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How to prove that

$$2\sin \frac{\pi}{14}+2\sin \frac{5\pi}{14}-2\sin \frac{3\pi}{14}=1$$

I absolutely have no idea other than using the property $\sin\bigl(\frac{\pi}{2}-a\bigr)=\cos a\;$

Quanto
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Dareal1 62
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  • Start by expanding $sin(2\pi/14 + \pi/14)$ – Ripi2 Nov 02 '19 at 20:04
  • If you accept that the individual $\sin$ values are all roots of $8x^3-4x^2-4x+1$ (e.g. https://math.stackexchange.com/questions/773131/prove-that-sin-frac-pi14-is-a-root-of-8x3-4x2-4x-1-0), then the result follows from Vieta's formulas directly. – Sil Nov 02 '19 at 22:54

2 Answers2

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Using the formula

we need $$S=2\cos3\pi/7+2\cos\pi/7-2\cos2\pi/7$$

As $2\pi/7+5\pi/7=\pi,\cos(\pi-x)=-\cos x$

$$S=2\sum_{r=0}^2\cos(2r+1)\pi/7$$

Use How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression?

lab bhattacharjee
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  • thank you! also, this expression is actually a result from another exercice wich might point out the fact that I need to rethink it... – Dareal1 62 Nov 02 '19 at 20:05
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With $a=\frac\pi7$ \begin{align} &\ 2\sin \frac{\pi}{14}+2\sin \frac{5\pi}{14}-2\sin \frac{3\pi}{14}\\ = & \ 2\cos a+ 2\cos 3a + 2\cos 5a \\ =&\ \frac{1}{\sin a}(2\sin a\cos a+ 2\sin a\cos 3a +2\sin a \cos 5a )\\ =&\ \frac{1}{\sin a}(-\sin 0a +\sin 2a -\sin 2a + \sin 4a - \sin 4a + \sin 6a)\\ = &\ \frac{\sin 6a}{\sin a} =1 \end{align}

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