If k is a positive integer then  $\sin \alpha +\sin \left(\pi +\alpha \right)+\sin \left(2\pi +\alpha \right)+\mathrm{...}$ + sin$\left(2k\pi +\alpha \right)=$

  $put k=1 then \sin \alpha +\sin (\pi +\alpha )+\sin (2\pi +\alpha )+=\sin \alpha -\sin \alpha +\sin \alpha =\sin \alpha$ 

  $$\begin{gathered} put k=2 then \sin \alpha +\sin (\pi +\alpha )+\sin (2\pi +\alpha )+\sin (3\pi +\alpha )+\sin (4\pi +\alpha ) \\ =\sin \alpha -\sin \alpha +\sin \alpha +\sin \alpha -\sin \alpha =\sin \alpha \end{gathered}$$

$\left(\left(or\right) we can use the formula \sin a +\sin \left(a+d\right) +\sin \left(a+2d\right)+.....+.\sin \left(a+\left(n-1\right)d\right)=\frac{\sin {\displaystyle \frac{nd}{2}}}{\sin \frac{d}{2}}\left(\sin \frac{\left(2a+\left(n-1\right)d\right)}{2}\right)\right)$