11.12. Find the relationship between $\arcsin \cos \arcsin x \quad$ and $\quad \arccos \sin \arccos x$.

## 11.3. Equations
11.12. Answer: $\arcsin \cos \arcsin x+\arccos \sin \arccos x=\frac{\pi}{2}$.

Let $\arcsin \cos \arcsin x=\alpha$ and $\arccos \sin \arccos x=\beta$. Then $0 \leqslant$ $\leqslant \alpha, \beta \leqslant \pi / 2$. Indeed, $0 \leqslant \cos \arcsin x \leqslant 1$, since $-\frac{\pi}{2} \leqslant$ $\leqslant \arcsin x \leqslant \frac{\pi}{2}$, and $0 \leqslant \sin \arccos x \leqslant 1$, since $0 \leqslant \arccos x \leqslant \pi$. Further, $\sin \alpha=\cos \arcsin x$, so $\arcsin x= \pm\left(\frac{\pi}{2}-\alpha\right)$ and $\sin \left( \pm\left(\frac{\pi}{2}-\alpha\right)\right)=$ $= \pm \cos \alpha ; \cos \beta=\sin \arccos x$, so $\arccos x=\frac{\pi}{2} \mp \beta$ and $x=$ $=\cos \left(\frac{\pi}{2} \mp \beta\right)= \pm \sin \beta$. From the fact that $\cos \alpha=\sin \beta(= \pm x)$, it follows that $\alpha+\beta=\frac{\pi}{2}$.