Finding the Sum of trigonometric series using complex numbers

Exercise:

Show that, $\sum\limits_{n=0}^{54}{\cos \left( \frac{n\pi }{11} \right)}=1$then deduce the value of $\sum\limits_{n=0}^{54}{\sin \left( \frac{n\pi }{11} \right)}$

Solution: Let ${{S}_{1}}=\sum\limits_{n=0}^{54}{\cos \left( \frac{n\pi }{11} \right)}$ & ${{S}_{2}}=\sum\limits_{n=0}^{54}{\sin \left( \frac{n\pi }{11} \right)}$ Put $L={{S}_{1}}+i{{S}_{2}}$ where $i\in \mathbb{C}$

So $L=1+\cos \left( \frac{\pi }{11} \right)+\cos \left( \frac{2\pi }{11} \right)+...+\cos \left( \frac{54\pi }{11} \right)+i\sin \left( \frac{\pi }{11} \right)+i\sin \left( \frac{2\pi }{11} \right)+...+i\sin \left( \frac{54\pi }{11} \right)$

$\Rightarrow L=1+\left( \cos \left( \frac{\pi }{11} \right)+i\sin \left( \frac{\pi }{11} \right) \right)+....+\left( \cos \left( \frac{54\pi }{11} \right)+i\sin \left( \frac{54\pi }{11} \right) \right)$

      $=1+{{e}^{i\frac{\pi }{11}}}+{{e}^{i\frac{2\pi }{11}}}+....+{{e}^{i\frac{54\pi }{11}}}=1+{{e}^{i\frac{\pi }{11}}}+{{\left( {{e}^{i\frac{\pi }{11}}} \right)}^{2}}+...+{{\left( {{e}^{i\frac{\pi }{11}}} \right)}^{54}}$

      $=\frac{1-{{\left( {{e}^{i\frac{\pi }{11}}} \right)}^{55}}}{1-{{e}^{i\frac{\pi }{11}}}}=\frac{1-{{e}^{i\frac{55\pi }{11}}}}{1-{{e}^{i\frac{\pi }{11}}}}=\frac{1-{{e}^{i5\pi }}}{1-{{e}^{i\frac{\pi }{11}}}}=\frac{1-{{\left( {{e}^{i\pi }} \right)}^{5}}}{1-{{e}^{i\frac{\pi }{11}}}}=\frac{2}{1-{{e}^{i\frac{\pi }{11}}}}$ $=\frac{2}{1-\cos \left( \frac{\pi }{11} \right)-i\sin \left( \frac{\pi }{11} \right)}$

Remark that $2{{\sin }^{2}}\theta =1-\cos 2\theta $ and $\sin 2\theta =2\sin \theta \cos \theta $

Here $2\theta =\frac{\pi }{11}\Rightarrow \theta =\frac{\pi }{22}$ Hence $1-\cos \left( \frac{\pi }{11} \right)=2{{\sin }^{2}}\left( \frac{\pi }{22} \right)$ and $\sin \left( \frac{\pi }{11} \right)=2\sin \left( \frac{\pi }{22} \right)\cos \left( \frac{\pi }{22} \right)$

So $L=\frac{2}{2{{\sin }^{2}}\left( \frac{\pi }{22} \right)-i2\sin \left( \frac{\pi }{22} \right)\cos \left( \frac{\pi }{22} \right)}=\frac{1}{-i\sin \left( \frac{\pi }{22} \right)\left[ i\sin \left( \frac{\pi }{22} \right)+\cos \left( \frac{\pi }{22} \right) \right]}$

Hence \[L=\frac{1}{-i\sin \left( \frac{\pi }{22} \right){{e}^{i\frac{\pi }{22}}}}=\frac{i{{e}^{-i\frac{\pi }{22}}}}{\sin \left( \frac{\pi }{22} \right)}=\frac{i}{\sin \left( \frac{\pi }{22} \right)}\left( \cos \left( \frac{\pi }{22} \right)-i\sin \left( \frac{\pi }{22} \right) \right)\]

$\Rightarrow L=\frac{\sin \left( \frac{\pi }{22} \right)}{\sin \left( \frac{\pi }{22} \right)}+i\frac{\cos \left( \frac{\pi }{22} \right)}{\sin \left( \frac{\pi }{22} \right)}=1+i\cot \left( \frac{\pi }{22} \right)$ thus ${{S}_{1}}=1\,\,\,\And \,\,{{S}_{2}}=\cot \left( \frac{\pi }{22} \right)\,$