Math 150 – Fall 2015
Section 8E
1 of 2
Section 8E – Inverse Sine Function
y = sin x
Recall, for a function to have an inverse, it must be one-to-one. For sin x to have an
inverse, we will restrict its domain to [− π2 , π2 ].
Definition. We define arcsin as the function with domain [−1, 1] and range [− π2 , π2 ]
such that
arcsin x = sin−1 x = y
if and only if
sin y = x
where y ∈ [− π2 , π2 ].
y = sin x, x ∈ [− π2 , π2 ]
y = arcsin x
Example 1. Fill in the following table:
y = sin x, x ∈ [− π2 , π2 ]
y = sin−1 x
Domain:
Range:
Note.
• When we have y = sin x, we think of x as an angle and y is the value of
sin at the angle x. The inverse x = arcsin y means that we are looking for an
angle x where y = sin x.
Math 150 – Fall 2015
Section 8E
2 of 2
• If we want to evaluate y = arcsin 0, there are many angles x where sin x = 0, such
as . . . , −2π, −π, 0, π, 2π, 3π . . . etc. The restriction tells us to only choose angles
in the interval [− π2 , π2 ] so arcsin 0 = 0.
• The function sin−1 x = arcsin x is the inverse of y = sin x. This is different from
−1
the reciprocal of sin x which is (sin x) = sin1 x = csc x.
Theorem. For the composition of y = sin x and y = arcsin−1 x, we have
sin sin−1 x = x for x ∈ [−1, 1]
and
h π πi
sin−1 (sin x) = x for x ∈ − ,
2 2
Example 2. Evaluate the following functions:
√ √
(a) If sin π4 = 22 , then sin−1 22 =
(b) sin−1
(c) sin−1
√ 3
2
=
√ − 2
2
=
=
(d) sin−1 sin 7π
6
(e) cos sin−1
−5
7
(f) cot sin−1
2
3
=
=