First we establish the following, as they will be needed later. {{begin-eqn}} {{eqn | o =        | r = \sin \paren {\alpha + n} \theta + \sin \paren {\alpha - n} \theta       | c =  }} {{eqn | r = 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}       | c = Prosthaphaeresis Formula for Sine plus Sine }} {{eqn | n = 1       | r = 2 \sin \alpha \theta \cos n \theta       | c = simplification }} {{end-eqn}} {{begin-eqn}} {{eqn | o =        | r = \sin \paren {\alpha + n} \theta - \sin \paren {\alpha - n} \theta       | c =  }} {{eqn | r = 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}       | c = Prosthaphaeresis Formula for Sine minus Sine }} {{eqn | n = 2       | r = 2 \cos \alpha \theta \sin n \theta       | c = simplification }} {{end-eqn}} {{begin-eqn}} {{eqn | o =        | r = \sum_{n \mathop = -m}^m \dfrac {\sin \paren {n + \alpha} \theta} {n + \alpha}       | c =  }} {{eqn | r = \dfrac {\sin \paren {0 + \alpha} \theta} {0 + \alpha} + \sum_{n \mathop = 1}^m \dfrac {\sin \paren {n + \alpha} \theta} {n + \alpha} + \sum_{n \mathop = -m}^{-1} \dfrac {\sin \paren {n + \alpha} \theta} {n + \alpha}       | c =  }} {{eqn | r = \dfrac {\sin \alpha \theta} \alpha + \sum_{n \mathop = 1}^m \dfrac {\sin \paren {n + \alpha} \theta} {n + \alpha} + \sum_{n \mathop = 1}^m \dfrac {\sin \paren {-n + \alpha} \theta} {-n + \alpha}       | c =  }} {{eqn | r = \dfrac {\sin \alpha \theta} \alpha + \sum_{n \mathop = 1}^m \paren {\dfrac {\sin \paren {\alpha + n} \theta} {\alpha + n} + \dfrac {\sin \paren {\alpha - n} \theta} {\alpha - n} }       | c =  }} {{eqn | r = \dfrac {\sin \alpha \theta} \alpha + \sum_{n \mathop = 1}^m \paren {\dfrac {\paren {\alpha - n} \sin \paren {\alpha + n} \theta + \paren {\alpha + n} \sin \paren {\alpha - n} \theta} {\alpha^2 - n^2} }       | c =  }} {{eqn | r = \dfrac {\sin \alpha \theta} \alpha + \sum_{n \mathop = 1}^m \paren {\dfrac {\alpha \sin \paren {\alpha + n} \theta - n \sin \paren {\alpha + n} \theta + \alpha \sin \paren {\alpha - n} \theta - n \sin \paren {\alpha - n} \theta} {\alpha^2 - n^2} }       | c =  }} {{eqn | r = \dfrac {\sin \alpha \theta} \alpha + \sum_{n \mathop = 1}^m \paren {\dfrac {\alpha \paren {\sin \paren {\alpha + n} \theta + \sin \paren {\alpha - n} \theta} - n \paren {\sin \paren {\alpha + n} \theta - \sin \paren {\alpha - n} \theta} } {\alpha^2 - n^2} }       | c =  }} {{eqn | r = \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} }       | c = from $(1)$ and $(2)$ }} {{end-eqn}} From Half-Range Fourier Cosine Series for $\cos \alpha x$ over $\openint 0 \pi$: :$\displaystyle \cos \alpha \theta \sim \frac {2 \alpha \sin \alpha \pi} \pi \paren {\frac 1 {2 \alpha^2} + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {\cos n \theta} {\alpha^2 - n^2} }$ From Half-Range Fourier Sine Series for $\sin \alpha x$ over $\openint 0 \pi$: :$\displaystyle \sin \alpha \theta \sim \frac {2 \sin \alpha \pi} \pi \paren {\sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {n \sin n \theta} {\alpha^2 - n^2} }$ {{help|can't seem to get anything to work}}