5. Let $S=\left\{(x, y) \mid x^{2}-y^{2}=\right.$ odd, $\left.x, y \in \mathbf{R}\right\}$, $T=\left\{(x, y) \mid \sin \left(2 \pi x^{2}\right)-\sin \left(2 \pi y^{2}\right)=\cos \left(2 \pi x^{2}\right)-\cos \left(2 \pi y^{2}\right), x, y \in \mathbf{R}\right\}$. Then
(A) $S \subset T$
( B ) $T \subset S$
(C) $S=T$
( D) $S \cap T=\Phi$
5.【Analysis and Solution】When $y^{2}=x^{2}+$ an odd number, it is easy to see that $\sin \left(2 \pi x^{2}\right)-\sin \left(2 \pi y^{2}\right)=\cos \left(2 \pi x^{2}\right)-\cos \left(2 \pi y^{2}\right)$ holds.
Therefore, when $(x, y) \in S$, it must belong to $T$, so $S \subseteq T$. Moreover, points $(x, y) \in T$ satisfying $x=y$ do not belong to $S$. Hence $S$ $\subset T$.