$\sin(\frac{\pi}{3})+\sin(\frac{2\pi}{3})+...+\sin(\frac{n\pi}{3})=2\sin(\frac{n\pi}{6})\sin(\frac{(n+1)\pi}{6})$ I need to prove that 
$\forall n\in\mathbb N:$
$$\sum_{i=1}^n \sin(\frac{i\pi}{3}) = 2\sin(\frac{n\pi}{6})\sin(\frac{(n+1)\pi}{6})$$
Using induction only leads me to proving quite complicated trig identities. I guess it also can be solved using reminders because sins on the LHS are equal to:
$$\frac{\sqrt3}{2},\frac{\sqrt3}{2},0,-\frac{\sqrt3}{2},-\frac{\sqrt3}{2},0$$
so per every 6th term they cancel out.
Could You give me any hint?
 A: You are trying to prove a theorem of the shape
$$f(1)+f(2)+\cdots+f(n)=g(n).\tag{1}$$
To do this you need to prove two things. One is
$$f(1)=g(1)\tag{2}$$
and the other is
$$g(n+1)-g(n)=f(n+1).\tag{3}$$
Once you have $(2)$ and $(3)$, then $(1)$ is evident, since 
$$f(1)+f(2)+\cdots+f(n)
=g(1)+(g(2)-g(1))+(g(3)-g(2))+\cdots+(g(n)-g(n-1))$$
etc.
A: Induction way: 
Assume that we have $$\sum_{k=0}^n \sin \frac{k\pi}{3} = 2\sin\frac{n\pi}{6}\sin\frac{(n+1)\pi}{6}.$$
We prove that 
$$\sum_{k=0}^{n+1} \sin \frac{k\pi}{3} = 2\sin\frac{(n+1)\pi}{6}\sin\frac{(n+2)\pi}{6}.$$
Indeed, one has $$LHS = \sum_{k=0}^n \sin \frac{k\pi}{3} + \sin\frac{(n+1)\pi}{3} = 2\sin\frac{n\pi}{6}\sin\frac{(n+1)\pi}{6} + 2\cos\frac{(n+1)\pi}{6}\sin\frac{(n+1)\pi}{6}$$
$$=2\sin \frac{(n+1)\pi}{6}\left(\sin\frac{n\pi}{6} + \cos\frac{(n+1)\pi}{6}\right) = 2\sin \frac{(n+1)\pi}{6}\left(\sin\frac{n\pi}{6} + \sin(\frac{\pi}{2}-\frac{(n+1)\pi}{6})\right)$$
$$=2\sin \frac{(n+1)\pi}{6}\left(2 \sin (\frac{\pi}{6})\cos(\frac{n\pi}{6}-\frac{\pi}{6})\right) = 2\sin \frac{(n+1)\pi}{6}\sin \frac{(n+2)\pi}{6}.$$
Shorter way
One has $$\sin \frac{\pi}{3}\sum_{k=0}^n \sin \frac{k\pi}{3} = \sum_{k=0}^n \sin \frac{k\pi}{3}\sin \frac{\pi}{3} =\sum_{k=0}^n\frac{1}{2}\left( \cos \frac{(k-1)\pi}{3} - \cos \frac{(k+1)\pi}{3}\right)$$ 
$$ = \frac{1}{2}\left(\cos\frac{-\pi}{3}+1-\cos\frac{n\pi}{3}-\cos\frac{(n+1)\pi}{3}\right) = \frac{1}{2}\left(2\cos^2\frac{\pi}{6}-2\cos\frac{(2n+1)\pi}{6}\cos\frac{\pi}{6}\right)$$
$$=\cos\frac{\pi}{6}\left(\cos\frac{\pi}{6}-\cos\frac{(2n+1)\pi}{6}\right)=2\sin\frac{\pi}{3}\sin\frac{n\pi}{6}\sin\frac{(n+1)\pi}{6}.$$
