Answer:
Option A
Explanation:
We have
$\cos \left(\frac{\pi}{7}\right)\cos \left(\frac{2\pi}{7}\right)\cos \left(\frac{4\pi}{7}\right)$
Let A =$\frac{\pi}{7}\Rightarrow \pi=7A$
= $\cos A \cos 2A \cos 4A= \cos A \cos 2A \cos 2^{2} A$
Now, we will use the formula
$\cos A \cos 2A \cos 2^{2} A.......\cos ^{n-1} A= \frac{\sin 2^{n} A}{2^{n} \sin A}$
$\therefore$ $\cos A \cos2A\cos 2^{2}A=\frac{\sin 2^{3} A}{2^{3} \sin A}$
=$\frac{ \sin 8 A}{8 \sin A}= \frac{ \sin(7A+A)}{8 \sin A}= \frac{\sin(\pi+A)}{8 \sin A}$
[$\because 7A=\pi]$
=$\frac{-\sin A}{8 \sin A}$ $[\because \sin(\pi+\theta)=-\sin \theta)$
= $-\frac{1}{8}$
