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

Question

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

Accepted Answer

$\frac{\sin^2(n\theta)}{\sin\theta}$

Factorial, logarithmic and trigonometric telescoping