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 graphs
3. Trigonometric functions of sum, difference, multiple and sub-multiple angles
Objectives:
1.
2.
3.
4.
5.
To define Inverse Trigonometric Functions
Define the Principle value of Inverse trigonometric Function
To find domain and range of Inverse Trigonometric Function
To state the properties of Inverse Trigonometric Function
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
a
1
a
1
a
1
b
2
b
2
b
2
c
3
c
3
c
3
d
d
Fig: -1
d
Fig:-2
Fig:-3
The association in Fig:-1 is not a function as dA is unassociated. The association in Fig:-2 is also not a
function as bA is not uniquely associated. The association in Fig:-3 is a function from A to B. (Why?)
Trigonometric Function: f:RR defined by y=f(x) =Sinx is a function as for every xR 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.
Q1: What are domain of definitions of (a) f(x) = tanx
Ans:


(a) R   x : x  (2n  1)
(b) f(x)=cosecx


, n  Z  , (Why ?)
2

(b) R  x : x  n , 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 
f ( x)  tan x is not well defined (i.e. does not exist). So,

2
,

 domain of tan x .
2
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)  x 2 . 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
2
4
-2
-4
If we try to visualize the inverse rule we get the following
R
R
4
2
-4
-2
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 f 1 : B  A exists and is one-one and onto. Then
f ( x)  y  f 1 ( y)  x where x  A and y  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
1
. This is just a notation given to write Inverse
sin y
Trigonometric Function.
Q3:
Let f : R  R be a function defined by f ( x)  sin x . Examine whether sin 1 x exists. Consider
graph of y  sin x
We can easily verify

3

2
 
 2 
 f  f
 etc. Further any y  R   1,1 is not an image.
3
3
 3 
Hence, the function sin 1 x does not exist, as f is not one-one and onto.
   
,
  1,1 defined by y  f ( x)  sin x . It is both one-one and
Now consider the function f : 
 2 2 
onto which can be visualized from the above graph. Hence, sin 1 x exists.
  3   
,
In fact, if we restrict the domain of the sin function to any one of the intervals 
,
2 
 2
       3 
 2 , 2  ,  2 , 2  ,
onto.
 3 5 
 2 , 2  .. etc., keeping the range  1,1 the function becomes one-one and
Then y  f ( x)  sin 1 x is a function whose domain is  1,1 and range is any one of the
  3           3   3 5 
,
intervals 
,
,
,
.. etc., [please recall that if f
,
,
,
2   2 2   2 2   2 2 
 2
f
1
1
exists, f and
interchange their domain and range.
  3        
,
Every branch of the graph of y  sin x corresponding to branches 
,
,
,
2   2 2 
 2
  3   3 5 
  
 2 , 2  ,  2 , 2  gives a one–one and onto function. The graph in the branch  2 , 2  called
the principle value branch.
We can also justify our answer from the chapter Trigonometric Equations and General Values.
f ( x)  sin x 
x
3
 2
7 8
has infinitely many solutions, x  ,
,
,
2
3
3 3
3
… etc.. The value

  

,
is called the principle value.
3  2 2 
We may say that sin 1
3 
  
 . Also value of sin 1 x  
,
is called the principle value.
2
3
 2 2 
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 Domain and range of corresponding inverse trigonometric
they are one-one & onto (Principal value branches)
functions
Function
Domain
Range
Function
Domain
Range
1. y  sin x
  
 2 , 2 
 1,1
1. y = sin 1 x
 1,1
  
 2 , 2 
2. y  cos x
0,  
 1,1
2. y = cos 1 x
 1,1
0,  
3. y  tan x
  
 2 , 2 
R
3. y  tan 1 x
R
  
 2 , 2 
     
 2 ,0    0, 2 
 ,1  1, 
 ,1  1, 
     
 2 ,0    0, 2 
4. y  cos ecx
4. y  cos ec 1 x
5. y  sec x
0,     
 ,1  1, 
0,  
R
5. y  sec 1 x
 ,1  1, 
2
6. y  cot x
0,     
2
6. y  cot 1 x
R
0,  
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 ecx and 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
 cos ec 1 x, x  1 or x  1
x
(b) cos 1
1
 sec 1 x, x  1 or x  1
x
(c) tan 1
1
 cot 1 x, x  0
x
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)    cos 1 x, x 1 [1,1]
(b) cot 1 ( x)    cot 1 x, x  R
(c) sec 1 ( x)    sec 1 x, | x | 1

4. (a) sin 1 x  cos 1 x 
(b) tan 1 x  cot 1 x 
2

2
(c) cos ec 1 x  sec 1 x 
, x  [1,1]
,xR

2
, | x | 1
5. (a) tan 1 x  tan 1 y  tan 1
(b) tan 1 x  tan 1 y  tan 1
x y
, xy  1
1  xy
x y
, xy  1
1  xy
Proof: Let tan 1 x   and tan 1 y  . Then tan   x and tan   y.
Now tan(   ) 
tan   tan 
x y

1  tan  tan  1  xy
x y
1  xy
     tan 1
 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 (
3
)
2
(e) cot 1 (1)
Ans:
(a) cos ec 1 (
2
3
) = sin 1 (
 3
)   ( say )
2
(c) tan 1 ( 3 ) (d) sec 1 (  2 )
then, sin  
 3
 3
 

, ] such that sin  
. Then the value of   [
is  
2 2
3
2
2
Hence, cos ec 1 (
Ans:
(b)
2
3

3
)=
(c)

3

3
Q. Find the value of sec{cos 1 (
Ans: Let cos 1 (
 cos  
(d)
3
4
(e)
3
4
3
)
2
3
) 
2
3

 
2
6
Hence, sec{cos 1 (
3

2
)  sec   sec 
2
6
3
Q. Prove that cos(sin 1 x)  1  x 2
Ans: Let sin 1 x  y
 sin y  x
Hence, cos(sin 1 x)  cos y  1  sin 2 y  1  x 2
Q. Prove the following:
(a) 2 sin 1 x  sin 1 2 x 1  x 2
(b) 2 cos 1 x  cos 1 (2 x 2  1)
Q. Prove the following:
(a) 2 tan 1 x  sin 1
2x
, | x | 1
1 x2
(b) 2 tan 1 x  cos 1
1 x2
,x  0
1 x2
(c) 2 tan 1 x  tan 1
2x
,1  x  1
1 x2
Ans:
(a) Let tan 1 x  y
 tan y  x Now sin 2 y 
 2 y  sin 1
2 tan y
2x

2
1  tan y 1  x 2
2x
1 x2
 2 tan 1 x  sin 1
2x
1 x2
Similar proof for (b) and (c) follow.
Q. Prove that
sin 1
12
4
63
 cos 1  tan 1

13
5
16
Ans: Let sin 1
12
4
63
 x, cos 1  y, tan 1
z
13
5
16
 sin x 
12
4
63
, cos y  , tan z 
13
5
16
 cos x 
5
3
12
3
, sin y  , tan x  and tan y 
13
5
5
4
12 3

tan x  tan y
5
4   63
Then tan( x  y ) 

12 3
1  tan x tan y
16
1 x
5 4
 tan( x  y )   tan z
 tan( x  y )  tan(  z ) or tan( x  y )  tan(   z )
 x  y  z or x  y    z
As, x, y, z are all positive, x+y  -z
Hence, x +y + z =   sin 1
12
4
63
 cos 1  tan 1
  proved.
13
5
16
Q. Solve the following equation:
(b) tan 1
(a) 2 tan 1 (cos x)  tan 1 2(cos ecx)
1 x 1
 tan 1 x
1 x 2
Ans: (a) 2 tan 1 (cos x)  tan 1 2(cos ecx)
2 cos x
2
)  tan 1 (
)
 tan 1 (
2
sin x
1  cos x

2 cos x
2

2
1  cos x sin x
 cos x. sin x  sin 2 x
 sin x(sin x  cos x)  0
As, sin x  0 ( why) , sin x  cos x  0
 tan x  1
x

4
, Ans.
Graded Assignment:
Assignment – I


1. Find the value of sin 1  sin
2 

3 
 3
  sec 1  2 
2. Write the value of sin  

 2 


3. Write the value of cos 1  cos


1
5
4. Evaluate tan  2 tan 1 
Assignment –II
2 
2 
1 
  sin  sin

3 
3 

Ans:

3
Ans:

3
Ans: 
Ans:
5
12
12
 15
cos x  sin
13
3
1. Simplify sin 1 

x

Ans: x  cos 1

 x 1 
1  x  1 
  tan 

 x2
 x2 4
2. Solve the equation tan 1 
3. Prove that sin 1
5
3
63
 cos 1  tan 1
13
5
16
4. Prove that tan 1
1
1
1 
 tan 1  tan 1 
2
5
8 4
5. Solve for x, sin 1 (1  x)  2 sin 1 x 


6. If sin  sin 1
Ans: 
12
13
1
2

2
1

 cos 1 x   1 , find value of x
5

Assignment – III
x
x 2 2 cos  y 2
1 4
 cos
  , prove that 2 
1. If cos
 2  sin 2 
a
6
ab
a
b
1
 1 x  1 x   1
1
   cos x
 1 x  1 x  4 2
2. Prove that tan 1 
3. If cos 1 x  cos 1 y  cos 1 z   , prove that x 2  y 2  z 2  2 xyz  1
1 x  1
1
  tan x  0, ( x  0)
1

x
2


4. Solve for x, tan 1 
a
a  2b
 1
 1
 cos 1   tan   cos 1  
b
b a
4 2
4 2
5. Prove that tan 
Courtesy: While drawing graph help from NCERT Text Book
Ans: x 
1
3