[Solution] Let $y=-x$, then $y \in\left(0, \frac{1}{2}\right)$, and
$$
\begin{array}{l}
a_{1}=\cos (\sin y \pi), \quad a_{2}=\sin (\cos y \pi), \quad a_{3}=\cos (1-y) \pi<0 . \\
\because \quad \sin y \pi+\cos y \pi=\sqrt{2} \sin \left(y \pi+\frac{\pi}{4}\right) \leqslant \sqrt{2} \leqslant \frac{\pi}{2}, \\
\therefore \quad 0<\cos y \pi<\frac{\pi}{2}-\sin y \pi<\frac{\pi}{2} .
\end{array}
$$

Thus $0<\sin (\cos y \pi)<\cos (\sin y \pi)$. Therefore, the correct choice is $(A)$.
9.26 Let $x \in\left(-\frac{1}{2}, 0\right)$, then $a_{1}=\cos (\sin x \pi), a_{2}=\sin (\cos x \pi)$, $a_{3}=\cos (x+1) \pi$ have the following size relationship:
(A) $a_{3}<a_{2}<a_{1}$.
(B) $a_{1}<a_{3}<a_{2}$.
(C) $a_{3}<a_{1}<a_{2}$.
(D) $a_{2}<a_{3}<a_{1}$.
(China High School Mathematics League, 1996)