Section 5.5 Inverse Trigonometric Functions and Their Graphs
Section 5.5 Inverse Trigonometric Functions and Their Graphs 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
1 , cos x
tan−1 x,
cot−1 x,
1 , tan x
1 , cot x
FUNCTION DOMAIN −1 sin x [−1, 1] −1 cos x [−1, 1] tan−1 x (−∞, +∞) cot−1 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]
FUNCTION sin−1 x cos−1 x tan−1 x cot−1 x sec−1 x csc−1 x
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]
EXAMPLES: −1
(a) sin
π π π [ π π] 1 = , since sin = 1 and ∈ − , . 2 2 2 2 2
( π) π π [ π π] (b) sin−1 (−1) = − , since sin − = −1 and − ∈ − , . 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 3 π π π [ π π] = , since sin = and ∈ − , . 2 3 3 2 3 2 2
√ (e) sin
−1
√ 2 2 π π π [ π π] = , since sin = and ∈ − , . 2 4 4 2 4 2 2
√ (f) sin
−1
EXAMPLES: −1
cos
π 0= , 2
tan−1 1 =
π , 4
−1
cos
1 = 0,
−1
−1
cos (−1) = π,
π tan−1 (−1) = − , 4
tan−1
√
cos
3=
π , 3
tan−1
EXAMPLES: Find sec−1 1, sec−1 (−1), and sec−1 (−2).
3
√ 3 2 π π −1 cos = , cos = 2 6 2 4 ( ) 1 π 1 π √ = , tan−1 − √ =− 6 6 3 3
1 π = , 2 3
√
−1
FUNCTION sin−1 x cos−1 x tan−1 x cot−1 x sec−1 x csc−1 x
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]
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 π = −1,
sec 0 = 1, and 0, π, Note that sec
4π = −2 3
4π [ π ) [ 3π ) ∈ 0, ∪ π, 3 2 2
2π is also −2, but 3 sec−1 (−2) 6=
since
sec
4π 3
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
FUNCTION sin−1 x cos−1 x tan−1 x cot−1 x sec−1 x csc−1 x
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]
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
EXAMPLES: Evaluate ( ) ( ( π) π) 7π (a) sin arcsin , arcsin sin , and arcsin sin . 6 6 6 ( ) ( ( π) π) 8π (b) sin arcsin , arcsin sin , and arcsin sin . 7 7 7 ( ( )) ( ) ( ) 2 2π 9π (c) cos arccos − , arccos cos , and arccos cos . 5 5 5 Solution: Since arcsin x is the inverse of the restricted sine function, we have sin(arcsin x) = x if x ∈ [−1, 1]
arcsin(sin x) = x if x ∈ [−π/2, π/2]
and
Therefore ( ( π) π π) (a) sin arcsin = arcsin sin = , but 6 6 6 ( ) ( ) 7π 1 π arcsin sin = arcsin − =− 6 2 6 or ) ( ( ( π) ( ( π) π π )) 7π = arcsin − sin = − arcsin sin =− = arcsin sin π + arcsin sin 6 6 6 6 6 ( ( π) π π) (b) sin arcsin = arcsin sin = , but 7 7 7 ) ( ( (π )) ( ( π) π) π 8π = arcsin sin +π = arcsin − sin = − arcsin sin =− arcsin sin 7 7 7 7 7 (c) Similarly, since arccos x is the inverse of the restricted cosine function, we have cos(arccos x) = x if x ∈ [−1, 1] and arccos(cos x) = x if x ∈ [0, π] ( ( )) ( ) 2 2π 2 2π Therefore cos arccos − , but = − and arccos cos = 5 5 5 5 ( ) ( ( ( π) π 9π π )) = arccos cos = arccos cos = arccos cos 2π − 5 5 5 5 5
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
1 , cos x
tan−1 x,
cot−1 x,
1 , tan x
1 , cot x
FUNCTION DOMAIN −1 sin x [−1, 1] −1 cos x [−1, 1] tan−1 x (−∞, +∞) cot−1 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]
FUNCTION sin−1 x cos−1 x tan−1 x cot−1 x sec−1 x csc−1 x
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]
EXAMPLES: −1
(a) sin
π π π [ π π] 1 = , since sin = 1 and ∈ − , . 2 2 2 2 2
( π) π π [ π π] (b) sin−1 (−1) = − , since sin − = −1 and − ∈ − , . 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 3 π π π [ π π] = , since sin = and ∈ − , . 2 3 3 2 3 2 2
√ (e) sin
−1
√ 2 2 π π π [ π π] = , since sin = and ∈ − , . 2 4 4 2 4 2 2
√ (f) sin
−1
EXAMPLES: −1
cos
π 0= , 2
tan−1 1 =
π , 4
−1
cos
1 = 0,
−1
−1
cos (−1) = π,
π tan−1 (−1) = − , 4
tan−1
√
cos
3=
π , 3
tan−1
EXAMPLES: Find sec−1 1, sec−1 (−1), and sec−1 (−2).
3
√ 3 2 π π −1 cos = , cos = 2 6 2 4 ( ) 1 π 1 π √ = , tan−1 − √ =− 6 6 3 3
1 π = , 2 3
√
−1
FUNCTION sin−1 x cos−1 x tan−1 x cot−1 x sec−1 x csc−1 x
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]
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 π = −1,
sec 0 = 1, and 0, π, Note that sec
4π = −2 3
4π [ π ) [ 3π ) ∈ 0, ∪ π, 3 2 2
2π is also −2, but 3 sec−1 (−2) 6=
since
sec
4π 3
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
FUNCTION sin−1 x cos−1 x tan−1 x cot−1 x sec−1 x csc−1 x
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]
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
EXAMPLES: Evaluate ( ) ( ( π) π) 7π (a) sin arcsin , arcsin sin , and arcsin sin . 6 6 6 ( ) ( ( π) π) 8π (b) sin arcsin , arcsin sin , and arcsin sin . 7 7 7 ( ( )) ( ) ( ) 2 2π 9π (c) cos arccos − , arccos cos , and arccos cos . 5 5 5 Solution: Since arcsin x is the inverse of the restricted sine function, we have sin(arcsin x) = x if x ∈ [−1, 1]
arcsin(sin x) = x if x ∈ [−π/2, π/2]
and
Therefore ( ( π) π π) (a) sin arcsin = arcsin sin = , but 6 6 6 ( ) ( ) 7π 1 π arcsin sin = arcsin − =− 6 2 6 or ) ( ( ( π) ( ( π) π π )) 7π = arcsin − sin = − arcsin sin =− = arcsin sin π + arcsin sin 6 6 6 6 6 ( ( π) π π) (b) sin arcsin = arcsin sin = , but 7 7 7 ) ( ( (π )) ( ( π) π) π 8π = arcsin sin +π = arcsin − sin = − arcsin sin =− arcsin sin 7 7 7 7 7 (c) Similarly, since arccos x is the inverse of the restricted cosine function, we have cos(arccos x) = x if x ∈ [−1, 1] and arccos(cos x) = x if x ∈ [0, π] ( ( )) ( ) 2 2π 2 2π Therefore cos arccos − , but = − and arccos cos = 5 5 5 5 ( ) ( ( ( π) π 9π π )) = arccos cos = arccos cos = arccos cos 2π − 5 5 5 5 5
