2.1.1 * Let $x \in\left(-\frac{1}{2}, 0\right)$, then the size relationship of $a_{1}=\cos (\sin x \pi), a_{2}=\sin (\cos x \pi), a_{3}=$ $\cos (x+1) \pi$ is ().
(A) $a_{3}<a_{2}<a_{1}$
(B) $a_{1}<a_{3}<a_{2}$
(C) $a_{3}<a_{1}<a_{2}$
(D) $a_{2}<a_{3}<a_{1}$
Given that $\cos (x+1) \pi<0$. Let $y=-x$, then $y \in\left(0, \frac{1}{2}\right)$, and $a_{1}=$ $\cos (\sin y \pi), a_{2}=\sin (\cos y \pi)$. Also, $\sin y \pi+\cos y \pi=\sqrt{2} \sin \left(y \pi+\frac{\pi}{4}\right) \leqslant \sqrt{2}<\frac{\pi}{2}$, so $0<\cos y \pi<\frac{\pi}{2}-\sin y \pi<\frac{\pi}{2}$. Therefore, $0<\sin (\cos y \pi)<\cos (\sin y \pi)$, hence the answer is A.

Comment The above solution uses the identity $a \sin x+b \cos x=\sqrt{a^{2}+b^{2}} \sin (x+\varphi)$, where $\tan \varphi=$ $\frac{b}{a}$, the terminal side of $\varphi$ passes through the point $(a, b)$, and $a \neq 0$.