# How do you find this product?

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

-
– lab bhattacharjee Jul 12 '14 at 9:09
These days, you can find most products on Amazon... ;-) – David Richerby Jul 12 '14 at 17:49
@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 '14 at 18:27

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: $$P = \sqrt{\frac{2015}{2^{2014}}} = \frac{\sqrt{2015}}{2^{1007}}.$$

-
Or more roughly put, $P \approx 0$. – Cole Johnson Jul 12 '14 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.

-
Any idea about how to prove this identity? – Pkwssis Jul 12 '14 at 8:07
@user157130 Good old fashioned induction is one option. – David H Jul 12 '14 at 8:30
@user157130 Also, see math.stackexchange.com/questions/8385/… – David H Jul 12 '14 at 8:52
Thanks! That was useful – Pkwssis Jul 12 '14 at 8:58