## Precalculus (6th Edition)

$\sin^{-1}2$ does not exist.
Solve for radians, then convert to degrees. $y=\sin^{-1}x$ Domain: $[-1, 1]$ Range: $[-\displaystyle \frac{\pi}{2}, \displaystyle \frac{\pi}{2}]$ Quadrants (unit circle): I and IV -------- $y$ is the number from $[-\displaystyle \frac{\pi}{2}, \displaystyle \frac{\pi}{2}]$ such that $\sin y=2.$ But, such a y does not exist, as sine can not be greater than 1. More directly, $2$ is not in the domain of $\sin^{-1}x$, so $\sin^{-1}2$ does not exist.