A curious identity, Part 2

We can rewrite this product as

$\displaystyle\dfrac{\prod_{k=1}^{2014}(w^{2k} + 1)}{\prod_{k=1}^{2014}w^{k}} = -\large\prod_{k=1}^{2014} (e^{\frac{i2\pi k}{2015}} + 1),$

since $\large\displaystyle\prod_{k=1}^{2014} w^{k} = w^{\sum_{k=1}^{2014} k} = w^{\frac{2014*2015}{2}} = e^{i\pi*1007} = -1.$

Now since the roots of the equation $z^{2015} - 1 = 0$ are of the form $\large e^{\frac{i2\pi k}{2015}}$ for integers $0 \le k \le 2014,$ , we see that

$z^{2015} - 1 = \large\displaystyle\prod_{k=0}^{2014} (z - e^{\frac{i2\pi k}{2015}}) = (z - 1)\prod_{k=1}^{2014} (z - e^{\frac{i2\pi k}{2015}}).$

Plugging $z = -1$ into this equation gives us that

$-2 = -2 * (-1)^{2014} * \large\displaystyle\prod_{k=1}^{2014} (1 + e^{\frac{i2\pi k}{2015}}) \Longrightarrow 1 = \prod_{k=1}^{2014} (1 + e^{\frac{i2\pi k}{2015}}).$

Thus the given expression is simply $\boxed{-1}.$

Now for the trigonometry. Note first that $e^{ik\theta} + e^{-ik\theta} = 2\cos(k\theta).$ So the given product can be written as

$2^{2014} \displaystyle\prod_{k=1}^{2014} \cos\left(\dfrac{k\pi}{2015}\right) = -1 \Longrightarrow \prod_{k=1}^{2015} \cos\left(\dfrac{k\pi}{2015}\right) = -\dfrac{1}{2^{2014}}.$

The general identity is $\displaystyle\prod_{k=1}^{n-1} \cos\left(\dfrac{k\pi}{n}\right) = \dfrac{\sin(\pi n/2)}{2^{n-1}}.$