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

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(D) Let $x = \cos 	heta$,where $	heta = \cos^{-1} x$. Since $x \in [-1, 1]$,$	heta \in [0, \pi]$.Substituting $x = \cos 	heta$ into the right-hand

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

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(D) Let $x = \cos 	heta$,where $	heta = \cos^{-1} x$. Since $x \in [-1, 1]$,$	heta \in [0, \pi]$.Substituting $x = \cos 	heta$ into the right-hand side:$\sin^{-1} (2 \cos 	heta \sqrt{1 - \cos^2 	heta}) = \sin^{-1} (2 \cos 	heta \sin 	heta) = \sin^{-1} (\sin 2 	heta)$.For this to equal $2 \cos^{-1} x = 2 	heta$,we require $\sin^{-1} (\sin 2 	heta) = 2 	heta$.This holds when $2 	heta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$,which implies $	heta \in [-\frac{\pi}{4}, \frac{\pi}{4}]$.Given $	heta \in [0, \pi]$,the intersection is $	heta \in [0, \frac{\pi}{4}]$.Since $	heta = \cos^{-1} x$,we have $0 \leq \cos^{-1} x \leq \frac{\pi}{4}$.Taking the cosine of all sides (noting that $\cos$ is a decreasing function in this interval):$\cos(0) \geq x \geq \cos(\frac{\pi}{4})$,which gives $1 \geq x \geq \frac{1}{\sqrt{2}}$ 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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