\(\text{a)}\) Find a closed form for the following Trigonometric Products\[ \bullet \prod_{r=1}^{n-1} \sin \left(k \cdot \dfrac{r\pi}{n} \right) \]

\[ \bullet \prod_{r=1}^{n-1} \cos \left(k \cdot \dfrac{r\pi}{n}\right) \]

Here \(k > 1\) is a positive integer.

\( \text{b)} \) Generalize for all complex \(k\)

This is a part of the set Formidable Series and Integrals.

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## Comments

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TopNewest@Otto Bretscher @Mark Hennings What are your thoughts on this?

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I'm at work now; just a quick reply. Those products are not hard to do with complex numbers. The sine product, obviously, is 0 when \(\gcd(k,n)\neq 1\), and it is \(\frac{n}{2^{n-1}}\) in absolute value when \(\gcd(n,k)=1\)... the roots of unity get permuted in that case. For the cosine it gets just a bit messier but it is also very doable.

You can play with my solution here to get those results.

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Okay. Sorry to disturb and thanks for the input. Btw, I had tried using roots of unity for cosine but was getting a contradiction (that was a year ago, I'll have to re-check my work now).

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here. It works the exact same way when \(\gcd(n,k)=1\). Now tackle the case \(\gcd(n,k)\neq 1\).

We know that roots of unity work for the cosine when \(k=1\)... look at my solutionLog in to reply

here but it got somewhat messy.

Hmm. I found my flaw. It's pretty straight forward with roots of unity. I was earlier committing a calculation mistake and getting a contradiction when I tried using roots of unity. I also tried using Chebyshev Polynomials on the lines of my solutionLog in to reply

Are we done with this, or do you want to pursue it further?

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In the sign product, the \(r\) should go from 1 to \(n-1\); the way it's written now the product is always 0 ;)

I will essentially be copying my solutions from here and here

If \(\gcd(k,n)\neq 1\) then the product is 0 as \(kr\) will be a multiple of \(n\) for some \(r\).

If \(\gcd(k,n)=1\) then

\[\prod_{r=1}^{n-1}2|\sin(kr\pi/n)|=\prod_{r=1}^{n-1}|e^{ikr\pi/n}-e^{-ikr\pi/n}|=\prod_{r=1}^{n-1}|1-e^{2ikr\pi/n}|=n\]

so that \[\prod_{r=1}^{n-1}|\sin(kr\pi/n)|=\frac{n}{2^{n-1}}\]

Likewise, if \(\gcd(k,n)=1\) and \(n\) is odd, then

\[\prod_{r=1}^{n}2|\cos(kr\pi/n)|=\prod_{r=1}^{n}|e^{ikr\pi/n}+e^{-ikr\pi/n}|=\prod_{r=1}^{n}|-1-e^{2ikr\pi/n}|=2\]

so that \[\prod_{r=1}^{n}|\cos(kr\pi/n)|=\frac{1}{2^{n-1}}\]

If \(k=pq\) where \(p=\gcd(k,n)\), let \(n=mp\). Then

\[\prod_{r=1}^{n}|\cos(kr\pi/n)|=\prod_{r=1}^{pm}|\cos(qr\pi/m)|=\left(\frac{1}{2^{m-1}}\right)^p=\frac{1}{2^{n-\gcd(n,k)}}\]

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other problem because there we were restricting ourselves to the \(r\) with \(\gcd(r,n)=1\).

Yes, of course, I wasn't using the cyclotomic polynomials in this problem either. I was using them for theLog in to reply

@Otto Bretscher Sorry to disturb again, but when I inspected my earlier work, I found that I was trying to generalize the result for all complex \(k\). I tried to use transformation of roots on the \(n^{th}\) roots of unity, but to no avail. Can this be done?

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