A) 0
B) 1
C) \[\frac{1}{2}\]
D) \[-\frac{1}{2}\]
Correct Answer: D
Solution :
We have, \[\cos \frac{2\pi }{7}+cos\frac{4\pi }{7}+cos\frac{6\pi }{7}\] \[=\,\,\frac{1}{2\sin \frac{\pi }{7}}\left[ 2\sin \frac{\pi }{7}\cos \frac{2\pi }{7}+2\sin \,\frac{\pi }{7}\cos \frac{4\pi }{7}+2\sin \frac{\pi }{7}\cos \frac{6\pi }{7} \right]\] \[=\,\,\frac{1}{2\sin \frac{\pi }{7}}\left[ \left( \sin \frac{3\pi }{7}-\sin \frac{\pi }{7} \right)+\left( \sin \frac{5\pi }{7}-\sin \frac{3\pi }{7} \right)+\left( \sin \frac{7\pi }{7}-\sin \frac{5\pi }{7} \right) \right]\] \[=\,\,\,-\frac{1}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ \because \,\,\,\sin \frac{7\pi }{7}=\sin \,\,\pi =0 \right]\]
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