Show that $\sin^{-1}(2x\sqrt{1-x^2}) = 2\sin^{-1}x$ for $-\frac{1}{\sqrt{2}} \leq x \leq \frac{1}{\sqrt{2}}$.

(N/A) Let $x = \sin \theta$. Then $\theta = \sin^{-1} x$.
Since $-\frac{1}{\sqrt{2}} \leq x \leq \frac{1}{\sqrt{2}}$,we have $-\frac{\pi}{4} \leq \theta \leq \frac{\pi}{4}$,which implies $-\frac{\pi}{2} \leq 2\theta \leq \frac{\pi}{2}$.
Now,consider the expression $\sin^{-1}(2x\sqrt{1-x^2})$.
Substituting $x = \sin \theta$,we get $\sin^{-1}(2\sin \theta \sqrt{1-\sin^2 \theta})$.
Using the identity $\cos^2 \theta = 1 - \sin^2 \theta$,we have $\sin^{-1}(2\sin \theta \sqrt{\cos^2 \theta}) = \sin^{-1}(2\sin \theta |\cos \theta|)$.
Since $-\frac{\pi}{4} \leq \theta \leq \frac{\pi}{4}$,$\cos \theta \geq 0$,so $|\cos \theta| = \cos \theta$.
Thus,the expression becomes $\sin^{-1}(2\sin \theta \cos \theta) = \sin^{-1}(\sin 2\theta)$.
Since $2\theta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$,$\sin^{-1}(\sin 2\theta) = 2\theta$.
Substituting back $\theta = \sin^{-1} x$,we get $2\sin^{-1} x$.