8.42 Let $f(x)=\arccos x+2 \arcsin x$, then $f^{-1}(x)$ is
(A) $\sin x, x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.
(B) $\cos x, x \in[0, \pi]$.
(C) $-\sin x, x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.
(D) $-\cos x, x \in[0, \pi]$.
(2nd "Hope Cup" National Mathematics Invitational Competition, 1991)
[Solution] If we let $y=f(x)$, then we have
$$
y=(\arcsin x+\arccos x)+\arcsin x=\frac{\pi}{2}+\arcsin x,
$$

Therefore,
$$
x=\sin \left(y-\frac{\pi}{2}\right)=-\cos y .
$$

Noting that $-\frac{\pi}{2} \leqslant \arcsin x \leqslant \frac{\pi}{2}$, we have $0 \leqslant \frac{\pi}{2}+\arcsin x \leqslant \pi$, thus
$$
f^{-1}(x)=-\cos x, \quad x \in[0, \pi] .
$$

Therefore, the answer is $(D)$.