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Chapter 2
INVERSE TRIGONOMETRIC
FUNCTIONS
2.1 Overview
2.1.1 Inverse function
Inverse of a function ‘ f ’ exists, if the function is one-one and onto, i.e, bijective.
Since trigonometric functions are many-one over their domains, we restrict their
domains and co-domains in order to make them one-one and onto and then find
their inverse. The domains and ranges (principal value branches) of inverse
trigonometric functions are given below:
Functions Domain Range (Principal value
branches)
y = sin–1 x [–1,1]
–π π , 2 2
y = cos
y = cosec–1 x R – (–1,1)
–π π , – {0} 2 2
y = sec–1 x R – (–1,1) [0,π] –
π
y = tan
–π π , 2 2
y = cot
Notes:
(i) The symbol sin
- x should not be confused with (sin x ) - . Infact sin - x is an
angle, the value of whose sine is x , similarly for other trigonometric functions.
(ii) The smallest numerical value, either positive or negative, of θ is called the
principal value of the function.
INVERSE TRIGONOMETRIC FUNCTIONS 19
(iii) Whenever no branch of an inverse trigonometric function is mentioned, we mean
the principal value branch. The value of the inverse trigonometic function which
lies in the range of principal branch is its principal value.
2.1.2 Graph of an inverse trigonometric function
The graph of an inverse trigonometric function can be obtained from the graph of
original function by interchanging x- axis and y -axis, i.e, if ( a , b ) is a point on the graph
of trigonometric function, then ( b , a ) becomes the corresponding point on the graph of
its inverse trigonometric function.
It can be shown that the graph of an inverse function can be obtained from the
corresponding graph of original function as a mirror image (i.e., reflection) along the
line y = x.
2.1.3 Properties of inverse trigonometric functions
1. sin–1^ (sin x ) = x :
x
^
cos
(cos x ) = x : x ∈[0, ]
tan
–π π , 2 2
x
cot
- (cot x ) = x : x ∈( 0, π)
sec
π [0, π] – 2
x
cosec–1(cosec x ) = x :
–π π , – {0} 2 2
x
2. sin (sin - x ) = x : x ∈[–1,1]
cos (cos–1^ x ) = x : x ∈[–1,1]
tan (tan–1^ x ) = x : x ∈ R
cot (cot
sec (sec
cosec (cosec
sin cosec x x
: x ∈ R – (–1,1)
cos sec x x
: x ∈ R – (–1,1)
INVERSE TRIGONOMETRIC FUNCTIONS 21
Solution If cos
=^ θ^ , then cos^ θ^ =^
Since we are considering principal branch, θ ∈ [0, π]. Also, since
0, θ being in
the first quadrant, hence cos
π
Example 2 Evaluate tan
–π sin 2
^
Solution tan–^
–π sin 2
^
^
= tan–^
π sin 2
^
= tan–1(–1) =
π
Example 3 Find the value of cos
13 π cos 6
Solution cos–^
13 π cos 6
= cos–1^ cos (2^ ) 6
π π+
–1 π cos cos 6
π .
Example 4 Find the value of tan
9 π tan 8
Solution tan–^
9 π tan 8
= tan–1^ tan 8
π π +
tan tan 8
π ^
π
Example 5 Evaluate tan (tan
Solution Since tan (tan–1 x ) = x, (^) ∀ x ∈ R, tan (tan–1(– 4) = – 4.
Example 6 Evaluate: tan
22 MATHEMATICS
Solution tan–1^3 – sec–1^ (– 2) = tan–1^3 – [π – sec–12]
cos 3 2 3 3 3
π (^) π π π − π+ (^) = − + = −
Example 7 Evaluate:
sin cos sin 2
^
Solution
–1 –1 3 –1 π sin cos sin sin cos 2 3
^ ^
–1 1 π sin 2 6
Example 8 Prove that tan(cot
- x ) = cot (tan - x ). State with reason whether the
equality is valid for all values of x.
Solution Let cot
or,
π tan – θ = 2
x
–1 π tan – θ 2
x =
So
–1 π^ π –1 – tan(cot ) tan θ cot – θ cot cot cot(tan ) 2 2
x x x
The equality is valid for all values of x since tan
- x and cot - x are true for x ∈ R.
Example 9 Find the value of sec
tan 2
^ y
Solution Let
tan =θ 2
y , where
π π θ , 2 2
. So, tanθ = 2
y ,
which gives
2 4 secθ= 2
Therefore,
2 –1 4 sec tan = secθ = 2 2
y^ +^ y
Example 10 Find value of tan (cos
- x ) and hence evaluate tan
cos 17
Solution Let cos–1 x = θ, then cos θ = x, where θ ∈ [0,π]
24 MATHEMATICS
Long Answer (L.A.)
Example 13 Prove that 2sin
π
Solution Let sin
= θ, then sinθ =
, where θ ∈ , 2 2
−π^ π
Thus tan θ =
, which gives θ = tan
Therefore, 2sin
= 2θ – tan
= 2 tan
–1 –
tan – tan (^9 ) 1– 16
= tan–^
tan 7 31
tan 24 17
7 31
π
Example 14 Prove that
cot–17 + cot–18 + cot–118 = cot–1 3
Solution We have
cot–17 + cot–18 + cot–1 18
= tan–^
(since cot–1^ x = tan–^
x
, if x > 0)
–1 –
tan tan (^1 1 ) 1 7 8
− ×
(since x. y =
INVERSE TRIGONOMETRIC FUNCTIONS 25
–1 3 –1^1
tan tan 11 18
tan 3 1 1 11 18
− ×
(since xy < 1)
tan 195
tan 3
= cot
Example 15 Which is greater, tan 1 or tan
Solution From Fig. 2.1, we note that tan x is an increasing function in the interval
−π π
, since 1 > 4
π ⇒ tan 1 > tan 4
π
. This gives
tan 1 > 1
⇒ tan 1 > 1 > 4
π
⇒ tan 1 > 1 > tan
Example 16 Find the value of
sin 2 tan cos (tan 3) 3
Solution Let tan
= x and tan
- 3 =^ y^ so that tan^ x^ =^
and tan y = 3.
Therefore,
sin 2 tan cos (tan 3) 3
= sin (2 x ) + cos y
2 tan 1
1 tan (^1) tan
x
x (^) y
( )
2
INVERSE TRIGONOMETRIC FUNCTIONS 27
⇒ sin (sin
x
π −^ −
⇒ 6 x = – cos (sin–1^6 3 x )
⇒ 6 x = –
2 1 108− x. Squaring, we get
36 x
2 = 1 – 108 x
2
⇒ 144 x
2 = 1 ⇒ x = ±
Note that x = –
is the only root of the equation as x =
does not satisfy it.
Example 20 Show that
2 tan–^
–1 sin^ cos tan .tan tan 2 4 2 cos sin
α (^) π β (^) α β (^) −^ = α +^ β
Solution L.H.S. =
2 2
2 tan .tan 2 4 2 tan
1 tan tan 2 4 2
α π β −
α (^) π β − (^) −
–1 – 2
since 2 tan tan 1
x x x
2
1 tan 2 2 tan 2 1 tan 2 tan
1 tan 2 1 tan 2 1 tan 2
β − α
β
β − α ^ − (^) β (^) +
2
2 tan. 1 tan 2 2 tan
1 tan tan 1 tan 2 2 2
α (^) β −
β^ α^ β +^ −^ −
28 MATHEMATICS
2
2 2 2
2 tan 1 tan 2 2 tan
1 tan 1 tan 2 tan 1 tan 2 2 2 2
α (^) β −
β^ α^ β^ α +^ −^ +^ +
2
2 2
2
2 2
2 tan 1 tan 2 2
1 tan 1 tan 2 2 tan
1 tan 2 tan 2 2
1 tan 1 tan 2 2
α β −
α β
α β −
α β
–1 sin^ cos tan cos sin
α β α +^ β
= R.H.S.
Objective type questions
Choose the correct answer from the given four options in each of the Examples 21 to 41.
Example 21 Which of the following corresponds to the principal value branch of tan–1?
(A) ,
π^ π −
(B) ,
π^ π −
(C) ,
π π −
Solution (A) is the correct answer.
Example 22 The principal value branch of sec–1^ is
(A) ,^ { }^0 2 2
π^ π − −
(B) [^ 0,^ ] 2
π π −
(C) (0, π) (D) , 2 2
π π −
30 MATHEMATICS
(C)
x
(D)
2 1 x
x
Solution (D) is the correct answer. Let sin
⇒ cosec θ =
x
⇒ cosec
2 θ = (^2)
x
⇒ 1 + cot
2 θ = (^2)
x
⇒ cotθ =
2 1 x
x
Example 27 If tan–1 x = 10
π
for some x ∈ R , then the value of cot–1 x is
(A)
π (B)
π (C)
π (D)
π
Solution (B) is the correct answer. We know tan
π
. Therefore
cot
π
- 10
π
⇒ cot–1 x = 2
π
- 10
π =
π .
Example 28 The domain of sin
(A) [0, 1] (B) [– 1, 1]
(C)
(D) [–2, 2]
Solution (C) is the correct answer. Let sin
- 2 x = θ so that 2 x = sin θ.
Now – 1 ≤ sin θ ≤ 1, i.e.,– 1 ≤ 2 x ≤ 1 which gives
− ≤ x ≤ (^).
Example 29 The principal value of sin
is
INVERSE TRIGONOMETRIC FUNCTIONS 31
(A)
π − (^) (B) 3
π − (^) (C)
π (D)
π .
Solution (B) is the correct answer.
sin sin – sin – sin sin – 2 3 3 3
(^) − π (^) π (^) π =^ =^ = ^ ^ ^
Example 30 The greatest and least values of (sin
2
2 are respectively
(A)
2 2 5 and 4 8
π π (B) and 2 2
π −π
(C)
2 2
and 4 4
π −π (D)
2
and 0 4
π .
Solution (A) is the correct answer. We have
(sin
2
2 = (sin
2
2 –1 – 2sin sin 4 2
x x
π (^) π − (^) −
= (^) ( )
(^2 ) –1 – sin 2 sin 4
x x
π − π +
= (^ )^
2 –1 2 – 2 sin sin 2 8
x x
(^) π π −^ +
(^2 )
2 sin 4 16
x
(^) π π (^) − (^) +
Thus, the least value is
2 2
2 i.e. 16 8
(^) π π
and the Greatest value is
(^2 )
2 2 4 16
(^) −π π (^) π (^) −^ +
i.e.
2 5
π .
Example 31 Let θ = sin–1^ (sin (– 600°), then value of θ is
INVERSE TRIGONOMETRIC FUNCTIONS 33
(A) [–1, 1] (B) [–1, π + 1]
(C) (^) ( – ∞ ∞, (^) ) (D) φ
Solution (A) is the correct answer. The domain of cos is R and the domain of sin
[–1, 1]. Therefore, the domain of cos x + sin
- x is R ∩ (^) [ –1,1], i.e., [–1, 1].
Example 35 The value of sin (2 sin
(A) .48 (B) .96 (C) 1.2 (D) sin 1.
Solution (B) is the correct answer. Let sin
- (.6) = θ, i.e., sin θ = .6.
Now sin (2θ) = 2 sinθ cosθ = 2 (.6) (.8) = .96.
Example 36 If sin–1^ x + sin–1^ y = 2
π , then value of cos–1^ x + cos–1^ y is
(A)
π (B) π (C) 0 (D)
π
Solution (A) is the correct answer. Given that sin
π .
Therefore,
–1 –
x y
π^ π^ π +^ =
⇒ cos
π .
Example 37 The value of tan
–1 3 –1^1
cos tan 5 4
is
(A)
(B)
(C)
(D)
Solution (A) is the correct answer. tan
–1 3 –1^1
cos tan 5 4
= tan
–1 4 –1^1
tan tan 3 4
34 MATHEMATICS
= tan tan
tan tan (^4 1 8 ) 1 3 4
=^ =
^
− ×
Example 38 The value of the expression sin [cot–1^ (cos (tan–1^ 1))] is
(A) 0 (B) 1 (C)
(D)
Solution (D) is the correct answer.
sin [cot
π )] = sin [cot
]=
sin sin 3 3
Example 39 The equation tan
has
(A) no solution (B) unique solution
(C) infinite number of solutions (D) two solutions
Solution (B) is the correct answer. We have
tan–1 x – cot–1 x = 6
π and tan–1 x + cot–1 x = 2
π
Adding them, we get 2tan
π
⇒ tan
π i.e., (^) x = 3.
Example 40 If (^) α ≤ 2 sin–1 x + cos–1 x (^) ≤β , then
(A) ,
−π π α = β = (^) (B) α = 0,β = π
(C)
−π π α = β= (^) (D) α = 0, β = 2 π
36 MATHEMATICS
7. Find the real solutions of the equation
( )
tan 1 sin 1
x x + + x + x + = .
8. Find the value of the expression sin ( )
2 tan cos tan 2 2 3
9. If 2 tan - (cos θ) = tan - (2 cosec θ), then show that θ =
where n is any integer.
10. Show that
–1 1 –1^1
cos 2 tan sin 4 tan 7 3
11. Solve the following equation (^) ( )
cos tan sin cot 4
x
Long Answer (L.A.)
12. Prove that
2 2 –1 –1 2
2 2
tan cos
1 – 1–^4
x x
x
x x
=^ +
13. Find the simplified form of
cos cos sin
x x
, where x ∈
^
14. Prove that
–1 8 –1 3 –1^77
sin sin sin
15. Show that
–1 5 –1 3 –1^63
sin cos tan
16. Prove that
tan tan sin
−
17. Find the value of
–1 1 –1^1
4 tan – tan
INVERSE TRIGONOMETRIC FUNCTIONS 37
18. Show that
tan sin
and justify why the other value
is ignored?
19. If a 1
, a 2
, a 3
,... ,a n
is an arithmetic progression with common difference d , then
evaluate the following expression.
–1 –1 –1 –
1 2 2 3 3 4 –
tan tan tan tan ... tan (^1 1 1 1) n n
d d d d
a a a a a a a a
+^ +^ +^ +
+^ +^ +^ +
Objective Type Questions
Choose the correct answers from the given four options in each of the Exercises from
20 to 37 (M.C.Q.).
20. Which of the following is the principal value branch of cos - x?
(A)
–π π , 2 2
(B) (0, π)
(C) [0, π] (D) (0, π) –
π
21. Which of the following is the principal value branch of cosec–1 x?
(A)
–π π , 2 2
(B) [0, π] –
π
(C)
–π π , 2 2
(D)
–π π , 2 2
22. If 3tan–1^ x + cot–1^ x = π, then x equals
(A) 0 (B) 1 (C) –1 (D)
23. The value of sin–
cos
^
^
is
(A)
3 π
(B)
π (C) 10
π (D)
π
