Finding the area under the inequality $\sin^2 \pi x + \sin^2 \pi y \le 1$ for $x,y \in [-1,1]$ 
Find the area under the inequality $$\sin^2 \pi x + \sin^2 \pi y \le 1 \text{ for } x,y \in [-1,1]$$

I coudn't do this problem without using a graphing calculator:

It's easy to see now that in each quadrant, the area is $1/2$ unit, so the total area would be $2$ units.
How would one do this without access to a graphing calculator? Looking at the graph, It looks like there is a pattern I am missing out on. One thought would be to make the implicit inequality explicit, and obtain
$$|\sin \pi y \le \cos \pi x|$$
but I still couldn't couldn't graph this manually.
 A: We have that
$$\sin^2 \pi x + \sin^2 \pi y = 1 \iff \sin^2 \pi x=\cos^2 \pi y \iff \sin \pi x=\pm\cos \pi y$$
and since by definition
$$
\begin{cases}
\sin A= \cos B \iff A=\frac \pi 2\pm B+2k\pi \\\\
\sin A= -\cos B=\cos (-B) \iff A=-\frac \pi 2\pm B+2k\pi
\end{cases}
$$
we obtain
$$
\begin{cases}
\pi x=\frac {3\pi} 2 \pm \pi y \iff x=\frac 32 \pm y\\\\
\pi x=\frac \pi 2 \pm \pi y \iff x=\frac 12 \pm y\\\\
\pi x=\frac \pi 2 \pm \pi y \iff x=\frac 12 \pm y\\\\
\pi x=-\frac {3\pi} 2  \pm \pi y \iff x=-\frac 32 \pm y
\end{cases}
$$
from here we can obtain the area under the inequality.
A: Use Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$
$$\sin^2\pi x\le\cos^2\pi y$$
$$\iff\cos\pi(x+y)\cos\pi(x-y)\ge0$$
Case$\#1:$
If $\cos\pi(x+y)\ge0,2n\pi-\dfrac\pi2\le\pi(x+y)\le2n\pi+\dfrac\pi2\iff2n-\dfrac12\le x+y\le2n-\dfrac12$
and then we need $\cos\pi(x-y)\ge0\iff 2m-\dfrac12\le x-y\le2m-\dfrac12$
where $m,n$ are integers
Case$\#2:$
$$\cos\pi(x+y)\le0,\cos\pi(x+y)\le0$$
A: We can write
$$ z(x,y)=\dfrac{\sin^2\pi x}{\cos^2\pi y} \ge 1$$
because $\le1$ is inadmissible.
The total area is partitioned into two regions in case of equality by two sets of straight lines having slopes $\pm 1$:
$$\pm x\pm y=\pm \dfrac{k}{2}$$
for integer$k$ which makes area vanish by virtue of symmetry wrt either axis.
