More fun in 2016, Part 21

$\displaystyle \prod_{k=1}^{2016} 2\left(1+\cos\left(\frac{(2k-1)\pi}{2016}\right) \right) \\ \displaystyle = 2^{2016} \prod_{k=1}^{1008} \left(1+\cos\left(\frac{(2k-1)\pi}{2016}\right) \right)^2 \quad \quad \small \color{#3D99F6}{\text{Since }\cos\left(\frac{(2016-2k-1)\pi}{2016}\right) =\cos\left(\frac{(2k-1)\pi}{2016}\right)} \\ \displaystyle = 2^{2016} \prod_{k=1}^{504} \left(1-\cos ^2 \left(\frac{(2k-1)\pi}{2016}\right) \right)^2 \quad \quad \small \color{#3D99F6}{\text{Since }\cos\left(\frac{(1008-2k-1)\pi}{2016}\right) =- \cos\left(\frac{(2k-1)\pi}{2016}\right)} \\ \displaystyle = 2^{2016} \prod_{k=1}^{504} \sin ^4 \left(\frac{(2k-1)\pi}{2016}\right) \\ \displaystyle = 2^{2016} \left( 2^{-1007} \right)^2 = 2^2 = \boxed{4} \quad \quad \small \color{#3D99F6}{\text{See Note.}}$

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$\color{#3D99F6}{\text{Note:}}$

From the identity (eq. 24 -- T. Drane, pers. comm., Apr. 19, 2006) :

$\begin{aligned} \prod_{k=1}^{2016} \sin \left(\frac{k \pi}{n}\right)& = 2^{1-n} n \\ \Rightarrow \left( \prod_{k=1}^{504} \sin \left(\frac{(2k-1)\pi}{2016}\right) \right)^2 & = \prod_{k=1}^{1008} \sin \left(\frac{(2k-1)\pi}{2016}\right) \\ & = \frac{\displaystyle \prod_{k=1}^{2015} \sin \left(\frac{k \pi}{2016}\right)}{\displaystyle\prod_{k=1}^{1007} \sin \left(\frac{k \pi}{1008}\right)} \\& = \frac{2^{-2015}\dot{}2016}{2^{-1007}\dot{}1008} = 2^{-1007} \end{aligned}$

Comrade Cheong has submitted a delightful solution that uses trigonometry alone. For the sake of variety, let me suggest a solution that makes use of roots of (negative) unity.

Considering the 2016th roots of $-1$ and grouping them in conjugate pairs, we can write $x^{2016}+1=\prod_{k=1}^{1008}\left(x-e^{\frac{(2k-1)\pi i}{2016}}\right)\left(x-e^{-\frac{(2k-1)\pi i}{2016}}\right)=\prod_{k=1}^{1008}\left(x^2-2x\cos\left(\frac{(2k-1)\pi}{2016}\right)+1\right)$

Evaluating this equation at $x=-1$ gives $2=\prod_{k=1}^{1008}2\left(1+\cos\left(\frac{(2k-1)\pi}{2016}\right)\right)$

This product comes out the same if we let $k$ run from 1009 to 2016, by the symmetry $\cos(2\pi-t)=\cos(t)$ , so that $P=2^2=\boxed{4}$

Alternatively, the result can be obtained quickly from the chebyshev polynomials of the first kind

$T_n(x)=2^{n-1}\prod_{k=1}^{n}\left(x-\cos\left(\frac{(2k-1)\pi}{2n}\right)\right)$

Again, evaluate at $x=-1$ and recall that $T_n(-1)=(-1)^n$ .