Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 7 - Trigonometric Identities and Equations - 7.5 Inverse Circular Functions - 7.5 Exercises: 11


Sine and cosine do not have a common interval on which BOTH are one-to-one, and BOTH have range $[-1,1]$.

Work Step by Step

Sine is not one-to-one on $[0, \pi]$ and the interval does not contain x-values with negative function values. So, the whole range is not covered on this interval. On the other hand, the interval $[0, \pi]$ satisfies the cosine function needs (values range from -1 to 1, and cosine is one-to-one here). The same problem will occur with cosine, when restricted to the interval $ [-\displaystyle \frac{\pi}{2}, \displaystyle \frac{\pi}{2}]$, which fits the sine function.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.