### Prove That,$\cos \frac{\pi }{7}-\cos \frac{{2\pi }}{7}+\cos \frac{{3\pi}}{7}=\frac{1}{2}$

L.H.S.=$\cos \frac{\pi }{7} - \cos \frac{{2\pi }}{7} + \cos \frac{{3\pi }}{7}$

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

= $\cos \frac{\pi }{7} + \cos \frac{{3\pi }}{7} + \cos \frac{{5\pi }}{7}$

= $\cos \left( {\frac{\pi }{7} + \frac{{3 - 1}}{2}.\frac{{2\pi }}{7}} \right).\frac{{\sin \left( {3.\frac{{2\pi /7}}{2}} \right)}}{{\sin \left( {\frac{{2\pi /7}}{2}} \right)}}$

= $\cos \left( {\frac{\pi }{7} + \frac{{2\pi }}{7}} \right).\frac{{\sin \frac{{3\pi }}{7}}}{{\sin \frac{\pi }{7}}}$

= $\frac{{\cos \frac{{3\pi }}{7}\sin \frac{{3\pi }}{7}}}{{\sin \frac{\pi }{7}}} = \frac{{2\cos \frac{{3\pi }}{7}\sin \frac{{3\pi }}{7}}}{{2\sin \frac{\pi }{7}}}$

= $\frac{{\sin \frac{{6\pi }}{7}}}{{2\sin \frac{\pi }{7}}} = \frac{{\sin \left( {\pi - \frac{\pi }{7}} \right)}}{{2\sin \frac{\pi }{7}}} = \frac{1}{2}$