prove that:
$$\sin \alpha+\sin2\alpha+\sin3\alpha+...+\sin N\alpha = \frac{\sin\frac{N\alpha}{2}\sin\frac{(N+1)\alpha}{2}}{\sin\frac{\alpha}{2}}$$ without $\sum$
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I would try induction! – Dr. Sonnhard Graubner May 02 '18 at 14:31
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Should it be a $N$ in your right formula? – Dr. Sonnhard Graubner May 02 '18 at 14:33
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yes N as any number – Hollycow May 02 '18 at 14:36