A) \[\frac{1}{2}\]
B) \[\frac{1}{4}\]
C) \[-\frac{1}{8}\]
D) \[\frac{1}{8}\]
Correct Answer: D
Solution :
We have,\[\cos \frac{\pi }{7}\cos \frac{4\pi }{7}\cos \frac{5\pi }{7}\] \[=\cos \frac{\pi }{7}\cos \frac{4\pi }{7}\cos \left( \pi -\frac{2\pi }{7} \right)\] \[=-\cos \frac{\pi }{7}\cos \frac{2\pi }{7}\cos \frac{4\pi }{7}\] \[[\because \,\,\cos (\pi -\theta )=-\cos \theta ]\] \[=\frac{1}{2\sin \frac{\pi }{7}}\left[ 2\sin \frac{\pi }{7}\cos \frac{\pi }{7}\cos \frac{2\pi }{7}\cos \frac{4\pi }{7} \right]\] \[=\frac{-1}{2\sin \frac{\pi }{7}}\left[ \sin \frac{2\pi }{7}\cos \frac{2\pi }{7}\cos \frac{4\pi }{7} \right]\] \[\left( \because \,\,2\sin \frac{\theta }{2}\cos \frac{\theta }{2}=\sin \theta \right)\] \[=\frac{-1}{{{2}^{2}}\sin \frac{\pi }{7}}\left[ 2\sin \frac{2\pi }{7}\cos \frac{2\pi }{7}\cos \frac{4\pi }{7} \right]\] \[=\frac{-1}{{{2}^{2}}\sin \frac{\pi }{7}}\left[ \sin \frac{4\pi }{7}\cos \frac{4\pi }{7} \right]\] \[=\frac{-1}{{{2}^{3}}\sin \frac{\pi }{7}}\left[ 2\sin \frac{4\pi }{7}\cos \frac{4\pi }{7} \right]\] \[=\frac{-1}{8\sin \frac{\pi }{7}}\left[ \sin \frac{8\pi }{7} \right]\] \[=\frac{\sin \left( \pi +\frac{\pi }{7} \right)}{8\sin \frac{\pi }{7}}\] \[=\sin \frac{\frac{\pi }{7}}{8\sin \frac{\pi }{7}}=\frac{1}{8}\] \[[\because \,\,\sin (\pi +\theta )=-\sin \theta ]\]
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