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)

$-\frac{1}{8}$

Answer 2

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