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)

Let ${\cos }^{-1}\frac{1}{8}=\theta where0<\theta <\pi$
$\Rightarrow \cos \left(\frac{1}{2}{\cos }^{-1}\frac{1}{8}\right)=\cos \frac{\theta }{2}$
Now ${\cos }^{-1}\frac{1}{8}=\theta \Rightarrow \cos \theta =\frac{1}{8}$
$\Rightarrow 2{\cos }^2\frac{\theta }{2}-1=\frac{1}{8}\Rightarrow \cos \frac{\theta }{2}=\frac{3}{4}$
$\left[\because 0<\frac{\theta }{2}<\frac{\pi }{2},\cos \frac{\pi }{2}\ne -\frac{3}{4}\right]$