Product of This Year?? (Part 1)

Using $\sin \theta = \sin(\pi - \theta)$ , It is quite easy to see that we are asked $\displaystyle \prod_{k=1}^{1007} \sin^2 \bigg(\frac{k \pi}{2015}\bigg)$

Let us generalize this problem(so that we can answer easily if the problem is asked again in 2016 :P)

We will prove that $\displaystyle \prod_{k=1}^{n} \sin^2 \bigg(\frac{k \pi}{2n+1}\bigg) = \frac{2n+1}{2^{2n}}$

We know that $x^2 - 2x \cos \theta + 1 = (x - e^{i \theta})(x-e^{-i \theta})$

Also, roots of $x^{2n+1} = 1$ are $\Large \displaystyle 1 , e^{\pm \frac{2i \pi}{2n+1}}, e^{\pm \frac{4i \pi}{2n+1}}, \dots e^{\pm \frac{2n i \pi}{2n+1}}$

Hence, $\large \displaystyle x^{2n+1} - 1 = (x-1) \prod_{k=1}^{n} \bigg(x -e^{ \frac{2k i \pi}{2n+1}} \bigg)\bigg(x - e^{ \frac{- 2k i \pi}{2n+1}} \bigg)$

Hence, $\displaystyle 1+x+ \dots x^{2n} = \prod_{k=1}^{n} \bigg(x^2 - 2x \cos \frac{2k \pi}{n} +1 \bigg)$

Replacing $x=1$ and using $\displaystyle 1 - \cos \theta = 2 \sin^2 \frac{\theta}{2}$ for $\displaystyle \theta = \frac{2k\pi}{n}$ , we have,

$2n+1 = \displaystyle \prod_{k=1}^{n} 4 \sin^2\frac{k \pi}{2n+1} = 2^{2n} \displaystyle \prod_{k=1}^{n} \sin^2 \frac{k \pi}{2n+1}$

Hence, $\displaystyle \prod_{k=1}^{n} \sin^2 \frac{k \pi}{2n+1} = \frac{2n+1}{2^{2n}}$

In our case, $n=1007$ , hence, answer is $\frac{2015}{2^{2014}}$ .

Thus, $a^2 - b^2 = (a-b)(a+b) = 4029$