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

$=\frac{1}{2\sin \left({\displaystyle \frac{\mathrm{\pi }}{7}}\right)}\left\{2\cos \left(\frac{2\pi }{7}\right)\sin \left(\frac{\mathrm{\pi }}{7}\right)+2\cos \left(\frac{4\pi }{7}\right)\sin \left(\frac{\mathrm{\pi }}{7}\right)+2\cos \left(\frac{6\pi }{7}\right)\sin \left(\frac{\mathrm{\pi }}{7}\right)\right\}$

$=\frac{1}{2\sin {\displaystyle \frac{\mathrm{\pi }}{7}}}\left\{\sin \frac{3\mathrm{\pi }}{7}-\sin \frac{\mathrm{\pi }}{7}+\sin \frac{5\mathrm{\pi }}{7}-\sin \frac{3\mathrm{\pi }}{7}+\sin \frac{7\mathrm{\pi }}{7}-\sin \frac{5\mathrm{\pi }}{7}\right\}$

Since, $2\cos A\sin B=\sin \left(A+B\right)-\sin \left(A-B\right)$

$=\frac{1}{2\sin {\displaystyle \frac{\mathrm{\pi }}{7}}}\left\{-\sin \frac{\mathrm{\pi }}{7}+\mathrm{sin\pi }\right\}$

$=\frac{-1}{2}, \left(\because \mathrm{sin\pi }=0\right)$