Sum of Sines of Arithmetic Sequence of Angles

Theorem

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

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

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


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

Its derivation can follow similar lines.