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

Answer

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.
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