Prove that $\sum_{r=0}^{n-1}\sin((2r+1)\theta)=\frac{\sin^2(n\theta)}{\sin\theta}$, where \(\sin\t

Question

Prove that $\sum_{r=0}^{n-1}\sin((2r+1)\theta)=\frac{\sin^2(n\theta)}{\sin\theta}$, where $\sin\theta\ne0$.

Accepted Answer

Use $2\sin((2r+1)\theta)\sin\theta=\cos(2r\theta)-\cos(2(r+1)\theta)$ and telescope.

Challenge problems