Theorem: Sum from -m to m of Sine of n + alpha of theta over n + alpha

For $0 < \theta < 2 \pi$:
:$\ds \sum_{n \mathop = -m}^m \dfrac {\sin \paren {n + \alpha} \theta} {n + \alpha} = \int_0^\theta \map \cos {\alpha \theta} \dfrac {\sin \paren {m + \frac 1 2} \theta \rd \theta} {\sin \frac 1 2 \theta}$

Proof:
We have:
{{begin-eqn}}
{{eqn | l = \sum_{n \mathop = -m}^m e^{i \paren {n + \alpha} \theta}
      | r = \sum_{n \mathop = -m}^m e^{i n \theta} e^{i \alpha \theta}
      | c = 
}}
{{eqn | r = e^{i \alpha \theta} e^{-i m \theta} \sum_{n \mathop = 0}^{2 m} e^{i n \theta}
      | c = 
}}
{{eqn | n = 1
      | r = e^{i \alpha \theta} e^{-i m \theta} \paren {\dfrac {e^{i \paren {2 m + 1} \theta} - 1} {e^{i \theta} - 1} }
      | c = Sum of Geometric Sequence
}}
{{eqn | r = e^{i \alpha \theta} e^{-i m \theta} \paren {\dfrac {e^{i \paren {2 m + 1} \theta / 2} \paren {e^{i \paren {2 m + 1} \theta / 2} - e^{-i \paren {2 m + 1} \theta / 2} } } {e^{i \theta / 2} \paren {e^{i \theta / 2} - e^{i \theta / 2} } } }
      | c = extracting factors
}}
{{eqn | r = e^{i \alpha \theta} \paren {\dfrac {e^{i \paren {2 m + 1} \theta / 2} - e^{-i \paren {2 m + 1} \theta / 2} } {e^{i \theta / 2} - e^{i \theta / 2} } }
      | c = Exponential of Sum and some algebra
}}
{{eqn | r = e^{i \alpha \theta} \frac {\map \sin {\paren {2 m + 2} \theta / 2} } {\map \sin {\theta / 2} }
      | c = Euler's Sine Identity
}}
{{eqn | ll= \leadsto
      | l = \sum_{n \mathop = -m}^m \paren {\cos \paren {n + \alpha} \theta + i \sin \paren {n + \alpha} \theta}
      | r = \paren {\map \cos {\alpha \theta} + i \map \sin {\alpha \theta} } \frac {\map \sin {\paren {m + \frac 1 2} \theta } } {\map \sin {\theta / 2} }
      | c = Euler's Formula and simplifying
}}
{{eqn | n = 2
      | ll= \leadsto
      | l = \sum_{n \mathop = -m}^m \cos \paren {n + \alpha} \theta
      | r = \map \cos {\alpha \theta} \frac {\map \sin {\paren {m + \frac 1 2} \theta } } {\map \sin {\theta / 2} }
      | c = equating real parts
}}
{{end-eqn}}
Note that the {{RHS}} at $(1)$ is not defined when $e^{i \theta} = 1$.
This happens when $\theta = 2 k \pi$ for $k \in \Z$.
For the given range of $0 < \theta < 2 \pi$ it is therefore seen that $(1)$ does indeed hold.
Then:
{{begin-eqn}}
{{eqn | l = \int_0^\theta \cos \paren {\alpha + n} \theta \rd \theta
      | r = \intlimits {\dfrac {\sin \paren {n + \alpha} \theta} {n + \alpha} } {\theta \mathop = 0} {\theta \mathop = \theta}
      | c = Primitive of $\cos a x$
}}
{{eqn | r = \paren {\dfrac {\sin \paren {n + \alpha} \theta} {n + \alpha} } - \paren {\dfrac {\sin \paren {n + \alpha} 0} {n + \alpha} }
      | c = 
}}
{{eqn | r = \dfrac {\sin \paren {n + \alpha} \theta} {n + \alpha}
      | c = Sine of Zero is Zero
}}
{{eqn | ll= \leadsto
      | l = \sum_{n \mathop = -m}^m \dfrac {\sin \paren {n + \alpha} \theta} {n + \alpha}
      | r = \sum_{n \mathop = -m}^m \int_0^\theta \cos \paren {\alpha + n} \theta \rd \theta
      | c = 
}}
{{eqn | r = \int_0^\theta \sum_{n \mathop = -m}^m \cos \paren {\alpha + n} \theta \rd \theta
      | c = Linear Combination of Definite Integrals
}}
{{eqn | r = \int_0^\theta \map \cos {\alpha \theta} \frac {\map \sin {\paren {m + \frac 1 2} \theta } } {\map \sin {\theta / 2} } \rd \theta
      | c = from $(2)$
}}
{{end-eqn}}
Hence the result.
{{qed}}