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.
 A: Solution without using trinometry, but only properties of roots of unity: 
(This proof has been refined after reviewing this excellent post here. I knew there had to be a more direct approach!:))
Let $\omega=e^{i2\pi/7}$, i.e. the primitive $7$th root of unity. By definition, 
$$1+\omega^1+\omega^2+\omega^3+\omega^4+\omega^5+\omega^6=0$$
Let 
$$\begin{align}
S&=\omega^1+\omega^2+\omega^4\\
\Rightarrow\quad 
S^2&=\omega^2+\omega^4+\omega^8+2(\omega^3+\omega^6+\omega^5)\\
&=2(\underbrace{\omega^1+\omega^2+\omega^3+\omega^4+\omega^5+\omega^6}_{=-1})-(\underbrace{\omega^1+\omega^2+\omega^4}_{=S})\\
&=-2-S\\
S^2+S+2&=0\\
S&=-\frac 12\pm\frac{\sqrt7}2i\\\\
\sin\frac{2\pi}7+\sin\frac{4\pi}7-\sin\frac{6\pi}7
&=\Im(\omega^1+\omega^2-\omega^3)\\
&=\Im(\omega^1+\omega^2+\omega^4)\\
&=\Im(S)\\
&=\frac{\sqrt7}2\quad\blacksquare
\end{align}$$
NB - The positive value is chosen as $\sin\frac {2\pi}7+\sin\frac {4\pi}7-\sin\frac {6\pi}7=\sin\frac {2\pi}7+\sin\frac {4\pi}7-\sin\frac {\pi}7>0$, since $\sin\frac \pi7<\sin\frac {2\pi}7$.

$\tiny\color{\lightgrey}{\text{Previous solution (now superceded) shown below.}}$
$$\tiny\color{lightgrey}{\begin{align}
\sin\frac {2\pi}7+\sin\frac {4\pi}7-\sin\frac {6\pi}7
&=\sin\theta+\sin 2\theta-\sin3\theta
\qquad\qquad (\theta=2\pi/7;\;\omega=e^{i\theta}=e^{i2\pi/7})\\
&=\frac 1{2i}\left[(\omega^1-\omega^{-1})+(\omega^2-\omega^{-2})-(\omega^3-\omega^{-3})\right]\\
&=\frac 1{2i}\left[(\omega^1-\omega^6)+(\omega^2-\omega^5)-(\omega^3-\omega^4)\right]\\
&=\frac 1{2i}\left[\underbrace{\omega^1+\omega^2+\omega^3+\omega^4+\omega^5+\omega^6}_{=-1}-2(\omega^3+\omega^5+\omega^6)\right]\\
&=-\frac 1{2i}\left[1+2(\omega^3+\omega^5+\omega^6)\right]\\
\left(\sin\frac {2\pi}7+\sin\frac {4\pi}7-\sin\frac {6\pi}7\right)^2
&=-\frac14\left[1+4(\omega^3+\omega^5+\omega^6)+4(\omega^3+\omega^5+\omega^6)^2\right]\\
&=-\frac 14\left[1+4(\omega^3+\omega^5+\omega^6)+4(\omega^6+\omega^{10}+\omega^{12}+2\omega^8+2\omega^{11}+2\omega^9)\right]\\
&=-\frac 14\left[1+4(\omega^3+\omega^5+\omega^6)+4(\omega^6+\omega^{3}+\omega^{5}+2\omega^1+2\omega^{3}+2\omega^2)\right]\\
&=-\frac 14\left[8(\underbrace{1+\omega^1+\omega^2+\omega^3+\omega^4+\omega^5+\omega^6}_{=0})-7\right]\\
&=\frac 74\\
\sin\frac {2\pi}7+\sin\frac {4\pi}7-\sin\frac {6\pi}7&=\frac{\sqrt7}{\;2}\quad\blacksquare
\end{align}}$$
A: Although it's not very elegant, you can do this calculation by brute force. We have:
$$
\begin{split} i(\sin(2\pi/7)+\sin(4\pi/7)-\sin(6\pi/7)&=\frac{1}{2}(e^{2\pi i/7}-e^{-2\pi i/7}+e^{4\pi i/7}-e^{-4\pi i/7}-e^{6\pi i/7}+e^{-6\pi i/7}) \\ &=\frac{1}{2}(e^{2\pi i/7}-e^{12\pi i/7}+e^{4\pi i/7}-e^{10\pi i/7}-e^{6\pi i/7}+e^{8\pi i/7}) \end{split}
$$
squaring this we obtain:
$$
\begin{split} && \frac{1}{4}(e^{2\pi i/7}-e^{12\pi i/7}+e^{4\pi i/7}-e^{10\pi i/7}-e^{6\pi i/7}+e^{8\pi i/7})^2 \\ =&& \frac{1}{4}(\underbrace{e^{4\pi i/7}+e^{10\pi i/7}+e^{8\pi i/7}+e^{6\pi i/7}+e^{12\pi i/7}+e^{2\pi i/7}}_{=-1}-2e^{2\pi i}+2e^{6\pi i/7}-2e^{12\pi i/7}-2e^{8\pi i/7}+2e^{10\pi i/7}-2e^{2\pi i/7}+2e^{8\pi i/7}+2e^{4\pi i/7}-2e^{6\pi i/7}-2e^{2\pi i} -2e^{10\pi i/7}+2e^{12\pi i/7}+2e^{2\pi i/7}-2e^{4\pi i/7}-2e^{2\pi i}) \\=&&\frac{1}{4}(-1-2-2-2)=\frac{-7}{4} \end{split}
$$
This gives us:
$$
\begin{split} &&(i(\sin(2\pi/7)+\sin(4\pi/7)-\sin(6\pi/7))^2=\frac{-7}{4} \\ &\implies& -(\sin(2\pi/7)+\sin(4\pi/7)-\sin(6\pi/7))^2=\frac{-7}{4} \\ &\implies& \sin(2\pi/7)+\sin(4\pi/7)-\sin(6\pi/7)=\frac{\sqrt{7}}{2}  \end{split}
$$
Unfortunately this approach is not very enlightening.
A: Since $ \sin\left(\frac{6\pi}7\right) = \sin\left(\frac{\pi}7\right) $, then the expression is positive because $ \sin\left(\frac{2\pi}7\right) - \sin\left(\frac{\pi}7\right) > 0 $.
Let $ I $ denote this expression, then $ I > 0 $. Consider
$ I^2 = \left[\underbrace{ \sin^2\left(\frac{2\pi}7\right) + \sin^2\left(\frac{4\pi}7\right) + \sin^2\left(\frac{6\pi}7\right)}_{J} \right] -  \left [ \underbrace{2\sin\left(\frac{2\pi}7\right)  \sin\left(\frac{4\pi}7\right) -  2\sin\left(\frac{2\pi}7\right)   \sin\left(\frac{6\pi}7\right) -  2\sin\left(\frac{4\pi}7\right)  \sin\left(\frac{6\pi}7\right)}_{K} \right ] $
By half angle formula, $\sin^2(A) = \frac12 (1-\cos(2A)) $, then $J = \frac32 - \frac12 \left( \cos\left(\frac{4\pi}7\right) + \cos\left(\frac{8\pi}7\right) + \cos\left(\frac{12\pi}7\right) \right) = \frac32 - \frac12\left( \cos\left(\frac{3\pi}7\right) + \cos\left(\frac{5\pi}7\right) + \cos\left(\frac{\pi}7\right)\right) $
$J = \frac32 + \frac12 \times \frac12 = \frac74$
By product to sum formula, $\cos(A) - \cos(B) = -2\sin\left( \frac{A+B}2\right)\sin\left( \frac{A-B}2\right) $, can you show that $ K = 0 $?
A: 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 $n=3$
