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

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

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