## Precalculus (6th Edition)

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