Question

If the value of $(\cos \frac{2\pi }{7}+\cos \frac{4\pi }{7}+\cos \frac{6\pi }{7})$ =p, then $2p$ is equal to____

Answer 1

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

Multiplying dividing by $2\sin \frac{\pi }{7}$

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

$2\cos A\sin B=\sin (A+B)-\sin (A-B)$

$\Rightarrow$ $\frac{\sin (\frac{\pi }{7}+\frac{2\pi }{7})-\sin (\frac{2\pi }{7}-\frac{\pi }{7})+\sin (\frac{4\pi }{7}+\frac{\pi }{7})-\sin (\frac{4\pi }{7}-\frac{\pi }{7})+\sin (\frac{6\pi }{7}+\frac{\pi }{7})-\sin (\frac{6\pi }{7}-\frac{\pi }{7})}{2\sin \frac{\pi }{7}}$

$\Rightarrow$ $\frac{\sin \frac{3\pi }{7}-\sin \frac{\pi }{7}+\sin \frac{5\pi }{7}-\sin \frac{3\pi }{7}+\sin \pi -\sin \frac{5\pi }{7}}{2\sin \frac{\pi }{7}}$

$\Rightarrow$ $\frac{-\sin \frac{\pi }{7}+\sin \pi }{2\sin \frac{\pi }{7}}$

$\Rightarrow$ $\frac{-\sin \frac{\pi }{7}}{2\sin \frac{\pi }{7}}$ [ Since, $\sin \pi =0$ ]

$\Rightarrow$ $\frac{-1}{2}$

So here, $p=\frac{-1}{2}$

$\therefore$ $2p=-1$