If 2 sin^{-1} x = sin^{-1}(2x sqrt{1-x^2}),th | vedclass

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(C) We know that $2 \sin^{-1} x = \sin^{-1}(2x \sqrt{1-x^2})$ holds true when $-\frac{1}{\sqrt{2}} \le x \le \frac{1}{\sqrt{2}}$.Let $x = \sin 	heta$

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$[\frac{-1}{\sqrt{2}}, \frac{1}{\sqrt{2}}]$

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(C) We know that $2 \sin^{-1} x = \sin^{-1}(2x \sqrt{1-x^2})$ holds true when $-\frac{1}{\sqrt{2}} \le x \le \frac{1}{\sqrt{2}}$.Let $x = \sin 	heta$,where $	heta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.Then $2 \sin^{-1}(\sin 	heta) = \sin^{-1}(2 \sin 	heta \sqrt{1-\sin^2 	heta}) = \sin^{-1}(2 \sin 	heta \cos 	heta) = \sin^{-1}(\sin 2	heta)$.For $2	heta = 2 \sin^{-1} x$ to be valid,$2	heta$ must lie in the principal value branch of $\sin^{-1}$,which is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.So,$-\frac{\pi}{2} \le 2	heta \le \frac{\pi}{2} \implies -\frac{\pi}{4} \le 	heta \le \frac{\pi}{4}$.Taking sine on all sides,$\sin(-\frac{\pi}{4}) \le \sin 	heta \le \sin(\frac{\pi}{4})$,which gives $-\frac{1}{\sqrt{2}} \le x \le \frac{1}{\sqrt{2}}$.Thus,$x \in [-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}]$.

If $2 \sin^{-1} x = \sin^{-1}(2x \sqrt{1-x^2})$,then $x \in$ . . . . . . .

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