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Chapter
2
INVERSE TRIGONOMETRIC
FUNCTIONS
Mathematics, in general, is fundamentally the science of
self-evident things
...
1 Introduction
In Chapter 1, we have studied that the inverse of a function
f, denoted by f –1, exists if f is one-one and onto
...
In Class XI, we
studied that trigonometric functions are not one-one and
onto over their natural domains and ranges and hence their
inverses do not exist
...
Besides, some elementary properties will also be discussed
...
D
...
The concepts of inverse trigonometric functions is also used in science and engineering
...
2 Basic Concepts
In Class XI, we have studied trigonometric functions, which are defined as follows:
sine function, i
...
, sine : R → [– 1, 1]
cosine function, i
...
, cos : R → [– 1, 1]
π
, n ∈ Z} → R
2
cotangent function, i
...
, cot : R – { x : x = nπ, n ∈ Z} → R
tangent function, i
...
, tan : R – { x : x = (2n + 1)
π
, n ∈ Z} → R – (– 1, 1)
2
cosecant function, i
...
, cosec : R – { x : x = nπ, n ∈ Z} → R – (– 1, 1)
secant function, i
...
, sec : R – { x : x = (2n + 1)
34
MATHEMATICS
We have also learnt in Chapter 1 that if f : X→Y such that f(x) = y is one-one and
onto, then we can define a unique function g : Y→X such that g (y) = x, where x ∈ X
and y = f (x), y ∈ Y
...
The function g is called the inverse of f and is denoted by f –1
...
Thus, g –1 = (f –1) –1 = f
...
If we restrict its domain to
2 , 2 , then it becomes one-one
and onto with range [– 1, 1]
...
, is one-one and its range is [–1, 1]
...
We denote the
inverse of sine function by sin–1 (arc sine function)
...
Corresponding to each such interval, we get a branch of the
−π π
function sin–1
...
When we refer
to the function sin–1, we take it as the function whose domain is [–1, 1] and range is
−π π
−π π
–1
2 , 2
...
In other words, if y = sin–1 x, then
2
2
sin y = x
...
Thus, the graph of sin–1 function can be obtained from the graph of original
function by interchanging x and y axes, i
...
, if (a, b) is a point on the graph of
sine function, then (b, a) becomes the corresponding point on the graph of inverse
INVERSE TRIGONOMETRIC FUNCTIONS
35
of sine function
...
The graphs of y = sin x and
y = sin–1 x are as given in Fig 2
...
The dark portion of the graph of
y = sin–1 x represent the principal value branch
...
e
...
This can be visualised by looking the graphs of y = sin x and
y = sin–1 x as given in the same axes (Fig 2
...
Fig 2
...
1 (ii)
Fig 2
...
If we restrict the domain of cosine function
to [0, π], then it becomes one-one and onto with range [–1, 1]
...
, is bijective with range as
[–1, 1]
...
We denote the inverse of the cosine function by cos–1 (arc cosine function)
...
Corresponding to each such interval, we get a branch of the
function cos–1
...
We write
cos–1 : [–1, 1] → [0, π]
...
The
graphs of y = cos x and y = cos–1 x are given in Fig 2
...
Fig 2
...
2 (ii)
Let us now discuss cosec–1x and sec–1x as follows:
1
Since, cosec x =
, the domain of the cosec function is the set {x : x ∈ R and
sin x
x ≠ nπ, n ∈ Z} and the range is the set {y : y ∈ R, y ≥ 1 or y ≤ –1} i
...
, the set
R – (–1, 1)
...
If we restrict the domain of cosec function to
π π
− 2 , 2 – {0}, then it is one to one and onto with its range as the set R – (– 1, 1)
...
, is bijective and its range is the set of all real numbers R – (–1, 1)
...
The
be any of the intervals , − {0} ,
2 2
2 2
2 2
−π π
function corresponding to the range , − {0} is called the principal value branch
2 2
of cosec–1
...
3 (i), (ii)
...
3 (i)
Fig 2
...
It means that sec (secant function) assumes
Also, since sec x =
all real values except –1 < y < 1 and is not defined for odd multiples of
restrict the domain of secant function to [0, π] – {
π
...
Actually, secant function restricted to any of the
−π
3π
π
intervals [–π, 0] – {
}, [0, π] – , [π, 2π] – {
} etc
...
Thus sec–1 can be defined as a function whose domain is R– (–1, 1) and
−π
π
3π
range could be any of the intervals [– π, 0] – {
}, [0, π] – { }, [π, 2π] – { } etc
...
π
The branch with range [0, π] – { } is called the principal value branch of the
2
function sec–1
...
4 (i), (ii)
...
4 (i)
Fig 2
...
It means that tan function
2
π
is not defined for odd multiples of
...
Actually, tangent function
−3 π −π −π π π 3 π
, , , , ,
restricted to any of the intervals
etc
...
Thus tan–1 can be defined as a function whose domain is R and
−3π −π −π π π 3π
, , , , , and so on
...
The branch with range ,
2 2
is called the principal value branch of the function tan–1
...
5 (i), (ii)
...
5 (i)
Fig 2
...
It means that cotangent function is not
defined for integral multiples of π
...
In fact, cotangent function restricted
to any of the intervals (–π, 0), (0, π), (π, 2π) etc
...
Thus
cot –1 can be defined as a function whose domain is the R and range as any of the
40
MATHEMATICS
intervals (–π, 0), (0, π), (π, 2π) etc
...
The function with range (0, π) is called the principal value branch of
the function cot –1
...
6 (i), (ii)
...
6 (i)
Fig 2
...
sin–1
:
[–1, 1]
→
π π
− 2 , 2
cos –1
:
[–1, 1]
→
[0, π]
cosec–1
:
R – (–1,1)
→
π π
− 2 , 2 – {0}
sec –1
:
R – (–1, 1) →
π
[0, π] – { }
2
tan–1
:
R
→
−π π
,
2 2
cot–1
:
R
→
(0, π)
INVERSE TRIGONOMETRIC FUNCTIONS
41
Note
1
1
...
In fact (sin x) –1 =
and
sin x
similarly for other trigonometric functions
...
Whenever no branch of an inverse trigonometric functions is mentioned, we
mean the principal value branch of that function
...
The value of an inverse trigonometric functions which lies in the range of
principal branch is called the principal value of that inverse trigonometric
functions
...
2
1
1
Solution Let sin–1
...
Then, sin y =
2
2
−π π
, and
We know that the range of the principal value branch of sin–1 is
2 2
1
1
π
π
sin =
...
Then,
3
π
2π
−1
π
cot y =
= − cot = cot π − = cot
3
3
3
3
We know that the range of principal value branch of cot –1 is (0, π) and
−1
2π
2 π −1
cot =
...
1
Find the principal values of the following:
1
1
...
cos–1 2
3
...
tan–1 ( − 3)
1
5
...
tan–1 (–1)
42
MATHEMATICS
2
7
...
cos–1 −
2
8
...
cosec–1 ( − 2 )
Find the values of the following:
1
1
1
2
11
...
cos–1 + 2 sin–1
2
2
13
...
tan–1
3 − sec−1 ( − 2 ) is equal to
(A) π
(B) −
π
3
(C)
π
3
(D)
2π
3
2
...
It may be mentioned here that these results are valid within the principal
value branches of the corresponding inverse trigonometric functions and wherever
they are defined
...
In fact, they will be valid only for some values of x for which
inverse trigonometric functions are defined
...
Let us recall that if y = sin–1x, then x = sin y and if x = sin y, then y = sin–1 x
...
We now prove
some properties of inverse trigonometric functions
...
(i) sin–1
(ii) cos–1
1
= cosec–1 x, x ≥ 1 or x ≤ – 1
x
1
= sec –1x, x ≥ 1 or x ≤ – 1
x
INVERSE TRIGONOMETRIC FUNCTIONS
1
= cot–1 x, x > 0
x
To prove the first result, we put cosec–1 x = y, i
...
, x = cosec y
1
Therefore
= sin y
x
1
Hence
sin–1 = y
x
1
or
sin–1
= cosec–1 x
x
Similarly, we can prove the other parts
...
(i) sin–1 (–x) = – sin–1 x, x ∈ [– 1, 1]
(ii) tan–1 (–x) = – tan–1 x, x ∈ R
(iii) cosec–1 (–x) = – cosec–1 x, | x | ≥ 1
Let sin–1 (–x) = y, i
...
, –x = sin y so that x = – sin y, i
...
, x = sin (–y)
...
3
...
e
...
(iii) tan–1
4
...
Then x = sin y = cos − y
2
π
π
−y =
− sin –1 x
Therefore
cos–1 x =
2
2
43
44
MATHEMATICS
Hence
sin–1 x + cos–1 x =
π
2
Similarly, we can prove the other parts
...
(i) tan–1 x + tan–1 y = tan–1
(ii) tan–1 x – tan–1 y = tan–1
x+ y
, xy < 1
1 – xy
x–y
, xy > – 1
1 + xy
Let tan–1 x = θ and tan–1 y = φ
...
6
...
Now
sin–1
2 tan y
2x
–1
2 = sin
1 + tan 2 y
1+ x
= sin–1 (sin 2y) = 2y = 2tan–1 x
INVERSE TRIGONOMETRIC FUNCTIONS
Also cos–1
1 − tan 2 y
1 − x2
= cos–1
= cos–1 (cos 2y) = 2y = 2tan–1 x
1 + tan 2 y
1 + x2
(iii) Can be worked out similarly
...
Example 3 Show that
1
1
≤ x≤
(i) sin–1 2x 1− x 2 = 2 sin–1 x, −
2
2
1
≤ x ≤1
(ii) sin–1 2x 1 − x2 = 2 cos–1 x,
2
Solution
(
)
(
)
(i) Let x = sin θ
...
We have
(
)
(
sin–1 2x 1− x 2 = sin–1 2sin θ 1 − sin 2 θ
)
= sin–1 (2sinθ cosθ) = sin–1 (sin2θ) = 2θ
= 2 sin–1 x
(ii) Take x = cos θ, then proceeding as above, we get, sin–1 ( 2 x 1 − x 2 ) = 2 cos–1 x
1
2
3
+ tan–1 = tan–1
2
11
4
Solution By property 5 (i), we have
Example 4 Show that tan–1
1 2
+
1
2
15
−1 3
–1 2 11
tan –1 + tan –1
L
...
S
...
H
...
1 2
4
2
11
20
1− ×
2 11
cos x − 3π
π
< x < in the simplest form
...
x −1
Solution Let x = sec θ, then
x2 − 1 =
sec2 θ − 1 = tan θ
INVERSE TRIGONOMETRIC FUNCTIONS
–1
Therefore, cot
1
= cot–1 (cot θ) = θ = sec–1 x, which is the simplest form
...
Then θ = tan–1 x
...
H
...
= tan –1
= tan
2
2
1− 3x
1 − 3tan θ
= tan–1 (tan3θ) = 3θ = 3tan–1 x = tan–1 x + 2 tan–1 x
2x
= L
...
S
...
2
Prove the following:
1 1
1
...
3cos–1 x = cos–1 (4x3 – 3x), x ∈ , 1
2
3
...
2
7
1
+ tan −1 = tan −1
11
24
2
2 tan −1
1
1
31
+ tan −1 = tan −1
2
7
17
Write the following functions in the simplest form:
1+ x 2 − 1
,x≠0
x
5
...
1 − cos x
tan −1
1 + cos x
47
,0
1
6
...
cos x − sin x
tan −1
, 0 < x< π
cos x + sin x
x2 −1
, |x | > 1
48
9
...
MATHEMATICS
x
tan −1
2
a − x2
, |x | < a
3a 2 x − x3
−a
a
2 , a > 0;
3
3
a − 3ax
Find the values of each of the following:
11
...
1
2x
1 − y2
tan sin –1
+ cos –1
, | x | < 1, y > 0 and xy < 1
2
1+ x 2
1 + y2
12
...
If sin sin
5
x −1
x +1 π
+ tan –1
= , then find the value of x
x−2
x+2 4
Find the values of each of the expressions in Exercises 16 to 18
...
If tan
16
...
19
...
3π
tan–1 tan
4
(C)
π
3
(D)
(C)
1
4
(D) 1
3
3
tan sin –1 + cot –1
5
2
7π
cos −1 cos is equal to
6
(A)
20
...
7π
6
(B)
5π
6
1
π
sin − sin −1 ( − ) is equal to
3
2
1
1
(A)
(B)
2
3
π
6
tan −1 3 − cot −1 ( − 3 ) is equal to
(A) π
(B) −
π
2
(C) 0
(D) 2 3
INVERSE TRIGONOMETRIC FUNCTIONS
Miscellaneous Examples
−1
Example 9 Find the value of sin (sin
3π
)
5
−1
Solution We know that sin −1 (sin x) = x
...
e
...
e
...
e
...
6
Since x = – 1 does not satisfy the equation, as the L
...
S
...
6
Miscellaneous Exercise on Chapter 2
Find the value of the following:
13π
cos –1 cos
6
Prove that
1
...
7π
tan–1 tan
6
4
...
cos –1
12
3
56
+ sin –1 = sin –1
13
5
65
3
...
4
12
33
cos –1 + cos –1 = cos –1
5
13
65
7
...
1
1
1
1 π
tan –1 + tan −1 + tan −1 + tan −1 =
5
7
3
8 4
63
5
3
= sin –1 + cos –1
16
13
5
52
MATHEMATICS
Prove that
9
...
1 + sin x + 1 − sin x x
π
x ∈ 0,
cot –1
1 + sin x − 1 − sin x = 2 ,
4
11
...
9π 9
1 9
2 2
− sin −1 = sin −1
8 4
3 4
3
π 1
1
–1
≤ x ≤ 1 [Hint: Put x = cos 2θ]
= − cos x , −
4 2
2
Solve the following equations:
13
...
tan –1
1− x 1
= tan –1 x, ( x > 0)
1+ x 2
15
...
sin–1 (1 – x) – 2 sin –1 x =
1 − x2
1 + x2
x
(D)
1 + x2
π
, then x is equal to
2
(A) 0,
17
...
In fact (sin x) –1 =
1
and
sin x
similarly for other trigonometric functions
...
For suitable values of domain, we have
y = sin–1 x ⇒ x = sin y
x = sin y ⇒ y = sin–1 x
sin (sin–1 x) = x
sin–1 (sin x) = x
sin–1
1
= cosec–1 x
x
cos–1 (–x) = π – cos–1 x
cos–1
1
= sec–1x
x
cot–1 (–x) = π – cot–1 x
tan–1
1
= cot–1 x
x
sec–1 (–x) = π – sec–1 x
53
54
MATHEMATICS
sin–1 (–x) = – sin–1 x
tan–1 (–x) = – tan–1 x
tan–1 x + cot–1 x =
π
2
cosec–1 (–x) = – cosec–1 x
sin–1 x + cos–1 x =
π
2
cosec–1 x + sec–1 x =
x+ y
tan–1 x + tan–1y = tan–1
1 − xy
2tan–1x = tan –1
π
2
2x
1 − x2
x−y
tan–1 x – tan–1 y = tan–1
2tan–1 x = sin–1
1 + xy
2x
1 − x2
= cos–1
1 + x2
1 + x2
Historical Note
The study of trigonometry was first started in India
...
D
...
D
...
D
...
D
...
All
this knowledge went from India to Arabia and then from there to Europe
...
In India, the predecessor of the modern trigonometric functions, known as
the sine of an angle, and the introduction of the sine function represents one of
the main contribution of the siddhantas (Sanskrit astronomical works) to
mathematics
...
D
...
A sixteenth century Malayalam work Yuktibhasa
contains a proof for the expansion of sin (A + B)
...
, were given by Bhaskara II
...
, for arc sin x, arc cos x, etc
...
W
...
C
...
He
is credited with the determination of the height of a great pyramid in Egypt by
measuring shadows of the pyramid and an auxiliary staff (or gnomon) of known
INVERSE TRIGONOMETRIC FUNCTIONS
55
height, and comparing the ratios:
H h
= = tan (sun’s altitude)
S s
Thales is also said to have calculated the distance of a ship at sea through
the proportionality of sides of similar triangles
...
—
—