$9 \cdot 24$ For any $\varphi \in\left(0, \frac{\pi}{2}\right)$, we have
(A) $\sin \sin \varphi\cos \varphi>\cos \cos \varphi$.
(C) $\sin \cos \varphi>\cos \varphi>\operatorname{cossin} \varphi$.
(D) $\sin \cos \varphi<\cos \varphi<\operatorname{cossin} \varphi$.
(China High School Mathematics League, 1982)
[Solution] In $\left(0, \frac{\pi}{2}\right)$, $\sin x, \cos x$ have the following properties:
(1) $\sin x \sin \varphi$, and from (2) we have $\cos \varphi < \cos \sin \varphi$.
Thus, $\sin \cos \varphi < \cos \varphi < \cos \sin \varphi$.
Therefore, the answer is $(D)$.