1. Let $S=\left\{(x, y) \mid x^{2}-y^{2}=\right.$ odd, $\left.x, y \in \mathbf{R}\right\}, T=\left\{(x, y) \mid \sin \left(2 \pi x^{2}\right)\right.$ $\left.-\sin \left(2 \pi y^{2}\right)-\cos \left(2 \pi x^{2}\right)-\cos \left(2 \pi y^{2}\right), x, y \in \mathbf{R}\right\}$, the relationship between sets $S$ and $T$ is $\qquad$ .
1. $S \subset T$.

When $x^{2}=y^{2}+(2 k-1)$ (where $k \in \mathbf{Z}$), it is easy to see that $\sin \left(2 \pi x^{2}\right)-\sin \left(2 \pi y^{2}\right)=\sin \left[2 \pi\left(y^{2}+\right.\right.$ $2 k-1)]-\sin \left(2 \pi y^{2}\right)=0$. Similarly, $\cos \left(2 \pi x^{2}\right)-\cos \left(2 \pi y^{2}\right)=0, \sin \left(2 \pi x^{2}\right)-\sin \left(2 \pi y^{2}\right)=$ $\cos \left(2 \pi x^{2}\right)-\cos \left(2 \pi y^{2}\right)$. When $(x, y) \in S$, it must be that $(x, y) \in T$; but when $x=y$, $(x, y) \in T$, while $(x, y) \notin S$, hence $S \subset T$.