Show that $\sin(\frac{x}{2})(\sin(x)+\sin(2x)+...+\sin(nx))=\sin(\frac{nx}{2})\sin(\frac{n+1}{2}x)$

Question

I want to show that $\sin(\frac{x}{2})(\sin(x)+\sin(2x)+...+\sin(nx))=\sin(\frac{nx}{2})\sin(\frac{n+1}{2}x)$

But I'm not quite sure how to start my proof. I tried to expand the left half of the equation but that didn't seem to be the right idea. Help appreciated!

Answer 1

Hint:

Since $\sin x\sin y=\frac{1}{2}\cos(x-y)-\frac{1}{2}\cos (x+y)$ we have $$\sin \left(\frac{x}{2}\right)\sum_{k=1}^n\sin \left(kx\right) =\sum_{k=1}^n\left\{\tfrac12\cos \left[(k-\tfrac{1}{2})x\right]-\tfrac12\cos \left[(k+\tfrac12)x\right]\right\}$$ Which is telescopic.