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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
π
π
sin x, − ≤ x ≤
2
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 ≤ π
DEFINITION: The inverse tangent function, denoted by tan−1 x (or arctan x), is deﬁned
to be the inverse of the restricted tangent function
π
π
tan x, − < x <
2
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 < π
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 ∈ [0, π/2) ∪ [π, 3π/2) or x ∈ [0, π/2) ∪ (π/2, π] 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 ∈ (0, π/2] ∪ (π, 3π/2] or x ∈ [−π/2, 0) ∪ (0, π/2] in some other textbooks
IMPORTANT: Do not confuse
sin−1 x,
with
cos−1 x,
1
,
sin x
1
,
cos x
tan−1 x,
cot−1 x,
1
,
tan x
1
,
cot x
FUNCTION
DOMAIN
−1
sin x
[−1, 1]
cos−1 x
[−1, 1]
−1
tan x
(−∞, +∞)
−1
cot x
(−∞, +∞)
−1
sec x
(−∞, −1] ∪ [1, +∞)
csc−1 x
(−∞, −1] ∪ [1, +∞)
2
sec−1 x,
1
,
sec x
csc−1 x
1
csc x
RANGE
[−π/2, π/2]
[0, π]
(−π/2, π/2)
(0, π)
[0, π/2) ∪ [π, 3π/2)
(0, π/2] ∪ (π, 3π/2]
Section 3.5 Inverse Trigonometric Functions
FUNCTION
sin−1 x
cos−1 x
tan−1 x
cot−1 x
sec−1 x
csc−1 x
2010 Kiryl Tsishchanka
DOMAIN
RANGE
[−1, 1]
[−π/2, π/2]
[−1, 1]
[0, π]
(−∞, +∞)
(−π/2, π/2)
(−∞, +∞)
(0, π)
(−∞, −1] ∪ [1, +∞) [0, π/2) ∪ [π, 3π/2)
(−∞, −1] ∪ [1, +∞) (0, π/2] ∪ (π, 3π/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 ± π) = − sin x
cos(x ± π) = − cos x
tan(x ± π) = tan x
sec(x ± π) = − sec x
csc(x ± π) = − csc x
cot(x ± π) = cot x
EXAMPLES:
(a) sin−1 1 =
π
π
π [ π π]
, since sin = 1 and ∈ − , .
2
2
2
2 2
( π)
π [ π π]
π
= −1 and − ∈ − , .
(b) sin−1 (−1) = − , since sin −
2
2
2
2 2
[ π π]
(c) sin−1 0 = 0, since sin 0 = 0 and 0 ∈ − , .
2 2
(d) sin−1
1
π
π
1
π [ π π]
= , since sin = and ∈ − , .
2
6
6
2
6
2 2
√
3
π
π
3
π [ π π]
= , since sin =
and ∈ − , .
2
3
3
2
3
2 2
√
(e) sin
−1
√
(f) sin
−1
√
2
π
π
2
π [ π π]
= , since sin =
and ∈ − , .
2
4
4
2
4
2 2
EXAMPLES:
π
cos−1 0 = ,
2
tan−1 1 =
π
,
4
cos−1 1 = 0,
cos−1 (−1) = π,
π
tan−1 (−1) = − ,
4
tan−1
√
cos−1
3=
π
,
3
tan−1
EXAMPLES: Find sec−1 1, sec−1 (−1), and sec−1 (−2).
3
√
3
π
2
π
cos−1
= , cos−1
=
2
6
2
4
(
)
1
π
1
π
√ = , tan−1 − √
=−
6
6
3
3
π
1
= ,
2
3
√
Section 3.5 Inverse Trigonometric Functions
FUNCTION
sin−1 x
cos−1 x
tan−1 x
cot−1 x
sec−1 x
csc−1 x
2010 Kiryl Tsishchanka
DOMAIN
RANGE
[−1, 1]
[−π/2, π/2]
[−1, 1]
[0, π]
(−∞, +∞)
(−π/2, π/2)
(−∞, +∞)
(0, π)
(−∞, −1] ∪ [1, +∞) [0, π/2) ∪ [π, 3π/2)
(−∞, −1] ∪ [1, +∞) (0, π/2] ∪ (π, 3π/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 ± π) = − sin x
cos(x ± π) = − cos x
tan(x ± π) = tan x
sec(x ± π) = − sec x
csc(x ± π) = − csc x
cot(x ± π) = cot x
EXAMPLES: Find sec−1 1, sec−1 (−1), and sec−1 (−2).
Solution: We have
sec−1 1 = 0,
sec−1 (−1) = π,
sec−1 (−2) =
since
sec 0 = 1,
and
sec π = −1,
sec
4π
3
4π
= −2
3
4π [ π ) [ 3π )
0, π,
∈ 0,
∪ π,
3
2
2
Note that sec
2π
is also −2, but
3
sec−1 (−2) 6=
since
2π
3
2π [ π ) [ 3π )
6∈ 0,
∪ π,
3
2
2
EXAMPLES: Find
tan−1 0
cot−1 0
cot−1 1
sec−1
4
√
2
csc−1 2
2
csc−1 √
3
Section 3.5 Inverse Trigonometric Functions
2010 Kiryl Tsishchanka
EXAMPLES: We have
tan−1 0 = 0,
cot−1 0 =
π
,
2
cot−1 1 =
π
,
4
sec−1
√
2=
π
,
4
csc−1 2 =
π
,
6
2
π
csc−1 √ =
3
3
(
)
( π)
8π
π)
EXAMPLES: Evaluate sin arcsin
, arcsin sin
, and arcsin sin
.
7
7
7
(
Solution: Since arcsin x is the inverse of the restricted sine function, we have
sin(arcsin x) = x if x ∈ [−1, 1]
(
π) π
sin arcsin
=
7
7
Therefore
but
arcsin(sin x) = x if x ∈ [−π/2, π/2]
and
(
8π
arcsin sin
7
)
(
π) π
arcsin sin
=
7
7
and
( (π
))
(
( π)
π)
π
= arcsin sin
+π
= arcsin − sin
= − arcsin sin
=−
7
7
7
7
(
2
EXAMPLES: Evaluate cot arcsin
5
)
(
2
and sec arcsin
5
)
.
Solution 1: We have
√
cos θ
± 1 − sin2 θ
cot θ =
=
sin θ
sin θ
Since −
sec θ =
1
1
= √
cos θ
± 1 − sin2 θ
π
π
2
≤ arcsin x ≤ , it follows that cos(arcsin x) ≥ 0. Therefore if θ = arcsin , then
2
2
5
√
1 − sin2 θ
1
cot θ =
and sec θ = √
sin θ
1 − sin2 θ
√
hence
(
2
cot arcsin
5
and
and
(
2
sec arcsin
5
(
2
1 − sin2 arcsin
5
(
)
2
sin arcsin
5
)
=
)
=√
1 − sin2
1
(
√
)
=
( )2
2
1−
√
5
21
=
2
2
5
5
1
)=√
( )2 = √21
2
2
arcsin
1−
5
5
2
2
Solution 2: Put θ = arcsin , so sin θ = . Then
5
5
√
(
(
)
)
2
21
2
5
cot arcsin
and sec arcsin
= cot θ =
= sec θ = √
5
2
5
21
(
)
EXAMPLES: Evaluate, if possible, cot sin−1 2 and sin (tan−1 2) .
5
!
5 !!!
!
!!
!
!! θ √
21
2
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
does
( 2 −1
) not exist, since 2 6∈ [−1, 1], that is, 2 is not in the domain of
−1
sin x. Therefore cot sin 2 does not exist.
We will evaluate sin (tan−1 2) in two diﬀerent ways:
Solution 1: We have
tan θ
sin θ = ± √
1 + tan2 θ
Since −π/2 < tan−1 x < π/2, it follows that cos (tan−1 x) > 0. Therefore if θ = tan−1 2, then
sin θ = √
hence
tan θ
1 + tan2 θ
(
)
tan (tan−1 2)
2
2
√
√
sin tan−1 2 = √
=
=
1 + 22
5
1 + tan2 (tan−1 2)
Solution 2: Put θ = tan−1 2 = tan−1
5
θ
√
2
2
, so tan θ = . Then
1
1
(
)
2
sin tan−1 2 = sin θ = √
5
(
EXAMPLES: Evaluate sin cot−1
(
1
−
2
))
(
and cos cot−1
6
(
1
1
−
2
))
.
2
Section 3.5 Inverse Trigonometric Functions
2010 Kiryl Tsishchanka
(
(
−1
EXAMPLES: Evaluate sin cot
1
−
2
))
(
(
−1
and cos cot
1
−
2
))
.
Solution 1: We have
sin θ = ± √
−1
Since 0 < cot
1 + cot2 θ
x < π, it follows that sin(cot
(
(
−1
sin cot
and
and
−1
sin θ = √
hence
1
1
−
2
1
1 + cot2 θ
=√
(
1
1 + cot2 θ
cos θ = √
(
1
2
1
−
2
)
, then
cot θ
1 + cot2 θ
)) = √
(
1
(
1+ −
2
)2 = √ 5
1
2
))
1
1
(
( ))
cot cot
−
−
1
1
2
2
cos cot−1 −
=√
=√
= −√
(
(
))
(
)
2
2
5
1
1
1 + cot2 cot−1 −
1+ −
2
2
−1
(
−1
Solution 2: Put θ = cot
and
(
−1
1 + cot2 cot−1 −
(
cot θ
x) > 0. Therefore if θ = cot
and
))
cos θ = ± √
)
1
−1
, so cot θ = − =
. Then
2
2
(
( ))
1
2
−1
sin cot
−
= sin θ = √
2
5
1
−
2
(
(
−1
cos cot
A
A
2
1
−
2
))
1
= cos θ = − √
5
A
A
A
A
A
A
A √
A 5
A
A
A
A
A
A
A
-1
7
Aθ
Section 3.5 Inverse Trigonometric Functions
2010 Kiryl Tsishchanka
THEOREM: We have
(a) (sin−1 u)0 = √
1
u0
1 − u2
(b) (cos−1 u)0 = − √
(c) (tan−1 u)0 =
(d) (cot−1 u)0 = −
1
u0
2
1−u
1
u0
1 + u2
1
(e) (sec−1 u)0 = √
u0
2
u u −1
1
u0
1 + u2
1
(csc−1 u)0 = − √
u0
2
u u −1
(f)
Proof:
(a) Let y = sin−1 u, then sin y = u. Therefore
0
(sin y) = u
Since −
0
0
cos y · y = u
=⇒
0
u0
y =
cos y
0
=⇒
π
π
−1
≤ sin
u} ≤ , it follows that cos y ≥ 0. Hence
|
{z
2
2
y
√
cos y =
1 − sin2 y = [sin y = u] =
√
1 − u2
y0 =
=⇒
u0
u0
=√
cos y
1 − u2
(b) Let y = cos−1 u, then cos y = u. Therefore
(cos y)0 = u0
=⇒
− sin y · y 0 = u0
=⇒
y0 = −
u0
sin y
−1
Since 0 ≤ cos
| {z u} ≤ π, it follows that sin y ≥ 0. Hence
y
sin y =
√
1 − cos2 y = [cos y = u] =
√
1 − u2
=⇒
y0 = −
u0
u0
= −√
sin y
1 − u2
(c) Let y = tan−1 u, then tan y = u. Therefore
(tan y)0 = u0
=⇒
sec2 y · y 0 = u0
=⇒
Note, that sec2 y = 1 + tan2 y = [tan y = u] = 1 + u2 . Hence
y0 =
u0
u0
=
sec2 y
1 + u2
8
y0 =
u0
sec2 y
Section 3.5 Inverse Trigonometric Functions
(a) (sin−1 u)0 = √
1
u0
2
1−u
(b) (cos−1 u)0 = − √
(c) (tan−1 u)0 =
2010 Kiryl Tsishchanka
(d) (cot−1 u)0 = −
1
u0
2
1−u
1
u0
2
1+u
1
(e) (sec−1 u)0 = √
u0
2
u u −1
1
u0
1 + u2
(f)
1
(csc−1 u)0 = − √
u0
u u2 − 1
(d) Let y = cot−1 u, then cot y = u. Therefore
(cot y)0 = u0
=⇒
− csc2 y · y 0 = u0
=⇒
y0 = −
u0
csc2 y
Note, that csc2 y = 1 + cot2 y = [cot y = u] = 1 + u2 . Hence
y0 = −
u0
u0
=
−
csc2 y
1 + u2
(e) Let y = sec−1 u, then sec y = u. Therefore
(sec y)0 = u0
sec y tan y · y 0 = u0
=⇒
=⇒
y0 =
u0
sec y tan y
−1
Since sec
| {z u} ∈ [0, π/2) ∪ [π, 3π/2), it follows that tan y ≥ 0. Hence
y
√
√
sec y tan y = sec y sec2 y − 1 = [sec y = u] = u u2 − 1
=⇒
u0
u0
√
y =
=
sec y tan y
u u2 − 1
0
(f) Let y = csc−1 u, then csc y = u. Therefore
(csc y)0 = u0
=⇒
− csc y cot y · y 0 = u0
=⇒
y0 = −
u0
csc y cot y
−1
Since csc
| {z u} ∈ (0, π/2] ∪ (π, 3π/2], it follows that cot y ≥ 0. Hence
y
√
√
csc y cot y = csc y csc2 y − 1 = [csc y = u] = u u2 − 1
EXAMPLES:
(a) Let f (x) = x tan−1 (1 − 2x). Find f 0 (x).
(b) Let f (x) = 2sin
(c) Let f (x) =
√
−1 (4x)
. Find f 0 (x).
sec−1 (1 − 3x). Find f 0 (x).
9
=⇒
u0
u0
√
y =−
=−
csc y cot y
u u2 − 1
0
Section 3.5 Inverse Trigonometric Functions
(a) (sin−1 u)0 = √
1
u0
1 − u2
(b) (cos−1 u)0 = − √
(c) (tan−1 u)0 =
2010 Kiryl Tsishchanka
(d) (cot−1 u)0 = −
1
u0
1 + u2
1
(e) (sec−1 u)0 = √
u0
2
u u −1
1
u0
2
1−u
1
u0
1 + u2
1
(csc−1 u)0 = − √
u0
2
u u −1
(f)
EXAMPLES:
(a) Let f (x) = x tan−1 (1 − 2x). Then
f 0 (x) = [x tan−1 (1 − 2x)]0 = x0 tan−1 (1 − 2x) + x[tan−1 (1 − 2x)]0
= 1 · tan−1 (1 − 2x) + x
= tan−1 (1 − 2x) + x
1
· (1 − 2x)0
2
1 + (1 − 2x)
1
· (−2)
1 + (1 − 2x)2
2x
1 + (1 − 2x)2
x
= tan−1 (1 − 2x) −
1 − 2x + 2x2
= tan−1 (1 − 2x) −
(b) Let f (x) = 2sin
−1 (4x)
. Then
[
]0
[
]0
−1
−1
f 0 (x) = 2sin (4x)
= 2sin (4x) ln 2 · sin−1 (4x)
= 2sin
= 2sin
−1 (4x)
−1 (4x)
ln 2 √
ln 2 √
−1 (4x)+2
2sin
= √
(c) Let f (x) =
√
1
1 − (4x)2
1
1 − (4x)2
· (4x)0
·4
ln 2
1 − (4x)2
sec−1 (1 − 3x). Then
1
f 0 (x) = [(sec−1 (1 − 3x))1/2 ]0 = (sec−1 (1 − 3x))−1/2 · [sec−1 (1 − 3x)]0
2
1
1
√
= (sec−1 (1 − 3x))−1/2
· (1 − 3x)0
2
2
(1 − 3x) (1 − 3x) − 1
1
1
√
· (−3)
= (sec−1 (1 − 3x))−1/2
2
(1 − 3x) (1 − 3x)2 − 1
=−
3
√
2(1 − 3x) 3x(3x − 2) sec−1 (1 − 3x)
10
```
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