11.2. Plot on the coordinate plane the set of points satisfying the inequality $\arcsin x+\arcsin y>\frac{\pi}{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_b09ddab305ed9897653ag-1.jpg?height=284&width=326&top_left_y=906&top_left_x=131)
Solution. Note that the given set is in the first quadrant, since if $x\frac{\pi}{2}-\arcsin x>\frac{\pi}{2}$. Therefore, we consider $x, y \in[0,1]$. We have $\arcsin y>\frac{\pi}{2}-\arcsin x \Leftrightarrow \sin (\arcsin y) \geq \sin \left(\frac{\pi}{2}-\arcsin x\right) \quad$ (the inequalities are equivalent, since for $\quad x \geq 0 \quad \frac{\pi}{2}-\arcsin x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ ). Thus, $y>\cos (\arcsin x)=\sqrt{1-x^{2}} \Leftrightarrow y^{2}>1-x^{2}$, (the inequalities are equivalent for $y \geq 0$ ). Therefore, we have the set in the square $[0,1] \times[0,1]$ outside the unit circle (see figure).