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

Let $x = \cos \theta$. Then $\theta = \cos ^{-1} x$.
Given the range $\frac{1}{\sqrt{2}} \leq x \leq 1$,we have $\cos \frac{\pi}{4} \leq \cos \theta \leq \cos 0$,which implies $0 \leq \theta \leq \frac{\pi}{4}$.
Thus,$0 \leq 2\theta \leq \frac{\pi}{2}$.
Now,substitute $x = \cos \theta$ into the expression:
$\sin ^{-1}(2 x \sqrt{1-x^{2}}) = \sin ^{-1}(2 \cos \theta \sqrt{1-\cos ^{2} \theta})$
$= \sin ^{-1}(2 \cos \theta \sin \theta)$
$= \sin ^{-1}(\sin 2 \theta)$
Since $0 \leq 2\theta \leq \frac{\pi}{2}$,$\sin ^{-1}(\sin 2 \theta) = 2 \theta$.
Substituting back $\theta = \cos ^{-1} x$,we get $2 \cos ^{-1} x$.
Hence,$\sin ^{-1}(2 x \sqrt{1-x^{2}}) = 2 \cos ^{-1} x$.