A) \[\frac{1}{2}\]
B) \[-\frac{1}{4}\]
C) \[\frac{1}{8}\]
D) \[-\frac{1}{8}\]
Correct Answer: D
Solution :
\[\cos \,\frac{\pi }{7}.\,\cos \,\frac{2\pi }{7}.\,\cos \frac{4\pi }{7}\] \[\Rightarrow \] \[\cos \,\,20.\frac{\pi }{7}.\cos {{2}^{1}}.\frac{\pi }{7}.\cos \,{{2}^{2}}.\frac{\pi }{7}\] \[\Rightarrow \] \[\frac{\sin \,{{2}^{3}}\,\left( \frac{\pi }{7} \right)}{{{2}^{3}}.\sin \frac{\pi }{7}}\] \[\left( \because \,\,\left\{ \begin{align} & \cos A.\cos 2A.\cos {{2}^{2}}A.....\cos \,{{2}^{n-1}}A \\ & =\frac{\sin \,{{2}^{n}}A}{{{2}^{n}}\,\sin \,A} \\ \end{align} \right\} \right)\] \[=\frac{\sin \,8\,\pi /7}{8.\,\sin \,\pi /7}=\frac{\sin \,(\pi +\pi /7)}{8.\sin \pi /7}\] \[=\frac{\sin \,8\,\pi /7}{8.\,\sin \,\pi /7}=\frac{\sin \,(\pi +\pi /7)}{8.\sin \pi /7}=\frac{-\sin \,\pi /7}{8.\,\sin \,\pi /7}\] \[=-1/8\]
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