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Inverse Trig Functions: For a function to have an inverse, it must be one-to-one – that is, it must pass the Horizontal Line test. The graph of the sine function does not pass this test.
If we restrict our domain to − π 2 ≤ x ≤ π 2 , we have:
- On the interval
[
π 2
π 2
]
, y = sin x is increasing.
- On the interval
[
− π 2 , π 2
]
, y = sin x takes on its full range of values, − 1 ≤ sin x ≤ 1.
- On the interval
[
π 2
π 2
]
, y = sin x is one-to-one.
Hence, on the restricted domain
[
− π 2 , π 2
]
, y = sin x has a unique inverse called the inverse sine function, denoted by y = arcsin x or y = sin−^1 x.
Note: The notation sin−^1 x is consistent with the notation f −^1 (x); here,
sin−^1 x 6 =
sin x. The inverse sine function is defined by y = arcsin x if and only if sin y = x,
where − 1 ≤ x ≤ 1 and − π 2 ≤^ y^ ≤^
π
- The domain of^ y^ = arcsin^ x^ is [−^1 ,^ 1], and the range in
[
π 2 ,
π 2
]
Ex: If possible, find the exact value of
arcsin
sin−^1 (−1) arcsin 3
Since sin
− π 4
for − π 2 ≤ y ≤ π 2 , we have arcsin
= − π 4
Since sin
π 2
= −1 for − π 2 ≤^ y^ ≤^
π 2 , we have sin
− (^1) (−1) = − π
Since − 1 ≤ sin x ≤ 1 for all real numbers, arcsin 3 has no solution. To define the inverse for the cosine function, we restrict our domain to 0 ≤ x ≤ π.
Definitions of the Inverse Trigonometric Functions: Function Domain Range y = arcsin x = sin−^1 x if and only if sin y = x − 1 ≤ x ≤ 1 − π 2 ≤^ y^ ≤^
π 2 y = arccos x = cos−^1 x if and only if cos y = x − 1 ≤ x ≤ 1 0 ≤ y ≤ π y = arctan x = tan−^1 x if and only if tan y = x −∞ ≤ x ≤ ∞ − π 2 < y < π 2
To obtain these graphs, you can either follow the process on page 196 by filling out a table, plotting those points, and then filling in the curve, or you can reflect the graph of your function (with its restricted domain) across the line y = x.
Ex: arccos
5 π 6 since cos
5 π 6
and 0 ≤ 5 π 6 ≤ π.
Ex: tan−^1
π 3 since tan
( (^) π 3
3 and − π 2
π 3
π 2
Inverse Property of Trigonometric Functions If − 1 ≤ x ≤ 1 and − π 2 ≤^ y^ ≤^
π 2 , then sin (arcsin x) = x and arcsin (sin y) = y. If − 1 ≤ x ≤ 1 and 0 ≤ y ≤ π, then cos (arccos x) = x and arccos (cos y) = y.
If x is a real number and − π 2 ≤^ y^ ≤^
π 2 , then tan (arctan x) = x and arctan (tan y) = y.
Ex: If possible, find the exact value.
cos−^1
cos
3 π 2
sin
arcsin
Note that − 3 π 2 is not in the interval [0, π], but the coterminal angle − 3 π 2
- 2π = π 2 is in the interval. So, we have
cos−^1
cos
− 3 π 2
= cos−^1
cos
( (^) π 2
= π 2
Since − 1 ≤
≤ 1, we have
sin
arcsin
Ex: Find the exact values of
tan
arccos
sin
arctan
