Question

If $-\frac{3\pi }{4}\le \left({\sin }^{-1}x-\frac{\pi }{4}\right)\le \frac{\pi }{4}$ for all $x\in \left[-1,1\right]$ then prove that $0\le {\left({\sin }^{-1}x-\frac{\pi }{4}\right)}^2\le \frac{9{\pi }^2}{16}$

Answer 1

$\begin{array}{c}-\frac{3\pi }{4}\le \left({\sin }^{-1}x-\frac{\pi }{4}\right)\le \frac{\pi }{4}\\ -\frac{3\pi }{4}\le A\le \frac{\pi }{4}\end{array}$
Break it into two equations
$-\frac{3\pi }{4}\le A\le \mathrm{0.....}\left(1\right)$ and $0\le A\le \frac{\pi }{4}.....\left(2\right)$
squaring on each side of eq$\left(1\right)$
$\frac{9{\pi }^2}{16}\le A^2\le \mathrm{0.....}\left(A\right)$
squaring on each side of eq$\left(2\right)$
$\frac{{\pi }^2}{16}\le A^2\le \mathrm{0.....}\left(B\right)$
From equation $A$ and $B$
$\frac{9{\pi }^2}{16}\le A^2\le 0$