Example 1 Let $x \geqslant y \geqslant z \geqslant \frac{\pi}{12}$, and $x+y+z=$ $\frac{\pi}{2}$. Find the maximum and minimum values of the product $\cos x \cdot \sin y \cdot \cos z$.
(1997, National High School Mathematics Competition)
Solution: From the given conditions, we have
$$
\begin{array}{l}
x=\frac{\pi}{2}-(y+z) \leqslant \frac{\pi}{2}-\left(\frac{\pi}{12}+\frac{\pi}{12}\right)=\frac{\pi}{3}, \\
\sin (x-y) \geqslant 0, \sin (y-z) \geqslant 0 .
\end{array}
$$

Thus,
$$
\begin{array}{l}
\cos x \cdot \sin y \cdot \cos z \\
=\frac{1}{2} \cos x \cdot[\sin (y+z)+\sin (y-z)] \\
\geqslant \frac{1}{2} \cos x \cdot \sin (y+z) \\
=\frac{1}{2} \cos ^{2} x \geqslant \frac{1}{2} \cos ^{2} \frac{\pi}{3}=\frac{1}{8},
\end{array}
$$

and the equality holds when $x=\frac{\pi}{3}, y=z=\frac{\pi}{12}$.
$\therefore$ The minimum value of $\cos x \cdot \sin y \cdot \cos z$ is $\frac{1}{8}$.
Also, $\cos x \cdot \sin y \cdot \cos z$
$$
\begin{aligned}
& =\frac{1}{2} \cos z \cdot[\sin (x+y)-\sin (x-y)] \\
\leqslant & \frac{1}{2} \cos z \cdot \sin (x+y)=\frac{1}{2} \cos ^{2} z \\
\leqslant & \frac{1}{2} \cos ^{2} \frac{\pi}{12}=\frac{1}{4}\left(1+\cos \frac{\pi}{6}\right)=\frac{2+\sqrt{3}}{8},
\end{aligned}
$$

and the equality holds when $x=y=\frac{5 \pi}{24}, z=\frac{\pi}{12}$.
$\therefore$ The maximum value of $\cos x \cdot \sin y \cdot \cos z$ is $\frac{2}{8}+\frac{\sqrt{3}}{8}$.