Answer:
Option D
Explanation:
We have ,
$\sum_{r=1}^{16} \left( \sin \frac{2r\pi}{17}+i\cos \frac{2r \pi}{17}\right)=i\sum_{r=1}^{16}\left( \cos \frac{2r\pi}{17}-i\sin\frac{2r \pi}{17}\right)$
=i$\sum_{r=1}^{16} e^{-\frac{i2r\pi}{17}}$
$=i\sum_{r=1}^{16} K^{r}$ [where, $K=e^{-\frac{2i \pi}{17}}$]
=$i\left[k\left(\frac{1-k^{16}}{1-k}\right)\right]= i\frac{(k-k^{17})}{(1-k)}$
=$\frac{i(k-1)}{(1-k)}$
$[\because k^{17}=e^{-2i\pi}=\cos (2\pi)-i\sin(2\pi)=1]$
=-i
