<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] 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 $?
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