Given that, ${\sin ^{ - 1}}x = 2{\sin ^{ - 1}}a$ We know, $- {\pi \over 2} \le {\sin ^{ - 1}}x \le {\pi \over 2}$ $\therefore$ $\,\,\,$ $- {\pi \over 2} \le 2{\sin ^{ - 1}}a \le {\pi \over 2}$ $\Rightarrow - {\pi \over 4} \le {\sin ^{ - 1}}a \le {\pi \over 4}$ $\Rightarrow \sin \left( { - {\pi \over 4}} \right) \le a \le \sin \left( {{\pi \over 4}} \right)$ $\Rightarrow - {1 \over {\sqrt 2 }} \le a \le {1 \over {\sqrt 2 }}$ $\therefore$ $\,\,\,\,\,\,\,\,\left| a \right| \le {1 \over {\sqrt 2 }}$