Explanation

(c)
$1 + \sum_{k = 0}^{14}\left\{ \cos{\frac{(2k + 1)}{15}\pi + i\sin{\frac{(2k + 1)}{15}\pi}} \right\}$
$= 1 + \sum_{k = 0}^{14}e^{i\frac{(2k + 1)}{15}\pi}$
$= 1 + \left( \alpha + \alpha^{3} + \alpha^{5} + ... + \alpha^{29} \right)\ (where\ \alpha = e^{i\pi/15})$
$= 1 + \alpha\left( \frac{1 - \alpha^{30}}{1 - \alpha^{2}} \right) = 1$ $\lbrack\because\alpha^{30} = e^{i2\pi = 1}\rbrack$