4.7
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/2. The
domain of y = arcsin x is [-1, 1] and the
range is [ -π/2 , π/2].
Remember from section 1.6 that for a function to have an
inverse function, it must be one-to-one (pass the Horizontal Line
Test). If you graph y = sin x if does not pass the Horizontal Line
Test. But if you take a portion of the graph, say between –π/2
and π/2, it does pass the test. Also,
1. On the interval [-π/2, π/2] the function y = sin x is increasing.
2. On the interval [-π/2, π/2] y = sin x take on its full range of
values -1 ≤ sin x ≤ 1.
3. On the interval [ -π/2, π/2] y = sin x is one-to-one.
Sine y = sin x is one-to-one it has a unique inverse function
called the inverse sine function denoted as y = arcsin x or
y = sin-1 x. Arcsin x means the angle (or arc) whose sine is x.
If possible, find the exact value
A) Arcsin (-1/2)
B) Sin-1
 3


 2 


C) Sin-1 2
Sketch a graph of y = arcsin x by hand.
Function
Domain
y = arcsin x
-1 ≤ x ≤ 1
y = arccos x
-1 ≤ x ≤ 1
y = arctan x
-∞ ≤ x ≤ ∞
Range
-π/2 ≤ y ≤ π/2
0≤y≤π
-π/2 ≤ y ≤ π/2
The graphs of these functions can be found on page 311.
Find the exact values.
A) arc cos  22 
B) Cos-1 (-1)
C) Arctan 0
D) Tan-1 (-1)
Use a calculator to approximate the value (if possible).
A) Arctan (-8.45)
B) Sin-1 (.2447)
C) Arccos 2
Remember from 1.6 that if a function has an inverse the
following properties must be true.
f(f-1(x)) = x
f-1(f(x)) = x
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.
If possible, find the exact value.
A) tan[arctan(-5)]
B)
Arcsin(sin(5π/3))
C) Cos(cos-1π)