Section 6.4 Inverse Trigonometric Functions and Right Triangles
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
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
