a b c d

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INVERSE TRIGONOMETRIC FUNCTIONS

Prerequisites:

1. Knowledge of function, their domain and range.2. One –One and many-one function, onto and into function and their graphs3. Trigonometric functions of sum, difference, multiple and sub-multiple angles

Objectives:

1. To define Inverse Trigonometric Functions2. Define the Principle value of Inverse trigonometric Function3. To find domain and range of Inverse Trigonometric Function4. To state the properties of Inverse Trigonometric Function5. To simplify expressions and solve problems involving Inverse Trigonometric Functions

Introduction:

The study of Inverse Trigonometric Function is the continuation of what we have learnt about trigonometric functions in class XI with the help of concept of inverse of a function.

Let us recall the concept of the function: A Rule f which associates every element of a non-empty set A to a unique (one and only one) element of set B is called a function from A to B.

Ex:

A B A B A B

Fig: -1 Fig:-2 Fig:-3

The association in Fig:-1 is not a function as dA is unassociated. The association in Fig:-2 is also not a function as bA is not uniquely associated. The association in Fig:-3 is a function from A to B. (Why?)

Trigonometric Function: f:RR defined by y=f(x) =Sinx is a function as for every xR there corresponds one and only one value of y (i.e. Sinx). Similarly Cosx , Tanx, Cosecx, Secx, Cotx are also trigonometric functions in its domain of definition.

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Q1: What are domain of definitions of (a) f(x) = tanx (b) f(x)=cosecx

Ans: (a) R−{x : x=(2n+1) π

2, n∈Z }

, (Why ?)

(b) R−{x : x=nπ ,n∈Z } ,(Why ?)

While defining domain and range simply we can say as being real function (where domain and range

both are subsets of R) the set of all those values of x for which f ( x )= y exists (is well defined and is

unique is called domain and set of values of y (i.e. set of images) forms range. Example: - At x= π2 ,

f ( x )=tan x is not well defined (i.e. does not exist). So,

π2∉

domain of tan x .

Now let us recall the idea of inverse of a function. A function f:A B is inevitable only

when it is both one and one and onto then the rule f−1

:B→A in which every y∈B is associated with

a unique x∈ A , for which f ( x )= y , is called inverse of f (such an x∈ A exists and is unique as f is both one-one and onto).

Let us recall the definition of One-One function: the function is One-One if different

elements of domain have different images in range. Example: consider f :R→R defined by

f ( x )=x2 . The function is not One-One as 2≠−2⇒ f (2 )= f (−2 ) .

Onto Function: A function f : A→B is Onto if every element of B (co-domain) is an

image. In above example −4∈R (Co-domain) is not an image. So, it is not an Onto function.

R f R

If we try to visualize the inverse rule we get the following

R R

Here both the condition of being a function fails as 4∈domain is not uniquely associated, whereas -4 is unassociated. Hence for existence of inverse, the function must be one-one and onto. That is, if

f : A→B is one-one and onto, then f1 :B→A exists and is one-one and onto. Then

f ( x )= y⇔f −1( y )=xwhere x∈ A andy∈B .

The inverse Trigonometric Functions are defined in the same way. Let us take a function

y=sin x ,x∈ R , y∈R . The inverse of this function is defined as x=sin−1 y (We may also write

y=sin−1 x , provided the function is both one-ne and onto).

Very important: sin−1 y does not mean

1sin y . This is just a notation given to write Inverse

Trigonometric Function.

Q3: Let f :R→R be a function defined byf ( x )=sin x . Examine whether sin−1 x exists. Consider

graph of y=sin x

We can easily verify

π3≠2π3

⇒ f ( π3 )= f ( 2π3 )etc. Further any y∈R−[−1,1 ] is not an image.

Hence, the function sin−1 x does not exist, as f is not one-one and onto.

Now consider the function f : [−π2 , π

2 ]→ [−1,1 ] defined by y= f ( x )=sin x . It is both one-one and

onto which can be visualized from the above graph. Hence, sin−1 x exists.

In fact, if we restrict the domain of the sin function to any one of the intervals[−3 π2,−π2 ] ,

[−π2 , π2 ] , [

π2, 3π2 ] , [

3 π2, 5π2 ] .. etc., keeping the range [−1,1 ] the function becomes one-one and

onto.

Then y= f ( x )=sin−1 x is a function whose domain is [−1,1 ] and range is any one of the

intervals [−3 π2,−π2 ] , [

−π2, π2 ] , [

π2, 3π2 ] , [

3 π2, 5π2 ] .. etc., [please recall that if f

−1exists, f and

f−1 interchange their domain and range.

Every branch of the graph of y=sin x corresponding to branches[−3 π2,−π2 ] , [

−π2, π2 ] ,

[ π2 , 3π2 ] , [3 π2, 5π2 ] gives a one–one and onto function. The graph in the branch [

−π2, π2 ]called

the principle value branch.

We can also justify our answer from the chapter Trigonometric Equations and General Values.

f ( x )=sin x=√32 has infinitely many solutions,

x= π3 ,

2π3 ,

7π3 ,

8π3 … etc.. The value

x=π3∈[−π2 , π

2 ] is called the principle value.

We may say that sin−1 √3

2=π3 . Also value of

sin−1 x∈[−π2 , π2 ] is called the principle value.

Similarly we restrict the domains of other trigonometric functions to make it invertible the table given below shows the domain and the range of the six basic inverse trigonometric function.

Domain and Range of trigonometric functions in which they are one-one & onto (Principal value branches)

Domain and range of corresponding inverse trigonometric functions

Function Domain Range Function Domain Range

1. y=sin x [−π2 , π2 ] [−1,1 ] 1. y = sin

−1 x [−1,1 ] [−π2 , π2 ]

2. y=cos x [0 , π ] [−1,1 ] 2. y = cos−1 x [−1,1 ] [0 , π ]

3. y= tan x ]−π2, π2

[ R 3. y=tan−1 x R ]−π

2, π2

[

4.y=cosecx [−π

2,0 [∪]0 , π

2] ]−∞ ,−1 ]∪[1 ,∞[ 4. y=cosec

−1 x ]−∞ ,−1 ]∪[1 ,∞[ [−π2,0 [∪]0 , π

2]

5.y=sec x [0 , π ]−{π2 } ]−∞ ,−1 ]∪[1 ,∞[ 5. y=sec−1 x ]−∞ ,−1 ]∪[1 ,∞[ [0 , π ]−{π2 }

6. y=cot x ]0 , π [ R 6. y=cot−1 x R ]0 , π [

The values of domain and range indicated for different trigonometric, inverse trigonometric functions can be observed from the graphs of trigonometric functions and their corresponding inverse trigonometric functions.

We can draw the graph of a inverse function of a function by taking image on line y = x.

Let us apply this to draw graph of y = sin−1 x .

2. The graph of the function y = cos x and y = cos−1 x

3. The graph of the function y = tanx and y = tan−1 x

4. The graph of the function y = cos ecxand y = cos ec−1 x

5. The graph of the function y = sec x and y = sec−1 x

6. The graph of the function y = cot x and y = cot−1 x

Note: Whenever no branch of an inverse function is mentioned, we consider the Principal value branch.

PROPERTIES OF INVERSE CIRCULAR FUNCTIONS

1. (a) sin−1 1

x= cosec−1 x , x≥1 or x≤−1

(b) cos−1 1

x= sec−1 x , x≥1 or x≤−1

(c) tan−1 1

x= cot−1 x , x>0

2. (a) sin−1 (−x )=−sin−1 x , x∈1 [−1,1 ]

(b) tan−1(−x )=−tan−1 x , x∈R

(c) cos ec−1(−x )=−cos ec−1 x , |x|≥1

3. (a) cos−1(−x )= π−cos−1x , x∈1 [−1,1 ]

(b) cot−1 (−x )= π−cot−1 x , x∈R

(c) sec−1(−x )= π−sec−1 x , |x|≥1

4. (a) sin−1 x+cos−1 x=π

2, x∈[−1,1 ]

(b) tan−1 x+cot−1 x=π

2, x∈R

(c) cos ec−1 x+sec−1 x= π

2,|x|≥1

5. (a) tan−1 x+ tan−1 y=tan−1 x+ y

1−xy, xy<1

(b) tan−1 x− tan−1 y= tan−1 x− y

1+xy, xy>−1

Proof: Let tan−1 x=θ and tan−1 y=φ . Then tan θ= x and tanφ= y .

Now tan(θ+φ )= tan θ+tanφ

1− tanθ tanφ= x+ y1−xy

⇒θ+φ=tan−1 x− y

1+xy

⇒tan−1 x+ tan−1 y=tan−1 x+ y

1−xy

Problems to review the concept:

Q. Find the Principal value of:

(a) cos ec−1(−2

√3) (b)

cos−1( √32

)(c) tan

−1(−√3) (d) sec−1(−√2)

(e) cot−1 (−1)

Ans:

(a)cos ec−1(−2

√3) = sin−1 (−√3

2)=θ( say )

then, sin θ=−√3

2 . Then the value of θ∈[−π

2, π2

]such that

sin θ=−√32 is

θ=−π3

Hence, cos ec−1(−2

√3)=

−π3

Ans: (b)

π3 (c)

−π3 (d)

3π4 (e)

3π4

Q. Find the value ofsec ¿¿

Ans: Let cos−1( √3

2)=θ

⇒ cosθ=√3

2⇒ θ=π

6

Hence, sec ¿¿

Q. Prove that cos (sin−1 x )=√1−x2

Ans: Let sin−1 x= y

⇒ sin y=x

Hence, cos (sin−1 x )=cos y=√1−sin2 y=√1−x2

Q. Prove the following:

(a) 2sin−1 x=sin−12x √1−x2

(b) 2cos−1 x=cos−1 (2 x2−1)

Q. Prove the following:

(a) 2 tan−1 x=sin−1 2x

1+x2,|x|≤1

(b) 2 tan−1 x=cos−1 1−x

2

1+x2, x≥0

(c) 2 tan−1 x= tan−1 2 x

1−x2,−1<x<1

Ans:

(a) Let tan−1 x= y

⇒ tan y=x Now sin 2 y= 2 tan y

1+ tan2 y= 2 x1+x2

⇒2 y=sin−1 2 x

1+x2

⇒2 tan−1 x=sin−1 2x

1+x2

Similar proof for (b) and (c) follow.

Q. Prove that

sin−11213

+cos−1 45+ tan−163

16=π

Ans: Let sin−112

13=x ,cos−1 4

5= y ,tan−163

16=z

⇒sin x=12

13,cos y=4

5,tan z=63

16

⇒cos x= 5

13,sin y=3

5,tan x=12

5and tan y=3

4

Then

tan( x+ y )=tan x+ tan y1− tan x tan y

=

125

+ 34

1−125x 34

=−6316

⇒ tan( x+ y )=− tan z

⇒ tan ( x+ y )=tan(−z ) or tan ( x+ y )=tan(π−z )

⇒ x+ y=−z or x+ y=π−z

As, x, y, z are all positive, x+y -z

Hence, x +y + z = ⇒sin−112

13+cos−1 4

5+ tan−163

16=π

proved.

Q. Solve the following equation:

(a) 2 tan−1 (cos x )=tan−12(cos ecx ) (b)

tan−1 1−x1+x

= 12tan−1 x

Ans: (a) 2 tan−1 (cos x )= tan−12(cos ecx )

⇒tan−1( 2cos x

1−cos2 x)=tan−1( 2

sin x)

⇒

2cos x1−cos2 x

= 2sin x

⇒ cos x .sin x=sin2 x

⇒ sin x (sin x−cos x )=0

As, sin x≠0 (why ) , sin x−cos x=0

⇒ tan x=1

⇒x= π4 , Ans.

Graded Assignment:

Assignment – I

1. Find the value of sin−1(sin 2π3 )

Ans:

π3

2. Write the value of sin−(−√3

2 )+sec−1 (−2 )Ans:

π3

3. Write the value of cos−1(cos 2 π3 )+sin−1 (sin 2π3 )

Ans: π

4. Evaluate tan(2 tan−1 1

5 ) Ans:

512

Assignment –II

1. Simplify sin−1 (153 cos x+1213 sin x) Ans:

x+cos−11213

2. Solve the equation tan−1( x−1x−2 )+ tan−1( x+1x+2 )=π4 Ans:

± 1√2

3. Prove that sin−1 5

13+cos−1 3

5=tan−163

16

4. Prove that tan−1 1

2+ tan−1 1

5+ tan−1 1

8= π4

5. Solve for x , sin−1(1−x )−2sin−1 x= π

2

6. If sin(sin−1 15+cos−1 x)=1 , find value of x

Assignment – III

1. If cos−1 x

a+cos−1 4

6=α ,

prove that

x2

a2−2cos α

ab+ y

2

b2=sin2α

2. Prove that tan−1[ √1+x−√1−x

√1+x+√1−x ]= π4−12cos−1x

3. If cos−1 x+cos−1 y+cos−1 z=π , prove that x

2+ y2+ z2+2xyz=1

4. Solve for x , tan−1 (1−x1+x )−12 tan−1 x=0 ,( x>0) Ans:

x= 1√3

5. Prove that tan( π4 + 1

2cos−1 a

b )+tan ( π4−12cos−1 a

b )= 2baCourtesy: While drawing graph help from NCERT Text Book