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Section 3.5 Inverse Trigonometric Functions 2010 Kiryl Tsishchanka
Inverse Trigonometric Functions
DEFINITION: The inverse sine function, denoted by sin??1 x (or arcsin x), is defined to be
the inverse of the restricted sine function
sin x; ??
2
 x  
2
DEFINITION: The inverse cosine function, denoted by cos??1 x (or arccos x), is defined to
be the inverse of the restricted cosine function
cos x; 0  x  
DEFINITION: The inverse tangent function, denoted by tan??1 x (or arctan x), is defined
to be the inverse of the restricted tangent function
tan x; ??
2
< x <

2
DEFINITION: The inverse cotangent function, denoted by cot??1 x (or arccot x), is defined
to be the inverse of the restricted cotangent function
cot x; 0 < x < 
1
Section 3.5 Inverse Trigonometric Functions 2010 Kiryl Tsishchanka
DEFINITION: The inverse secant function, denoted by sec??1 x (or arcsec x), is defined to
be the inverse of the restricted secant function
sec x; x 2 [0; =2) [ [; 3=2)
[
or x 2 [0; =2) [ (=2; ] in some other textbooks
]
DEFINITION: The inverse cosecant function, denoted by csc??1 x (or arccsc x), is defined
to be the inverse of the restricted cosecant function
csc x; x 2 (0; =2] [ (; 3=2]
[
or x 2 [??=2; 0) [ (0; =2] in some other textbooks
]
IMPORTANT: Do not confuse
sin??1 x; cos??1 x; tan??1 x; cot??1 x; sec??1 x; csc??1 x
with
1
sin x
;
1
cos x
;
1
tan x
;
1
cot x
;
1
sec x
;
1
csc x
FUNCTION DOMAIN RANGE
sin??1 x [??1; 1] [??=2; =2]
cos??1 x [??1; 1] [0; ]
tan??1 x (??1;+1) (??=2; =2)
cot??1 x (??1;+1) (0; )
sec??1 x (??1;??1] [ [1;+1) [0; =2) [ [; 3=2)
csc??1 x (??1;??1] [ [1;+1) (0; =2] [ (; 3=2]
2
Section 3.5 Inverse Trigonometric Functions 2010 Kiryl Tsishchanka
FUNCTION DOMAIN RANGE
sin??1 x [??1; 1] [??=2; =2]
cos??1 x [??1; 1] [0; ]
tan??1 x (??1;+1) (??=2; =2)
cot??1 x (??1;+1) (0; )
sec??1 x (??1;??1] [ [1;+1) [0; =2) [ [; 3=2)
csc??1 x (??1;??1] [ [1;+1) (0; =2] [ (; 3=2]
sin(??x) = ??sin x cos(??x) = cos x tan(??x) = ??tan x
csc(??x) = ??csc x sec(??x) = sec x cot(??x) = ??cot x
sin(x  ) = ??sin x cos(x  ) = ??cos x tan(x  ) = tan x
sec(x  ) = ??sec x csc(x  ) = ??csc x cot(x  ) = cot x
EXAMPLES:
(a) sin??1 1 =

2
; since sin

2
= 1 and

2
2
[
??
2
;

2
]
:
(b) sin??1(??1) = ??
2
; since sin
(
??
2
)
= ??1 and ??
2
2
[
??
2
;

2
]
:
(c) sin??1 0 = 0; since sin 0 = 0 and 0 2
[
??
2
;

2
]
:
(d) sin??1 1
2
=

6
; since sin

6
=
1
2
and

6
2
[
??
2
;

2
]
:
(e) sin??1
p
3
2
=

3
; since sin

3
=
p
3
2
and

3
2
[
??
2
;

2
]
:
(f) sin??1
p
2
2
=

4
; since sin

4
=
p
2
2
and

4
2
[
??
2
;

2
]
:
EXAMPLES:
cos??1 0 =

2
; cos??1 1 = 0; cos??1(??1) = ; cos??1 1
2
=

3
; cos??1
p
3
2
=

6
; cos??1
p
2
2
=

4
tan??1 1 =

4
; tan??1(??1) = ??
4
; tan??1
p
3 =

3
; tan??1 p1
3
=

6
; tan??1
(
??p1
3
)
= ??
6
EXAMPLES: Find sec??1 1; sec??1(??1); and sec??1(??2):
3
Section 3.5 Inverse Trigonometric Functions 2010 Kiryl Tsishchanka
FUNCTION DOMAIN RANGE
sin??1 x [??1; 1] [??=2; =2]
cos??1 x [??1; 1] [0; ]
tan??1 x (??1;+1) (??=2; =2)
cot??1 x (??1;+1) (0; )
sec??1 x (??1;??1] [ [1;+1) [0; =2) [ [; 3=2)
csc??1 x (??1;??1] [ [1;+1) (0; =2] [ (; 3=2]
sin(??x) = ??sin x cos(??x) = cos x tan(??x) = ??tan x
csc(??x) = ??csc x sec(??x) = sec x cot(??x) = ??cot x
sin(x  ) = ??sin x cos(x  ) = ??cos x tan(x  ) = tan x
sec(x  ) = ??sec x csc(x  ) = ??csc x cot(x  ) = cot x
EXAMPLES: Find sec??1 1; sec??1(??1); and sec??1(??2):
Solution: We have
sec??1 1 = 0; sec??1(??1) = ; sec??1(??2) =
4
3
since
sec 0 = 1; sec  = ??1; sec
4
3
= ??2
and
0; ;
4
3
2
[
0;

2
)
[
[
;
3
2
)
Note that sec
2
3
is also ??2; but
sec??1(??2) 6=
2
3
since
2
3
62
[
0;

2
)
[
[
;
3
2
)
EXAMPLES: Find
tan??1 0 cot??1 0 cot??1 1 sec??1
p
2 csc??1 2 csc??1 p2
3
4
Section 3.5 Inverse Trigonometric Functions 2010 Kiryl Tsishchanka
EXAMPLES: We have
tan??1 0 = 0; cot??1 0 =

2
; cot??1 1 =

4
; sec??1
p
2 =

4
; csc??1 2 =

6
; csc??1 p2
3
=

3
EXAMPLES: Evaluate sin
(
arcsin

7
)
; arcsin
(
sin

7
)
; and arcsin
(
sin
8
7
)
:
Solution: Since arcsin x is the inverse of the restricted sine function, we have
sin(arcsin x) = x if x 2 [??1; 1] and arcsin(sin x) = x if x 2 [??=2; =2]
Therefore
sin
(
arcsin

7
)
=

7
and arcsin
(
sin

7
)
=

7
but
arcsin
(
sin
8
7
)
= arcsin
(
sin
(
7
+ 
))
= arcsin
(
??sin

7
)
= ??arcsin
(
sin

7
)
= ??
7
EXAMPLES: Evaluate cot
(
arcsin
2
5
)
and sec
(
arcsin
2
5
)
:
Solution 1: We have
cot  =
cos 
sin 
=

?
1 ?? sin2 
sin 
and sec  =
1
cos 
=
1

?
1 ?? sin2 
Since ??
2
 arcsin x  
2
; it follows that cos(arcsin x)  0: Therefore if  = arcsin
2
5
; then
cot  =
?
1 ?? sin2 
sin 
and sec  =
1 ?
1 ?? sin2 
hence
cot
(
arcsin
2
5
)
=
?
1 ?? sin2
(
arcsin
2
5
)
sin
(
arcsin
2
5
) =
?
1 ??
(
2
5
)2
2
5
=
p
21
2
and
sec
(
arcsin
2
5
)
=
1 ?
1 ?? sin2
(
arcsin
2
5
) =
1 ?
1 ??
(
2
5
)2
=
p5
21
Solution 2: Put  = arcsin
2
5
; so sin  =
2
5
: Then
cot
(
arcsin
2
5
)
= cot  =
p
21
2
and sec
(
arcsin
2
5
)
= sec  =
p5
21
!!!!!!!!!!

2
p
21
5
EXAMPLES: Evaluate, if possible, cot
(
sin??1 2
)
and sin (tan??1 2) :
5
Section 3.5 Inverse Trigonometric Functions 2010 Kiryl Tsishchanka
EXAMPLES: Evaluate, if possible, cot
(
sin??1 2
)
and sin (tan??1 2) :
We first note that sin??1 2 does not exist, since 2 62 [??1; 1]; that is, 2 is not in the domain of
sin??1 x. Therefore cot
(
sin??1 2
)
does not exist.
We will evaluate sin (tan??1 2) in two different ways:
Solution 1: We have
sin  = p tan 
1 + tan2 
Since ??=2 < tan??1 x < =2; it follows that cos (tan??1 x) > 0: Therefore if  = tan??1 2; then
sin  =
p tan 
1 + tan2 
hence
sin
(
tan??1 2
)
=
tan (tan??1 2) ?
1 + tan2 (tan??1 2)
=
p 2
1 + 22
=
p2
5
Solution 2: Put  = tan??1 2 = tan??1 2
1
; so tan  =
2
1
: Then
sin
(
tan??1 2
)
= sin  =
p2
5
EXAMPLES: Evaluate sin
(
cot??1
(
??1
2
))
and cos
(
cot??1
(
??1
2
))
:





















2
1
p
5
6
Section 3.5 Inverse Trigonometric Functions 2010 Kiryl Tsishchanka
EXAMPLES: Evaluate sin
(
cot??1
(
??1
2
))
and cos
(
cot??1
(
??1
2
))
:
Solution 1: We have
sin  = p 1
1 + cot2 
and cos  = p cot 
1 + cot2 
Since 0 < cot??1 x < ; it follows that sin(cot??1 x) > 0: Therefore if  = cot??1
(
??1
2
)
; then
sin  =
p 1
1 + cot2 
and cos  =
p cot 
1 + cot2 
hence
sin
(
cot??1
(
??1
2
))
=
1 ?
1 + cot2
(
cot??1
(
??1
2
)) =
1 ?
1 +
(
??1
2
)2
=
p2
5
and
cos
(
cot??1
(
??1
2
))
=
cot
(
cot??1
(
??1
2
))
?
1 + cot2
(
cot??1
(
??1
2
)) =
??1
? 2
1 +
(
??1
2
)2
= ??p1
5
Solution 2: Put  = cot??1
(
??1
2
)
; so cot  = ??1
2
=
??1
2
: Then
sin
(
cot??1
(
??1
2
))
= sin  =
p2
5
and
cos
(
cot??1
(
??1
2
))
= cos  = ??p1
5
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A

p
5
-1
2
7
Section 3.5 Inverse Trigonometric Functions 2010 Kiryl Tsishchanka
THEOREM: We have
(a) (sin??1 u)0 =
p 1
1 ?? u2
u0 (d) (cot??1 u)0 = ?? 1
1 + u2 u0
(b) (cos??1 u)0 = ??p 1
1 ?? u2
u0 (e) (sec??1 u)0 =
1
u
p
u2 ?? 1
u0
(c) (tan??1 u)0 =
1
1 + u2 u0 (f) (csc??1 u)0 = ?? 1
u
p
u2 ?? 1
u0
Proof:
(a) Let y = sin??1 u; then sin y = u: Therefore
(sin y)0 = u0 =) cos y
y0 = u0 =) y0 =
u0
cos y
Since ??
2
 sin??1 u | {z }
y
 
2
; it follows that cos y  0: Hence
cos y =
?
1 ?? sin2 y = [sin y = u] =
p
1 ?? u2 =) y0 =
u0
cos y
=
u0
p
1 ?? u2
(b) Let y = cos??1 u; then cos y = u: Therefore
(cos y)0 = u0 =) ??sin y
y0 = u0 =) y0 = ?? u0
sin y
Since 0  cos??1 u | {z }
y
 ; it follows that sin y  0: Hence
sin y =
?
1 ?? cos2 y = [cos y = u] =
p
1 ?? u2 =) y0 = ?? u0
sin y
= ?? u0
p
1 ?? u2
(c) Let y = tan??1 u; then tan y = u: Therefore
(tan y)0 = u0 =) sec2 y
y0 = u0 =) y0 =
u0
sec2 y
Note, that sec2 y = 1 + tan2 y = [tan y = u] = 1 + u2: Hence
y0 =
u0
sec2 y
=
u0
1 + u2
8
Section 3.5 Inverse Trigonometric Functions 2010 Kiryl Tsishchanka
(a) (sin??1 u)0 =
p 1
1 ?? u2
u0 (d) (cot??1 u)0 = ?? 1
1 + u2 u0
(b) (cos??1 u)0 = ??p 1
1 ?? u2
u0 (e) (sec??1 u)0 =
1
u
p
u2 ?? 1
u0
(c) (tan??1 u)0 =
1
1 + u2 u0 (f) (csc??1 u)0 = ?? 1
u
p
u2 ?? 1
u0
(d) Let y = cot??1 u; then cot y = u: Therefore
(cot y)0 = u0 =) ??csc2 y
y0 = u0 =) y0 = ?? u0
csc2 y
Note, that csc2 y = 1 + cot2 y = [cot y = u] = 1 + u2: Hence
y0 = ?? u0
csc2 y
= ?? u0
1 + u2
(e) Let y = sec??1 u; then sec y = u: Therefore
(sec y)0 = u0 =) sec y tan y
y0 = u0 =) y0 =
u0
sec y tan y
Since sec??1 u | {z }
y
2 [0; =2) [ [; 3=2); it follows that tan y  0: Hence
sec y tan y = sec y
?
sec2 y ?? 1 = [sec y = u] = u
p
u2 ?? 1 =) y0 =
u0
sec y tan y
=
u0
u
p
u2 ?? 1
(f) Let y = csc??1 u; then csc y = u: Therefore
(csc y)0 = u0 =) ??csc y cot y
y0 = u0 =) y0 = ?? u0
csc y cot y
Since csc??1 u | {z }
y
2 (0; =2] [ (; 3=2]; it follows that cot y  0: Hence
csc y cot y = csc y
?
csc2 y ?? 1 = [csc y = u] = u
p
u2 ?? 1 =) y0 = ?? u0
csc y cot y
= ?? u0
u
p
u2 ?? 1
EXAMPLES:
(a) Let f(x) = x tan??1(1 ?? 2x): Find f0(x):
(b) Let f(x) = 2sin??1(4x): Find f0(x):
(c) Let f(x) =
?
sec??1(1 ?? 3x): Find f0(x):
9
Section 3.5 Inverse Trigonometric Functions 2010 Kiryl Tsishchanka
(a) (sin??1 u)0 =
p 1
1 ?? u2
u0 (d) (cot??1 u)0 = ?? 1
1 + u2 u0
(b) (cos??1 u)0 = ??p 1
1 ?? u2
u0 (e) (sec??1 u)0 =
1
u
p
u2 ?? 1
u0
(c) (tan??1 u)0 =
1
1 + u2 u0 (f) (csc??1 u)0 = ?? 1
u
p
u2 ?? 1
u0
EXAMPLES:
(a) Let f(x) = x tan??1(1 ?? 2x): Then
f0(x) = [x tan??1(1 ?? 2x)]0 = x0 tan??1(1 ?? 2x) + x[tan??1(1 ?? 2x)]0
= 1
tan??1(1 ?? 2x) + x
1
1 + (1 ?? 2x)2

(1 ?? 2x)0
= tan??1(1 ?? 2x) + x
1
1 + (1 ?? 2x)2

(??2)
= tan??1(1 ?? 2x) ?? 2x
1 + (1 ?? 2x)2
= tan??1(1 ?? 2x) ?? x
1 ?? 2x + 2x2
(b) Let f(x) = 2sin??1(4x): Then
f0(x) =
[
2sin??1(4x)
]0
= 2sin??1(4x) ln 2

[
sin??1(4x)
]0
= 2sin??1(4x) ln 2
1 ?
1 ?? (4x)2

(4x)0
= 2sin??1(4x) ln 2
1 ?
1 ?? (4x)2

4
=
2sin??1(4x)+2 ln 2 ?
1 ?? (4x)2
(c) Let f(x) =
?
sec??1(1 ?? 3x): Then
f0(x) = [(sec??1(1 ?? 3x))1=2]0 =
1
2
(sec??1(1 ?? 3x))??1=2
[sec??1(1 ?? 3x)]0
=
1
2
(sec??1(1 ?? 3x))??1=2 1
(1 ?? 3x)
?
(1 ?? 3x)2 ?? 1

(1 ?? 3x)0
=
1
2
(sec??1(1 ?? 3x))??1=2 1
(1 ?? 3x)
?
(1 ?? 3x)2 ?? 1

(??3)
= ?? 3
2(1 ?? 3x)
?
3x(3x ?? 2) sec??1(1 ?? 3x)
10