Multiplying First Quadrant Cosines: 2016 Of Them

$x = \displaystyle \prod \limits^{2016}_{n=1}\cos \left( \frac{n\pi }{2017} \right)$ $y = \displaystyle\prod \limits^{2016}_{n=1}\sin\left( \frac{n\pi }{2017} \right)$ $x. y =\displaystyle\prod \limits^{2016}_{n=1}\sin\left( \frac{n\pi }{2017} \right) \cos \left( \frac{n\pi }{2017} \right)$ $x. y = \displaystyle\prod_{n=1}^{2016}\frac{1}{2}\sin\left( \frac{2n\pi}{2017} \right) = \bigg(\frac{1}{2}\bigg)^{2016}[\sin \left( \frac{2\pi }{2017} \right) \sin \left( \frac{4\pi }{2017} \right) \cdots \sin \left( \frac{14\pi }{2017} \right) \sin \left( \frac{16\pi }{2017} \right) \sin \left( \frac{18\pi }{2017} \right) \cdots \sin \left( \frac{4032\pi }{2017} \right)]$ $\color{#3D99F6}{\boxed{\color{#D61F06}{\boxed{\theta+\alpha=2\pi \rightarrow \sin(\theta)=-\sin(2\pi-\alpha)}}}}$ $x. y = \displaystyle\prod_{n=1}^{2016}\frac{1}{2}\sin\left( \frac{2n\pi}{2017} \right) = \bigg(\frac{1}{2}\bigg)^{2016}\times(-1)^{1008}[\displaystyle\prod \limits^{2016}_{n=1}\sin\left( \frac{n\pi }{2017} \right) ]$ $x. y =\left( \frac{1}{2} \right) ^{2016}\times\left( -1\right) ^{1008}. y$ $x =\left( \frac{1}{2^{2016}} \right)=(\frac{1}{A^B})$ $\Large\color{#3D99F6}{\boxed{\color{forestgreen}{\boxed{A+B=2018}}}}$

simply we can use the identity $P=\large \prod_{k=1}^{n-1} \cos\left(\frac{k\pi}{n}\right) = \frac{\sin(\pi n/2)}{2^{n-1}}$ now put $n=2017$ to get $P=\frac {1}{2^{2016}}$ So $\Large\color{#624F41}{\boxed{A+B=2018}}$