I have run across this interesting identity that I am unable to verify.
$$\sin(nx)=2^{n-1}\prod_{k=0}^{n-1}\sin\left(\frac{k\pi}n+x\right)$$
Can anyone provide a hint as how one would prove this?
This identity was found on the Wolfram site1.
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That "identity" fails for $n=1$. – Mark Viola Jan 12 '17 at 21:14
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1You want to start at $k=0$, not $1$. – Robert Israel Jan 12 '17 at 21:15
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Duplicate of this and this. – Lucian Jan 12 '17 at 21:53
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@lucian Not quite duplicates. The appearance of the $x$ makes things a bit more complicated. – Mark Viola Jan 12 '17 at 22:23
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2Similar question: Proof of $\sin nx=2^{n-1}\prod_{k=0}^{n-1} \sin\left( x + \frac{k\pi}{n} \right)$. Found using Approach0. – Martin Sleziak Jan 13 '17 at 04:32
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This solution herein is inspired by a comment left by @user1952009 on another posted solution. That solution is elegant in that it relies on only $(i)$ Euler's formula to write
$$\sin(nx)=\frac{e^{inx}-e^{-inx}}{2i}\tag 1$$
and $(ii)$ the $n$ roots of unity, $z=e^{-i2k \pi/n}$ for $k=0,\dots,n-1$, of the equation $z^n=1$ to write
$$z^n-1=\prod_{k=0}^{n-1}(z-e^{-i2\pi k/n})\tag 2$$
Proceeding, we find
$$\begin{align} \sin(nx)&=\frac{e^{inx}-e^{-inx}}{2i}&\\\\ &=\frac{e^{-inx}}{2i}\left(e^{i2nx}-1\right)\\\\ &=\frac{e^{-inx}}{2i}\prod_{k=0}^{n-1}(e^{i2x}-e^{-i2\pi k/n})\\\\ &=\frac{e^{-inx}}{2i}\prod_{k=0}^{n-1}\left(\color{blue}{e^{ix}}\color{red}{e^{-i\pi k/n}}\color{green}{(2i)}\sin\left(x+\frac{k \pi}{n}\right)\right)\\\\ &=\frac{e^{-inx}}{2i} \color{blue}{e^{inx}} \color{red}{e^{-i(\pi/n)n(n-1)/2}}\color{green}{(2i)^n}\prod_{k=0}^{n-1}\sin\left(x+\frac{k \pi}{n}\right)\\\\ &=2^{n-1}\prod_{k=0}^{n-1}\sin\left(x+\frac{k \pi}{n}\right) \end{align}$$
as was to be shown!
Mark Viola
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I failed to understand how you reached from $$\frac{e^{-inx}}{2i}\left(e^{i2nx}-1\right)\\$$ to $$\frac{e^{-inx}}{2i}\prod_{k=0}^{n-1}(e^{i2x-e^{-i2\pi k/n}})\\$$ and subsequent steps Can you please explain a bit. – Navin Feb 21 '17 at 14:48
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@navinstudent The equations in your comment have not been formatted – Mark Viola Feb 21 '17 at 14:50
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Yes I see that . please do also clarify subsequent step.It really puzzled me. – Navin Feb 21 '17 at 15:01
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Subsequent steps use straightforward arithmetic. Are you familiar with Euler's Formula in $(1)? – Mark Viola Feb 21 '17 at 15:02
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I think you meant $$\frac{e^{-inx}}{2i}\prod_{k=0}^{n-1}(e^{i2x-e^{-i2\pi k/n}})\\$$ to be $$\frac{e^{-inx}}{2i}\prod_{k=0}^{n-1}(e^{i2x}-e^{-i2\pi k/n})\\$$.I am sorry I am not getting how you reached the next step despite mant tries. – Navin Feb 21 '17 at 15:37
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@navinstudent Yes, of course. Thanks for catching the typographical error. – Mark Viola Feb 21 '17 at 15:41
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Start by factoring $$e^{i2x}-e^{-i2\pi k/n}=e^{ix}e^{-i\pi k/n}\left(\cdot\right)$$What do you get? – Mark Viola Feb 21 '17 at 15:43
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@DDD4C4U The term $(2i)^n$ did not "disappear." Note $(2i)^n=2^n e^{in\pi/2}$. The note the term in red-colored font. – Mark Viola Jun 19 '20 at 13:59
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One more question, You have to wrote $$ z^n -1 = \prod_{k=0}^{k=n-1} (z- e^{ \frac{-2\pi k}{n}) $$
or,
$$ z^n -1 = \prod_{k=0}^{k=n-1} (z- e^{ \frac{2\pi k}{n})$$. Now why did u prefer first instead of second? I tried using second way but I got a different result for same expresison
– Clemens Bartholdy Jun 19 '20 at 14:12 -
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@DDD4C4U Those expressions, as far as I can understand them as written, are identical. – Mark Viola Jun 19 '20 at 17:50
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@MarkViola: unfortunately this was deleted. Could you help me undelete it (2 more votes)? No problem if you can't. – Mittens Dec 13 '22 at 20:41
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You may prove such identity through Herglotz trick, for instance. Both the RHS and the LHS are entire functions in the complex plane of order $1$, having simple zeroes at every point of $\frac{\pi}{n}\mathbb{Z}$ and only there. Additionally, they both are $\frac{2\pi}{n}$-periodic functions. It follows that such identity holds as soon as it holds for some $x\not\in\frac{\pi}{n}\mathbb{Z}$, like $x=\frac{\pi}{2n}$. In such a case, it is straightforward to prove through De Moivre's identity $\sin(x)=\frac{e^{ix}-e^{-ix}}{2}$.
Since $\sin(\pi z)=\frac{\pi}{\Gamma(z)\Gamma(1-z)}$, you may also prove it through the multiplication theorem for the $\Gamma$ function.
Jack D'Aurizio
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Perhaps you mean $\sin z=\frac{\pi}{\Gamma\left(\frac{z}{\pi}\right)\Gamma\left(1-\frac{z}{\pi}\right)}$. – Poder Rac Jul 26 '20 at 09:28
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The right side is odd and periodic with period $2\pi/n$. Repeated use of the identities for $\sin(a) \sin(b)$ and $\cos(a) \sin(b)$ must give you a sum of terms of the form $a_j \sin(j x)$ and $b_j \cos(j x)$ with $0 \le j \le n$, but the oddness and periodicity implies that the terms for $j < n$ and the $\cos$ terms cancel, leaving a multiple of $\sin(nx)$. Now all you have to do is show that the constant is $1$...
Robert Israel
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