Question

The value of $\cos \left(\frac{1}{2}{\cos }^{-1}\frac{1}{8}\right)$ is equal to

• A. $\frac{3}{4}$
• B. $-\frac{3}{4}$
• C. $\frac{1}{16}$
• D. $\frac{1}{4}$

Answer 1

Answer: (A)

$\cos \left(\frac{1}{2}{\cos }^{-1}\frac{1}{8}\right)=x$ (let)
$\Rightarrow {\cos }^{-1}\frac{1}{8}=2{\cos }^{-1}x$
$={\cos }^{-1}\left(2x^2-1\right)$
$\Rightarrow \frac{1}{8}=2x^2-1$
$\Rightarrow 2x^2=\frac{1}{8}+1=\frac{9}{8}$
$\Rightarrow x^2=\frac{9}{16}$
$\Rightarrow x=\sqrt{\frac{9}{16}}=\frac{3}{4}$
$[\because 0\le x\le 1$]