Question

Find $\cos \frac{2\pi }{7}+\cos \frac{4\pi }{7}+\cos \frac{6\pi }{7}$

• A. $\frac{1}{2}$
• B. $-\frac{1}{2}$
• C. $\frac{2}{3}$
• D. $\frac{1}{3}$

Answer 1

The correct option is B $-\frac{1}{2}$

Let $s=\cos \frac{2\pi }{7}+\cos \frac{4\pi }{7}+\cos \frac{6\pi }{7}s=(\cos \frac{2\pi }{7}+\cos \frac{4\pi }{7}+\cos \frac{6\pi }{7})\frac{2\sin \frac{\pi }{7}}{2\sin \frac{\pi }{7}}s=\frac{1}{2\sin \frac{\pi }{7}}[2\sin \frac{\pi }{7}\cos \frac{2\pi }{7}+2\sin \frac{\pi }{7}\cos \frac{4\pi }{7}+2\sin \frac{\pi }{7}\cos \frac{6\pi }{7}]s=\frac{1}{2\sin \frac{\pi }{7}}\left[\right(\sin \frac{3\pi }{7}-\sin \frac{\pi }{7})+(\sin \frac{5\pi }{7}-\sin \frac{3\pi }{7})+(\sin \frac{7\pi }{7}-\cos \frac{5\pi }{7})]$

since $2\sin A\cos B=\sin (A+B)+\sin (A-B)s=\frac{1}{2\sin \frac{\pi }{7}}[-\sin \frac{\pi }{7}+0]=\frac{-1}{2}$