Question

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

Answer 1

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, $\cos ^{-1} \frac{1}{8}=\theta \Rightarrow \cos \theta=\frac{1}{8}$
$\Rightarrow 2 \cos ^{2} \frac{\theta}{2}-1=\frac{1}{8}$
$\Rightarrow \cos ^{2} \frac{\theta}{2}=\frac{9}{16} \Rightarrow \cos \frac{\theta}{2}=\frac{3}{4}$
$\Rightarrow 4 \cos \frac{\theta}{2}=3$
${\left[\because 0<\frac{\theta}{2}<\frac{\pi}{4}, \text{ so} \cos \frac{\theta}{2} \neq-\frac{3}{4}\right]}$