Answer
$-45^0$
Work Step by Step
The value of $\theta$ must be in the interval $[-90^o, 90^o]$.
Note that $\sin{45^o} = \frac{\sqrt{2}}{2}$.
Since $\sin{-\theta} = -\sin{\theta}$,
Then $\sin{(-45^o)}=-\sin{45^o} = -\frac{\sqrt{2}}{2}$
Thus, $\arcsin{(-\frac{\sqrt2}{2})}=-45^o$
