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Closed Form of Generalized Trigonometric Products

\(\text{a)}\) 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.

Note by Ishan Singh
5 months ago

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@Otto Bretscher @Mark Hennings What are your thoughts on this? Ishan Singh · 5 months ago

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@Ishan Singh 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. Otto Bretscher · 5 months ago

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@Otto Bretscher 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). Ishan Singh · 5 months ago

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@Ishan Singh No problem at all. I will write the solutions up in the evening unless somebody else does the work ;) Otto Bretscher · 5 months ago

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@Ishan Singh Try it yourself! If you use the complex expressions for sin and cosine, the computations are quite straightforward, and fun. Otto Bretscher · 5 months ago

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@Otto Bretscher I'll try it again, maybe I did some calculation mistake when I last tried it a year ago. Ishan Singh · 5 months ago

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@Ishan Singh We know that roots of unity work for the cosine when \(k=1\)... look at my solution here. It works the exact same way when \(\gcd(n,k)=1\). Now tackle the case \(\gcd(n,k)\neq 1\). Otto Bretscher · 5 months ago

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@Otto Bretscher 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 solution here but it got somewhat messy. Ishan Singh · 5 months ago

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@Ishan Singh I love Chebyshev polynomials, but here roots of unity seem like the way to go.

Are we done with this, or do you want to pursue it further? Otto Bretscher · 5 months ago

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@Otto Bretscher POST SOLUTION PLEASE! Pi Han Goh · 5 months ago

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@Pi Han Goh Finding the signs is a straightforward but tedious book keeping issue; let's just go for the absolute values for now since I have limited time.

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

\[\prod_{r=1}^{n}|\cos(kr\pi/n)|=\prod_{r=1}^{pm}|\cos(qr\pi/m)|=\left(\frac{1}{2^{m-1}}\right)^p=\frac{1}{2^{n-\gcd(n,k)}}\] Otto Bretscher · 5 months ago

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@Otto Bretscher (+1) Nice solution! My solution is essentially the same as yours, except instead of using cyclotomic polynomials, if \(\gcd (k,n) =1\) , note that \(\displaystyle \prod_{r=1}^{n-1} \sin \left(\dfrac{rk \pi}{n} \right) = (-1)^{n-1} \prod_{r=1}^{n-1} \sin \left(\dfrac{r \pi}{n} \right) = (-1)^{n-1} \dfrac{n}{2^{n-1}} \) since the set \(rk \pmod {n} \) is exhaustive for \(1 \leq r \leq n-1\). Similarly for \(\cos\) Ishan Singh · 5 months ago

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@Ishan Singh Yes, of course, I wasn't using the cyclotomic polynomials in this problem either. I was using them for the other problem because there we were restricting ourselves to the \(r\) with \(\gcd(r,n)=1\). Otto Bretscher · 5 months ago

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@Otto Bretscher @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? Ishan Singh · 5 months ago

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@Ishan Singh It's a nice sunny spring day here, and I don't feel like thinking about messy stuff like this right now. Maybe on a rainy day... ;) Otto Bretscher · 5 months ago

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@Otto Bretscher Yeah sure :) Thanks for the help. Ishan Singh · 5 months ago

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@Pi Han Goh I have solved it for both cases. I'll post solution by night. Ishan Singh · 5 months ago

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