We know that $e^{ix}=\cos{x}+i\sin{x}$

So, the summation of $\cos{x}$ and $\sin{x}$ would be the summation of real and imaginary parts of $e^{ix}$ respectively.

For the summation of $\sin{x}$ , $\displaystyle \sum_{x=0}^n \sin{x}=\displaystyle \sum_{x=0}^n \Im(e^{ix})=\Im \Bigg(\displaystyle \sum_{x=0}^n e^{ix}\Bigg)$ Clearly, this summation is a geometric one, so $\Im \Bigg(\displaystyle \sum_{x=0}^n e^{ix}\Bigg)=\Im \Bigg(\frac{e^{i(n+1)}-1}{e^i-1}\Bigg)$ $=\Im \Bigg(\frac{e^\frac{i(n+1)}{2}}{e^\frac{i}{2}} \cdot \frac{e^\frac{i(n+1)}{2}-e^{-\frac{i(n+1)}{2}}}{e^\frac{i}{2}-e^{-\frac{i}{2}}}\Bigg)$ $=\frac{\sin(\frac{n+1}{2})}{\sin(\frac{1}{2})}\Im(e^{\frac{in}{2}})$ $=\frac{\sin(\frac{n+1}{2}) \sin(\frac{n}{2})}{\sin(\frac{1}{2})}$ Doing the same for the summation of $\cos{x}$ gives $\displaystyle \sum_{x=0}^n \cos{x}=\Re \Bigg(\displaystyle \sum_{x=0}^n e^{ix}\Bigg)=\frac{\sin(\frac{n+1}{2}) \cos(\frac{n}{2})}{\sin(\frac{1}{2})}$ $\frac{\displaystyle \sum_{x=0}^{2016} \sin{x}}{\displaystyle \sum_{x=0}^{2016} \cos{x}}=\frac{\sin(\frac{1}{2} \times 2016^{\circ})}{\cos(\frac{1}{2} \times 2016^{\circ})}=\tan(1008^{\circ})$ $=\tan(-72^\circ)=-\sqrt{5-2\sqrt{5}}$

above expression terminates to :

-(sin37+ sin38+sin39...........sin216)/cos37cos38......cos179)

Which further terminates to -cot(18)

Hence answer=10

We can use formula transformation sum of trigonometric functions sin and cos to product: $\frac{(\sin(2016°)+\sin(0°))+(\sin(2015°)+\sin(1°))+\dots+(\sin(1012°) + \sin(1014°)) +( \sin(1013°)}{(\cos(2016°) + \cos(0°)) + (\cos(2015°) + \cos(1°)) +\dots+ (\cos(1012°) + \cos(1014°)) + \cos(1013°)}=$ $=\frac{(2 \sin(1008°)[\cos(1008°) + \cos(1007°) +\dots+\ cos(1°) + 1]}{(2\cos(1008°)[\cos(1008°) + \cos(1008°) +\dots+ \cos(1°) +1]}=\tan(1008°)$