A) \[1\]
B) \[-1\]
C) \[\frac{1}{2}\]
D) \[-\frac{1}{2}\]
Correct Answer: D
Solution :
\[\therefore \]\[\cos \frac{2\pi }{7}+\cos \frac{4\pi }{7}+\cos \frac{6\pi }{7}\] \[=\frac{1}{2\sin \frac{\pi }{7}}\times \]\[\left[ 2\sin \frac{\pi }{7}\cos \frac{2\pi }{7}+2\sin \frac{\pi }{7}\cos \frac{4\pi }{7}+2\sin \frac{\pi }{7}\cos \frac{6\pi }{7} \right]\] \[=\frac{1}{2\sin \frac{\pi }{7}}\times \]\[\left[ \sin \frac{3\pi }{7}-\sin \frac{\pi }{7}+\sin \frac{5\pi }{7}-\sin \frac{3\pi }{7}+\sin \pi -\frac{5\pi }{7} \right]\] \[=-\frac{1}{2}\]
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