Question

The value of $\cos \frac{\pi }{7}.\cos \frac{2\pi }{7}.\cos \frac{4\pi }{7}$ is equal to

• A. $\frac{1}{2}$
• B. $-\frac{1}{4}$
• C. $\frac{1}{8}$
• D. $-\frac{1}{8}$

Answer 1

Answer: (D)

$\cos \frac{\pi }{7}.\cos \frac{2\pi }{7}.\cos \frac{4\pi }{7}$
$\Rightarrow$ $\cos 20.\frac{\pi }{7}.\cos 2^1.\frac{\pi }{7}.\cos 2^2.\frac{\pi }{7}$
$\Rightarrow$ $\frac{\sin 2^3\left(\frac{\pi }{7}\right)}{2^3.\sin \frac{\pi }{7}}$
$\left(\because \left\{\begin{array}{rl}& \cos A.\cos 2A.\cos 2^{2}A.....\cos 2^{n-1}A\\ & =\frac{\sin 2^{n}A}{2^n\sin A}\end{array}\right\}\right)$
$=\frac{\sin 8\pi /7}{8.\sin \pi /7}=\frac{\sin (\pi +\pi /7)}{8.\sin \pi /7}$
$=\frac{\sin 8\pi /7}{8.\sin \pi /7}=\frac{\sin (\pi +\pi /7)}{8.\sin \pi /7}=\frac{-\sin \pi /7}{8.\sin \pi /7}$
$=-1/8$