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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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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!

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  • $\begingroup$ 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. $\endgroup$ – Navin Feb 21 '17 at 14:48
  • $\begingroup$ @navinstudent The equations in your comment have not been formatted $\endgroup$ – Mark Viola Feb 21 '17 at 14:50
  • $\begingroup$ @navinstudent Look at Equation $(2)$. Let $z=e^{i2x}$ $\endgroup$ – Mark Viola Feb 21 '17 at 14:59
  • $\begingroup$ Yes I see that . please do also clarify subsequent step.It really puzzled me. $\endgroup$ – Navin Feb 21 '17 at 15:01
  • $\begingroup$ Subsequent steps use straightforward arithmetic. Are you familiar with Euler's Formula in $(1)? $\endgroup$ – Mark Viola Feb 21 '17 at 15:02
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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.

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  • $\begingroup$ Perhaps you mean $\sin z=\frac{\pi}{\Gamma\left(\frac{z}{\pi}\right)\Gamma\left(1-\frac{z}{\pi}\right)}$. $\endgroup$ – Poder Rac Jul 26 '20 at 9:28
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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 $1$...

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