4. (5th "Hope Cup" Senior High School Competition) The functions $f_{1}(x)=\left|\sin \frac{x}{2}\right| \cdot\left|\cos \frac{x}{2}\right|, f_{2}(x)=\sin \frac{2 x}{3} +\cos \frac{2 x}{3}, f_{3}(x)=\arccos (\sin x)$ have the smallest positive periods $T_{1}, T_{2}, T_{3}$, respectively, then
A. $T_{1}<T_{2}<T_{3}$
B. $T_{3}<T_{2}<T_{1}$
C. $T_{1}<T_{3}<T_{2}$
D. $T_{3}<T_{1}<T_{2}$
4. C $f_{1}(x)=\left|\sin \frac{x}{2}\right| \cdot\left|\cos \frac{x}{2}\right|=\frac{1}{2}|\sin x|$, the smallest positive period $T_{1}=\pi ; f_{2}(x)=\sin \frac{2 x}{3}+\cos \frac{2 x}{3}$ $=\sqrt{2} \sin \left(\frac{2 x}{3}+\frac{\pi}{4}\right)$, the smallest positive period $T_{2}=3 \pi; f_{3}(x)=\arccos (\sin x)$ has the smallest positive period $T_{3}=2 \pi$. Therefore, $T_{1}<$ $T_{3}<T_{2}$. Hence, the answer is C.