Example: Let $x \in \mathbf{R}$, compare $f(x)=\cos (\cos x)$ with $g(x)=\sin (\sin x)$.
Analyzing these two functions, both are composite functions of trigonometric functions. Therefore, we still use the properties of trigonometric functions, such as monotonicity, periodicity, and boundedness, to solve the problem.

Solution: Since \( f(x+2\pi) = f(x) \) and \( g(x+2\pi) = g(x) \), we only need to consider the relationship between \( f(x) \) and \( g(x) \) when \( x \in [-\pi, \pi] \).
When \( x \in [-\pi, 0] \), \( f(x) > 0 \) and \( g(x) \leq 0 \), so \( f(x) > g(x) \).
When \( x \in [0, \pi] \), \( \sin(\sin x) = \cos\left(\frac{\pi}{2} - \sin x\right) \) and \( \cos(\cos x) = \cos|\cos x| \). Both \( \frac{\pi}{2} - \sin x \) and \( |\cos x| \) belong to the decreasing interval of the cosine function, \([0, \pi]\), and by
\[ |\cos x| + \sin x = \sqrt{1 + 2 \sin x |\cos x|} = \sqrt{1 + |\sin 2x|} \leq \sqrt{2} < \frac{\pi}{2} \]
we know:
\[ |\cos x| < \frac{\pi}{2} - \sin x \]
Therefore, \( \sin(\sin x) < \cos(\cos x) \)
which means \( g(x) < f(x) \)
In summary, \( g(x) < f(x) \)