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Math Help - [SOLVED] Trigonometric identity

  1. #1
    Moo
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    [SOLVED] Trigonometric identity

    Hi,

    I've came accross this question in another forum and have no idea (yet) on how to get it ^^
    Prove that :

    \prod_{k=0}^{2p-1} \frac{\cos\left(\tfrac \mu 2-\tfrac{k\pi}{2p}\right)}{\cos\left(\mu-\tfrac{k\pi}{2p}\right)}=\frac{\sin(p\mu)}{\sin(2p  \mu)}

    I don't know what to use, it's university level, that's why I put it in calculus

    (I don't know if it's possible, but no induction...that would not be funny )
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  2. #2
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    Quote Originally Posted by Moo View Post
    \prod_{k=0}^{2p-1} \frac{\cos\left(\tfrac \mu 2-\tfrac{k\pi}{2p}\right)}{\cos\left(\mu-\tfrac{k\pi}{2p}\right)} =\frac{\sin(p\mu)}{\sin(2p\mu)}
    This looks like a question on complex numbers. Let \alpha = e^{\pi i/p} (a (2p)th root of unity). Then -\alpha is also a primitive (2p)th root of unity, and the roots of \lambda^{2p}=1 can be taken to be \{-\alpha^k\}_{k=0}^{2p-1}. Therefore \lambda^{2p}-1 = \prod_{k=0}^{2p-1}(\lambda+\alpha^k).

    Now let z = e^{i\mu/2} and \omega = e^{\pi i/(2p)}. Then \frac{\sin(p\mu)}{\sin(2p\mu)} = \frac{\frac1{2i}(z^{2p}-z^{-2p})}{\frac1{2i}(z^{4p}-z^{-4p})} = \frac{z^{2p}(z^{4p}-1)}{z^{8p}-1}. Factorise the numerator using the result in the previous paragraph with \lambda=z^2 (and \alpha=\omega^2), and the denominator with \lambda=z^4. This gives

    \frac{\sin(p\mu)}{\sin(2p\mu)} = \frac{z^{2p}\prod(z^2+\omega^{2k})} {\prod(z^4+\omega^{2k})} = \prod\frac{z\omega^{-k}+ z^{-1}\omega^k}{z^2\omega^{-k}+ z^{-2}\omega^k} (the products going from k=0 to 2p1).

    But z\omega^{-k}+ z^{-1}\omega^k = 2\cos\bigl(\tfrac\mu2-\tfrac{k\pi}{2p}\bigr) and z^2\omega^{-k}+ z^{-2}\omega^k = 2\cos\bigl(\mu-\tfrac{k\pi}{2p}\bigr), so that gives the result.
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