Precalculus (6th Edition)

$\displaystyle \csc^{-1}(-\frac{1}{2})$ does not exist
$y=\csc^{-1}x$ Domain: $(-\infty, -1]\cup[1, \infty)$ Range: $[-\displaystyle \frac{\pi}{2},0)\cup(0,\frac{\pi}{2}]$ --------------- $-\displaystyle \frac{1}{2}\not\in(-\infty, -1]\cup[1, \infty)$. So, $-\displaystyle \frac{1}{2}$ is not in the domain of $\csc^{-1}x$, meaning that it is not in the range of $\csc x.$ (There is no y from $[-\displaystyle \frac{\pi}{2},0)\cup(0,\frac{\pi}{2}]$ such that $\displaystyle \csc y=\frac{1}{\sin y}=-\frac{1}{2},$ as this would mean that $\sin y=-2 < -1)$ $\displaystyle \csc^{-1}(-\frac{1}{2})$ does not exist