Example 6. Find the inverse function of the function $y=\sin x, x \in[-\pi$, $-\frac{\pi}{2}$].
Given $\because \quad x \in\left[-\pi,-\frac{\pi}{2}\right]$,
$$
\therefore \quad y \in[-1,0] \text {. }
$$

According to the definition of the inverse sine, if $A \in\left(-\frac{\pi}{2}\right.$, $\left.\frac{\pi}{2}\right]$, and $\sin A=N, \quad|N| \leqslant 1$, then $\arcsin N=A$. Therefore, the problem must be transformed to solve it.
$$
\begin{array}{l}
\because \quad-\pi \leqslant s \leqslant-\frac{\pi}{2}, \\
\therefore \quad 0 \leqslant \theta=x+\pi \leqslant \frac{\pi}{2} . \\
\text { Substituting } y=\sin x=\sin (\theta-\pi)=-\sin \theta, \\
\because 0 \in\left[0, \frac{\pi}{2}\right],
\end{array}
$$
$\therefore$ Solving inversely, we get $0=\arcsin (-y)$. That is, $x+\pi=\arcsin (-y)$.

Therefore, by swapping $x$ and $y$, we get $y=-\pi-\arcsin x$, $x \in[-1,0]$, which is the required inverse function.