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 February 20th, 2017, 11:22 PM #1 Senior Member   Joined: Jul 2011 Posts: 400 Thanks: 15 Product Finding $$\displaystyle \tan \left(\frac{\pi}{14}\right)\cdot \tan \left(\frac{3\pi}{14}\right)\cdot \tan \left(\frac{5\pi}{14}\right)$$
 March 5th, 2017, 04:12 AM #2 Newbie     Joined: Jun 2016 From: Hong Kong Posts: 20 Thanks: 2 I give up for proving,check out List of trigonometric identities $\displaystyle \prod_{k=0}^{n-1} \sin(x+\frac{k\pi}{n})=\frac{\sin(nx)}{2^{n-1}}$ $\displaystyle \prod_{k=1}^n \tan(\frac{(2k-1)\pi}{4n+2})=\frac{1}{\sqrt{2n+1}}$ $\displaystyle \tan(\frac{\pi}{14}) \tan(\frac{3\pi}{14}) \tan(\frac{5\pi}{14})=\frac{1}{\sqrt{7}}$ Thanks from panky
 March 5th, 2017, 07:17 AM #3 Global Moderator   Joined: Dec 2006 Posts: 18,962 Thanks: 1606 \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 $\tan(\pi/14)\tan(3\pi/14)\tan(5\pi/14) = 1/\sqrt7$. Thanks from greg1313 and panky

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