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I have seen this identity on Wolfram mathworld and in a comment to another similar trigonometric proof: Prove that $\prod_{k=1}^{n-1}\sin\frac{k \pi}{n} = \frac{n}{2^{n-1}}$

I can't seem to find a proof to this identity online anywhere. What method is used to prove this?

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    $\begingroup$ with $q = e^{2inx}$ : $\displaystyle \sin(nx) = \frac{e^{i n x}-e^{-inx}}{2 i } = \frac{q^{n/2}-q^{-n/2}}{2 i } = \frac{q^{-n/2}}{2 i }(q^{n}-1) = \frac{q^{-n/2}}{2 i } \prod_{k=0}^{n-1} (q-e^{2 i \pi k / n})$ $\endgroup$ – reuns Jun 5 '16 at 13:50
  • $\begingroup$ @user1952009 You should post the comment as a really solid solution! Well done. -Mark $\endgroup$ – Mark Viola Jan 13 '17 at 15:00
  • $\begingroup$ reuns solution in the comment above is very elegant. A couple of notes: First a typo: $q=e^{2ix}$. Second, one can take the norm of both sides. Thirdly, for real $\alpha$ and $\beta$ we have $|e^{i \alpha}- e^{i \beta}|=|e^{i(\alpha-\beta)/2}- e^{i(\beta-\alpha)/2}|= 2 |\sin(\alpha/2 -\beta/2)|$. By exchanging $x$ with $-x$, signs work out. $\endgroup$ – Chris Judge Dec 26 '17 at 1:46
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There is a nice argument based on Weierstrass products.

The sine function has its (simple) zeroes at $\pi\mathbb{Z}$ and

$$\frac{\sin x}{x}=\prod_{m\geq 1}\left(1-\frac{x^2}{n^2\pi^2}\right)\tag{1}$$ holds. The $\sin(n x)$ function has its zeroes at $\frac{\pi}{n}\mathbb{Z}=\left(\pi Z\right)\cup\left(\pi \mathbb{Z}+\frac{1}{n}\right)\cup\ldots\cup\left(\pi\mathbb{Z}+\frac{n-1}{n}\right)$, so by separating the zeroes according to their residue class $\pmod{\pi}$ and using $(1)$ and $$ \prod_{k=1}^{n-1}\sin\frac{k\pi}{n}=\frac{2n}{2^n}\tag{2}$$ your identity easily follows.

With a similar argument, you may check that your RHS and LHS have the same value at $x=0$ and by applying $\frac{d}{dz}\log(\cdot)$ to both sides you get two meromorphic functions with the same Eisenstein series, by Herglotz' trick.

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  • $\begingroup$ do you think you can reverse the process, and use my solution to prove the Weierstrass factorization of $\sin x$ ? $\endgroup$ – reuns Jun 5 '16 at 14:29
  • $\begingroup$ @user1952009: of course, it is possible to use exactly that approach to prove $(1)$, by proving $(2)$ and your claim by other means (roots of unity and Viète's theorem, for instance). $\endgroup$ – Jack D'Aurizio Jun 5 '16 at 14:35
  • $\begingroup$ sorry I didn't get your idea, once we proved the OP formula, what do you do next ? (knowing where are the zeros of $\sin$ and that they are of order $1$, but not using the Liouville theorem or such things) $\endgroup$ – reuns Jun 5 '16 at 14:40
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    $\begingroup$ @DurgeshTiwari: $(2)$ is a well-known identity. You may easily prove it by recalling that $2i \sin\theta = e^{i\theta}-e^{-i\theta}$. $\endgroup$ – Jack D'Aurizio Feb 22 '17 at 12:26
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    $\begingroup$ @DurgeshTiwari: you may write both the RHS and the LHS of OP's identity as infinite products, and they turn out to be the same infinite product. Have a look at Herglotz' trick - math.stackexchange.com/questions/581162/… $\endgroup$ – Jack D'Aurizio Feb 22 '17 at 12:33

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