Answer
$-\dfrac{\pi}{2}$
Work Step by Step
We are given the expression: $\sin^{-1} \left(-1\right)$
$y=\sin^{-1} x$ is the value of $y$ so that $\sin y=x$, where $-\dfrac{\pi}{2}\leq y\leq\dfrac{\pi}{2}$.
So $y=\sin^{-1} \left(-1\right)$ is the value of $y$ so that $\sin y=-1$, where $-\dfrac{\pi}{2}\leq y\leq \dfrac{\pi}{2}$
From the unit circle we have:
$x=\sin \left(-\dfrac{\pi}{2}\right)$
Therefore the value of $y$ for which $x=-1$ in the interval $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$ is $-\dfrac{\pi}{2}$, so we got:
$\sin^{-1} \left(-1\right)=-\dfrac{\pi}{2}$
