I would appreciate if somebody could help me with the following problem. How can we find the product

$$ \prod_{n=1}^{999}\sin\left(\frac{n \pi}{1999}\right)$$


closed as off-topic by user21820, Claude Leibovici, The Phenotype, TheSimpliFire, Parcly Taxel Feb 18 '18 at 11:22

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  • $\begingroup$ No idea how to do this exactly. If a decimal approximation is an acceptable answer you could use Sage. $\endgroup$ – Jim Mar 8 '13 at 6:24
  • $\begingroup$ @wildcow: whats source ? $\endgroup$ – M.H Mar 8 '13 at 6:36
  • $\begingroup$ source : high school problem $\endgroup$ – Young Mar 8 '13 at 11:18
  • $\begingroup$ @Young do you attend the NUS high school? This looks like a SMO problem. $\endgroup$ – anegligibleperson Jun 7 '13 at 9:53
  • $\begingroup$ Using complex numbers and eulers formula seem apparent $\endgroup$ – Torsten Hĕrculĕ Cärlemän Jul 15 '13 at 11:25

$$\prod_{k=1}^{n-1}\sin\left(\frac{k\pi}{n}\right)=\frac{n}{2^{n-1}}$$ Here, let $$x=\prod_{k=1}^{999}\sin\left(\frac{k\pi}{1999}\right) \tag{1}$$

Since $\sin t=\sin (\pi-t)$, therefore,

$$x=\prod_{k=1}^{999}\sin\left(\frac{(1999-k)\pi}{1999}\right)=\prod_{k=1000}^{1998}\sin\left(\frac{k\pi}{1999}\right) \tag{2}$$

Multiplying equation $(1)$ by equation $(2)$ gives,

$$x^2=\prod_{k=1}^{1998}\sin\left(\frac{k\pi}{1999}\right)=\frac{1999}{2^{1998}}$$ $$\implies x=\frac{\sqrt{1999}}{2^{999}}$$

We took $x>0$ because all angles are in $1^{st}$ and $2^{nd}$ quadrant.

  • $\begingroup$ Thank you so much for your concern. But how can I prove the first indentity: $$\prod_{k=1}^{n}\sin\left(\frac{k\pi}{n}\right)=\frac{n}{2^{n-1}}$$ $\endgroup$ – Young Mar 8 '13 at 6:51
  • 2
    $\begingroup$ That's pretty neat! $\endgroup$ – Inceptio Mar 8 '13 at 6:51
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    $\begingroup$ Can you please explain how u got $\prod_{k=1}^{n-1}\sin\big(\frac{k\pi}{n}\big)=\frac{n}{2^{n-1}}$ $\endgroup$ – Phani Raj Mar 8 '13 at 6:53
  • $\begingroup$ @PhaniRaj: I updated my answer. Have a look. $\endgroup$ – Aang Mar 8 '13 at 6:58
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    $\begingroup$ @Avatar Nice answer! $\endgroup$ – Start wearing purple Jul 15 '13 at 11:17

I have an even more precise bound: the product is less than

$$(999)! \left(\frac{\pi}{1999}\right)^{999}$$

Use Stirling approximation to get that this in turn is less than

$$ \sqrt{999 e} \left ( \frac{999 \pi}{1999 e}\right )^{999} \approx 4 \times 10^{-237}$$


The first digits of your number are:


To get this estimate notice that all terms of the product are less than $1$ and that $|\sin(x)| \le |x|$.

To be more precise your number is between $0$ and $10^{-80}$.

  • 4
    $\begingroup$ Hows this helpful? $\endgroup$ – Inceptio Mar 8 '13 at 6:34
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    $\begingroup$ @Inceptio: I know more digits of this number than of $\pi$... $\endgroup$ – Emanuele Paolini Mar 8 '13 at 6:38

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