Algebraic values of sine at sevenths of the circle At the end of a calculation it turned out that I wanted to know the value of
$$\sin(2\pi/7) + \sin(4\pi/7) - \sin(6\pi/7).$$
Since I knew the answer I was supposed to get, I was able to work out that the the above equals $\sqrt{7}/2$, and I can confirm this numerically. How would I prove this? I suspect I want to use the seventh roots of unity in some way but I am not sure how to proceed.

Using Prosthaphaeresis Formulas, 
$$\sin2x+\sin4x-\sin6x=2\sin2x\cos2x-(2\sin2x\cos4x)$$
$$=2\sin2x(\cos2x-\cos4x)=4\sin2x\sin3x\sin x$$
Now from this, $\sin(2n+1)x=(2n+1)\sin x+\cdots+2^{2n}(-1)^n\sin^{2n+1}x$
If $\sin(2n+1)x=0,(2n+1)x=m\pi$ where $m$ is any integer
$x=\dfrac{m\pi}{2n+1}$ where $m\equiv0,\pm1,\pm2,\cdots,\pm n\pmod{2n+1}$
So, the roots of $$2^{2n}(-1)^nt^{2n+1}x+\cdots+(2n+1)t=0$$ are $\sin\dfrac{m\pi}{2n+1}$ where $m\equiv0,\pm1,\pm2,\cdots,\pm n\pmod{2n+1}$
So, the roots of $$2^{2n}(-1)^nt^{2n}x+\cdots+2n+1=0$$ are $\sin\dfrac{m\pi}{2n+1}$ where $m\equiv\pm1,\pm2,\cdots,\pm n\pmod{2n+1}$
$\implies\prod_{r=-n}^n\sin\dfrac{m\pi}{2n+1}=\dfrac{2n+1}{2^{2n}(-1)^n}$
Now as $\sin(-y)=-\sin y,$
$\implies\prod_{r=1}^n(-1)^n\sin^2\dfrac{m\pi}{2n+1}=\dfrac{2n+1}{2^{2n}(-1)^n}$
As $0<\dfrac{m\pi}{2n+1}<\dfrac\pi2$ for $1\le m\le n,$
$\implies\prod_{r=1}^n\sin\dfrac{m\pi}{2n+1}=\dfrac{\sqrt{2n+1}}{2^n}$
Here $n=3$
