How to evaluate $\sum_{k=1} ^{n-1} \frac{\sin (k\theta)}{\sin \theta}$ How to evaluate $$\sum_{k=1} ^{n-1} \frac{\sin (k\theta)}{\sin \theta}$$
Any help ? I tried to use difference method. But I'm not getting there.  
 A: HINT:
Use the Prosthaphaeresis Formula,
$$\cos((k-1)\theta)-\cos((k+1)\theta)=2\sin(k\theta)\sin(\theta)$$
and then sum the resulting telescoping series.
A: For a different approach:
$$\sum_{k=1}^{n-1}\frac{\sin(k\theta)}{\sin(\theta)}=\frac1{\sin(\theta)}\sum_{k=0}^{n-1}\sin(k\theta)$$
$\Im$ means the imaginary part.
$$=\frac1{\sin(\theta)}\Im\sum_{k=0}^{n-1}(\cos(k\theta)+i\sin(k\theta))$$
$$=\frac1{\sin(\theta)}\Im\sum_{k=0}^{n-1}e^{k\theta i}$$
$$=\frac1{\sin(\theta)}\Im\frac{1-e^{n\theta i}}{1-e^{\theta i}}$$
$$=\frac1{\sin(\theta)}\Im\frac{1-\cos(n\theta)-i\sin(n\theta)}{1-\cos(\theta)-i\sin(\theta)}$$
$$=\frac1{\sin(\theta)}\Im\frac{1-\cos(n\theta)-i\sin(n\theta)}{1-\cos(\theta)-i\sin(\theta)}\cdot\frac{1-\cos(\theta)+i\sin(\theta)}{1-\cos(\theta)+i\sin(\theta)}$$
$$=\frac1{\sin(\theta)}\Im\frac{(1-\cos(n\theta)-i\sin(n\theta))(1-\cos(\theta)+i\sin(\theta))}{(1-\cos(\theta))^2+\sin^2(\theta)}$$
$$=\frac1{\sin(\theta)}\Im\frac{(1-e^{n\theta i})(1-e^{-\theta i})}{(1-\cos(\theta))^2+\sin^2(\theta)}$$
$$=\frac1{\sin(\theta)}\Im\frac{1-e^{n\theta i}-e^{-\theta i}+e^{(n-1)\theta i}}{2(1-\cos(\theta))}$$
$$=\frac1{\sin(\theta)}\Im\frac{1-\cos(n\theta)-i\sin(n\theta)-\cos(\theta)+i\sin(\theta)+\cos((n-1)\theta)+i\sin((n-1)\theta)}{2(1-\cos(\theta))}$$
$$=\frac1{\sin(\theta)}\frac{-\sin(n\theta)+\sin(\theta)+\sin((n-1)\theta)}{2(1-\cos(\theta))}$$
You might be able to simplify that last line, but it's a well known solution.
