<s>[INST] 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.
 [/INST] 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 $\sin\frac \pi7<\sin\frac {2\pi}7$.

$\tiny\color{\lightgrey}{\text{Previous solution (now superceded) shown below.}}$
$$\tiny\color{lightgrey}{\begin{align}
\sin\frac {2\pi}7+\sin\frac {4\pi}7-\sin\frac {6\pi}7
&=\sin\theta+\sin 2\theta-\sin3\theta
\qquad\qquad (\theta=2\pi/7;\;\omega=e^{i\theta}=e^{i2\pi/7})\\
&=\frac 1{2i}\left[(\omega^1-\omega^{-1})+(\omega^2-\omega^{-2})-(\omega^3-\omega^{-3})\right]\\
&=\frac 1{2i}\left[(\omega^1-\omega^6)+(\omega^2-\omega^5)-(\omega^3-\omega^4)\right]\\
&=\frac 1{2i}\left[\underbrace{\omega^1+\omega^2+\omega^3+\omega^4+\omega^5+\omega^6}_{=-1}-2(\omega^3+\omega^5+\omega^6)\right]\\
&=-\frac 1{2i}\left[1+2(\omega^3+\omega^5+\omega^6)\right]\\
\left(\sin\frac {2\pi}7+\sin\frac {4\pi}7-\sin\frac {6\pi}7\right)^2
&=-\frac14\left[1+4(\omega^3+\omega^5+\omega^6)+4(\omega^3+\omega^5+\omega^6)^2\right]\\
&=-\frac 14\left[1+4(\omega^3+\omega^5+\omega^6)+4(\omega^6+\omega^{10}+\omega^{12}+2\omega^8+2\omega^{11}+2\omega^9)\right]\\
&=-\frac 14\left[1+4(\omega^3+\omega^5+\omega^6)+4(\omega^6+\omega^{3}+\omega^{5}+2\omega^1+2\omega^{3}+2\omega^2)\right]\\
&=-\frac 14\left[8(\underbrace{1+\omega^1+\omega^2+\omega^3+\omega^4+\omega^5+\omega^6}_{=0})-7\right]\\
&=\frac 74\\
\sin\frac {2\pi}7+\sin\frac {4\pi}7-\sin\frac {6\pi}7&=\frac{\sqrt7}{\;2}\quad\blacksquare
\end{align}}$$
</s>