Question

Prove that : $\cos \frac{2\pi }{7}+\cos \frac{4\pi }{7}+\cos \frac{6\pi }{7}=-\frac{1}{2}$.

Answer 1

Multiply both sides with $\sin \left(\frac{\pi }{7}\right)$

$\sin \left(\frac{\pi }{7}\right)\cos \left(2\frac{\pi }{7}\right)+\sin \left(\frac{\pi }{7}\right)\cos \left(4\frac{\pi }{7}\right)+\sin \left(\frac{\pi }{7}\right)\cos \left(6\pi \frac{\pi }{7}\right)$

$=\frac{-1}{2}.\sin \left(\frac{\pi }{7}\right)$

Then, because $\sin A.\cos B=\frac{1}{2}\left[\sin \right(A-B)+\sin (A+B\left)\right]$

$\therefore \left(\frac{1}{2}\right)\left[\sin \right(\frac{\pi }{7})-(\frac{2\pi }{7})+\sin (\frac{\pi }{7}+2\frac{\pi }{7}\left)\right]+\left(\frac{1}{2}\right)\left[\sin \right(\frac{\pi }{7}-\frac{4\pi }{7})+\sin (\frac{\pi }{7}+\frac{4\pi }{7}\left)\right]+\left(\frac{1}{2}\right[\sin (\frac{\pi }{7}-6\frac{\pi }{7})+\sin (\frac{\pi }{7}+6\frac{\pi }{7})\left]\right)$

$=\frac{-\sin \frac{\pi }{7}}{2}$

We'll divide by $\left(\frac{1}{2}\right)$ both sides

$-\sin \left(\frac{\pi }{7}\right)+3\sin \left(\frac{\pi }{7}\right)-3\sin \left(\frac{\pi }{7}\right)+5\sin \left(\frac{\pi }{7}\right)-\sin \left(\frac{5\pi }{7}\right)+\sin \left(7\frac{\pi }{7}\right)=-\sin \frac{\pi }{7}$

Cancelling the same terms yields to

$-\sin \left(\frac{\pi }{7}\right)+\sin \left(\pi \right)=\sin \left(\frac{\pi }{7}\right)$

But $\sin \left(\pi \right)=0$

$\therefore -\sin \left(\frac{\pi }{7}\right)=-\sin \left(\frac{\pi }{7}\right)$

$LHS=RHS$

$\therefore \cos \frac{2\pi }{7}+\cos \frac{4\pi }{7}+\cos \frac{6\pi }{7}=\frac{-1}{2}$