Question

$2si{n}^{-1}\sqrt{\frac{1-x}{2}}=$

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

Answer 1

Answer: (C)

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