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How can one show that for any angle $\theta$ such that $0<\theta<2\pi/5$ the following equation is true?

$\sqrt{1-\cos\theta} +\sqrt{1-\cos(\theta-\frac{2\pi}{5})}+\sqrt{1-\cos(\theta+\frac{4\pi}{5})}=\sqrt{1-\cos(\theta+\frac{2\pi}{5})}+\sqrt{1-\cos(\theta-\frac{4\pi}{5})}$

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Note that $$\sqrt{1 - \cos(2x)} = \sqrt{2}\,|\sin(x)|.$$ So for $x \in (0, \frac{\pi}{5})$ we have to show $$\sin(x) - \sin(x - \frac{\pi}{5}) + \sin(x + \frac{2\pi}{5}) - \sin(x + \frac{\pi}{5}) + \sin(x - \frac{2 \pi}{5}) = 0.$$ Using that $\sin(x + 2\pi) = \sin(x)$ and $\sin(x + \pi) = -\sin(x)$ this can be rewritten as $$\sin(x) + \sin(x + \frac{4\pi}{5}) + \sin(x + \frac{2\pi}{5}) + \sin(x + \frac{6\pi}{5}) + \sin(x + \frac{8 \pi}{5}) = 0.$$ Now the left hand side is the sum of the $y$-coordinates of a regular pentagon centred at the origin and tilted at an angle $x$. Since the vertices of such a pentagon sum up to the zero vector the sum of the $y$-coordinates equals $0$.

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    $\begingroup$ In the first equality, I think you mean $2|\sin(\theta)|$. $\endgroup$ – Ian Mateus Nov 17 '13 at 19:03
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    $\begingroup$ @IanMateus Thanks, $\sqrt{2}$ even. $\endgroup$ – WimC Nov 17 '13 at 19:04

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