Sum of Sines of Arithmetic Sequence of Angles/Formulation 2

Theorem
Let $\alpha \in \R$ be a real number such that $\alpha \ne 2 \pi k$ for $k \in \Z$.

Then:

Proof
From Sum of Complex Exponentials of i times Arithmetic Sequence of Angles: Formulation 2:


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

From Euler's Formula, this can be expressed as:


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

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