289. Find the relationship between $\arcsin \cos \arcsin x$ and $\arccos \sin \arccos x$
289. Let

$$
\arcsin \cos \operatorname{arsin} x=\alpha
$$

The angle $\alpha$ is within the range $0 \leqslant \alpha \leqslant \frac{\pi}{2}$, because $0 \leqslant \cos \arcsin x \leqslant 1$ (since $-\frac{\pi}{2} \leqslant \arcsin x \leqslant \frac{\pi}{2}$). Further, $\sin \alpha=\cos \arcsin x$; therefore,

$$
\arcsin x= \pm\left(\frac{\pi}{2}-\alpha\right) \text { and } x=\sin \left[ \pm\left(\frac{\pi}{2}-\alpha\right)\right]= \pm \cos \alpha
$$

Similarly, if $\arccos \sin \arccos x=\beta$, then $0 \leqslant \beta \leqslant \frac{\pi}{2}$ (since $0 \leqslant \sin \arccos x \leqslant 1$, as $0 \leqslant \arccos x \leqslant \pi$) and $\cos \beta=\sin \arccos x_{\mathbf{i}}$; therefore,

$$
\arccos x=\frac{\pi}{2} \mp \beta \text { and } x=\cos \left(\frac{\pi}{2} \mp \beta\right)= \pm \sin \beta
$$

From the fact that $\cos \alpha=\sin \beta(= \pm x)$, we conclude:

$$
\alpha+\beta=\arcsin \cos \arcsin x+\arccos \sin \arccos x=\frac{\pi}{2}
$$