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Page 1
The inverse of a function B A f ? : exists if f is one-one onto i.e., a bijection and is given by
x y f y x f ? ? ?
?
) ( ) (
1
.
Consider the sine function with domain R and range [–1, 1]. Clearly this function is not a
bijection and so it is not invertible. If we restrict the domain of it in such a way that it becomes
one–one, then it would become invertible. If we consider sine as a function with domain
?
?
?
?
?
?
?
2
,
2
? ?
and co-domain [–1, 1], then it is a bijection and therefore, invertible. The inverse of
sine function is defined as x x ? ? ?
?
? ? sin sin
1
, where
?
?
?
?
?
?
? ?
2
,
2
? ?
? and ] 1 , 1 [ ? ? x .
Properties of Inverse Trigonometric Functions.
(1) Meaning of inverse function
(i) x ? ? sin ? ? ?
?
x
1
sin (ii) x ? ? cos ? ? ?
?
x
1
cos (iii) x ? ? tan ? ? ?
?
x
1
tan
(iv) x ? ? cot ? ? ?
?
x
1
cot (v) x ? ? sec ? ? ?
?
x
1
sec (vi) x ? ? cosec ? ? ?
?
x
1
cosec
(2) Domain and range of inverse functions
(i) If , sin x y ? then , sin
1
x y
?
? under certain condition.
; 1 sin 1 ? ? ? y but x y ? sin . 1 1 ? ? ? ? x
Again,
2
1 sin
?
? ? ? ? ? y y and
2
1 sin
?
? ? ? y y .
Keeping in mind numerically smallest angles or real numbers.
2 2
? ?
? ? ? ? y
These restrictions on the values of x and y provide us with the domain and range for the
function x y
1
sin
?
? .
i.e., Domain : ] 1 , 1 [ ? ? x
Range:
?
?
?
?
?
?
? ?
2
,
2
? ?
y
(ii) Let x y ? cos , then x y
1
cos
?
? , under certain conditions 1 cos 1 ? ? ? y
? 1 1 ? ? ? x
? ? ? ? ? y y 1 cos
0 1 cos ? ? ? y y
? ? ? ? y 0 {as cos x is a decreasing function in [ ? , 0 ];
hence 0 cos cos cos ? ? y ?
These restrictions on the values of x and y provide us the domain and range for the function
x y
1
cos
?
? .
Y
y = cos
–
1
x
O
(–1,
?/2)
X
(1, 0)
Y
(1, ?/2)
y = sin
–
1
x
O
(–1, –
?/2)
X
Page 2
The inverse of a function B A f ? : exists if f is one-one onto i.e., a bijection and is given by
x y f y x f ? ? ?
?
) ( ) (
1
.
Consider the sine function with domain R and range [–1, 1]. Clearly this function is not a
bijection and so it is not invertible. If we restrict the domain of it in such a way that it becomes
one–one, then it would become invertible. If we consider sine as a function with domain
?
?
?
?
?
?
?
2
,
2
? ?
and co-domain [–1, 1], then it is a bijection and therefore, invertible. The inverse of
sine function is defined as x x ? ? ?
?
? ? sin sin
1
, where
?
?
?
?
?
?
? ?
2
,
2
? ?
? and ] 1 , 1 [ ? ? x .
Properties of Inverse Trigonometric Functions.
(1) Meaning of inverse function
(i) x ? ? sin ? ? ?
?
x
1
sin (ii) x ? ? cos ? ? ?
?
x
1
cos (iii) x ? ? tan ? ? ?
?
x
1
tan
(iv) x ? ? cot ? ? ?
?
x
1
cot (v) x ? ? sec ? ? ?
?
x
1
sec (vi) x ? ? cosec ? ? ?
?
x
1
cosec
(2) Domain and range of inverse functions
(i) If , sin x y ? then , sin
1
x y
?
? under certain condition.
; 1 sin 1 ? ? ? y but x y ? sin . 1 1 ? ? ? ? x
Again,
2
1 sin
?
? ? ? ? ? y y and
2
1 sin
?
? ? ? y y .
Keeping in mind numerically smallest angles or real numbers.
2 2
? ?
? ? ? ? y
These restrictions on the values of x and y provide us with the domain and range for the
function x y
1
sin
?
? .
i.e., Domain : ] 1 , 1 [ ? ? x
Range:
?
?
?
?
?
?
? ?
2
,
2
? ?
y
(ii) Let x y ? cos , then x y
1
cos
?
? , under certain conditions 1 cos 1 ? ? ? y
? 1 1 ? ? ? x
? ? ? ? ? y y 1 cos
0 1 cos ? ? ? y y
? ? ? ? y 0 {as cos x is a decreasing function in [ ? , 0 ];
hence 0 cos cos cos ? ? y ?
These restrictions on the values of x and y provide us the domain and range for the function
x y
1
cos
?
? .
Y
y = cos
–
1
x
O
(–1,
?/2)
X
(1, 0)
Y
(1, ?/2)
y = sin
–
1
x
O
(–1, –
?/2)
X
i.e. Domain: ] 1 , 1 [ ? ? x
Range : ] , 0 [ ? ? y
(iii) If x y ? tan , then , tan
1
x y
?
? under certain conditions.
Here, R x R y ? ? ? tan ,
2 2
tan
? ?
? ? ? ? ? ? ? ? ? y y
Thus, Domain R x ? ;
Range ?
?
?
?
?
?
? ?
2
,
2
? ?
y
(iv) If , cot x y ? then x y
1
cot
?
?
under certain conditions, ; cot R x R y ? ? ?
? ? ? ? ? ? ? ? ? y y 0 cot
These conditions on x and y make the function, x y ? cot one-one
and onto so that the inverse function exists. i.e., x y
1
cot
?
? is
meaningful.
? Domain : R x ?
Range : ) , 0 ( ? ? y
(v) If , sec x y ? then , sec
1
x y
?
? where 1 | | ? x and
2
, 0
?
? ? ? ? y y
Here, Domain: ) 1 , 1 ( ? ? ? R x
Range:
?
?
?
?
?
?
? ?
2
] , 0 [
?
? y
(vi) If x y ? cosec , then x y
1
cosec
?
?
Where 1 | | ? x and 0 ,
2 2
? ? ? ? y y
? ?
Here, Domain ) 1 , 1 ( ? ? ? R
Range } 0 {
2
,
2
?
?
?
?
?
?
?
? ?
? ?
Function Domain (D) Range (R)
x
1
sin
?
1 1 ? ? ? x or ] 1 , 1 [ ?
2 2
?
?
?
? ? ? or
?
?
?
?
?
?
?
2
,
2
? ?
x
1
cos
?
1 1 ? ? ? x or ] 1 , 1 [ ? ? ? ? ? 0 or ] , 0 [ ?
x
1
tan
?
? ? ? ?? x i.e., R x ? or
) , ( ? ??
2 2
?
?
?
? ? ? or ?
?
?
?
?
?
?
2
,
2
? ?
x
1
cot
?
? ? ? ?? x i.e., R x ? or ? ? ? ? 0 or ) , 0 ( ?
y
O ?
x ?
y = tan
–
1
x
y = – ?/2
y =
? ?2
x
y = sec
–
1
x
y = ?/2
( –1, ?)
O (1,0
)
x
y = cosec
–
1
x
(1,
?/2)
y
O ? (–1, –
?)
y = ?
O
x ?
(0,
?/2)
y = cot
–
1
x
Page 3
The inverse of a function B A f ? : exists if f is one-one onto i.e., a bijection and is given by
x y f y x f ? ? ?
?
) ( ) (
1
.
Consider the sine function with domain R and range [–1, 1]. Clearly this function is not a
bijection and so it is not invertible. If we restrict the domain of it in such a way that it becomes
one–one, then it would become invertible. If we consider sine as a function with domain
?
?
?
?
?
?
?
2
,
2
? ?
and co-domain [–1, 1], then it is a bijection and therefore, invertible. The inverse of
sine function is defined as x x ? ? ?
?
? ? sin sin
1
, where
?
?
?
?
?
?
? ?
2
,
2
? ?
? and ] 1 , 1 [ ? ? x .
Properties of Inverse Trigonometric Functions.
(1) Meaning of inverse function
(i) x ? ? sin ? ? ?
?
x
1
sin (ii) x ? ? cos ? ? ?
?
x
1
cos (iii) x ? ? tan ? ? ?
?
x
1
tan
(iv) x ? ? cot ? ? ?
?
x
1
cot (v) x ? ? sec ? ? ?
?
x
1
sec (vi) x ? ? cosec ? ? ?
?
x
1
cosec
(2) Domain and range of inverse functions
(i) If , sin x y ? then , sin
1
x y
?
? under certain condition.
; 1 sin 1 ? ? ? y but x y ? sin . 1 1 ? ? ? ? x
Again,
2
1 sin
?
? ? ? ? ? y y and
2
1 sin
?
? ? ? y y .
Keeping in mind numerically smallest angles or real numbers.
2 2
? ?
? ? ? ? y
These restrictions on the values of x and y provide us with the domain and range for the
function x y
1
sin
?
? .
i.e., Domain : ] 1 , 1 [ ? ? x
Range:
?
?
?
?
?
?
? ?
2
,
2
? ?
y
(ii) Let x y ? cos , then x y
1
cos
?
? , under certain conditions 1 cos 1 ? ? ? y
? 1 1 ? ? ? x
? ? ? ? ? y y 1 cos
0 1 cos ? ? ? y y
? ? ? ? y 0 {as cos x is a decreasing function in [ ? , 0 ];
hence 0 cos cos cos ? ? y ?
These restrictions on the values of x and y provide us the domain and range for the function
x y
1
cos
?
? .
Y
y = cos
–
1
x
O
(–1,
?/2)
X
(1, 0)
Y
(1, ?/2)
y = sin
–
1
x
O
(–1, –
?/2)
X
i.e. Domain: ] 1 , 1 [ ? ? x
Range : ] , 0 [ ? ? y
(iii) If x y ? tan , then , tan
1
x y
?
? under certain conditions.
Here, R x R y ? ? ? tan ,
2 2
tan
? ?
? ? ? ? ? ? ? ? ? y y
Thus, Domain R x ? ;
Range ?
?
?
?
?
?
? ?
2
,
2
? ?
y
(iv) If , cot x y ? then x y
1
cot
?
?
under certain conditions, ; cot R x R y ? ? ?
? ? ? ? ? ? ? ? ? y y 0 cot
These conditions on x and y make the function, x y ? cot one-one
and onto so that the inverse function exists. i.e., x y
1
cot
?
? is
meaningful.
? Domain : R x ?
Range : ) , 0 ( ? ? y
(v) If , sec x y ? then , sec
1
x y
?
? where 1 | | ? x and
2
, 0
?
? ? ? ? y y
Here, Domain: ) 1 , 1 ( ? ? ? R x
Range:
?
?
?
?
?
?
? ?
2
] , 0 [
?
? y
(vi) If x y ? cosec , then x y
1
cosec
?
?
Where 1 | | ? x and 0 ,
2 2
? ? ? ? y y
? ?
Here, Domain ) 1 , 1 ( ? ? ? R
Range } 0 {
2
,
2
?
?
?
?
?
?
?
? ?
? ?
Function Domain (D) Range (R)
x
1
sin
?
1 1 ? ? ? x or ] 1 , 1 [ ?
2 2
?
?
?
? ? ? or
?
?
?
?
?
?
?
2
,
2
? ?
x
1
cos
?
1 1 ? ? ? x or ] 1 , 1 [ ? ? ? ? ? 0 or ] , 0 [ ?
x
1
tan
?
? ? ? ?? x i.e., R x ? or
) , ( ? ??
2 2
?
?
?
? ? ? or ?
?
?
?
?
?
?
2
,
2
? ?
x
1
cot
?
? ? ? ?? x i.e., R x ? or ? ? ? ? 0 or ) , 0 ( ?
y
O ?
x ?
y = tan
–
1
x
y = – ?/2
y =
? ?2
x
y = sec
–
1
x
y = ?/2
( –1, ?)
O (1,0
)
x
y = cosec
–
1
x
(1,
?/2)
y
O ? (–1, –
?)
y = ?
O
x ?
(0,
?/2)
y = cot
–
1
x
) , ( ? ??
x
1
sec
?
1 , 1 ? ? ? x x or ) , 1 [ ] 1 , ( ? ? ? ??
? ?
?
? ? ? ? 0 ,
2
or
?
?
?
?
?
?
? ?
?
?
?
?
?
?
? ?
,
2 2
, 0
x
1
cosec
?
1 , 1 ? ? ? x x or ) , 1 [ ] 1 , ( ? ? ? ??
2 2
, 0
?
?
?
? ? ? ? ? or
?
?
?
?
?
?
? ?
?
?
?
?
?
?
2
, 0 0 ,
2
? ?
(3) ? ? ?
?
) (sin sin
1
, Provided that
2 2
?
?
?
? ? ? , ? ? ?
?
) (cos cos
1
, Provided that
? ? ? ? 0
? ? ?
?
) (tan tan
1
, Provided that
2 2
?
?
?
? ? ? , ? ? ?
?
) (cot cot
1
, Provided that ? ? ? ? 0
? ? ?
?
) (sec sec
1
, Provided that
2
0
?
? ? ? or ? ?
?
? ?
2
, ) cosec ( cosec
1
? ? ?
?
Provided that 0
2
? ? ? ?
?
or
2
0
?
? ? ?
(4) , ) sin(sin
1
x x ?
?
Provided that 1 1 ? ? ? x , , ) cos(cos
1
x x ?
?
Provided that 1 1 ? ? ? x
tan , ) (tan
1
x x ?
?
Provided that ? ? ? ? ? x , ) cot(cot
1
x x ?
?
Provided that
? ? ? ? ? x
, ) sec(sec
1
x x ?
?
Provided that 1 ? ? ? ? ? x or ? ? ? x 1
, ) cosec ( cosec
–1
x x ? Provided that 1 ? ? ? ? ? x or ? ? ? x 1
(5) x x
1 1
sin ) ( sin
? ?
? ? ? x x
1 1
cos ) ( cos
? ?
? ? ? ? , x x
1 1
tan ) ( tan
? ?
? ? ?
x x
1 1
cot ) ( cot
? ?
? ? ? ? x x
1 1
sec ) ( sec
? ?
? ? ? ? x x
–1 1
cosec ) ( cosec ? ? ?
?
(6)
2
cos sin
1 1
?
? ?
? ?
x x , for all ] 1 , 1 [ ? ? x
2
cot tan
1 1
?
? ?
? ?
x x , for all R x ?
2
cosec sec
1 - 1
?
? ?
?
x x , for all ) , 1 [ ] 1 , ( ? ? ? ?? ? x
Important Tips
? Here; x x x
1 1 1
tan , cosec , sin
? ? ?
belong to I and IV Quadrant.
? Here; x x x
1 1 1
cot , sec , cos
? ? ?
belong to I and II Quadrant.
? I Quadrant is common to all the inverse functions.
? III Quadrant is not used in inverse function.
? IV Quadrant is used in the clockwise direction i.e., 0
2
? ? ? y
?
(7) Principal values for inverse circular functions
Principal values for 0 ? x Principal values for 0 ? x
2
sin 0
1
?
? ?
?
x 0 sin
2
1
? ? ?
?
x
?
2
cos 0
1
?
? ?
?
x
?
?
? ?
?
x
1
cos
2
I
IV
– ? ? ? ?
?/2
I
0 ? ?
II
Page 4
The inverse of a function B A f ? : exists if f is one-one onto i.e., a bijection and is given by
x y f y x f ? ? ?
?
) ( ) (
1
.
Consider the sine function with domain R and range [–1, 1]. Clearly this function is not a
bijection and so it is not invertible. If we restrict the domain of it in such a way that it becomes
one–one, then it would become invertible. If we consider sine as a function with domain
?
?
?
?
?
?
?
2
,
2
? ?
and co-domain [–1, 1], then it is a bijection and therefore, invertible. The inverse of
sine function is defined as x x ? ? ?
?
? ? sin sin
1
, where
?
?
?
?
?
?
? ?
2
,
2
? ?
? and ] 1 , 1 [ ? ? x .
Properties of Inverse Trigonometric Functions.
(1) Meaning of inverse function
(i) x ? ? sin ? ? ?
?
x
1
sin (ii) x ? ? cos ? ? ?
?
x
1
cos (iii) x ? ? tan ? ? ?
?
x
1
tan
(iv) x ? ? cot ? ? ?
?
x
1
cot (v) x ? ? sec ? ? ?
?
x
1
sec (vi) x ? ? cosec ? ? ?
?
x
1
cosec
(2) Domain and range of inverse functions
(i) If , sin x y ? then , sin
1
x y
?
? under certain condition.
; 1 sin 1 ? ? ? y but x y ? sin . 1 1 ? ? ? ? x
Again,
2
1 sin
?
? ? ? ? ? y y and
2
1 sin
?
? ? ? y y .
Keeping in mind numerically smallest angles or real numbers.
2 2
? ?
? ? ? ? y
These restrictions on the values of x and y provide us with the domain and range for the
function x y
1
sin
?
? .
i.e., Domain : ] 1 , 1 [ ? ? x
Range:
?
?
?
?
?
?
? ?
2
,
2
? ?
y
(ii) Let x y ? cos , then x y
1
cos
?
? , under certain conditions 1 cos 1 ? ? ? y
? 1 1 ? ? ? x
? ? ? ? ? y y 1 cos
0 1 cos ? ? ? y y
? ? ? ? y 0 {as cos x is a decreasing function in [ ? , 0 ];
hence 0 cos cos cos ? ? y ?
These restrictions on the values of x and y provide us the domain and range for the function
x y
1
cos
?
? .
Y
y = cos
–
1
x
O
(–1,
?/2)
X
(1, 0)
Y
(1, ?/2)
y = sin
–
1
x
O
(–1, –
?/2)
X
i.e. Domain: ] 1 , 1 [ ? ? x
Range : ] , 0 [ ? ? y
(iii) If x y ? tan , then , tan
1
x y
?
? under certain conditions.
Here, R x R y ? ? ? tan ,
2 2
tan
? ?
? ? ? ? ? ? ? ? ? y y
Thus, Domain R x ? ;
Range ?
?
?
?
?
?
? ?
2
,
2
? ?
y
(iv) If , cot x y ? then x y
1
cot
?
?
under certain conditions, ; cot R x R y ? ? ?
? ? ? ? ? ? ? ? ? y y 0 cot
These conditions on x and y make the function, x y ? cot one-one
and onto so that the inverse function exists. i.e., x y
1
cot
?
? is
meaningful.
? Domain : R x ?
Range : ) , 0 ( ? ? y
(v) If , sec x y ? then , sec
1
x y
?
? where 1 | | ? x and
2
, 0
?
? ? ? ? y y
Here, Domain: ) 1 , 1 ( ? ? ? R x
Range:
?
?
?
?
?
?
? ?
2
] , 0 [
?
? y
(vi) If x y ? cosec , then x y
1
cosec
?
?
Where 1 | | ? x and 0 ,
2 2
? ? ? ? y y
? ?
Here, Domain ) 1 , 1 ( ? ? ? R
Range } 0 {
2
,
2
?
?
?
?
?
?
?
? ?
? ?
Function Domain (D) Range (R)
x
1
sin
?
1 1 ? ? ? x or ] 1 , 1 [ ?
2 2
?
?
?
? ? ? or
?
?
?
?
?
?
?
2
,
2
? ?
x
1
cos
?
1 1 ? ? ? x or ] 1 , 1 [ ? ? ? ? ? 0 or ] , 0 [ ?
x
1
tan
?
? ? ? ?? x i.e., R x ? or
) , ( ? ??
2 2
?
?
?
? ? ? or ?
?
?
?
?
?
?
2
,
2
? ?
x
1
cot
?
? ? ? ?? x i.e., R x ? or ? ? ? ? 0 or ) , 0 ( ?
y
O ?
x ?
y = tan
–
1
x
y = – ?/2
y =
? ?2
x
y = sec
–
1
x
y = ?/2
( –1, ?)
O (1,0
)
x
y = cosec
–
1
x
(1,
?/2)
y
O ? (–1, –
?)
y = ?
O
x ?
(0,
?/2)
y = cot
–
1
x
) , ( ? ??
x
1
sec
?
1 , 1 ? ? ? x x or ) , 1 [ ] 1 , ( ? ? ? ??
? ?
?
? ? ? ? 0 ,
2
or
?
?
?
?
?
?
? ?
?
?
?
?
?
?
? ?
,
2 2
, 0
x
1
cosec
?
1 , 1 ? ? ? x x or ) , 1 [ ] 1 , ( ? ? ? ??
2 2
, 0
?
?
?
? ? ? ? ? or
?
?
?
?
?
?
? ?
?
?
?
?
?
?
2
, 0 0 ,
2
? ?
(3) ? ? ?
?
) (sin sin
1
, Provided that
2 2
?
?
?
? ? ? , ? ? ?
?
) (cos cos
1
, Provided that
? ? ? ? 0
? ? ?
?
) (tan tan
1
, Provided that
2 2
?
?
?
? ? ? , ? ? ?
?
) (cot cot
1
, Provided that ? ? ? ? 0
? ? ?
?
) (sec sec
1
, Provided that
2
0
?
? ? ? or ? ?
?
? ?
2
, ) cosec ( cosec
1
? ? ?
?
Provided that 0
2
? ? ? ?
?
or
2
0
?
? ? ?
(4) , ) sin(sin
1
x x ?
?
Provided that 1 1 ? ? ? x , , ) cos(cos
1
x x ?
?
Provided that 1 1 ? ? ? x
tan , ) (tan
1
x x ?
?
Provided that ? ? ? ? ? x , ) cot(cot
1
x x ?
?
Provided that
? ? ? ? ? x
, ) sec(sec
1
x x ?
?
Provided that 1 ? ? ? ? ? x or ? ? ? x 1
, ) cosec ( cosec
–1
x x ? Provided that 1 ? ? ? ? ? x or ? ? ? x 1
(5) x x
1 1
sin ) ( sin
? ?
? ? ? x x
1 1
cos ) ( cos
? ?
? ? ? ? , x x
1 1
tan ) ( tan
? ?
? ? ?
x x
1 1
cot ) ( cot
? ?
? ? ? ? x x
1 1
sec ) ( sec
? ?
? ? ? ? x x
–1 1
cosec ) ( cosec ? ? ?
?
(6)
2
cos sin
1 1
?
? ?
? ?
x x , for all ] 1 , 1 [ ? ? x
2
cot tan
1 1
?
? ?
? ?
x x , for all R x ?
2
cosec sec
1 - 1
?
? ?
?
x x , for all ) , 1 [ ] 1 , ( ? ? ? ?? ? x
Important Tips
? Here; x x x
1 1 1
tan , cosec , sin
? ? ?
belong to I and IV Quadrant.
? Here; x x x
1 1 1
cot , sec , cos
? ? ?
belong to I and II Quadrant.
? I Quadrant is common to all the inverse functions.
? III Quadrant is not used in inverse function.
? IV Quadrant is used in the clockwise direction i.e., 0
2
? ? ? y
?
(7) Principal values for inverse circular functions
Principal values for 0 ? x Principal values for 0 ? x
2
sin 0
1
?
? ?
?
x 0 sin
2
1
? ? ?
?
x
?
2
cos 0
1
?
? ?
?
x
?
?
? ?
?
x
1
cos
2
I
IV
– ? ? ? ?
?/2
I
0 ? ?
II
2
tan 0
1
?
? ?
?
x 0 tan
2
1
? ? ?
?
x
?
2
cot 0
1
?
? ?
?
x
?
?
? ?
?
x
1
cot
2
2
sec 0
1
?
? ?
?
x ?
?
? ?
?
x
1
sec
2
2
cosec 0
1
?
? ?
?
x
0 cosec
2
1
? ? ?
?
x
?
Thus ,
6 2
1
sin
1
?
? ?
?
?
?
?
?
?
not
3
2
2
1
cos ;
6
5
1
? ?
? ?
?
?
?
?
?
?
?
not
3
4 ?
;
3
) 3 ( tan
1
?
? ? ?
?
not
3
2 ?
;
4
3
) 1 ( cot
1
?
? ?
?
not
4
?
? etc.
Note : ? x x x
1 1 1
tan , cos , sin
? ? ?
are also written as arc x sin , arc x cos and arc x tan respectively.
? It should be noted that if not otherwise stated only principal values of inverse
circular functions are to be considered.
(8) Conversion property : Let, y x ?
?1
sin ? y x sin ? ? ?
?
?
?
?
?
?
x
y
1
cosec ? ?
?
?
?
?
?
?
x
y
1
cosec
1 –
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ? ?
? ? ? ? ?
x
x
x
x
x
x
x x
1
cosec
1
1
sec
1
cot
1
tan 1 cos sin
1 –
2
1
2
1
2
1 2 1 1
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
?
?
?
?
?
?
?
?
?
? ? ?
? ? ? ? ?
2
1
2
1 – 1
2
1 2 1 1
1
cot
1
1
cosec
1
sec
1
tan 1 sin cos
x
x
x
x x
x
x x
?
?
?
?
?
?
?
?
?
? ? ? ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ? ? ? ? ?
x
x
x
x
x x
x
x
2
1 2 1 1
2
1
2
1 1
1
cosec 1 sec
1
cot
1
1
cos
1
sin tan
Note : ? x
x
1 1
cosec
1
sin
? ?
? ?
?
?
?
?
?
, for all ) , 1 [ ] 1 , ( ? ? ?? ? x
? , sec
1
cos
1 1
x
x
? ?
? ?
?
?
?
?
?
for all ) , 1 [ ] 1 , ( ? ? ?? ? x
?
?
?
?
? ? ?
?
? ?
?
?
?
?
?
?
?
?
0 for , cot
0 for , cot 1
tan
1
1
1
x x
x x
x ?
(9) General values of inverse circular functions: We know that if ? ? ?is the smallest angle whose
sine is x, then all the angles whose sine is x can be written as , ) 1 ( ?
n
nx ? ? where ,..... 2 , 1 , 0 ? n
Therefore, the general value of x
1
sin
?
can be taken as ? ?
n
n ) 1 ( ? ? . The general value of x
1
sin
?
is
denoted by x
1
sin
?
.
Thus, we have
2 2
and sin if 1, 1 , 1) ( sin
1
p
a
p
x a x a n p x
n
? ? ? ? ? ? ? ? ? ?
?
Similarly, general values of other inverse circular functions are given as follows:
x
1
? ? ?–
x
2
Page 5
The inverse of a function B A f ? : exists if f is one-one onto i.e., a bijection and is given by
x y f y x f ? ? ?
?
) ( ) (
1
.
Consider the sine function with domain R and range [–1, 1]. Clearly this function is not a
bijection and so it is not invertible. If we restrict the domain of it in such a way that it becomes
one–one, then it would become invertible. If we consider sine as a function with domain
?
?
?
?
?
?
?
2
,
2
? ?
and co-domain [–1, 1], then it is a bijection and therefore, invertible. The inverse of
sine function is defined as x x ? ? ?
?
? ? sin sin
1
, where
?
?
?
?
?
?
? ?
2
,
2
? ?
? and ] 1 , 1 [ ? ? x .
Properties of Inverse Trigonometric Functions.
(1) Meaning of inverse function
(i) x ? ? sin ? ? ?
?
x
1
sin (ii) x ? ? cos ? ? ?
?
x
1
cos (iii) x ? ? tan ? ? ?
?
x
1
tan
(iv) x ? ? cot ? ? ?
?
x
1
cot (v) x ? ? sec ? ? ?
?
x
1
sec (vi) x ? ? cosec ? ? ?
?
x
1
cosec
(2) Domain and range of inverse functions
(i) If , sin x y ? then , sin
1
x y
?
? under certain condition.
; 1 sin 1 ? ? ? y but x y ? sin . 1 1 ? ? ? ? x
Again,
2
1 sin
?
? ? ? ? ? y y and
2
1 sin
?
? ? ? y y .
Keeping in mind numerically smallest angles or real numbers.
2 2
? ?
? ? ? ? y
These restrictions on the values of x and y provide us with the domain and range for the
function x y
1
sin
?
? .
i.e., Domain : ] 1 , 1 [ ? ? x
Range:
?
?
?
?
?
?
? ?
2
,
2
? ?
y
(ii) Let x y ? cos , then x y
1
cos
?
? , under certain conditions 1 cos 1 ? ? ? y
? 1 1 ? ? ? x
? ? ? ? ? y y 1 cos
0 1 cos ? ? ? y y
? ? ? ? y 0 {as cos x is a decreasing function in [ ? , 0 ];
hence 0 cos cos cos ? ? y ?
These restrictions on the values of x and y provide us the domain and range for the function
x y
1
cos
?
? .
Y
y = cos
–
1
x
O
(–1,
?/2)
X
(1, 0)
Y
(1, ?/2)
y = sin
–
1
x
O
(–1, –
?/2)
X
i.e. Domain: ] 1 , 1 [ ? ? x
Range : ] , 0 [ ? ? y
(iii) If x y ? tan , then , tan
1
x y
?
? under certain conditions.
Here, R x R y ? ? ? tan ,
2 2
tan
? ?
? ? ? ? ? ? ? ? ? y y
Thus, Domain R x ? ;
Range ?
?
?
?
?
?
? ?
2
,
2
? ?
y
(iv) If , cot x y ? then x y
1
cot
?
?
under certain conditions, ; cot R x R y ? ? ?
? ? ? ? ? ? ? ? ? y y 0 cot
These conditions on x and y make the function, x y ? cot one-one
and onto so that the inverse function exists. i.e., x y
1
cot
?
? is
meaningful.
? Domain : R x ?
Range : ) , 0 ( ? ? y
(v) If , sec x y ? then , sec
1
x y
?
? where 1 | | ? x and
2
, 0
?
? ? ? ? y y
Here, Domain: ) 1 , 1 ( ? ? ? R x
Range:
?
?
?
?
?
?
? ?
2
] , 0 [
?
? y
(vi) If x y ? cosec , then x y
1
cosec
?
?
Where 1 | | ? x and 0 ,
2 2
? ? ? ? y y
? ?
Here, Domain ) 1 , 1 ( ? ? ? R
Range } 0 {
2
,
2
?
?
?
?
?
?
?
? ?
? ?
Function Domain (D) Range (R)
x
1
sin
?
1 1 ? ? ? x or ] 1 , 1 [ ?
2 2
?
?
?
? ? ? or
?
?
?
?
?
?
?
2
,
2
? ?
x
1
cos
?
1 1 ? ? ? x or ] 1 , 1 [ ? ? ? ? ? 0 or ] , 0 [ ?
x
1
tan
?
? ? ? ?? x i.e., R x ? or
) , ( ? ??
2 2
?
?
?
? ? ? or ?
?
?
?
?
?
?
2
,
2
? ?
x
1
cot
?
? ? ? ?? x i.e., R x ? or ? ? ? ? 0 or ) , 0 ( ?
y
O ?
x ?
y = tan
–
1
x
y = – ?/2
y =
? ?2
x
y = sec
–
1
x
y = ?/2
( –1, ?)
O (1,0
)
x
y = cosec
–
1
x
(1,
?/2)
y
O ? (–1, –
?)
y = ?
O
x ?
(0,
?/2)
y = cot
–
1
x
) , ( ? ??
x
1
sec
?
1 , 1 ? ? ? x x or ) , 1 [ ] 1 , ( ? ? ? ??
? ?
?
? ? ? ? 0 ,
2
or
?
?
?
?
?
?
? ?
?
?
?
?
?
?
? ?
,
2 2
, 0
x
1
cosec
?
1 , 1 ? ? ? x x or ) , 1 [ ] 1 , ( ? ? ? ??
2 2
, 0
?
?
?
? ? ? ? ? or
?
?
?
?
?
?
? ?
?
?
?
?
?
?
2
, 0 0 ,
2
? ?
(3) ? ? ?
?
) (sin sin
1
, Provided that
2 2
?
?
?
? ? ? , ? ? ?
?
) (cos cos
1
, Provided that
? ? ? ? 0
? ? ?
?
) (tan tan
1
, Provided that
2 2
?
?
?
? ? ? , ? ? ?
?
) (cot cot
1
, Provided that ? ? ? ? 0
? ? ?
?
) (sec sec
1
, Provided that
2
0
?
? ? ? or ? ?
?
? ?
2
, ) cosec ( cosec
1
? ? ?
?
Provided that 0
2
? ? ? ?
?
or
2
0
?
? ? ?
(4) , ) sin(sin
1
x x ?
?
Provided that 1 1 ? ? ? x , , ) cos(cos
1
x x ?
?
Provided that 1 1 ? ? ? x
tan , ) (tan
1
x x ?
?
Provided that ? ? ? ? ? x , ) cot(cot
1
x x ?
?
Provided that
? ? ? ? ? x
, ) sec(sec
1
x x ?
?
Provided that 1 ? ? ? ? ? x or ? ? ? x 1
, ) cosec ( cosec
–1
x x ? Provided that 1 ? ? ? ? ? x or ? ? ? x 1
(5) x x
1 1
sin ) ( sin
? ?
? ? ? x x
1 1
cos ) ( cos
? ?
? ? ? ? , x x
1 1
tan ) ( tan
? ?
? ? ?
x x
1 1
cot ) ( cot
? ?
? ? ? ? x x
1 1
sec ) ( sec
? ?
? ? ? ? x x
–1 1
cosec ) ( cosec ? ? ?
?
(6)
2
cos sin
1 1
?
? ?
? ?
x x , for all ] 1 , 1 [ ? ? x
2
cot tan
1 1
?
? ?
? ?
x x , for all R x ?
2
cosec sec
1 - 1
?
? ?
?
x x , for all ) , 1 [ ] 1 , ( ? ? ? ?? ? x
Important Tips
? Here; x x x
1 1 1
tan , cosec , sin
? ? ?
belong to I and IV Quadrant.
? Here; x x x
1 1 1
cot , sec , cos
? ? ?
belong to I and II Quadrant.
? I Quadrant is common to all the inverse functions.
? III Quadrant is not used in inverse function.
? IV Quadrant is used in the clockwise direction i.e., 0
2
? ? ? y
?
(7) Principal values for inverse circular functions
Principal values for 0 ? x Principal values for 0 ? x
2
sin 0
1
?
? ?
?
x 0 sin
2
1
? ? ?
?
x
?
2
cos 0
1
?
? ?
?
x
?
?
? ?
?
x
1
cos
2
I
IV
– ? ? ? ?
?/2
I
0 ? ?
II
2
tan 0
1
?
? ?
?
x 0 tan
2
1
? ? ?
?
x
?
2
cot 0
1
?
? ?
?
x
?
?
? ?
?
x
1
cot
2
2
sec 0
1
?
? ?
?
x ?
?
? ?
?
x
1
sec
2
2
cosec 0
1
?
? ?
?
x
0 cosec
2
1
? ? ?
?
x
?
Thus ,
6 2
1
sin
1
?
? ?
?
?
?
?
?
?
not
3
2
2
1
cos ;
6
5
1
? ?
? ?
?
?
?
?
?
?
?
not
3
4 ?
;
3
) 3 ( tan
1
?
? ? ?
?
not
3
2 ?
;
4
3
) 1 ( cot
1
?
? ?
?
not
4
?
? etc.
Note : ? x x x
1 1 1
tan , cos , sin
? ? ?
are also written as arc x sin , arc x cos and arc x tan respectively.
? It should be noted that if not otherwise stated only principal values of inverse
circular functions are to be considered.
(8) Conversion property : Let, y x ?
?1
sin ? y x sin ? ? ?
?
?
?
?
?
?
x
y
1
cosec ? ?
?
?
?
?
?
?
x
y
1
cosec
1 –
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ? ?
? ? ? ? ?
x
x
x
x
x
x
x x
1
cosec
1
1
sec
1
cot
1
tan 1 cos sin
1 –
2
1
2
1
2
1 2 1 1
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
?
?
?
?
?
?
?
?
?
? ? ?
? ? ? ? ?
2
1
2
1 – 1
2
1 2 1 1
1
cot
1
1
cosec
1
sec
1
tan 1 sin cos
x
x
x
x x
x
x x
?
?
?
?
?
?
?
?
?
? ? ? ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ? ? ? ? ?
x
x
x
x
x x
x
x
2
1 2 1 1
2
1
2
1 1
1
cosec 1 sec
1
cot
1
1
cos
1
sin tan
Note : ? x
x
1 1
cosec
1
sin
? ?
? ?
?
?
?
?
?
, for all ) , 1 [ ] 1 , ( ? ? ?? ? x
? , sec
1
cos
1 1
x
x
? ?
? ?
?
?
?
?
?
for all ) , 1 [ ] 1 , ( ? ? ?? ? x
?
?
?
?
? ? ?
?
? ?
?
?
?
?
?
?
?
?
0 for , cot
0 for , cot 1
tan
1
1
1
x x
x x
x ?
(9) General values of inverse circular functions: We know that if ? ? ?is the smallest angle whose
sine is x, then all the angles whose sine is x can be written as , ) 1 ( ?
n
nx ? ? where ,..... 2 , 1 , 0 ? n
Therefore, the general value of x
1
sin
?
can be taken as ? ?
n
n ) 1 ( ? ? . The general value of x
1
sin
?
is
denoted by x
1
sin
?
.
Thus, we have
2 2
and sin if 1, 1 , 1) ( sin
1
p
a
p
x a x a n p x
n
? ? ? ? ? ? ? ? ? ?
?
Similarly, general values of other inverse circular functions are given as follows:
x
1
? ? ?–
x
2
1 1 , 2 cos
1
? ? ? ? ?
?
x n x ? ? ; If x ? ? cos , ? ? ? ? 0
, tan
1
? ? ? ?
?
n x R x ? ; If , tan x ? ?
2 2
?
?
?
? ? ?
, cot
1
? ? ? ?
?
n x R x ? ; If x ? ? cot , ? ? ? ? 0
? ? ? ?
?
n x 2 sec
1
, 1 ? x or 1 ? ? x ; If
2
and 0 , sec
?
? ? ? ? ? ? ? x
1 , ) 1 ( cosec
1
? ? ? ?
?
x n x
n
? ? or 1 ? ? x ; If 0 and
2 2
, cosec ? ? ? ? ? x x
?
?
?
?
Example: 1 The principal value of
?
?
?
?
?
?
?
?
?
?
2
3
sin
1
is [Roorkee 1992]
(a)
3
2 ?
? (b)
3
?
? (c)
3
4 ?
(d)
8
5 ?
Solution: (b)
3 3
sin sin
1
? ?
? ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ? ?
?
2
sin
2
1
? ?
x ?
Example: 2 ? ?
?
)] 30 [sec( sec
1 o
[MP PET 1992]
(a) –
o
60 (b)
o
30 ? (c)
o
30 (d)
o
150
Solution: (c)
o o o
30 ) 30 (sec sec )] 30 [sec( sec
1 1
? ? ?
? ?
.
Example: 3 The principal value of ?
?
?
?
?
?
?
3
5
sin sin
1
?
is [MP PET 1996]
(a)
3
5 ?
(b)
3
5 ?
? (c)
3
?
? (d)
3
4 ?
Solution: (c)
3 2
3
sin
3
5
sin sin
1 1
? ?
? ?
?
?
?
?
?
?
?
?
? ? ?
?
?
?
?
?
? ?
.
Example: 4 The principal value of
?
?
?
?
?
?
?
?
?
?
?
?
?
3
2
sin sin
1
?
is [IIT 1986]
(a)
3
2 ?
? (b)
3
2 ?
(c)
3
4 ?
(d) None of these
Solution: (d) The principal vlaue of )]
3
2
[sin( sin
1
?
? ?
?
=
3 3
sin sin
1
? ?
? ?
?
?
?
?
?
?
.
Example: 5 Considering only the principal values, if
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
2
1
cot sin ) tan(cos
1 1
x , then x is equal to [IIT 1991; AMU 2001]
(a)
5
1
(b)
5
2
(c)
5
3
(d)
3
5
Solution: (d) Put
2
1
cot
2
1
cot
1
? ? ? ?
?
?
?
?
?
?
? ?
? .
5
2
sin ? ? Put ? ?
?
x
1
cos then ? cos ? x
Also ?
3
5
cos ,
5
2
tan ? ? ? ? ? ? x .
Example: 6 If )] 600 ( [sin sin
1
? ? ?
?
? , then one of the possible value of ? is [Kerala (Engg.) 2002]
Read More
FAQs on Detailed Notes: Inverse Trigonometry Functions - JEE
| 1. What are the basic properties of inverse trigonometry functions? |  |
| 2. How do we find the values of inverse trigonometry functions? | |
Ans. To find the values of inverse trigonometry functions, we use the properties of these functions along with trigonometric identities. We can also use the unit circle or right triangle to determine the values of inverse trigonometry functions. Additionally, we can use the graphs of these functions to find their values.
| 3. What is the relationship between trigonometry functions and their inverses? | |
Ans. The relationship between trigonometry functions and their inverses is that they are reciprocals of each other. This means that the inverse trigonometry functions "undo" the actions of the trigonometry functions. For example, the sine function and its inverse, arcsine function, are reciprocals of each other.
| 4. How do we solve equations involving inverse trigonometry functions? | |
Ans. To solve equations involving inverse trigonometry functions, we can use the properties of these functions to simplify the equations. We can also use trigonometric identities to rewrite the equations in a more manageable form. It is essential to be familiar with the properties of inverse trigonometry functions to solve such equations effectively.
| 5. What are some common applications of inverse trigonometry functions in real life? | |
Ans. Some common applications of inverse trigonometry functions in real life include finding angles in navigation, engineering, physics, and computer graphics. Inverse trigonometry functions are used to calculate angles and distances in various practical scenarios where trigonometry is involved.
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