Discussion Forum

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

Answer:

Explanation:

Let  $\cos^{-1}\frac{1}{8}=\theta, where 0<\theta<\frac{\pi}{2}$

 $\Rightarrow \frac{1}{2}\cos^{-1}\frac{1}{8}=\frac{1}{2}\theta,$

$\Rightarrow \cos(\frac{1}{2}\cos^{-1}\frac{1}{8})=\cos\frac{1}{2}\theta,$

Now, $\cos^{-1}\frac{1}{8}=\theta\Rightarrow\cos\theta=\frac{1}{8}$

$\Rightarrow 2\cos^{2}\frac{\theta}{2}-1=\frac{1}{8}$

$\Rightarrow \cos^{2}\frac{\theta}{2}=\frac{9}{16}\Rightarrow\cos\frac{\theta}{2}=\frac{3}{4}$

$[\because 0<\frac{\theta}{2}<\frac{\pi}{4},so \cos\frac{\theta}{2}\neq-\frac{3}{4}]$