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Thread: An Identity of Trig

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    An Identity of Trig

    For $\displaystyle n\geq 2, $ Show that $\displaystyle \cot(\frac{\pi}{2n})\cot(\frac{2\pi}{2n})\cot(\fra c{3\pi}{2n})\cdots\cot(\frac{(n-1)\pi}{2n})=1$ .(hint: Use complex numbers.)
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  2. #2
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    Quote Originally Posted by bigli View Post
    For $\displaystyle n\geq 2, $ Show that $\displaystyle \cot(\frac{\pi}{2n})\cot(\frac{2\pi}{2n})\cot(\fra c{3\pi}{2n})\cdots\cot(\frac{(n-1)\pi}{2n})=1$ .(hint: Use complex numbers.)
    All you need is the trig identity $\displaystyle \sin\left( \frac{\pi}{2} - x\right) = \cos x$. If you write $\displaystyle \cot \theta = \frac{\cos \theta}{\sin \theta}$ and reverse the product of the denominator terms everything cancels. Let me write out a few to demonstrate.

    $\displaystyle
    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}}
    $ and each term equals one. For example the first term


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

    $\displaystyle
    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

    $\displaystyle \frac{\cos \frac{3\pi}{8}}{\sin \frac{\pi}{8}} = 1 $ since $\displaystyle \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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