An Identity of Trig

**bigli** · December 12th 2009, 03:58 AM

For $n\geq 2,$ Show that $\cot(\frac{\pi}{2n})\cot(\frac{2\pi}{2n})\cot(\fra c{3\pi}{2n})\cdots\cot(\frac{(n-1)\pi}{2n})=1$ .(hint: Use complex numbers.)

**Danny** · December 12th 2009, 06:31 AM

Originally Posted by bigli

For $n\geq 2,$ Show that $\cot(\frac{\pi}{2n})\cot(\frac{2\pi}{2n})\cot(\fra c{3\pi}{2n})\cdots\cot(\frac{(n-1)\pi}{2n})=1$ .(hint: Use complex numbers.)

All you need is the trig identity $\sin\left( \frac{\pi}{2} - x\right) = \cos x$ . If you write $\cot \theta = \frac{\cos \theta}{\sin \theta}$ and reverse the product of the denominator terms everything cancels. Let me write out a few to demonstrate.

$<br /> n= 3\;\;\frac{\cos \frac{\pi}{6}}{\sin \frac{\pi}{6}} \cdot \frac{\cos \frac{2\pi}{6}}{\sin \frac{2\pi}{6}} = \frac{\cos \frac{\pi}{6}}{\sin \frac{2\pi}{6}} \cdot \frac{\cos \frac{2\pi}{6}}{\sin \frac{\pi}{6}}<br />$ and each term equals one. For example the first term

$\frac{\cos \frac{\pi}{6}}{\sin \frac{2\pi}{6}} = 1$ since $\sin \left(\frac{\pi}{2} - \frac{\pi}{6}\right) = \sin \frac{2\pi}{6} = \cos \frac{\pi}{6}$ from the identity.

$<br /> n= 4\;\;\frac{\cos \frac{\pi}{8}}{\sin \frac{\pi}{8}} \cdot \frac{\cos \frac{2\pi}{8}}{\sin \frac{2\pi}{8}} \cdot \frac{\cos \frac{3\pi}{8}}{\sin \frac{3\pi}{8}} = \frac{\cos \frac{\pi}{8}}{\sin \frac{3\pi}{8}} \cdot \frac{\cos \frac{2\pi}{8}}{\sin \frac{2\pi}{8}} \cdot \frac{\cos \frac{3\pi}{8}}{\sin \frac{\pi}{8}}$

or the last term

$\frac{\cos \frac{3\pi}{8}}{\sin \frac{\pi}{8}} = 1$ since $\cos \frac{3\pi}{8} = \sin \left(\frac{\pi}{2} - \frac{3\pi}{8}\right) = \sin \frac{\pi}{8}$ from the identity.

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