We have,
$$
\sum_{r=1}^{16}\left(\sin \frac{2 r \pi}{17}+i \cos \frac{2 r \pi}{17}\right)=i \sum_{r=1}^{16}\left(\cos \frac{2 r \pi}{17}-i \sin \frac{2 r \pi}{17}\right)
$$

$$
\begin{aligned}
& =i \sum_{r=1}^{16} e^{-\frac{i 2 r \pi}{17}} \\
& \left.=i \sum_{r=1}^{16} K^r \quad \text { [where, } K=e^{-\frac{2 i \pi}{17}}\right] \\
& =i\left[k\left(\frac{1-k^{16}}{1-k}\right)\right]=i \frac{\left(k-k^{17}\right)}{(1-k)} \\
& =\frac{i(k-1)}{(1-k)} \quad\left[\because K^{17}=e^{-2 i \pi}=\cos (2 \pi)-i \sin 2 \pi=1\right] \\
& =-i \quad
\end{aligned}
$$