$2^{\sin(x) + \cos(y)} = 1$ , $16^{\sin^2(x) + \cos^2(y)} = 4$ (system of equations) My progress so far:
$2^{\sin(x) + \cos(y)} = 1$ 
$16^{\sin^2(x) + \cos^2(y)} = 4$
||||||||||||||||||||
$\sin(x) + \cos(y) = 0$
$\sin^2(x) + \cos^2(y) = \frac12$
I think I'm on the right track, but I'm not sure how to continue. Thanks in advance to anyone who helps.

Suppose $\sin x + \cos y = 0$. Then
$$\sin ^2x + \cos ^2y = (\sin x + \cos y) ^2 -2\sin x\cos y = -2\sin x\cos y. $$
Thus we have left to consider
$$\begin{cases} \sin x + \cos y = 0 \\ \sin x\cos y = -\frac{1}{4} \end{cases} $$
Can you take it from here?
