This Problem Was Made In 2016

New Solution: (credits to those who commented!)

Let $a=x+1$ and $b=y+1$ . We see that $\begin{cases} a^3+2016a=\sin^2\theta - \frac{1}{2} \\ b^3+2016b=\cos^2\theta - \frac{1}{2} \end{cases}$ Adding both, we have $a^3+b^3+2016(a+b)=0$ .

We can factor this further to get $(a+b)(a^2-ab+b^2+2016)=0$ , so we have either $a+b=x+y+2=0$ or $a^2-ab+b^2+2016=0$ .

We express $a^2-ab+b^2+2016$ in the form $(a-b)^2+ab+2016$ , then we must have $a=b$ and $ab=-2016$ which clearly has no solution.

We can also express $a^2-ab+b^2+2016$ in the form $\dfrac{a^2+b^2+(a-b)^2}{2}+2016$ to see that $a^2-ab+b^2+2016\geq2016$ .

Thus, we find our solution in $a+b=x+y+2=0$ . We have $x+y=\boxed{-2}$ .

Original (Wrong) Solution:

We have $\sin^2\theta-\frac{1}{2}=-\left(\cos^2\theta-\frac{1}{2}\right)$ .

We factor the $L.H.S.$ of each equation as $\begin{cases} (x+1)(x^2+2x+2017)=\sin^2\theta - \frac{1}{2} \\ (y+1)(y^2+2y+2017)=\cos^2\theta - \frac{1}{2} \end{cases}$

Adding both, we have $(x+1)(x^2+2x+2017)+(y+1)(y^2+2y+2017)=0$ . Since $x^2+2x+2017>\left(x+1\right)^2\geq0$ and $y^2+2y+2017>\left(y+1\right)^2\geq0$ , we must have $x=-1$ and $y=-1$ . Thus $x+y=\boxed{-2}$ .