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February 21st, 2017, 12:22 AM   #1
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Finding $$\displaystyle \tan \left(\frac{\pi}{14}\right)\cdot \tan \left(\frac{3\pi}{14}\right)\cdot \tan \left(\frac{5\pi}{14}\right)$$
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March 5th, 2017, 05:12 AM   #2
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I give up for proving,check out List of trigonometric identities

$\displaystyle \prod_{k=0}^{n-1} \sin(x+\frac{k\pi}{n})=\frac{\sin(nx)}{2^{n-1}}$

$\displaystyle \prod_{k=1}^n \tan(\frac{(2k-1)\pi}{4n+2})=\frac{1}{\sqrt{2n+1}}$

$\displaystyle \tan(\frac{\pi}{14}) \tan(\frac{3\pi}{14}) \tan(\frac{5\pi}{14})=\frac{1}{\sqrt{7}}$
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March 5th, 2017, 08:17 AM   #3
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$\displaystyle \begin{align*}8\sin(\pi/14)\cos(\pi/14)\sin(3\pi/14)\sin(5\pi/14) &= 4\sin(2\pi/14)\sin(3\pi/14)\sin(5\pi/14) \\
&= 2\sin(2\pi/14)(\cos(2\pi/14) - \cos(8\pi/14)) \\
&= 2\sin(2\pi/14)\cos(2\pi/14) - 2\sin(2\pi/14)\cos(8\pi/14) \\
&= \sin(4\pi/14) - \sin(10\pi/14) + \sin(6\pi/14) \\
&= \cos(\pi/14)\end{align*}$
so $8\sin(\pi/14)\sin(3\pi/14)\sin(5\pi/14) = 1$.

$\displaystyle \begin{align*}8\cos(\pi/14)\cos(3\pi/14)\cos(5\pi/14) &= 4\cos(3\pi/14)\cos(6\pi/14) + 4\cos(3\pi/14)\cos(4\pi/14) \\
&= 2\cos(9\pi/14) + 2\cos(3\pi/14) + 2\cos(7\pi/14) + 2\cos(\pi/14) \\
&= 2\cos(\pi/14) + 2\cos(3\pi/14) - 2\cos(5\pi/14)\end{align*}$

After squaring and simplifying, one gets $(8\cos(\pi/14)\cos(3\pi/14)\cos(5\pi/14))^2 = 7$.

Hence $\tan(\pi/14)\tan(3\pi/14)\tan(5\pi/14) = 1/\sqrt7$.
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