\(\text{a)}\) 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.

## Comments

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TopNewest@Otto Bretscher @Mark Hennings What are your thoughts on this? – Ishan Singh · 7 months, 2 weeks ago

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You can play with my solution here to get those results. – Otto Bretscher · 7 months, 2 weeks ago

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– Ishan Singh · 7 months, 2 weeks ago

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).Log in to reply

– Otto Bretscher · 7 months, 2 weeks ago

No problem at all. I will write the solutions up in the evening unless somebody else does the work ;)Log in to reply

– Otto Bretscher · 7 months, 2 weeks ago

Try it yourself! If you use the complex expressions for sin and cosine, the computations are quite straightforward, and fun.Log in to reply

– Ishan Singh · 7 months, 2 weeks ago

I'll try it again, maybe I did some calculation mistake when I last tried it a year ago.Log in to reply

here. It works the exact same way when \(\gcd(n,k)=1\). Now tackle the case \(\gcd(n,k)\neq 1\). – Otto Bretscher · 7 months, 2 weeks ago

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. – Ishan Singh · 7 months, 2 weeks ago

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? – Otto Bretscher · 7 months, 2 weeks ago

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– Pi Han Goh · 7 months, 2 weeks ago

POST SOLUTION PLEASE!Log in to reply

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 · 7 months, 2 weeks ago

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– Ishan Singh · 7 months, 1 week ago

(+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 \(\cos\)Log in to reply

other problem because there we were restricting ourselves to the \(r\) with \(\gcd(r,n)=1\). – Otto Bretscher · 7 months, 1 week ago

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? – Ishan Singh · 7 months, 1 week ago

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– Otto Bretscher · 7 months, 1 week ago

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... ;)Log in to reply

– Ishan Singh · 7 months, 1 week ago

Yeah sure :) Thanks for the help.Log in to reply

– Ishan Singh · 7 months, 2 weeks ago

I have solved it for both cases. I'll post solution by night.Log in to reply