If $\mathrm{a}=\frac{\mathrm{\pi }}{18}$, then $\mathrm{cosa}+\cos 2\mathrm{a}+....+\cos 18\mathrm{a}=$

$$\begin{gathered} Recall that \cos ( \pi - A ) = - \cos A \\ hence, \cos \left( \frac{17 \pi }{18}\right) = \cos \left(\pi -\frac{ \pi }{18} \right) = - \cos \left(\frac{ \pi }{18}\right) \\ and hence \cos \left(\frac{17 \pi }{18}\right) + \cos \frac{\pi }{18}= 0 \\ or \cos \alpha + \cos 17 \alpha = 0 \\ Similarly \cos 2 \alpha + \cos 16 \alpha = 0 \\ \cos 3 \alpha + \cos 15 \alpha = 0 \\ \cos 4 \alpha + \cos 14 \alpha = 0 \\ ... ................... \\ \cos 8 \alpha + \cos 10 \alpha = 0 \\ as \cos 9 \alpha = \cos ( \pi 2 ) = 0 \\ and \cos 18 \alpha = \cos \pi = - 1 \\ Hence \cos \alpha + \cos 2 \alpha + ... + \cos 18 \alpha = - 1 \\ But as \cos 0 \alpha = \cos 0 = 1 , \\ we have \\ \cos 0 \alpha + \cos 1 \alpha + \cos 2 \alpha + ... + \cos 18 \alpha = 0 \\ and \cos 1 \alpha + \cos 2 \alpha + ... + \cos 18 \alpha = - 1 \end{gathered}$$