1. ## [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 )

2. Originally Posted by Moo
$\prod_{k=0}^{2p-1} \frac{\cos\left(\tfrac \mu 2-\tfrac{k\pi}{2p}\right)}{\cos\left(\mu-\tfrac{k\pi}{2p}\right)} =\frac{\sin(p\mu)}{\sin(2p\mu)}$
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 2p–1).

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.