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

Let $c{\mathrm{os}}^{-1}(1/8)=\theta$, where$0<\theta <\pi$. then

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

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

$\begin{array}{r}\Rightarrow {\cos }^2 \frac{\theta }{2}=\frac{9}{16}\Rightarrow \cos \frac{\theta }{2}=\frac{3}{4}\\ \left[\because 0<\frac{\theta }{2}<\frac{\pi }{2},\mathrm{so} \cos \frac{\theta }{2}\ne -\frac{3}{4}\right]\end{array}$