value of

Now, let x =

sec x =sec ()

Here range of principle value of sec is [0,]

x = [0,]

Hence for all values of x in range [0,] ,the value of

is

sec x =sec (2) (sec ()= sec () )

sec x =sec () (sec ()= sec )

x =

To Find: The value of

Now, let x =

cosec x =cosec ()

Here range of principle value of cosec is [-]

x = [-]

Hence for all values of x in range [-] ,the value of

is

cosec x =cosec () (cosec ()= cosec () )

cosec x =cosec () (cosec ()= cosec())

x = -

To Find: The value of

Now, let x =

tan x =tan ()

Here range of principle value of tan is []

x = []

Hence for all values of x in range [] ,the value of

is

tan x =tan () (tan ()= tan () )

tan x =tan () (tan ()= tan())

x =

To Find: The value of

Now, let x =

x = ()

x = ( = )

x =

x = =

To Find: The value ofsin()

Now, let x = sin()

x = sin () ()

x = 1 (

Given: x 0

To Find: The value ofcos()

Now, let x = cos()

x = cos () ()

x = 0 (

To Find: The value of sin()

Now, let x =

cos x =

Now ,sin x =

=

=

x = =

Therefore,

sin() = sin()

Let , Y= sin()

=

Y =

To Find: The value of

Here,consider ()

=

Now,consider

Since here the principle value of sine lies in range [] and since []

=

=

=

Therefore,

= +

=

=

To Find: The value of

Let , x =

x = – [-] ()

x = – [ -]

x = – []

x = -

To Find: The value of

Now, let x =

x= +2() (cos ()=and sin ()=)

x= +

x=

To Find: The value of

Now, let x =

x = + [-] + [-] ()

x = + [-] + [- ]

x = + -

x =

To Find: The value of tan(2 - )

Consider , tan(2 - ) =tan( - )

()

= tan( - )

= tan( - )

= tan( - )) (tan()=1)

= tan()

( - =

= tan()

tan(2 - ) =

To Find: The value of tan ( )

Let , x =

cos x =

Now, tan ( ) becomes

tan ( )= tan (x) =tan

=

=

=

=

tan ( ) =

To Find: The value of sin()

Let, x =

cos x =

Now , sin() becomes sin (x)

Since we know that sin x =

=

sin() = Sin x =

To Find: The value of cos()

Let x =

tan x =

tan x = =

We know that by pythagorus theorem ,

(Hypotenuse )² = (opposite side )² + (adjacent side )²

Therefore, Hypotenuse = 5

cos x = =

Since here x = hence cos() becomes cos x

Hence , cos() = cos x =

To Find: The value of of sin}

Let, x = sin}

x = sin} ()

x = sin)

x = sin) = sin) = 1

To Find: The value of sin()

Let x =

cos x =

Therefore sin() becomes sin(),i.e sin ()

We know that sin () =

=

=

sin () =

To Find: The value of

Let , x =

x = (sin ()=)

x =

x = = = (cos ()=)

Given: = x

To Find: The value of sin x

Since , x =

cot x = =

By pythagorus theroem ,

(Hypotenuse )² = (opposite side )² + (adjacent side )²

Therefore, Hypotenuse =

sin x = =

To Find: The value of +

Let , x = +

x = - + 2 [] ()

x = - () + 2 []

x = - () + 2 []

x = - +

x =

Tag:

To Find: The value of +

Let , x = +

x = - + ()

()

x = - + ()

x = - +

x =

To Find: The value of cot ()

Let , x = cot ()

x = cot () ()

x = 0

To Find: The value of +

Let , x = +

Since we know that + =

+ = =

To Find: The value of +

Let , x = +

Since we know that + =

+ = = =

To Find: The value of 2 i.e, +

Let , x = +

Since we know that + =

+ = =

To Find: The value of cos (2)

Let , x = cos (2)

x = cos (+ )

Since we know that + =

+ = =

x = cos ()

Now , let y =

tan y =

By pythagorus theroem ,

(Hypotenuse )² = (opposite side )² + (adjacent side )²

Therefore, Hypotenuse = 5

cos ()=cos y =

To Find: The value of sin (2)

Let , x = sin(2)

We know that =

x = sin( = sin( =

To Find: The value of sin (2)

Let , x =

sin x =

We know that ,cos x =

=

=

Now since, x = ,hence sin (2) becomes sin(2x)

Here, sin(2x)= 2 sin x cos x

=2

=

To Find: The value of = -

Now , = - ()

Since we know that - =

+ = =

=

x =

To Find: The value of + =

Since we know that + =

+ =

=

=

Here since + =

=

= ()

=

=

x = 0

Given:+ =

To Find: The value of +

Since we know that+ =

= -

Similarly = -

Now consider + = - + -

= – []

= -

=

To Find: The value of +

Since we know that + =

+ =

=

=

Since the principle value of tan lies in the range [0,]

=

Given: + =

To Find: The value of x

Here + = can be written as

= -

Since we know that - =

= - =

=

=

x =

Given: + =

To Find: The value of x

Since we know that + =

+ =

=

Now since + =

+ = ()

=

= 1

6 + 5x -1 =0

x = or x= -1

To Find: The value of tan {}

Let x =

cos x = =

By pythagorus theroem ,

(Hypotenuse )² = (opposite side )² + (adjacent side )²

Therefore , opposite side = 3

tan x= =

x =

Now tan {} = tan {}

Since we know that + =

tan {} = tan ()

= tan ()

=

To Find: The value of

Now can be written in terms of tan inverse as

=

Since we know that + =

=

=

= =

To Find: The range of

Here,the inverse function is given by y =

The graph of the function y = can be obtained from the graph of

Y = sin x by interchanging x and y axes.i.e, if (a,b) is a point on Y = sin x then (b,a) is

The point on the function y =

Below is the Graph of range of

From the graph, it is clear that the range of is restricted to the interval

[]

To Find: The range of

Here,the inverse function is given by y =

The graph of the function y = can be obtained from the graph of

Y = cos x by interchanging x and y axes.i.e, if (a,b) is a point on Y = cos x then (b,a) is the point on the function y =

Below is the Graph of the range of

From the graph, it is clear that the range of is restricted to the interval

[]

To Find: The range of tan^-1 x

Here,the inverse function is given by y =

The graph of the function y = can be obtained from the graph of

Y = tan x by interchanging x and y axes.i.e, if (a,b) is a point on Y = tan x then (b,a) is the point on the function y =

Below is the Graph of the range of

From the graph, it is clear that the range of is restricted to any of the intervals like [] , [] , [] and so on. Hence the range is given by

().

To Find:The range of

Here,the inverse function is given by y =

The graph of the function y = can be obtained from the graph of

Y = sec x by interchanging x and y axes.i.e, if (a,b) is a point on Y = sec x then (b,a) is the point on the function y =

Below is the Graph of the range of

From the graph, it is clear that the range of is restricted to interval

[0,] – {}

To Find: The range of

Here,the inverse function is given by y =

The graph of the function y = can be obtained from the graph of

Y = cosec x by interchanging x and y axes.i.e, if (a,b) is a point on Y = cosec x then (b,a) is the point on the function y =

Below is the Graph of the range of

From the graph it is clear that the range of is restricted to interval

[] – {0}

To Find: The Domain of

Here,the inverse function of cos is given by y =

The graph of the function y = can be obtained from the graph of

Y = cos x by interchanging x and y axes.i.e, if (a,b) is a point on Y = cos x then (b,a) is the point on the function y =

Below is the Graph of the domain of

From the graph, it is clear that the domain of is [-1,1]

To Find: The Domain of

Here,the inverse function is given by y =

The graph of the function y = can be obtained from the graph of

Y = sec x by interchanging x and y axes.i.e, if (a,b) is a point on Y = sec x then (b,a) is the point on the function y =

Below is the Graph of the domain of

From the graph, it is clear that the domain of is a set of all real numbers excluding -1 and 1 i.e, R – [-1,1]