Sum of Sines of Arithmetic Sequence of Angles

Theorem

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

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

Also presented as
This result is also seen presented in the form:


 * $\displaystyle \sum_{k \mathop = 1}^n \cos \paren {\theta + k \alpha} = \frac {\sin \paren {n \alpha / 2} } {\sin \paren {\alpha / 2} } \sin \paren {\theta + \frac {n + 1} 2 \alpha}$

Its derivation can follow similar lines.