Question

The value of $4\cos \left(\frac{1}{2}{\left(\cos \right)}^{-1}\frac{1}{8}\right)$ is equal to

Answer 1

Answer: 3

Let ${\cos }^{-1}\frac{1}{8}=\theta$, where $0<\theta <\frac{\pi }{2}$.
$\Rightarrow \frac{1}{2}{\cos }^{-1}\frac{1}{8}=\frac{\theta }{2}$
$\Rightarrow \cos \left(\frac{1}{2}(\cos {)}^{-1}\frac{1}{8}\right)=\cos \left(\frac{\theta }{2}\right)$
Now, ${\cos }^{-1}\frac{1}{8}=\theta \Rightarrow \cos \theta =\frac{1}{8}$
$\Rightarrow 2{\cos }^2\frac{\theta }{2}-1=\frac{1}{8}$
$\Rightarrow {\cos }^2\frac{\theta }{2}=\frac{9}{16}\Rightarrow \cos \frac{\theta }{2}=\frac{3}{4}$
$\Rightarrow 4\cos \frac{\theta }{2}=3$
$\left[\because 0<\frac{\theta }{2}<\frac{\pi }{4},\text{ so}\cos \frac{\theta }{2}\ne -\frac{3}{4}\right]$