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

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

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

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

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

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

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

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

### Sum of the coefficients in the expansion of $(x+y)^n$ ....

If the sum of the coefficients in the expansion of $(x+y)^n$ is 4096, then the greatest coefficient in the expansion is _ _ _ _ . Solution $C_0 + C_1 + C_2 + C_3 + ......................... + C_n =4096$ $\therefore 2^n = 4096 =2^{12}$ $\Rightarrow n = 12$ Greatest coefficient = ${}^{12}{C_6} = 924$