Download INVERSE TRIGONOMETRIC FUNCTIONS

Inverse Trigonometric Functions
MODULE - IV
Functions
Notes
18
INVERSE TRIGONOMETRIC FUNCTIONS
In the previous lesson, you have studied the definition of a function and different kinds of functions.
We have defined inverse function.
Let us briefly recall :
Let f be a one-one onto function from A to B.
Let y be an arbitary element of B. Then, f being
onto, ∃ an element x ∈ A such that f ( x ) = y .
Also, f being one-one, then x must be unique. Thus
for each y ∈ B , ∃ a unique element x ∈ A such
that f ( x ) = y . So we may define a function,
denoted by f −1 as f −1 : B → A
f −1 ( y ) = x ⇔ f ( x ) = y
∴
The above function f −1 is called the inverse of f. A function is invertiable if and only if f is
one-one onto.
In this case the domain of f −1 is the range of f and the range of f −1 is the domain f.
Let us take another example.
We define a function : f : Car → Registration No.
If we write,
g : Registration No. → Car, we see that the domain of f is range of g and the
range of f is domain of g.
So, we say g is an inverse function of f, i.e., g = f −1 .
In this lesson, we will learn more about inverse trigonometric function, its domain and range, and
simplify expressions involving inverse trigonometric functions.
OBJECTIVES
After studying this lesson, you will be able to :
l
define inverse trigonometric functions;
MATHEMATICS
141
Inverse Trigonometric Functions
MODULE - IV
Functions
l
l
l
l
l
Notes
state the condition for the inverse of trigonometric functions to exist;
define the principal value of inverse trigonometric functions;
find domain and range of inverse trigonometric functions;
state the properties of inverse trigonometric functions; and
simplify expressions involving inverse trigonometric functions.
EXPECTED BACKGROUND KNOWLEDGE
l
l
Knowledge of function and their types, domain and range of a function
Formulae for trigonometric functions of sum, difference, multiple and sub-multiples of
angles.
18.1 IS INVERSE OF EVERY FUNCTION POSSIBLE ?
Take two ordered pairs of a function ( x1, y ) and ( x 2 , y )
If we invert them, we will get ( y , x1 ) and ( y , x2 )
This is not a function because the first member of the two ordered pairs is the same.
Now let us take another function :
 sin π , 1

,

2 
π 1 


π 3
 sin ,
 and  sin ,

4 2


3 2 
Writing the inverse, we have
 3
π
1,sin π   1 ,sin π 
,sin 

, 
 and 
4

2  2
 2
3
which is a function.
Let us consider some examples from daily life.
f : Student → Score in Mathematics
Do you think f −1 will exist ?
It may or may not be because the moment two students have the same score, f −1 will cease to be
a function. Because the first element in two or more ordered pairs will be the same. So we
conclude that
every function is not invertible.
Example 18.1 If f : R → R defined by f (x) = x 3 + 4 . What will be f −1 ?
Solution : In this case f is one-to-one and onto both.
⇒ f is invertible.
Let
∴
y = x3 + 4
y − 4 = x3 ⇒ x
3= y
4−
So f −1 , inverse function of f i.e., f −1 ( y ) = 3 y − 4
142
MATHEMATICS
Inverse Trigonometric Functions
The functions that are one-to-one and onto will be invertible.
Let us extend this to trigonometry :
Take y = sin x. Here domain is the set of all real numbers. Range is the set of all real numbers
lying between −1 and 1, including −1 and 1 i.e. −1 ≤ y ≤1 .
We know that there is a unique value of y for each given number x.
In inverse process we wish to know a number corresponding to a particular value of the sine.
y = sin x =
Suppose
sin x = sin
⇒
MODULE - IV
Functions
Notes
1
2
π
5π
13π
= sin
= sin
= ....
6
6
6
x may have the values as
π 5π 13π
= ....
,
,
6 6
6
Thus there are infinite number of values of x.
y = sin x can be represented as
 π 1   5π 1 
 ,  ,  ,  ,....
6 2  6 2
The inverse relation will be
 1 π   1 5π 
 , ,  ,
 ,....
2 6  2 6 
It is evident that it is not a function as first element of all the ordered pairs is
1
, which contradicts
2
the definition of a function.
Consider y = sin x, where x ∈ R (domain) and y ∈ [−1, 1] or −1 ≤y ≤1 which is called range.
This is many-to-one and onto function, therefore it is not invertible.
Can y = sin x be made invertible and how? Yes, if we restrict its domain in such a way that it
becomes one-to-one and onto taking x as
(i)
−
π
π
≤ x≤ ,
2
2
y ∈ [−1, 1]
or
(ii)
3π
5π
≤x≤
2
2
y ∈ [−1, 1]
or
(iii)
−
5π
3π
≤x ≤−
2
2
y ∈ [−1, 1]
etc.
Now consider the inverse function y = sin−1 x .
We know the domain and range of the function. We interchange domain and range for the
inverse of the function. Therefore,
MATHEMATICS
143
Inverse Trigonometric Functions
MODULE - IV
Functions (i)
(ii)
(iii)
Notes
π
π
≤y ≤
2
2
x ∈ [−1, 1]
or
3π
5π
≤ y≤
2
2
x ∈ [−1, 1]
or
x ∈ [−1, 1]
etc.
−
−
5π
3π
≤ y≤−
2
2
Here we take the least numerical value among all the values of the real number whose sine is x
which is called the principle value of sin−1 x .
For this the only case is −
π
π
≤ y ≤ . Therefore, for principal value of y = sin−1 x , the domain
2
2
is [−1, 1] i.e. x∈ [−1, 1] and range is −
π
π
≤y ≤ .
2
2
Similarly, we can discuss the other inverse trigonometric functions.
Function
Domain
Range
(Principal value)
1.
y = sin−1 x
[−1, 1]
− π , π 
 2 2 
2.
y = cos −1 x
[−1, 1]
[0, π ]
3.
y = tan −1 x
R
− π, π 


 2 2
4.
y = cot −1 x
R
[0, π ]
5.
y = sec−1 x
x ≥ 1 or x ≤ − 1
 π  π 
0, 2  ∪  2 , π


6.
y = cosec−1 x
x ≥ 1 or x ≤ − 1
 π   π
 − 2 , 0  ∪  0, 2 


18.2 GRAPH OF INVERSE TRIGONOMETRIC FUNCTIONS
18.2 Gr
y = sin−1 x
144
y = cos −1 x
MATHEMATICS
Inverse Trigonometric Functions
MODULE - IV
Functions
Notes
y = tan −1 x
y = cot −1 x
y = sec −1 x
y = cosec −1 x
Fig. 18.2
Example 18.2 Find the principal value of each of the following :
 1 
−1  1 
(i) sin −1 
 (ii) cos  − 2 
 2
Solution : (i) Let
sin θ =
or
 1 
(iii) tan −1  −
3 

−1  1 
sin 
=θ
 2
1
π
= sin   or
2
4 
θ=
π
4
(ii) Let cos −1  − 1  = θ
 2
⇒
cos θ = −
1
π

 2π 
= cos  π −  = cos   or
2

3
 3 
1 
(iii) Let tan −1  −
=θ
3 

⇒
MATHEMATICS
−
or
θ=−
θ=
2π
3
1
π
= tan θ or tan θ = tan  − 
3
 6
π
6
145
Inverse Trigonometric Functions
MODULE - IV
Functions
Example 18.3
Find the principal value of each of the following :
(a)
1 
(i) cos −1 

 2
(b)
Find the value of the following using the principal value :
(ii) tan −1 (−1)


sec  cos −1 3 

2 
Notes
−1  1 
Solution : (a) (i) Let cos 
 = θ , then
 2
1
= cos θ
2
⇒
θ=
π
4
or
cos θ = cos
or
π
tan θ= tan  − 
 4
or
π
cos θ = cos  
6
π
4
(ii) Let tan −1 (−1) = θ , then
−1 = tan θ
θ=−
⇒
π
4


(b) Let cos −1  3  = θ , then
 2 
3
= cos θ
2
π
θ=
6
⇒
∴


2
π
sec  cos −1 3  = sec θ = sec   =
6
3

2 
Example 18.4
Simplify the following :
(ii) cot [ cosec−1 x ]
(i) cos [sin −1 x ]
Solution : (i) Let sin −1 x = θ
⇒
∴
x = sin θ
cos [sin −1 x ] = cos θ = 1 − sin 2 θ = 1 − x 2
(ii) Let cosec−1x = θ
⇒
146
x = cosec θ
MATHEMATICS
Inverse Trigonometric Functions
MODULE - IV
Functions
cot θ = cosec θ− 1
2
Also
= x 2 −1
CHECK YOUR PROGRESS 18.1
1.
2.
Find the principal value of each of the following :
(a)
 3
cos −1 

 2 
(b)
cosec −1 ( − 2 )
(d)
tan −1 ( − 3 )
(e)
cot −1 (1)

3
(c) sin−1  −

 2 
Evaluate each of the following :

π
−1 2 

(b) cosec−1  cosec  (c) cos  cosec

3

4

 −1 1 
(a) cos  cos


3
(d) tan (sec −1 2 )
3.
Notes
(e) cosec  cot −1 ( − 3 )
Simplify each of the following expressions :
(a) sec ( tan −1 x )
x
(b) tan  cosec−1  (c) cot (cosec −1 x 2 )

2
−1 2
(d) cos ( cot x )
(e) tan (sin −1 ( 1 − x ))
18.3 PROPERTIES OF INVERSE TRIGONOMETRIC FUNCTIONS
Property 1
sin −1 (sin θ ) = θ , −
π
π
≤θ≤
2
2
Solution : Let sin θ = x
⇒
θ = sin −1 x
= sin −1 (sin θ ) = θ
Also
sin (sin −1 x ) = x
Similarly, we can prove that
(i)
cos −1 ( cosθ ) = θ , 0 ≤θ≤π
(ii)
π
π
tan −1 ( tan θ ) = θ, − < θ <
2
2
Property 2
−1
−1  1 
(i) cosec x = sin  
x 
MATHEMATICS
1
(ii) cot −1 x = tan −1  
x
147
Inverse Trigonometric Functions
MODULE - IV
Functions (iii)
1
sec −1 x = cos −1  
x
Solution : (i) Let cosec −1 x = θ
⇒
x = cosec θ
Notes
(ii) Let
⇒
cot −1 x = θ
x = cot θ
⇒
1
= sin θ
x
⇒
1
= tan θ
x
∴
1 
θ = sin−1  
x 
⇒
1
θ = tan −1  
x
∴
1
cot −1 x = tan −1  
x
⇒
(iii)
⇒
1 
cosec −1 x = sin−1  
x 
sec −1 x = θ
x = sec θ
∴
1
= cos θ
x
∴
1
sec −1 x = cos −1  
x
Property 3
or
(i) sin −1 (− x ) = − sin −1 x
1
θ = cos −1  
x
(ii) tan −1 ( − x) = − tan −1 x
(iii) cos −1 ( − x ) = π −cos −1 x
−1
Solution : (i) Let sin (− x ) = θ
⇒
− x = sin θ
or
x = − sin θ = sin ( −θ)
∴
− θ = sin −1 x
or
θ = − sin −1 x
or
x = − tan θ = tan ( −θ)
θ = − tan −1 x or
tan −1 ( − x) = − tan −1 x
or
sin −1 (− x ) = − sin −1 x
(ii) Let tan −1 ( − x ) = θ
⇒
− x = tan θ
∴
(iii) Let cos −1 ( −x ) = θ
⇒
− x = cos θ
∴
cos −1 x = π − θ
∴
cos −1 ( −x ) = π − cos −1 x
Property 4.
(iii)
148
x = − cos θ = cos (π − θ )
or
−1
−1
(i) sin x + cos x =
π
2
cosec −1 x + sec −1 x =
π
2
(ii)
tan −1 x + cot −1 x =
π
2
MATHEMATICS
Inverse Trigonometric Functions
Soluton : (i) sin −1 x + cos −1 x =
MODULE - IV
Functions
π
2
π

Let sin −1 x = θ ⇒ x = sin θ = cos  − θ 
2

π

cos −1 x =  − θ 
2

or
⇒
θ + cos −1 x =
(ii) Let cot −1 x = θ
or
sin −1 x + cos −1 x =
π
2
π

⇒ x = cot θ = tan  − θ 
2

tan −1 x =
∴
or
π
2
Notes
π
−θ
2
cot −1 x + tan −1 x =
or
θ + tan −1 x =
π
2
or
θ + sec−1 x =
π
2
π
2
(iii) Let cosec −1 x = θ
π

x = cosec θ = sec  − θ 
2

⇒
sec −1 x =
∴
⇒ cosec −1 x + sec −1 x =
Property 5
π
−θ
2
π
2
x+y 
(i) tan −1 x + tan −1 y = tan −1 

 1 − xy 
x−y 
(ii) tan −1 x − tan −1 y =tan −1 

 1 + xy 
Solution : (i) Let tan −1 x = θ , tan −1 y = φ ⇒ x = tan θ , y = tan φ
x+y 
We have to prove that tan −1 x + tan −1 y = tan −1 

 1 − xy 
By substituting that above values on L.H.S. and R.H.S., we have
−1  tan θ + tan φ 
L.H.S. = θ + φ and R.H.S. = tan 

1 − tan θ tan φ 
= tan −1  tan (θ + φ ) = θ + φ = L.H.S.
∴ The result holds.
Simiarly (ii) can be proved.
MATHEMATICS
149
Inverse Trigonometric Functions
MODULE - IV
Functions
Property 6
2
 2x 
−1 1 − x 
−1  2x 
2tan −1 x = sin −1 
=
cos
=
tan


1 + x 2 
1 + x2 
1 − x 2 
(i)
(ii)
(iii)
(iv)
Let x = tan θ
Substituting in (i), (ii), (iii), and (iv) we get
Notes
2tan −1 x = 2 tan −1 ( tan θ ) = 2 θ
......(i)
−1  2x 
−1  2 tan θ 
sin 
= sin 

2 
 1+ x 
 1 + tan 2 θ 
−1  2 tan θ 
= sin 

 sec2 θ 
= sin −1 ( 2sinθ cos θ )
= sin −1 ( sin2 θ ) = 2 θ
.....(ii)
2
 1− x2 
−1  1 − tan θ 
cos −1 
=
cos



 1+ x2 
 1 + tan 2 θ 
 cos 2 θ − sin 2 θ 
= cos− 1 

 cos 2 θ + sin 2 θ 
= cos −1 (cos 2 θ − sin 2 θ )
= cos− 1 ( cos2θ ) = 2 θ
......(iii)
−1  2x 
−1  2tan θ 
tan 
= tan 

2 
 1− x 
 1 − tan 2 θ 
= tan −1 ( tan 2 θ ) = 2 θ
.....(iv)
From (i), (ii), (iii) and (iv), we get
2

 2x 
−
1 1− x
2tan −1 x = sin −1 
=
cos

 =tan

1 + x2 
 1+ x2 
1−
2x 


1 − x 2 
Property 7
(i)
sin
−1
x = cos
−1
(
1− x
2
) = tan − 1 

2 
 1− x 
x
1 
−1 
= sec 
2 
 1− x 
 1− x2 

= cot 
 x 
−1
1 
= cosec−1  
x 
150
MATHEMATICS
Inverse Trigonometric Functions
cos −1 x = sin −1 (
(ii)
MODULE - IV
Functions
 1− x2 

1 − x 2 ) = tan − 1 
 x 
1 
−1 
= cosec 
2 
 1−x 
Notes
x 
−1 
= cot 
2 
 1− x 
 1
= sec− 1  
 x
Proof : Let sin −1 x = θ ⇒ sin θ = x
(i) cos θ = 1 − x 2 , tan θ =
∴
sin
−1
x = θ = cos
(
−1
x
, sec θ =
1 − x2
1− x
2
) = tan −1 
1
1− x2
, cot θ =
and cosec θ =
x
x
1− x2
1


 1− x 
x
2
1 
−1 
= sec 
2 
 1− x 
 1− x2 

= cot −1 
 x 
1
= cosec −1  
x
(ii) Let cos −1 x = θ ⇒
∴
and
x = cos θ
sin θ = 1 − x ,
2
cosec θ =
tan θ =
1− x2
,
x
sec θ =
1
,
x
cot θ =
x
1− x2
1
1− x2
cos −1 x = sin −1 ( 1 − x 2 )
 1− x2 

= tan 
 x 
−1
1 
−1 
= cosec 
2 
 1− x 
1
= sec−1  
x
MATHEMATICS
151
Inverse Trigonometric Functions
MODULE - IV
Functions
Notes
Property 8
(i)
sin −1 x + sin −1 y = sin − 1  x 1− y2 + y 1− x2 
(ii)
cos −1 x + cos −1 y = cos −1  xy − 1 − x 2 1 − y 2 
(iii)
sin −1 x − sin
(iv)
cos −1 x − cos
−1
y =sin
−1
y =cos
1− 
2
2
 x 1 −y −y 1 x− 
1− 
2
2
 xy + 1 −x 1 −y 
Proof (i) : Let x = sin θ , y = sin φ , then
L.H.S. = θ + φ
R.H.S. = sin −1 (sin θ cos φ + cos θ sin φ )
= sin −1 sin (θ + φ ) = θ + φ
∴
L.H.S. = R.H.S.
x = cos θ and y = cos φ
(ii) Let
L.H.S. = θ + φ
R.H.S. = cos −1 (cos θ cos φ − sin θ sin φ)
= cos −1  cos ( θ + φ ) = θ + φ
∴
L.H.S. = R.H.S.
(iii) Let x = sin θ , y = sin φ
L.H.S. = θ − φ
R.H.S. = sin −1  x 1− y 2 − y 1− x2 
= sin −1 sin θ 1 − sin 2 φ − sin φ 1 − sin 2 θ 
= sin −1 [ sin θcos φ− cos θsin φ]
= sin −1 sin (θ − φ ) = θ − φ
∴
L.H.S. = R.H.S.
x = cos θ , y = cos φ
L.H.S = θ − φ
(iv) Let
∴
R.H.S. = cos −1 [cos θ cos φ + sin θ sin φ]
= cos −1  cos (θ − φ ) = θ − φ
∴
152
L.H.S. = R.H.S.
MATHEMATICS
Inverse Trigonometric Functions
Example 18.5
MODULE - IV
Functions
Evaluate :
3
5

cos sin −1 + sin −1 

5
13 
 3
−1  5 
Soluton : Let sin−1   = θ and sin   = φ , then
 13 
 5
⇒
sin θ =
3
5
and sin φ =
5
13
cos θ =
4
12
and cos φ =
5
13
Notes
∴ The given expression becomes cos [θ + φ ]
= cos θ cos φ − sin θ sin φ
=
4 12 3 5
33
⋅ − ⋅
=
5 13 5 13
65
Example 18.6 Prove that
1
 1
 2
tan −1   + tan −1   = tan −1  
7
 13 
9
Solution : Applying the formula :
 x+y 
tan −1 x + tan −1 y = tan −1 
 , we have
 1 − xy 
 1 1 
+
 20 
2
−1  1 
−1  1 
− 1  7 13 
tan   + tan   = tan 
= tan −1   = tan −1  

7
 13 
 90 
9
 1 − 1 × 1 
 7 13 
Example 18.7 Prove that
 4
 12 
 33 
cos −1   + cos −1   = cos −1  
5
 13 
 65 
Applying the property
(
)
cos −1 x + cos −1 y = cos −1 xy − 1 − x 2 1 − y 2 , we have
 4 12
16
144 
 4
 12 
cos −1   + cos −1   = cos −1  × − 1 −
1−

5
 13 
25
169 
 5 13
 33 
= cos −1  
 65 
MATHEMATICS
153
Inverse Trigonometric Functions
MODULE - IV
Functions
Example 18.8 Prove that
tan −1
Solution : Let
x=
1
 1− x 
cos −1 

2
 1+ x 
x = tan θ then
2
1
1
−1  1 − tan θ 
cos
= cos−1 ( c o s 2 θ )
L.H.S. = θ and R.H.S. =

2 
2
 1 + tan θ  2
Notes
∴
1
= ×2θ=θ
2
L.H.S. = R.H.S.
Example 18.9 Solve the equation
 1− x  1
−1
tan −1 
 = tan x , x > 0
 1+ x  2
Solution : Let x = tan θ , then
 1 − tan θ  1
−1
tan −1 
 = tan ( tan θ)
 1 + tan θ  2
⇒
 π
 1
tan −1  tan  − θ   = θ
 2
 4
π
1
−θ= θ
4
2
⇒
π 2 π
× =
4 3 6
⇒
θ=
∴
 π 1
x = tan   =
6
3
Example 18.10 Show that
 1+ x2 + 1− x2
tan −1 

2
2
 1 + x − 1− x
 π 1
 = + cos −1 ( x 2 )
 4 2

Solution : Let x 2 = cos2θ , then
2 θ = cos −1 ( x 2 )
θ=
⇒
1
cos −1 x 2
2
Substituting x 2 = c o s 2 θ in L.H.S. of the given equation, we have
 1 + x 2 + 1− x 2
tan 

2
2
 1 + x − 1− x
−1
154

 1 + cos2θ + 1 − c o s 2θ 
 = tan −1 

 1 + cos2θ − 1 − c o s 2θ 



MATHEMATICS
Inverse Trigonometric Functions
= tan
MODULE - IV
Functions
−1 
2cos θ+ 2sin θ 


 2cos θ− 2sin θ 
 cos θ + sin θ 
= tan −1 

 cos θ − sin θ 
 1 + tan θ 
= tan −1 

 1 − tan θ 
Notes
 π

= tan −1  tan  + θ  

 4
=
π
+θ
4
=
π 1
+ cos −1 ( x 2 )
4 2
CHECK YOUR PROGRESS 18.2
1.
2.
Evaluate each of the following :
(a)
π
 1 
sin  − sin −1  − 
 2 
3
(c)
2 
1
2x
−1 1 − y
tan  sin −1
+
cos

2
1+ x2
1 + y2 
(d)
1

tan  2tan −1 

5
(b)
(e)
cot ( tan −1 α + cot −1 α )
1 π

tan  2tan −1 − 

5 4
If cos −1 x + cos −1 y = β , prove that
x 2 − 2xycos β + y 2 = sin 2 β
3.
−1
−1
−1
If cos x + cos y + cos z = π , prove that
x 2 + y 2 + z 2 + 2xyz = 1
4.
Prove each of the following :
1
2
π
+ sin − 1
=
5
5 2
(a)
sin −1
(c)
4
3
27
cos −1 + tan −1 = tan −1
(d)
5
5
11
MATHEMATICS
(b)
4
5
16 π
sin −1 + sin −1 + sin −1
=
5
13
65 2
tan −1
1
1
1 π
+ tan −1 + tan −1 =
2
5
8 4
155
Inverse Trigonometric Functions
MODULE - IV
Functions
LET US SUM UP
LET US SUM UP
l
Notes
l
l
Inverse of a trigonometric function exists if we restrict the domain of it.
π
π
(i) sin −1 x = y if siny = x where −1 ≤ x ≤ 1, − ≤ y ≤
2
2
(ii) cos −1 x = y if cos y = x where −1 ≤ x ≤ 1, 0 ≤ y ≤π
π
π
(iii) tan −1 x = y if tan y = x where x ∈ R, − < y <
2
2
−
1
(iv) cot x = y if cot y = x where x ∈ R , 0 < y < π
π
π
(v) sec −1 x = y if sec y = x where x ≥ 1, 0≤ y<
or x ≤ − 1, < y ≤ π
2
2
π
(vi) cosec −1 x = y if cosec y = x where x ≥ 1, 0 < y ≤
2
π
x ≤ −1, − ≤ y < 0
or
2
Graphs of inverse trigonometric functions can be represented in the given intervals by
interchanging the axes as in case of y = sin x, etc.
Properties :
(i)
(ii)
−1
sin −1 (sin θ ) = θ , tan −1 ( tan θ ) = θ, tan (tan −1 θ ) = θ and sin (sin θ ) = θ
1
1 
−1
−1  1 
cosec −1 x = sin−1   , cot −1 x = tan −1   , sec x = cos  
x
x
x 
(iii) sin −1 (− x ) = − sin −1 x , tan −1 ( − x) = − tan −1 x , cos −1 ( −x ) = π − cos −1 x
π
π
π
−1
−1
−1
−1
−1
−1
(iv) sin x + cos x = , tan x + cot x = , cosec x + sec x =
2
2
2
(v)
 x+y 
tan −1 x + tan −1 y = tan −1 
, tan −1 x − tan

 1 − xy 
−1
y =tan
x−y 
 1 + xy 


−
1
2 

(vi) 2tan −1 x = sin −1  2x  = cos−1  1 − x  = tan−1  2x 
 1+ x2 
 1 + x2 
 1 − x2 
(vii) sin −1 x = cos −1
(
1 −x 2
 1
= sec−1 
 1− x2
)
=tan
−1 
x 

2 
 1−x 

−1  1 − x 2
 = cot 

 x

−1  1 
 = cosec  
x

−1
−1
−1
2
2
(viii) sin x ± sin y = sin  x 1− y ± y 1− x 
(ix) cos −1 x ± cos −1 y = cos 1−  xy m y 1 −x 2 1 −y2 
156
MATHEMATICS
Inverse Trigonometric Functions
MODULE - IV
Functions
SUPPORTIVE WEB SITES
l
l
http://www.wikipedia.org
http://mathworld.wolfram.com
Notes
TERMINAL EXERCISE
1.
2.
3.
Prove each of the following :
(a)
 3
 8
 77 
1
1 1
 3
sin −1   + sin −1   = sin − 1   (b) tan −1   + tan −1   = cos− 1  
 5
 17 
 85 
4
9 2
 5
(c)
 4
 3
 27 
cos −1   + tan −1   = tan −1  
5
 5
 11 
Prove each of the following :
(a)
1
 1
 23 
2tan −1   + tan −1   = tan −1   (b)
2
 5
 11 
(c)
1
 1
1
tan −1   + tan − 1   = tan − 1  
8
 5
 3
1
tan −1   + 2tan
2
−1  1 
  = tan
 3
1−
2
(a) Prove that 2sin −1 x = sin − 1 ( 2x 1− x2 )
(b) Prove that 2cos −1 x = cos −1 ( 2x 2 − 1)
4.
5.
 1− x
(c) Prove that cos −1 x = 2sin −1 
2

Prove the following :

−1  1 + x 
 = 2cos 

2 


(a)
 cosx  π x
tan −1 
= −
 1 + sin x  4 2
(c)
 ab + 1 
−1  bc + 1 
−1  ca + 1 
cot −1 
 + cot 
 + cot 
= 0
 a −b 
 b− c 
 c−a 
(b)
 cosx − sin x  π
tan −1 
= −x
 cosx + sin x  4
Solve each of the following :
π
4
(a)
tan −1 2x + tan −1 3x =
(c)
1  π
cos −1 x + sin −1  x  =
2  6
MATHEMATICS
(b) 2tan −1 ( cosx ) = tan −1 ( 2cosecx)
−1
−1
(d) cot x − cot (x + 2 ) =
π
,x>0
12
157
Inverse Trigonometric Functions
MODULE - IV
Functions
ANSWERS
CHECK YOUR PROGRESS 18.1
1.
(a)
π
6
(b) −
(a)
1
3
(b)
Notes
2.
π
4
(c) −
π
4
(c)
π
3
1
2
(d) −
π
3
(a) 1 + x 2
(b)
x −4
2
π
4
(e) −2
(d) 1
x2
2
3.
(e)
(c)
x4 −1
(d)
x4 +1
(e)
1− x
x
(e) −
7
17
CHECK YOUR PROGRESS 18.2
1.
(a) 1
(b) 0
(c)
x+y
1 − xy
(d)
5
12
(d)
3
TERMINAL EXERCISE
5.
158
(a)
1
6
(b)
π
4
(c) ±1
MATHEMATICS