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

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

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

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

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

The first digits of your number are:

0.00


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 $10^{-80}$.

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

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

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

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