Inverse-Trigonometric-Functions.pdf

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Complete Study Guide & Notes On
INVERSE TRIGONOMETRIC FUNCTIONS
The essence of Mathematics is not to make simple things complicated,
but to make complicated things simple!
IMPORTANT TERMS, DEFINITIONS & RESULTS
 A list of formulae for Trigonometric Functions (class XI) has been added for
reference at the end of this formulae guide.
01. Basic Introduction: A function : A B
f  is said to be invertible if f is bijective (i.e., one-one
and onto). The inverse of the function f is denoted by : B A
f  such that 1
( )
f y x

 if ( )
f x y
 ,
A, B
x y
   . As trigonometric functions are many-one so, their inverse doesn’t exist. But they
become one-one onto by restricting their domains. Therefore, inverse of trigonometric functions are
defined with restricted domains. In fact, in the discussion below we have used all the restrictions
required so that the inverse of the concerned trigonometric functions do exist. If these restrictions are
removed, the terms will represent inverse trigonometric relations and not the functions. Note that
the inverse trigonometric functions are also called as Inverse Circular Functions.
02. List of Formulae and their proofs for Inverse Trigonometric Functions:
A. a)    
1 1 1
1,1
sin cosec ,
   

 
 
 
x
x
x b)      
1 1 1
cosec sin , , 1 1,
   
 
 
     
x
x
x
c)    
1 1 1
1,1
cos sec ,
   

 
 
 
x
x
x d)      
1 1 1
sec cos , , 1 1,
   
 
 
     
x
x
x
e)  
1
1
1
1
1
π
cot , 0
tan
cot , 0



  
 

  

 
   
  




x
x
x
x
x
f)  
1
1
1
1
1
π
tan , 0
cot
tan , 0



  
 

  

 
   
  




x
x
x
x
x
PROOF a) Let 1
( )
sin θ


x then, sinθ  x
1
cosecθ

x
= 1 1
θ cosec  
 
 

x
=
1 1 1
sin cosec
   
   
 
x
x
[H.P.]
Other results can be proved in the same way!
B. a)    
1 1
sin sin , 1,1
x x x
 
     b)    
1 1
cos π cos , 1,1
 
    
x x x
c)  
1 1
tan tan , R
x x x
 
    d)  
1 1
| |
cosec cosec , 1
 

   x
x x
e)  
1 1
| |
sec π sec , 1
 
   
x
x x f)  
1 1
cot π cot , R
 
   
x x x
PROOF b) Let  
1
cos θ

 
x then, cosθ  x
  1
cosθ cos π θ cos π θ


       
x
x x
 
1 1 1
cos
θ π cos π cos
  
 
    
x x x [H.P.]
Similarly, we can prove other results!
C. a)  
1 π π
2 2
sin sin ,

 
  x
x x b)  
1
0 π
cos cos ,

 
 x
x x
c)  
1 π π
2 2
tan tan ,

 
  x
x x d)  
1 π π
, 0
2 2
cosec cosec ,

  
  x x
x x
e)  
1 π
0 π,
2
sec sec ,

  
 x x
x x f)  
1
0 π
cot cot ,

 
 x
x x
A Formulae Guide By OP Gupta (Indira Award Winner)

Inverse Trigonometric Functions By OP Gupta (INDIRA AWARD Winner, Elect. & Comm. Engineering)
2
PROOF a) Let    
1
sin θ ...


x i
then,  
sinθ ...
 ii
x
Substituting the value of x in (i) from (ii) we get,
1
sin sinθ θ

 (Replacing θ by x)
1
sin sin

 
x x [H.P.]
For other results we can proceed similarly!
D. a)  
1 1 π
sin cos , 1,1
2
 
   
x x x
b) 1 1 π
tan cot , R
2
 
  
x x x
c) 1 1
| | 1 or 1
π
cosec sec , 1 . .,
2
 
  
  
x x
x x i e x
PROOF a) Let 1
( )
sin θ


x then, sinθ  x
1 1
π π π
2 2 2
cos θ θ cos cos θ
 
 
 
 
 
      
x x x
1 1 π
, 1 1
2
sin cos
 
   
 
x x
x [H.P.]
Similarly, proceed for other results!
E. a) 1 1 1 2 2
sin sin sin 1 1
    
    
 
x y x y y x
b) 1 1 1 2 2
cos cos cos 1 1
    
   
 

x y xy x y
c)
1
1 1 1
1
, 1
, 0, 0, 1
, 0, 0, 1
tan
1
tan tan π tan
1
π tan
1

  

  

  
 

  

  
  
 

  
   
 
  




  


 

xy
x y xy
x y xy
x y
xy
x y
x y
xy
x y
xy
d)
1
1 1 1
1
, 1
, 0, 0, 1
, 0, 0, 1
tan
1
tan tan π tan
1
π tan
1

  

  
 
  
 

  

   
  
 

  
    
 
  




  


 

xy
x y xy
x y xy
x y
xy
x y
x y
xy
x y
xy
e) 1 1
1 1
1
tan tan tan tan
 
   
  
 
  
 
  
x y z xyz
z
xy yz zx
x y
PROOF a) Let 1
sin θ


x and 1
sin β


y . Then, sinθ  x and sinβ  y .
Now ( ) sinθcosβ cosθsinβ
sin θ β  

2 2
sinθ 1 sin β 1 sin θsinβ
    
( )
sin θ β
 + 2 2
1 1
   
x y y x
1 2 2
θ β sin 1 1
  
     
 
x y y x
1 1 1 2 2
sin sin sin 1 1
    
     
 
x y x y y x [H.P.]

A Complete Formulae Guide  Compiled By OP Gupta (M.+91-9650350480 | +91-9718240480)
3
b) 1 1 1 2 2
cos cos cos 1 1
    
   
 

x y xy x y
Do yourself. Proceed as in (a).
c) Let 1
tan θ


x and 1
tan β


y . Then, tanθ  x and tanβ  y .
Now  
tan θ tanβ
tan
1 tanθtanβ 1
θ β
 
 
 

x y
xy
1
θ β tan
1
  

    

 
x y
xy
1 1 1
tan tan tan
1
    

    

 
x y
x y
xy
For 0

x , 1
tan
x will be a positive angle and for 0

y , 1
tan
y will also be a positive angle.
Therefore, LHS of (c) will be a positive angle and hence RHS should also be a positive angle.
Case I When 0, 0
 
x y and 1

xy then
1


x y
xy
is positive.
So, 1
tan
1
  

 

 
x y
xy
will be a positive angle.
Hence, 1 1 1
tan tan tan
1
    

   

 
x y
x y
xy
[H.P.]
Case II When 0, 0
 
x y and 1

xy then
1


x y
xy
is negative.
So, 1
tan
1
  

 

 
x y
xy
will be a negative angle. Therefore we add π to make it positive and balanced.
Hence, 1 1 1
tan tan π tan
1
    

    

 
x y
x y
xy
[H.P.]
Case III When 0, 0
 
x y and 1

xy then
1


x y
xy
is positive.
So, 1 1
tan tan
 

x y will be a negative angle and 1
tan
1
  

 

 
x y
xy
will be a positive angle. Therefore to
balance it we will be adding – π .
Hence, 1 1 1
tan tan π tan
1
    

     

 
x y
x y
xy
[H.P.]
d) Let 1
tan θ


x and 1
tan β


y . Then, tanθ  x and tanβ  y .
Now  
tan θ tanβ
tan
1 tanθtanβ 1
θ β
x y
xy
 
 
 

1
θ β tan
1
x y
xy
  

    

 
1 1 1
tan tan tan
1
x y
x y
xy
    

    

 
…(A)
Case I When 0, 0
 
x y and 1
xy   then
1
x y
xy


is positive (or negative depending upon the
absolute value of the angles x and y). Also if 0, 0
 
x y then,
π
θ,β 0,
2
 
 
 
. So
π π
θ β ,
2 2
 
  
 
 
i.e.,
1 1 π π
tan tan ,
2 2
x y
   
  
 
 
.

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Hence, 1 1 1
tan tan tan
1
x y
x y
xy
    

   

 
, 1
xy   [By using (A) [H.P.]
Case II When 0, 0
x y
  and 1
xy   then
1
x y
xy


is negative. Also if 0, 0
x y
  then,
π π
θ 0, ,β ,0
2 2
   
  
   
   
i.e.,
π π
θ 0, , β 0,
2 2
   
  
   
   
. So θ β (0,π)
  i.e., 1 1
tan tan (0,π)
x y
 
  .
As 1
tan
1
x y
xy
  

 

 
is a negative angle. Therefore we add π in RHS of (A) to make it positive and
balanced.
Hence, 1 1 1
tan tan π tan
1
x y
x y
xy
    

    

 
, 0, 0
x y
  and 1
xy   [H.P.]
Case III When 0, 0
x y
  and 1
xy   then
1
x y
xy


is positive. Also if 0, 0
x y
  then,
π π
θ ,0 ,β 0,
2 2
   
  
   
   
i.e.,
π π
θ ,0 , β ,0
2 2
   
    
   
   
. So θ β ( π,0)
   i.e., 1 1
tan tan ( π,0)
x y
 
   .
So, 1 1
tan tan
x y
 
 will be a negative angle and 1
tan
1
x y
xy
  

 

 
will be a positive angle. Therefore to
balance it we will be adding – π in RHS of (A).
Hence, 1 1 1
tan tan π tan
1
x y
x y
xy
    

     

 
, 0, 0
x y
  and 1
xy   [H.P.]
F. a) 2
1 1 2
2 tan sin , | | 1
1
   
 
 

 
x
x x
x
b)
2
1
2
1 1
2 tan cos , 0
1
   

 
 

 
x
x x
x
c) 1
2
1 2
2 tan tan , 1 1
1
   
   
 

 
x
x x
x
PROOF a) Let 1
tan θ


x then, tanθ  x .
As 2 2
2tanθ 2
sin2θ sin2θ
1 tan θ 1
  
 
x
x
i.e., 1
2 2
1 1
2 2
2θ sin 2tan sin
1 1
    
    
 
 
x x
x
x x
[H.P.]
Other results can also be proved in the same way!
03. Principal Value: Numerically smallest angle is known as the principal value.
Finding the principal value: For finding the principal value, following algorithm can be followed–
STEP1– Firstly, draw a trigonometric circle and mark the quadrant in which the angle may lie.
STEP2– Select anticlockwise direction for 1st
and 2nd
quadrants and clockwise direction for 3rd
and
4th
quadrants.
STEP3– Find the angles in the first rotation.
STEP4– Select the numerically least (magnitude wise) angle among these two values. The angle thus
found will be the principal value.
STEP5– In case, two angles one with positive sign and the other with the negative sign qualify for
the numerically least angle then, it is the convention to select the angle with positive sign as
principal value.
The principal value is never numerically greater than  .

5
04. Table demonstrating domains and ranges of Inverse Trigonometric functions:
Discussion about the range of inverse circular functions
other than their respective principal value branch
We know that the domain of sine function is the set of real numbers and range is the closed
interval [–1, 1]. If we restrict its domain to
3π π
,
2 2
 
 
 
 
,
π π
,
2 2
 

 
 
,
π 3π
,
2 2
 
 
 
etc. then, it
becomes bijective with the range [–1, 1]. So, we can define the inverse of sine function in
each of these intervals. Hence, all the intervals of sin–1
function, except principal value
branch (here except of
π π
,
2 2
 

 
 
for sin–1
function) are known as the range of sin–1
other
than its principal value branch. The same discussion can be extended for other inverse
circular functions. (Refer Q16 in Mathematicia Vol.1 By OP Gupta)
05. To simplify inverse trigonometrical expressions, following substitutions can be considered:
Inverse Trigonometric Functions i.e., ( )
f x Domain/ Values of x Range/ Values of ( )
f x
1
sin
x [ 1, 1]
 π π
,
2 2
 

 
 
1
cos
x [ 1, 1]
 [0, π]
1
cosec
x R ( 1, 1)
  π π
, {0}
2 2
 
 
 
 
1
sec
x R ( 1, 1)
  π
[0, π]
2
 
  
 
1
tan
x R π π
,
2 2
 

 
 
1
cot
x R (0, π)
Expression Substitution
2 2 2 2
or
a x a x
  tanθ or cotθ
 
x a x a
2 2 2 2
or
a x a x
  sinθ or cosθ
 
x a x a
2 2 2 2
or
x a x a
  secθ or cosecθ
 
x a x a
or
a x a x
a x a x
 
 
cos2θ

x a
2 2 2 2
2 2 2 2
or
a x a x
a x a x
 
 
2 2
cos 2θ

x a
or
x a x
a x x


2 2
sin θ or cos θ
 
x a x a
or
x a x
a x x


2 2
tan θ or cot θ
 
x a x a

6
Note the followings and keep them in mind:
 The symbol 1
sin x

is used to denote the smallest angle whether positive or negative, such
that the sine of this angle will give us x. Similarly 1 1 1 1
, , , ,
cos x tan x cosec x sec x
   
and
1
cot x

are defined.
 You should note that 1
sin x

can be written as arcsinx . Similarly other Inverse Trigonometric
Functions can also be written as arccosx, arctanx, arcsecx etc.
 Also note that sin x
1
(and similarly other Inverse Trigonometric Functions) is entirely
different from 1
( )
sin x 
. In fact, 1
sin x

is the measure of an angle in Radians whose sine is x
whereas 1
( )
sin x 
is
1
sin x
(which is obvious as per the laws of exponents).
 Keep in mind that these inverse trigonometric relations are true only in their domains i.e.,
they are valid only for some values of ‘x’ for which inverse trigonometric functions are well
defined!
Check out NCERT Textbook Part I for the Graphs of Inverse Trigonometric Functions.
 Trigonometric Formulae (Only For Reference):
 Relation between trigonometric ratios
a)
sinθ
tan θ
cosθ
 b)
1
tan θ
cot θ
 c) tan θ.cotθ 1

d)
cosθ
cot θ
sinθ
 e)
1
cosecθ
sin θ
 f)
1
secθ
cosθ

 Trigonometric identities
a) 2 2
sin θ cos θ 1
 
b) 2 2
sec θ 1 tan θ
 
c) 2 2
cosec θ 1 cot θ
 
 Addition / subtraction formulae & some related
results
a)  
sin sin cos cos sin
A B A B A B
  
b)  
cos cos cos sin sin
A B A B A B
  
    2 2
2 2
) cos cos cos sin
cos sin
c A B A B A B
B A
   
 
    2 2
2 2
) sin sin sin sin
cos cos
d A B A B A B
B A
   
 
e)  
tan tan
tan
1 tan tan
A B
A B
A B

 

f)  
cot cot 1
cot
cot cot
B A
A B
B A
 


 Multiple angle formulae involving 2A & 3A
a)sin2 2sin cos
A A A

b)
2 2
sin 2sin cos
A A
A 
c) 2 2
cos2 cos sin
A A A
 
d) 2 2
cos cos sin
2 2
A A
A  
e) 2
cos2 2cos 1
A A
 
f) 2
2cos 1 cos2
A A
 
g) 2
cos2 1 2sin
A A
 
h) 2
2sin 1 cos2
A A
 
i) 2
2tan
sin2
1 tan
A
A
A


j)
2
2
1 tan
cos2
1 tan
A
A
A




7
k) 2
2 tan
tan 2
1 tan
A
A
A


l) 3
sin3 3sin 4sin
A A A
 
m) 3
cos3 4cos 3cos
A A A
 
n)
3
2
3tan tan
tan 3
1 3tan
A A
A
A



 Transformation of sums / differences into products
& vice-versa
a) sin sin 2sin cos
2 2
C D
C D
C D

 

b)sin sin 2cos sin
2 2
C D
C D
C D

 

c)cos cos 2cos cos
2 2
C D
C D
C D

 

d)cos cos 2sin sin
2 2
C D C D
C D
 
  
e)    
sin sin
2sin cos A B A B
A B    
f)    
2 sin sin
cos sin A B A B
A B    
g)    
cos cos
2cos cos A B A B
A B    
h)    
cos cos
2sin sin A B A B
A B    
 Relations in Different Measures of Angle
  
π
Angle in Radian Measure Angle in Degree Measure
180
= ×
  
180
Angle in Degree Measure Angle in Radian Measure
π
= ×
 θ (in radian measure)
l
r

Also followings are of importance as well:
 o
1Right angle 90
  o
1 = 60 , 1 = 60
  
  
.
o
1 = = 0.01745 radians Approx
180

 o
1 radian = 57 17 45 or 206265 seconds
  .
 General Solutions
a) sin sin ( 1 ) ,
n
x y x n y where n Z

      .
b) cos cos 2 ,
x y x n y where n Z

     .
c) tan tan ,
x y x n y where n Z

     .
 Relation in Degree & Radian Measures
Angles in Degree o
0 o
30 o
45 o
60 o
90 o
180 o
270 o
360
Angles in Radian
c
0
c
6

 
 
 
c
4

 
 
 
c
3

 
 
 
c
2

 
 
 
 c

c
3
2

 
 
 
 c
2
 In actual practice, we omit the exponent ‘c’ and instead of writing  c
we simply write  and
similarly for others.
 For the values of Trigonometric Ratios at Standard Angles (i.e, o o o o o
0 ,30 ,45 ,60 and 90 ), check the
following page.
For the complete discussion of the Trigonometric Functions, please refer to the
FORMULAE GUIDE Of TRIGONOMETRIC FUNCTIONS of Class XI.

8
 Trigonometric Ratio of Standard Angles
Degree / Radian  
 o
0 o
30 o
45 o
60 o
90
T – Ratios 
 0
π
6
π
4
π
3
π
2
sin 0 1
2
1
2
3
2
1
cos 1 3
2
1
2
1
2
0
tan 0 1
3
1 3 
cosec  2 2
2
3
1
sec 1
2
3
2 2 
cot  3 1
1
3
0
 Trigonometric Ratios of Allied Angles
Angles  

π
θ
2

π
θ
2
 π θ
 π θ
 3π
θ
2

3π
θ
2
 2π θ
 OR
θ

2π θ

T- Ratios  

sin cosθ cosθ sinθ  sinθ  cosθ  cosθ  sinθ sinθ
cos sinθ  sinθ  cosθ  cosθ  sinθ sinθ cosθ cosθ
tan cot θ  cot θ tanθ
 tan θ cot θ  cot θ  tan θ tan θ
cot tan θ  tan θ  cot θ cot θ tan θ  tan θ  cot θ cot θ
sec cosecθ  cosecθ  secθ  secθ  cosecθ cosecθ secθ secθ
cosec secθ secθ cosecθ  cosecθ  secθ  secθ  cosecθ cosecθ
 Domain and Range of Trigonometric Functions
T- Functions 
 Domain Range
sin x R [ 1, 1]

cos x R [ 1, 1]

tan x { R : (2 1)π 2, Z}
x x n n
    R
cot x { R : π, Z}
x x n n
   R
cosec x { R : π, Z}
x x n n
   R ( 1, 1)
 
sec x { R : (2 1)π 2, Z}
x x n n
    R ( 1, 1)
 
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theopgupta@gmail.com
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School/Institution, Place, Contact No. (if you wish)

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