Sum of Sines of Arithmetic Sequence of Angles

Theorem

 * $\displaystyle \sum_{k \mathop = 0}^n \sin \left({\theta + k \alpha}\right) = \frac {\sin \left({\alpha \left({n+1}\right) / 2}\right)} {\sin \left({\alpha / 2}\right)} \sin \left({\theta + \frac {n \alpha} 2}\right)$

Proof
Equating imaginary parts:
 * $\displaystyle \sum_{k \mathop = 0}^n \sin \left({\theta + k \alpha}\right) = \frac {\sin \left({\alpha \left({n+1}\right) / 2}\right)} {\sin \left({\alpha / 2}\right)} \sin \left({\theta + \frac {n \alpha} 2}\right)$