7.22 If $S=\left\{(x, y) \mid x^{2}-y^{2}=\right.$ odd, $\left.x, y \in \mathrm{R}\right\}$, and $T=\left\{(x, y) \mid \sin \left(2 \pi x^{2}\right)-\sin \left(2 \pi y^{2}\right)=\cos \left(2 \pi x^{2}\right)-\cos \left(2 \pi y^{2}\right)\right.$, where $x, y \in \mathrm{R}\}$. Then
(A) $S \subset T$.
(B) $T \subset S$.
(C) $S=T$.
(D) $S \cap T=\varnothing$.
(China High School Mathematics League, 1991)
[Solution 1] When $x^{2}-y^{2}$ is odd, 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 $(x, y) \in T$ that satisfy $x=y$ do not belong to $S$. Thus, $S \subset T$. Hence, the answer is $(A)$.
[Solution 2] Take $A(0,0)$ and $B(1,0)$, clearly $A \in T$, and $A \notin S$; also $B \in S$, and $B \in T$, which rules out $(B),(C),(D)$. Hence, the answer is $(A)$.