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Is there a way to find the exact value of the product $$P=\displaystyle\prod_{n=1}^{1007} \sin {\left(\dfrac{n\pi}{2015}\right)}$$

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These days, you can find most products on Amazon... ;-) –  David Richerby Jul 12 at 17:49
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@DavidRicherby Stuck on a test because you can't find the product? Well, with Amazon's (brand new) Yesterday Shipping™, you'll already have the product! ;) –  Cole Johnson Jul 12 at 18:27

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up vote 7 down vote accepted

By symmetry, your product is just: $$ P = \sqrt{\prod_{n=1}^{2014}\sin\frac{n \pi}{2015}}=\sqrt{\frac{1}{2^{2014}}\prod_{n=1}^{2014}\left(\exp\left(\frac{2\pi i n}{2015}\right)-1\right)},$$ where the innermost product is the product of the roots of the polynomial: $$ \frac{(x+1)^{2015}-1}{x}.$$ Hence, by Vieta's theorem: $$ P = \sqrt{\frac{2015}{2^{2014}}} = \frac{\sqrt{2015}}{2^{1007}}.$$

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Or more roughly put, $P \approx 0$. –  Cole Johnson Jul 12 at 18:23

For all positive integers $n\in\mathbb{N}$, the following finite product identity holds:

$$\prod_{k=1}^{\lfloor\frac{n-1}{2}\rfloor}\sin{\left(\frac{\pi\,k}{n}\right)}=2^{\frac{1-n}{2}}\,\sqrt{n}~.$$

Since you have the good fortune that the pair of integers $1007$ and $2015$ satisfy the necessary arithmetic relationship, $\lfloor\frac{2015-1}{2}\rfloor=1007$, the above product identity applies.

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Any idea about how to prove this identity? –  Pkwssis Jul 12 at 8:07
    
@user157130 Good old fashioned induction is one option. –  David H Jul 12 at 8:30
    
@user157130 Also, see math.stackexchange.com/questions/8385/… –  David H Jul 12 at 8:52
    
Thanks! That was useful –  Pkwssis Jul 12 at 8:58

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