1. Let the sets be
$$
\begin{aligned}
S= & \left\{(x, y) \left\lvert\, x-\frac{1}{2} y^{2}\right. \text { is odd, } x, y \in \mathbf{R}\right\}, \\
T= & \left\{(x, y) \mid \sin (2 \pi x)-\sin \left(\pi y^{2}\right)=\right. \\
& \left.\cos (2 \pi x)-\cos \left(\pi y^{2}\right), x, y \in \mathbf{R}\right\} .
\end{aligned}
$$

Then the relationship between $S$ and $T$ is ( ).
(A) $S \subset T$
(B) $T$ ■
(C) $S=T$
(D) $S \cap T=\varnothing$
$-1 .(\mathrm{A})$.
When $x=\frac{1}{2} y^{2}+$ odd number, it is obvious that
$$
\sin (2 \pi x)-\sin \left(\pi y^{2}\right)=\cos (2 \pi x)-\cos \left(\pi y^{2}\right)
$$

holds.
Therefore, when $(x, y) \in S$, $(x, y) \in T$. Thus $S \subseteq T$.
Moreover, points $(x, y) \in T$ that satisfy $x=\frac{1}{2} y^{2}$ do not belong to $S$, hence $S \subset T$.