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: 7

Answer

a. $[-1, 1] $ b. $[-\displaystyle \frac{\pi}{2}, \displaystyle \frac{\pi}{2}]$ c. increasing d. because $-2$ is not in the range of $\sin x$

Work Step by Step

See figure 14 on p.699 (or the table on page 703 ) $y=\sin^{-1}x$ ($y$ is the number from $[-\displaystyle \frac{\pi}{2}, \displaystyle \frac{\pi}{2}]$ for which $\sin y=x$) Domain: $[-1, 1] $ Range:$ [-\displaystyle \frac{\pi}{2}, \displaystyle \frac{\pi}{2}]$ Quadrants (unit circle): I and IV. Figure 14: $\sin^{-1}x$ is increasing. For part (d), the domain of the inverse function equals the range of the function. There is no y for which $\sin y=-2$, so $\sin^{-1}(-2) $is not defined. a. $[-1, 1] $ b. $[-\displaystyle \frac{\pi}{2}, \displaystyle \frac{\pi}{2}]$ c. increasing d. because $-2$ is not in the range of $\sin x$
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