Question

$2\,sin^{-1}\sqrt{\frac{1-x}{2}} =$

• A. $cos^{-1}\, x$
• B. $2\,cos^{-1}\sqrt{\frac{1+x}{2}}$
• C. Both $(a)$ and $(b)$
• D. None of these

Answer 1

Answer: (C)

Let $cos^{-1}\, x = y$. Then, $x = cos\, y$.
$\therefore 2\,sin^{-1}\sqrt{\frac{1-x}{2}}$
$= 2\,sin^{-1}\left\{\sqrt{\frac{1-cos\,y}{2}}\right\}$
$= 2\,sin^{-1}\left\{sin \frac{y}{2}\right\} = 2 \left(\frac{y}{2}\right) = y$
and, $2\,cos^{-1}\sqrt{\frac{1-x}{2}}$
$= 2\,cos^{-1}\left\{\sqrt{\frac{1+ cos\,y}{2}}\right\}$
$= 2\,cos^{-1}\left\{cos \frac{y}{2}\right\} = 2\left(\frac{y}{2}\right) = y$
Hence, $2\,cos^{-1}\,x = 2\,sin^{-1}\sqrt{\frac{1-x}{2}}$
$= 2\,cos^{-1} \sqrt{\frac{1+x}{2}}$