#### Answer

$-45^{o}$

#### Work Step by Step

Solve for radians, then convert to degrees.
$y=\sin^{-1}x =\arcsin x$
Domain: $[-1, 1] $
Range: $[-\displaystyle \frac{\pi}{2}, \displaystyle \frac{\pi}{2}]$
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$y$ is the number from $[-\displaystyle \frac{\pi}{2}, \displaystyle \frac{\pi}{2}]$
such that $\displaystyle \sin y=-\frac{\sqrt{2}}{2}.$
In quadrant I, $\displaystyle \sin(\frac{\pi}{4})=\frac{\sqrt{2}}{2}.$
In quadrant IV,
$\displaystyle \sin(-\frac{\pi}{4})=-\frac{\sqrt{2}}{2}\qquad$and$\displaystyle \quad -\frac{\pi}{4}\in[-\frac{\pi}{2}, \displaystyle \frac{\pi}{2}]$,
so
$y =-\displaystyle \frac{\pi}{4}$
To convert radians to degrees, multiply y with $\displaystyle \frac{180^{o}}{\pi}$
$\displaystyle \theta=-\frac{\pi}{4}\cdot\frac{180^{o}}{\pi}=-45^{o}$