Intriguing Products, Part 2

The Chebyshev polynomial of the first kind $T_n(x)$ has leading term $2^{n-1} x^n$ and zeros $\cos \tfrac{(2k-1) \pi}{2n}$ for $1 \le k \le n$ . Thus $T_n(x) \; = \; 2^{n-1} \prod_{k=1}^n \left( x - \cos\tfrac{(2k-1)\pi}{2n}\right) \; = \; \tfrac12\prod_{k=1}^n \left(2x - 2\cos\tfrac{(2k-1)\pi}{2n}\right)$ and so the product of the question is equal to $\prod_{k=1}^{2016} \left(\sqrt{2} - 2\cos\tfrac{(2k-1)\pi}{2\times2016}\right) \; = \; 2T_{2016}\big(\tfrac{1}{\sqrt{2}}\big) \;.$ Since $T_n(\cos\theta) \; =\; \cos n\theta \;,$ we deduce that $T_n\big(\tfrac{1}{\sqrt{2}}\big) \; = \; \cos \tfrac14n\pi \;,$ and hence $T_{2016}\big(\tfrac{1}{\sqrt{2}}\big) = 1$ , making the answer equal to $\boxed{2}$ .

There are other quick ways to do this, for example

$\prod_{k=1}^{2016}\left(\sqrt{2}-2\cos\left(\frac{(2k-1)\pi}{2\times 2016}\right)\right)$ $=\prod_{k=1}^{1008}\left(2-4\cos^2\left(\frac{(2k-1)\pi}{2\times 2016}\right)\right)=2^{1008}\prod_{k=1}^{1008}\left(\cos\left(\frac{(2k-1)\pi}{2016}\right)\right)=\boxed{2}$

where we first (and last) use $\cos(\pi-x)=-\cos(x)$ , then $1-2\cos^2(x)=-\cos(2x)$ , and finally the last of the "Identities without Variables" listed here , with $n=504$ .