Intriguing Product

The Chebyshev polynomial of the second kind $U_n(x)$ has leading term $2^n x^n$ and zeros $\cos \tfrac{k \pi}{n+1}$ for $1 \le k \le n$ . Thus $U_n(x) \; = \; 2^n \prod_{k=1}^n \left( x - \cos\tfrac{k\pi}{n+1}\right) \; = \; \prod_{k=1}^n \left(2x - 2\cos\tfrac{k\pi}{n+1}\right)$ and so the product of the question is equal to $\prod_{k=0}^{2016} \left(1 - 2\cos\tfrac{k\pi}{2017}\right) \; = \; -\prod_{k=1}^{2016} \left(1 - 2\cos\tfrac{k\pi}{2017}\right) \; =\; -U_{2016}(\tfrac12) \;.$ Since $U_n(\cos\theta) \; =\; \frac{\sin(n+1)\theta}{\sin\theta} \;,$ we deduce that $U_n(\tfrac12) \; = \; \frac{\sin \tfrac13(n+1)\pi}{\sin \tfrac13\pi} \;,$ and hence $U_{2016}(\tfrac12) = 1$ , making the answer equal to $\boxed{-1}$ .