Section 6.8, The Inverse Trigonometric Functions and Their
Derivatives
Homework: 6.8 #1-73 odds
Since all trigonometric functions are 2π-periodic, they are not one-to-one. Therefore, in order to
define inverse functions, we need to restrict the domain.
For sine and cosine, we restrict their domains to [−π/2, π/2] and [0, π ], respectively. Then,
x= sin−1y⇔y= sin x, −π
2≤x≤π
2
x= cos−1y⇔y= cos x, 0≤x≤π
For tangent and secant, the domains are restricted to (−π/2, π/2) and [0, π/2) ∪(π/2, π], respectively.
Furthermore,
x= tan−1y⇔y= tan x, −π
2<x<π
2
x= sec−1y⇔y= sec x, 0≤x≤π, x 6=π
2
This book has no need to define csc−1or cot−1. Also, the domain of secant is sometimes restricted
in different ways.
There is alternate notation for inverse trigonometric functions: arcsin x= sin−1x, arccos x=
cos−1x, and arctan x= tan−1x.
Examples
Calculate each of the following:
1. sin−11
2
Since sin(π/6) = 1/2, sin−11
2=π
6.
2. arccos(−√2
2)
Since cos(3π/4) = −√2/2, arccos(−√2
2) = 3π
4.
3. arccos(cos(5π
4))
Since 5π/4 is not in the range for arccos, we can use that cos( 5π
4) = cos(3π
4). Therefore,
arccos(cos(5π
4)) = 3π
4.
4. tan(tan−1(1))
tan(tan−1(1)) = 1 since tan and tan−1are inverse functions.
5. cos(sec−1(2))
We can use that secx= 1/cos x, so cos(sec−1(2)) = cos(cos−1(1/2)) = 1/2.
6. cos sin−12
3
We want the value of the cosine for the angle whose sine is 2/3. Drawing a right triangle
with hypotenuse 1, we get that the two legs have lengths 2/3 and p1−(2/3)2=√5/3, so
cos sin−12
3=√5
3.
There are many identities that can be used for inverse trigonometric functions. Some of them are:
Theorem A
1. sin(cos−1x) = √1−x2
2. cos(sin−1x) = √1−x2
1
