# Trigonometric sum of a ratio of two sine functions

Evaluate $$\sum^{13}_{k=1}\frac{\sin (30^\circ k +45^\circ)}{\sin(30^\circ(k-1)+45^\circ)}.$$

Put $30^\circ = \alpha, 45^\circ = \beta$. Then \begin{align} S:=&\sum^{13}_{k=1}\frac{\sin (\alpha k+\beta)}{\sin(\alpha(k-1)+\beta)} = \sum^{13}_{k=1}\frac{\sin(\alpha(k-1)+\beta+\alpha)}{\sin(\alpha(k-1)+\beta)}\\ &= \sum^{13}_{k=1}\frac{\sin(\alpha(k-1)+\beta)\cos \alpha+\cos(\alpha(k-1)+\beta)\sin \alpha}{\sin(\alpha(k-1)+\beta)}\\ &= \cos \alpha\sum^{13}_{k=1}1+\sin \alpha\sum^{13}_{k=1}\cot(\alpha(k-1)+\beta) \\ &= \frac{\sqrt{3}}{2}\cdot 13+\frac{1}{2}\sum^{13}_{k=1}\cot(\alpha(k-1)+\beta). \end{align} Could some help me to solve it, thanks.

You are on the right track. Note that $\cot(x)=\tan(90^{\circ}-x)$ and $\tan(x)=\tan(x+180^{\circ})$. Hence \begin{align}\sum_{k=1}^{13}&\cot((k-1)30^\circ+45^\circ)=\sum_{k=0}^{12}\tan(45^\circ-k30^\circ)\\ &=1+\sum_{k=1}^{6}\tan(45^\circ-k30^\circ)+\sum_{k=1}^{6}\tan(45^\circ-(k+6)30^\circ)\\ &=1+2\sum_{k=1}^{6}\tan(45^\circ-k30^\circ)\\ &=1+2\left(\tan(15^\circ)+\tan(-15^\circ)+\tan(-45^\circ) +\tan(-75^\circ) +\tan(-105^\circ) +\tan(-135^\circ)\right)\\ &=1+2\left(-\tan(45^\circ) -\tan(75^\circ) +\tan(-180^\circ+75^\circ)+\tan(-180^\circ+45^\circ)\right)\\ &=1. \end{align}