Explanation

(c)
of $\sum_{k = 1}^{10}\left( \sin{\frac{2k\pi}{11} + i\cos\frac{2k\pi}{11}} \right) = i\ \sum_{k = 1}^{10}\left( \cos{\frac{2k\pi}{11} - i\sin\frac{2k\pi}{11}} \right)$
$= i\ \sum_{k = 1}^{10}\left( e^{- \frac{2k\pi}{11}i} \right)\$
$= i\ \sum_{k = 1}^{10}r^{k}\ where\ r = e^{\frac{- 2\pi i}{11}}$
$= i(r + r^{2} + r^{3} + ...r^{10})$
$= \frac{i.r(r^{10} - 1)}{r - 1}$
$= i\left( \frac{r^{11} - r}{r - 1} \right)$
$= i\left( \frac{1 - r}{r - 1} \right)$ $\because\ r^{11} = e^{- 2i\pi} = 1$
$= - i$