Fun with 2018 #3

$\begin{aligned} \cos \left(\frac \pi{2018}\right) & = \prod_{n=0}^\infty \left(1-\frac a{(2018(2n+1))^2} \right) & \small \color{#3D99F6} \text{Let }s = \frac {\sqrt a}{2018} \\ & = \prod_{n=0}^\infty \left(1-\frac {s^2}{(2n+1)^2} \right) \\ & = \left(1-\frac {s^2}{1^2} \right) \left(1-\frac {s^2}{3^2} \right) \left(1-\frac {s^2}{5^2} \right) \left(1-\frac {s^2}{7^2} \right) \cdots \\ & = \frac {\left(1-\frac {s^2}{1^2} \right) \left(1-\frac {s^2}{2^2} \right) \left(1-\frac {s^2}{3^2} \right) \left(1-\frac {s^2}{4^2} \right) \cdots}{\left(1-\frac {s^2}{2^2} \right) \left(1-\frac {s^2}{4^2} \right) \left(1-\frac {s^2}{6^2} \right) \left(1-\frac {s^2}{8^2} \right) \cdots} \\ & = \frac {\left(1-\frac {s^2}{1^2} \right) \left(1-\frac {s^2}{2^2} \right) \left(1-\frac {s^2}{3^2} \right) \left(1-\frac {s^2}{4^2} \right) \cdots}{\left(1-\frac {(s/2)^2}{1^2} \right) \left(1-\frac {(s/2)^2}{2^2} \right) \left(1-\frac {(s/2)^2}{3^2} \right) \cdots} \\ & = \frac {\prod_{n=1}^\infty \left(1-\frac {s^2}{n^2}\right)} {\prod_{n=1}^\infty \left(1-\frac {(s/2)^2}{n^2}\right)} & \small \color{#3D99F6} \text{By identity: } s \pi \prod_{n=1}^\infty \left(1-\frac {s^2}{n^2}\right) = \sin (s\pi) \\ & = \frac {\sin (s\pi)}{2\sin \left(\frac {s\pi}2\right)} = \frac {2\sin \left(\frac {s\pi}2\right)\cos \left(\frac {s\pi}2\right)}{2\sin \left(\frac {s\pi}2\right)} \\ & = \cos \left(\frac {s\pi}2\right) = \cos \left(\frac {\sqrt a\pi}{4036} \right) \\ \implies \frac {\sqrt a}{4036} & = \frac 1{2018} \\ a & = \boxed{4} \end{aligned}$

Reference for $\displaystyle s \pi \prod_{n=1}^\infty \left(1-\frac {s^2}{n^2}\right) = \sin (s\pi)$ see equation 22 .