If $\mathrm{z}=\cos 6^0+\mathrm{i} \sin 6^0, \mathrm{then} \sum _{\mathrm{n}=1}^{20}\mathrm{Im}\left({\mathrm{z}}^{2\mathrm{n}-1}\right)=$

$\begin{array}{rcl}\mathrm{z}& =& \cos 6^0+\mathrm{i} \sin 6^0\end{array}$

$\begin{array}{rcl} & & \\ \sum _{\mathrm{n}=1}^{20} {\mathrm{I}}_{\mathrm{m}} \left({\mathrm{z}}^{2\mathrm{n}-1}\right)& =& \sum _{\mathrm{n}=1}^{20} {\mathrm{I}}_{\mathrm{m}}{\left(\cos 6^0+\mathrm{i} \sin 6^0\right)}^{2\mathrm{n}-1}\\ & =& \sum _{\mathrm{n}=1}^{20} {\mathrm{I}}_{\mathrm{m}}\left(\cos (2\mathrm{n}-1)6^0+\mathrm{i} \sin (2\mathrm{n}-1)6^0\right)\\ & =& \sum _{\mathrm{n}=1}^{20} \sin (2\mathrm{n}-1) 6^0\\ & =& \sin 6^0+\sin {18}^0+\sin {30}^0+...\sin {234}^0\\ & =& \frac{\sin 20\left(6^0\right)}{\sin 6^0} \sin \frac{(6^0+{234}^0)}{2}=\frac{{\sin }^2{120}^0}{\sin {60}^0}\\ & =& \frac{3}{4\sin 6^0}\end{array}$