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13 Invnerse Trignometric
Functions
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13. Inverse Trignometric Functions eLearn.Punjab 13. Inverse Trignometric Functions eLearn.Punjab

 -p p 
13.1 Introduction
restrict the domain of y = sinx to the interval  ,  , then the restricted function y = sinx,
 2 2
p p
- ≤ x ≤ is called the principal sine function; which is now one-to-one and hence will
We have been inding the values of trigonometric functions for given measures
of the angles. But in the application of trigonometry, the problem has also been the other 2 2
way round and we are required to ind the measure of the angle when the value of its
have an inverse as shown in igure 2.
trigonometric function is given. For this purpose, we need to have the knowledge of inverse
This inverse function is called the inverse sin function and is written as sin-1x or arc
trigonometric functions.
sinx.
In chapter 2, we have discussed inverse functions. We learned that only a one-to-
The Inverse sine Function is deined by:
one function will have an inverse. If a function is not one-to-one, it may be possible to restrict
y = sin-1x , if and only if x = sin y.
its domain to make it one-to-one so that its inverse can be found.
p p
where - ≤ y≤ and - 1 ≤ x ≤ 1
In this section we shall deine the inverse trigonometric functions.
2 2
13.2 The Inverse sine Function: Here y is the angle whose sine is x. The domain of the function

p p
The graph of y = sinx, -T< x < +T, is shown in the igure 1.
y = sin-1x is - 17 x 7 1, its range is - ≤ y≤
2 2

The graph of y = sin-1x is obtained by relecting the restricted portion of the graph of
y = sinx about the line y = x as shown in igure 3.
We notice that the graph of y = sinx is along the x - axis whereas the graph of y = sin-1x
is along the y - axis.

Note: It must be remembered that sin-1x ≠ (sin x)-1.

(i) sin -1 (ii) sin -1 (- )

3 1
Example 1: Find the value of
2 2

3
Solution: (i) We want to ind the angle y, whose sine is
2

p p
=
⇒ sin y -, ≤ ≤
3
y

p
We observe that every horizontal line between the lines y = 1 and y= -1 intersects the 2 2 2
graph ininitly many times. It follows that the sine function is not one-to-one.However, if we ⇒ y=
3

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We observe that every horizontal line between the lines y = 1 and y = -1 intersects the
3 p
)=
graph ininitly many times. It follows that the cosine function is not one-to-one. However, if
sin -1 (
2 3 we restrict the domain of y = cosx to the interval [0, p], then the restricted function y = cosx,
We want to ind the angle y whose sine is - 0 7x7p is called the principal cosine function; which is now one-to-one and hence will
1
(ii)

p p
2 have an inverse as shown in igure 5.
⇒ sin y =- , - ≤ y≤
1 This inverse function is called the inverse cosine function and is written as cos-1x or arc

p
2 2 2
cosx.
∴ - y= The Inverse Cosine Function is deined by:

p
6
y = cos-1x, if and only if x= cos y.
∴ sin -1 (- ) =
- where 0 7y7p and -17x 7 1.
1

Here y is the angle whose cosine is x . The domain of the function y = cos-1x is -17x 71
2 6

and its range is 07y7p .

13.3 The Inverse Cosine Function: The graph of y = cos-1x is obtained by relecting the restricted portion of the graph of
y = cos x about the line y = x as shown in igure 6.
The graph of y = cosx, -T< x < +T, is shown in the igure 4. We notice that the graph of y = cos x is along the x - axis whereas the graph of y = cos-1x
is along the y - axis .

Note: It must be remembered that cos-1x ≠ (cosx)-1

cos -1 1 cos -1 (- )
1
Example 2: Find the value of (i) (ii)
2

Solution: (i) We want to ind the angle y whose cosine is 1

⇒=
cos y 1, 0≤ y ≤p

⇒ y=
0

∴ cos -1 1 =
0

We want to ind the angle y whose cosine is -

1
(ii)
2

⇒ cos y =- , 0≤ y ≤p
1
2

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2p -p p
∴ y= function y = tanx < x < , is called the Principal tangent function; which is now one-to-
3 2 2
one and hence will have an inverse as shown in igure 8.
2p
∴ cos -1 (- ) =
-
1 This inverse function is called the inverse tangent function and is written as tan-1x or
2 3 arc tanx.
The Inverse Tangent Function is deined by:
13.4 Inverse Tangent Function: y = tan-1x , if and only if x = tan y.

p p
where - < y< and - ∞ < x < +∞ .
2 2

Here y is the angle whose tangent is x. The domain of the function y = tan-1x is -T< x <

p p
+T and its range is - < y<
2 2

The graph of y = tan-1x is obtained by relecting the restricted portion of the graph of
y = tanx about the line y = x as shown in igure 9.
We notice that the graph of y = tanx is along the x - axis whereas the graph of y = tanx
is along the y- axis.

Note: It must be remembered that tan-1x ≠ (tanx)-1 .

Example 3: Find the value of (i) tan -1 1 (ii) tan -1 (- 3)

Solution: (i) We want to find the angle y, whose tangent is 1

p p
⇒=
tan y 1, - < y<

p
2 2
⇒ y=

p
4
The graph of y = tanx, -T< x < +T, is shown in the igure 7. ∴ tan -1 1 =
We observe that every horizontal line between the lines y = 1 and y = -1 intersect the graph 4
ininitly many times. It follows that the tangent function is not one-to-one. (ii) We want to ind the angle y whose tangent is - 3
-p p p p
However, if we restrict the domain of y = Tanx to the interval < x < , then the restricted ⇒ tan y =- 3 - <y <
2 2 2 2
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which is one-to-one and has an inverse.

2p
∴ y= The Inverse Cosecant Function is deined by:
y = csc -1 x, if and only if x = csc y
2p
3
∴ tan -1 (- 3) =
p p
3 where - ≤ y≤ , y ≠ 0 and x ≥ 1
2 2
13.5 Inverse Cotangent, Secant and Cosecant Functions
The students should draw the graph of y = csc -1 by taking the relection of y = cscx in
These inverse functions are not used frequently and most of the calculators do not the line y = x. This is left an exercise for them.
even have keys for evaluating them. However, we list their deinitions as below:
i) Inverse Cotangent function: Note. While discusspjpihe Inverse Trigonometric Functions, we have seen that there are in
y = cotx, where 0 7x7p is called the Principal Cotangent Function, which is one-to- general, no inverses of Trigonometric Functions, but restricting their domain to
one and has an inverse. principal Functions, we have made them as functions.
The inverse cotangent function is deined by:
y = cot-1x , if and only if x = coty
Where 0 < y < p and - ∞ < x < +∞ 13.6 Domains and Ranges of Principal Trigonometric Function
The students should draw the graph of y = cot-1 x by taking the relection of y = cotx in and Inverse Trigonometric Functions.
the line y = x. This is left as an exercise for them.
ii) Inverse Secant function From the discussion on the previous pages we get the following table showing domains

p
and ranges of the Principal Trigonometric and Inverse Trigonometric Functions.
y = sec x, where 0 ≤ x ≤ p and x ≠ is called the Principal Secant Function, which is
2
Functions Domain Range
one-to-one and has an inverse.
y = sin x -p p -1 ≤ x ≤ 1
≤x≤
The Inverse Secant Function is deined by:
y = sec-1x. if and only if x = secy 2 2

p -1 ≤ x ≤ 1 -p p
where 0 ≤ y ≤ p , y≠ and x ≥ 1 y = sin -1 x ≤x≤
2 2 2
y = cos x 0≤ x ≤p -1 ≤ x ≤ 1
The students should draw the graph of y = sec-1x by taking the relection of y = secx in
the line y = x. This is left an exercise for them,
y = cos -1 x -1 ≤ x ≤ 1 0≤ x ≤p
iii) Inverse Cosecant Function

p p y = tan x -p p (-∞, ∞) or ℜ
y = csc x, where - ≤ y≤ and x ≠ 0 is called the Principal Cosecant Function, <x<
2 2 2 2

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sin a = ⇒ a = sin -1
-p p
5 5
(-∞, ∞) or ℜ
Thus
y = tan -1 x <x<
13 13
cos -1 = sin -1
2 2 12 5
Hence
y = cot x 0< x <p (-∞, ∞) or ℜ
13 13

0< x <p
Example 5: Find the value of
y = cot x
-1 (-∞, ∞) or ℜ
sin (cos -1 cos (tan -1 0) sec [sin -1 (- )] Solution:
3 1
i) ) ii) iii)
y = sec x p y ≤ -1 or y ≥ 1
[0, p ], x ≠
2 2
3
2 i) we irst ind the value of y, whose cosine is
2

y = sec x x ≥ -1 or x ≤ 1 p = 0≤ y ≤p
[0, p ], y ≠
3
-1 cos y ,

p
2
⇒ y=
2

y = csc x p p y ≤ -1 or y ≥ 1
[- , ], x ≠ 0
6
p
⇒ (cos -1 =
2 2 3
)
2 6
y = csc -1 x x ≤ -1 or x ≥ 1 p p p 1
[- , ], y ≠ 0 ∴ sin(cos -1 = )=
3
sin
2 2 2 6 2
ii) we irst ind the value of y, whose tangent is 0

p p
Example 4: Show that cos -1 sin -1 tan= - < y<
12 5
y 0,
13 13 2 2
Solution: Let cos -1 a
=
⇒ = a ⇒ y=
12 12
cos 0
⇒ =
13 13
(tan -1 0) 0
 12  ∴ =
cos(tan -1 0) =
∴ sin a = ± 1 - cos a = ± 1 -  
2
cos0 1
 13 
2

we irst ind the value of y, whose sine is -

1
iii)
= ± 1-
p p
144 2
sin y =- , - ≤ y≤
169 1
169 - 144
±= ±= ±= p
25 5 2 2 2
169 169 13 ⇒ -y =

p
6
a cos a is +ve and domain of a is [0, p], in which sine is +ve. ⇒ sin -1 (- ) =
1
-
2 6
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∴ sec[sin -1 (- )] =  -2   1 
cot -1 (-1) cosec -1   sin -1  - 
1 2
 3  2
vii) viii) ix)
2 3
Example: 6 Prove that the inverse trigonometric functions satisfy the following identities:
Without using table/ Calculator show that:
p p
2.
sin -1 x = cos -1 x cos--1 x sin -1 x = -
= sin -1 =
i)
tan -1 2cos -1 sin -1
and 5 5 4 24
i) ii)
p p
2 2
tan -1 x = cot--1 x = -
12 13 5 25
cot -1 x tan -1 x
=
ii)
cos -1 cot -1
and 4 4
iii)
p p
2 2
sec -1 x = csc--1 x = -
5 3
iii) csc -1 x and sec -1 x
2 2 3. Find the value of each expression:
Proof:
Consider the right triangle given in the igure Angles a and b are acute and complementary.  1   1  3
cos  sin -1  sec  cos -1  tan  cos -1 
 2  2
i) ii) iii)
p  
a +b =
csc ( tan -1 (-1) ) tan ( tan -1 (-1) )
⇒
2
 1 
sec  sin -1 (- ) 
p p  2 
2 iv) v) vi)
⇒ a =- b and b =- a ...(i)
sin ( tan -1 (-1) )
 1   1 
sin  sin -1 ( )  tan  sin -1 (- ) 
p
2 2
 2   2 
vii) viii) ix)
Now sin a= sin( - b=
) cos b= x (say)
2

=
∴ a sin -1 x = and b cos -1 x 13.7 Addition and Subtraction Formulas
Thus from (i) we have:

p p
sin -1 x = cos --1 x = -
1) Prove that:
cos -1 x sin -1 x
sin -1 A + sin
= -1
sin -1 ( A 1 - B 2 + B 1 - A2 )
and
2 2 B
In a similar way, we can derive the identities (ii) and (iii). Proof: Let sin -1 A = x ⇒ sin x = A
and sin -1 B = y ⇒ sin y =B
= ± 1 - sin 2 x = ± 1 - A2
Exercise 13.1
Now cos x

 p p
In sinx = A, domain =  - ,  , in which
1. Evaluate without using tables / calculator:
 2 2
 3
sin -1 (-1)
Cosine is +ve,
sin -1 (1) cos -1  
∴ cos x = -
i) ii) iii)
 2  1 A2
 1  1  1  Similarly, cos y= 1 - B2
tan -1  -  cos -1   tan -1  
 3 2  3
iv) vi)
Now sin( x += sin x cos y + cos x sin y
v)
y)
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= A 1 - B 2 + B 1 - A2 A+ A
tan -1 A + tan -1 A = tan -1
⇒ x=
+y sin -1 ( A 1 - B 2 - B 1 - A2 ) ∴ 1 - A2
⇒ 2 tan -1 A = tan -1
sin -1 A + sin
= B sin -1 ( A 1 - B 2 + B 1 - A2 )
2A
-1
1 - A2
In a similar way, we can prove that
Exercise 13.2
2) sin A - sin
-1
= B sin ( A 1 - B - B 1 - A )
-1 -1 2 2

3) cos -1 A + cos= B cos -1 ( AB - (1 - A 2 )(1 - B 2 ))

-1 Prove the following:

4) cos A - cos=
-1
B cos ( AB + (1 - A )(1 - B ))
-1 -1 sin -1 + sin -1 = cos -1 tan -1 + tan -1 =tan -1
5 7 253 1 1 9
2 2 1. 2.
13 25 325 4 5 19
5) Prove that:
 12 
A+ B = sin -1 =
=
13 
2 tan -1 : Let tan -1
2 12 2
tan -1 A + tan -1 B =
3.
tan -1
Hint x.and shown sin 2 x
1 - AB
3 13 3

Let tan -1 A = x ⇒ tan x = A p

tan -1 = 2cos -1 sin -1 + cot -1 3 =
Proof: 120 12 1
tan B = y
-1
⇒ tan y =B
4. 5.
and 119 13 5 4

tan x + tan y A+ B
tan(=
x + y) = sin -1 + sin -1 = - sin -1 =
1 - tan x tan y 1 - AB sin -1 sin -1 cos -1
Now 6.
3 8 77
7.
77 3 15
5 17 85 85 5 17
A+ B
⇒ x+ y =tan -1
1 - AB 8. cos -1
63
+ 2 tan -1 =
1
sin -1
3
65 5 5
A+ B
∴ tan -1 A + tan B =
tan -1
1 - AB 8 p
tan -1 + tan -1 - tan -1 =
3 3
9.
4 5 19 4
In a similar way, we can prove that

A- B  
tan -1 A - tan -1 B =  Hint : First add tan 4 + tan 5 and then proceed 
-1 3 -1 3
tan -1
1 + AB
6)

16 p
sin -1 + sin -1 + sin -1 =
Cor. Putting A - B in 4 5
A+ B
10.
tan -1 A + tan -1B =
5 13 65 2
tan -1
1 - AB
, we get

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tan -1 + tan -1 = tan -1 + tan -1

1 5 1 1
11.

1 p
11 6 3 2
2 tan -1 + tan -1 =
1
12.
3 7 4

13. Show that cos(sin -1 x) = 1 - x2

14. Show that sin(2cos -1 x=
) 2 x 1 - x2
15. Show that cos(2sin -1 x) = 1 - 2 x2
16. Show that tan -1 (- x) - =
tan -1 x
17. Show that sin -1 (- x)- =
sin -1 x
18. Show that cos -1 (- x) p cos -1 x
-=

tan(sin -1 x) =
x
1 - x2
19. Show that

Given that x = sin -1 , ind the values of following trigonometric functions: sinx, cosx,
1
20.
2

tanx, cotx, secx and cscx.

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