# prove the product $\sin^2\frac{(k-j)\pi}n$

$n\geq2$, prove that the product $$\prod_{1 \leq j<k\leq n \atop \gcd(j,n)=\gcd(k,n)=1}4 \sin^2\frac{(k-j)\pi}{n}=\dfrac{n^{\varphi(n)}}{\prod\limits_{p\mid n, p\; prime}{p^{\frac{\varphi(n)}{p-1}}}}$$ where $\varphi(n)$ is the Euler's totient function

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What have you tried? What is the problem (specific)? –  Mixxiphoid Aug 6 '13 at 8:16
Using the imperative mood might not endear you to potential answerers. –  user17794 Aug 6 '13 at 8:19
Zero personal input after 25+ questions asked on the site is rather curious, no? –  Did Aug 6 '13 at 9:03
Personal input $\ne$ adding wanted formula. –  Did Aug 6 '13 at 10:09
I don't think it's a homework question –  ziang chen Aug 6 '13 at 22:18

Let $\Phi_n(x)$ be the $n$th cyclotomic polynomial. So we have:

$$\Phi_n(x)=\prod_{\substack{1\le k\le n\\\gcd(k,n)=1}}(x-e^{\frac{2k\pi i}{n}})$$

The roots of $\Phi_n(x)$ are exactly the primitive $n$th roots of unity, so its discriminant is:

$$D(\Phi_n(x))=\prod_{\substack{1\le j<k\le n\\\gcd(j,n)=\gcd(k,n)=1}}(e^{\frac{2k\pi i}{n}}-e^{\frac{2j\pi i}{n}})^2$$

Now see if you can show the identity:

$$|e^{\frac{2k\pi i}{n}}-e^{\frac{2j\pi i}{n}}|^2=4\sin^2\left(\frac{(k-j)\pi}{n}\right)$$

It follows that your desired product is the absolute value of the discriminant of the $n$th cyclotomic polynomial.

Edit: To address the edit to your question, after an internet search, I've learned it is actually true that

$$|D(\Phi_n(x))|=\frac{n^{\phi(n)}}{\prod_{p|n}p^{\frac{\phi(n)}{p-1}}}$$

so that this answer is consistent with what you want to prove. However, the calculation of the discriminant of the cyclotomic polynomial seems fairly complicated.

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