Discussion Forum

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

Answer:

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