Theorem with its proof to find the sum of sine series

Theorem:

Let $a,d\in \mathbb{R}$ , $d\ne 0$ and let $n\in {{\mathbb{Z}}^{+}}$ then we have

          $\sum\limits_{k=0}^{n-1}{\sin \left( a+kd \right)}=\frac{\sin \left( \frac{nd}{2} \right)}{\sin \frac{d}{2}}\sin \left( a+\frac{\left( n-1 \right)d}{2} \right)$

 and   $\sum\limits_{k=0}^{n-1}{\cos \left( a+kd \right)=\frac{\sin \left( \frac{nd}{2} \right)}{\sin \frac{d}{2}}\cos \left( a+\frac{\left( n-1 \right)d}{2} \right)}$

Proof: Let $C=\sum\limits_{k=0}^{n-1}{\cos \left( a+kd \right)}$ multiply by $2\sin \left( \frac{d}{2} \right)$ both sides to get

$2C\sin \left( \frac{d}{2} \right)=\sum\limits_{k=0}^{n-1}{2\cos \left( a+kd \right)\sin \left( \frac{d}{2} \right)}$

Remember that $\sin \left( p+q \right)-\sin \left( p-q \right)=2\cos p\sin q$

So $\sum\limits_{k=0}^{n-1}{2\cos \left( a+kd \right)\sin \left( \frac{d}{2} \right)=}\sum\limits_{k=0}^{n-1}{\sin \left( a+kd+\frac{d}{2} \right)-\sin \left( a+kd-\frac{d}{2} \right)}$

                                 $=\sum\limits_{k=0}^{n-1}{\sin \left( a+\frac{\left( 2k+1 \right)d}{2} \right)-\sin \left( a+\frac{\left( 2k-1 \right)d}{2} \right)}$

Thus $2C\sin \left( \frac{d}{2} \right)=\sum\limits_{k=0}^{n-1}{\sin \left( a+\frac{\left( 2k+1 \right)d}{2} \right)-\sin \left( a+\frac{\left( 2k-1 \right)d}{2} \right)}$

                     $=\sum\limits_{k=0}^{n-1}{\sin \left( a+\frac{\left( 2k+1 \right)d}{2} \right)-\sum\limits_{k=0}^{n-1}{\sin \left( a+\frac{\left( 2k-1 \right)d}{2} \right)}}$

$=\left\{ \sin \left( a+\frac{d}{2} \right)+\sin \left( a+\frac{3d}{2} \right)+\sin \left( a+\frac{5d}{2} \right)+....+\sin \left( a+\frac{\left( 2n-2+1 \right)d}{2} \right) \right\}$


 $-\left\{ \sin \left( a-\frac{d}{2} \right)+\sin \left( a+\frac{d}{2} \right) \right.+\sin \left( a+\frac{3d}{2} \right)+\sin \left( a+\frac{5d}{2} \right)+.......+\sin \left. \left( a+\frac{\left( 2n-2-1 \right)d}{2} \right) \right\}$

$=\sin \left( a+\frac{\left( 2n-1 \right)d}{2} \right)-\sin \left( a-\frac{d}{2} \right)$  (Telescope series canceled easily)

\[=2\cos \left( a+\frac{\left( n-1 \right)d}{2} \right)\sin \left( \frac{nd}{2} \right)\]

Thus $\sum\limits_{k=0}^{n-1}{\cos \left( a+kd \right)=\frac{\sin \left( \frac{nd}{2} \right)}{\sin \left( \frac{d}{2} \right)}\cos \left( a+\frac{\left( n-1 \right)d}{2} \right)}$