# Sum over Integers of Sine of n + alpha of theta over n + alpha

## Theorem

For $0 < \theta < 2 \pi$:

$\ds \sum_{n \mathop \in \Z} \dfrac {\map \sin {n + \alpha} \theta} {n + \alpha} = \pi$

## Proof

First we establish the following, as they will be needed later.

 $\ds$  $\ds \map \sin {\alpha + n} \theta + \map \sin {\alpha - n} \theta$ $\ds$ $=$ $\ds 2 \map \sin {\dfrac {\paren {\alpha + n} \theta + \paren {\alpha - n} \theta} 2} \map \cos {\dfrac {\paren {\alpha + n} \theta - \paren {\alpha - n} \theta} 2}$ Prosthaphaeresis Formula for Sine plus Sine $\text {(1)}: \quad$ $\ds$ $=$ $\ds 2 \sin \alpha \theta \cos n \theta$ simplification

 $\ds$  $\ds \map \sin {\alpha + n} \theta - \map \sin {\alpha - n} \theta$ $\ds$ $=$ $\ds 2 \map \cos {\dfrac {\paren {\alpha + n} \theta + \paren {\alpha - n} \theta} 2} \map \sin {\dfrac {\paren {\alpha + n} \theta - \paren {\alpha - n} \theta} 2}$ Prosthaphaeresis Formula for Sine minus Sine $\text {(2)}: \quad$ $\ds$ $=$ $\ds 2 \cos \alpha \theta \sin n \theta$ simplification

 $\ds$  $\ds \sum_{n \mathop = -m}^m \dfrac {\map \sin {n + \alpha} \theta} {n + \alpha}$ $\ds$ $=$ $\ds \dfrac {\map \sin {0 + \alpha} \theta} {0 + \alpha} + \sum_{n \mathop = 1}^m \dfrac {\map \sin {n + \alpha} \theta} {n + \alpha} + \sum_{n \mathop = -m}^{-1} \dfrac {\map \sin {n + \alpha} \theta} {n + \alpha}$ $\ds$ $=$ $\ds \dfrac {\sin \alpha \theta} \alpha + \sum_{n \mathop = 1}^m \dfrac {\map \sin {n + \alpha} \theta} {n + \alpha} + \sum_{n \mathop = 1}^m \dfrac {\map \sin {-n + \alpha} \theta} {-n + \alpha}$ $\ds$ $=$ $\ds \dfrac {\sin \alpha \theta} \alpha + \sum_{n \mathop = 1}^m \paren {\dfrac {\map \sin {\alpha + n} \theta} {\alpha + n} + \dfrac {\map \sin {\alpha - n} \theta} {\alpha - n} }$ $\ds$ $=$ $\ds \dfrac {\sin \alpha \theta} \alpha + \sum_{n \mathop = 1}^m \paren {\dfrac {\paren {\alpha - n} \map \sin {\alpha + n} \theta + \paren {\alpha + n} \map \sin {\alpha - n} \theta} {\alpha^2 - n^2} }$ $\ds$ $=$ $\ds \dfrac {\sin \alpha \theta} \alpha + \sum_{n \mathop = 1}^m \paren {\dfrac {\alpha \map \sin {\alpha + n} \theta - n \map \sin {\alpha + n} \theta + \alpha \map \sin {\alpha - n} \theta - n \map \sin {\alpha - n} \theta} {\alpha^2 - n^2} }$ $\ds$ $=$ $\ds \dfrac {\sin \alpha \theta} \alpha + \sum_{n \mathop = 1}^m \paren {\dfrac {\alpha \paren {\map \sin {\alpha + n} \theta + \map \sin {\alpha - n} \theta} - n \paren {\map \sin {\alpha + n} \theta - \map \sin {\alpha - n} \theta} } {\alpha^2 - n^2} }$ $\ds$ $=$ $\ds \dfrac {\sin \alpha \theta} \alpha + \sum_{n \mathop = 1}^m \paren {\dfrac {2 \alpha \sin \alpha \theta \cos n \theta - 2 n \cos \alpha \theta \sin n \theta} {\alpha^2 - n^2} }$ from $(1)$ and $(2)$
$\ds \cos \alpha \theta \sim \frac {2 \alpha \sin \alpha \pi} \pi \paren {\frac 1 {2 \alpha^2} + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {\cos n \theta} {\alpha^2 - n^2} }$
$\ds \sin \alpha \theta \sim \frac {2 \sin \alpha \pi} \pi \paren {\sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {n \sin n \theta} {\alpha^2 - n^2} }$