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{align} & \cos A.\cos 2A.\cos {{2}^{2}}A.....\cos \,{{2}^{n-1}}A \\ & =\frac{\sin \,{{2}^{n}}A}{{{2}^{n}}\,\sin \,A} \\ \end{align} \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 $