2 cos ^{-1} x = sin ^{-1} left( 2 x sqrt{1 - | vedclass

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(C) Let $\cos ^{-1} x = y$,then $x = \cos y$. Since the range of $\cos ^{-1} x$ is $[0, \pi]$,we have $0 \leq y \leq \pi$.The given equation is $2y =

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$\frac{1}{\sqrt{2}} \leq x \leq 1$

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(C) Let $\cos ^{-1} x = y$,then $x = \cos y$. Since the range of $\cos ^{-1} x$ is $[0, \pi]$,we have $0 \leq y \leq \pi$.The given equation is $2y = \sin ^{-1} (2 \cos y \sin y) = \sin ^{-1} (\sin 2y)$.For $\sin ^{-1} (\sin 2y) = 2y$,we must have $-\frac{\pi}{2} \leq 2y \leq \frac{\pi}{2}$,which implies $-\frac{\pi}{4} \leq y \leq \frac{\pi}{4}$.Combining $0 \leq y \leq \pi$ and $-\frac{\pi}{4} \leq y \leq \frac{\pi}{4}$,we get $0 \leq y \leq \frac{\pi}{4}$.Since $y = \cos ^{-1} x$,we have $0 \leq \cos ^{-1} x \leq \frac{\pi}{4}$.Taking cosine on all sides (noting that cosine is a decreasing function in $[0, \pi]$),we get $\cos(0) \geq x \geq \cos(\frac{\pi}{4})$.Thus,$1 \geq x \geq \frac{1}{\sqrt{2}}$,or $\frac{1}{\sqrt{2}} \leq x \leq 1$.

$2 \cos ^{-1} x = \sin ^{-1} \left( 2 x \sqrt{1 - x^2} \right)$ is valid for all values of $x$ satisfying

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