Section 6.4 Inverse Trigonometric Functions and Right
Triangles
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
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
FUNCTION
sin−1 x
cos−1 x
tan−1 x
cot−1 x
sec−1 x
csc−1 x
1
,
cos x
tan−1 x,
cot−1 x,
1
,
tan x
1
,
cot x
DOMAIN
[−1, 1]
[−1, 1]
(−∞, +∞)
(−∞, +∞)
(−∞, −1] ∪ [1, +∞)
(−∞, −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]
FUNCTION
sin−1 x
cos−1 x
tan−1 x
cot−1 x
sec−1 x
csc−1 x
DOMAIN
[−1, 1]
[−1, 1]
(−∞, +∞)
(−∞, +∞)
(−∞, −1] ∪ [1, +∞)
(−∞, −1] ∪ [1, +∞)
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]
EXAMPLES:
(a) sin
−1
π
π
π h π πi
1 = , since sin = 1 and ∈ − , .
2
2
2
2 2
π
π
π h π πi
(b) sin−1 (−1) = − , since sin −
= −1 and − ∈ − , .
2
2
2
2 2
h π πi
(c) sin−1 0 = 0, since sin 0 = 0 and 0 ∈ − , .
2 2
(d) sin−1
(e) sin
(f) sin
−1
−1
π
π
1
π h π πi
1
= , since sin = and ∈ − , .
2
6
6
2
6
2 2
√
√
π
π
π h π πi
3
3
= , since sin =
and ∈ − , .
2
3
3
2
3
2 2
√
√
2
2
π
π
π h π πi
= , 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) = π,
cos
−1
π
tan−1 (−1) = − ,
4
tan−1
√
3=
−1
π
,
3
√
3
2
π
π
−1
cos
= , cos
=
2
6
2
4
π
π
1
1
√ = , tan−1 − √
=−
6
6
3
3
1
π
= ,
2
3
tan−1
−1
EXAMPLES: Find sec−1 1, sec−1 (−1), and sec−1 (−2).
Solution: We have
sec−1 1 = 0,
sec−1 (−1) = π,
since
sec 0 = 1,
sec π = −1,
and
sec−1 (−2) =
4π
= −2
3
3π sec
4π h π h
0, π,
∈ 0,
∪ π,
3
2
2
2π
2π
is also −2, but sec−1 (−2) 6=
, since
Note that sec
3
3
2π h π h 3π 6∈ 0,
∪ π,
3
2
2
3
4π
3
√
π
π
8π
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]
Therefore
and
π π
=
sin arcsin
7
7
but
8π
arcsin sin
7
arcsin(sin x) = x if x ∈ [−π/2, π/2]
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
and
sec θ =
2
and sec arcsin
5
.
Solution 1: We have
cos θ
±
cot θ =
=
sin θ
Since −
hence
1 − sin2 θ
sin θ
1
1
= p
cos θ
± 1 − sin2 θ
π
2
π
≤ arcsin x ≤ , it follows that cos(arcsin x) ≥ 0. Therefore if θ = arcsin , then
2
2
5
p
1 − sin2 θ
1
cot θ =
and sec θ = p
sin θ
1 − sin2 θ
2
cot arcsin
5
and
p
2
sec arcsin
5
=
s
=s
1
=s
2
1 − sin arcsin
5
2
sin arcsin
5
2
s
1 − sin2 arcsin
2
5
=
2
2
1−
√
5
21
=
2
2
5
5
1
2 = √21
2
1−
5
2
2
Solution 2: Put θ = arcsin , so sin θ = . Then
5
5
√
2
2
5
21
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) .
4
5 ✦✦✦
✦
✦
✦✦
✦✦
θ√
21
✦
2
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 6∈ [−1, 1], that is, 2 is not in the domain of
−1
−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 = p
=√
=√
2
−1
1 + 22
5
1 + tan (tan 2)
2
2
Solution 2: Put θ = tan−1 2 = tan−1 , so tan θ = . Then
1
1
EXAMPLES: Evaluate sin cot
−1
1
−
2
and cos cot
5
−1
✁
5✁
✁
✁
✁
✁θ
2
sin tan−1 2 = sin θ = √
5
√
✁
✁
1
−
2
.
1
✁
✁
2
EXAMPLES: Evaluate sin cot
−1
1
−
2
and cos cot
−1
1
−
2
.
Solution 1: We have
sin θ = ± √
Since 0 < cot
−1
and
1 + cot2 θ
1
−1
cos θ = ± √
1 + cot θ
1 + cot2 θ
x) > 0. Therefore if θ = cot
and
2
cot θ
cos θ = √
−1
1
−
, then
2
cot θ
1 + cot2 θ
1
1
2
1
−1
−
sin cot
=s
=√
=s
2
2
5
1
1
1 + cot2 cot−1 −
1+ −
2
2
1
1
cot cot
−
−
1
1
2
2
cos cot−1 −
=s
=s
= −√
2
2
5
1
1
1 + cot2 cot−1 −
1+ −
2
2
Solution 2: Put θ = cot
and
and
x < π, it follows that sin(cot
sin θ = √
hence
1
−1
−1
1
−1
, so cot θ = − =
. Then
2
2
1
2
−1
−
= sin θ = √
sin cot
2
5
1
−
2
cos cot
−1
❆
❆
2
❆
❆
1
−
2
❆
❆
❆
❆
1
= cos θ = − √
5
❆ √
❆ 5
❆
❆
❆
❆
-1
6
❆
❆
❆
❆θ
EXAMPLE: A 40-ft ladder leans against a building. If the base of the ladder is 6 ft from the
base of the building, what is the angle formed by the ladder and the building?
Solution: First we sketch a diagram as in the Figure below (left). If θ is the angle between the
ladder and the building, then
6
= 0.15
sin θ =
40
Now we use sin−1 to find θ:
θ = sin−1 (0.15) ≈ 8.6◦
EXAMPLE: A lighthouse is located on an island that is 2 mi off a straight shoreline (see the
Figure above on the right). Express the angle formed by the beam of light and the shoreline
in terms of the distance d in the figure.
Solution: From the figure we see that
tan θ =
2
d
Taking the inverse tangent of both sides, we get
2
tan (tan θ) = tan
d
2
−1
θ = tan
d
−1
−1
7