Example 2 (2005 Daoxi College Entrance Examination Question) Given the vectors $\boldsymbol{a}=\left(2 \cos \frac{x}{2}, \tan \left(\frac{x}{2}+\frac{\pi}{4}\right)\right), \boldsymbol{b}=$ $\left(\sqrt{2} \sin \left(\frac{x}{2}+\frac{\pi}{4}\right), \tan \left(\frac{x}{2}-\frac{\pi}{4}\right)\right)$, let $f(x)=\boldsymbol{a} \cdot \boldsymbol{b}$. Find the maximum value of the function $f(x)$, its smallest positive period, and the monotonic intervals of $f(x)$ on $\left[0, \frac{\pi}{2}\right]$.
Analysis: The key to solving this example is to transform the expression into one of the simple trigonometric function forms listed above. Based on the structural characteristics of this example, consider the methods of converting compound angles to single angles and converting tangents to sines and cosines.

Solution: 
$$
f(x) = \boldsymbol{a} \cdot \boldsymbol{b} = 2 \sqrt{2} \cos \frac{x}{2} \sin \left(\frac{x}{2} + \frac{\pi}{4}\right) + \tan \left(\frac{x}{2} - \frac{\pi}{4}\right) \tan \left(\frac{x}{2} - \frac{\pi}{4}\right)
$$
$$
\begin{array}{l}
= 2 \sqrt{2} \cos \frac{x}{2} \left(\frac{\sqrt{2}}{2} \sin \frac{x}{2} + \frac{\sqrt{2}}{2} \cos \frac{x}{2}\right) + \frac{1 + \tan \frac{x}{2}}{1 - \tan \frac{x}{2}} \cdot \frac{\tan \frac{x}{2} - 1}{1 + \tan \frac{x}{2}} \\
= 2 \sin \frac{x}{2} \cos \frac{x}{2} + 2 \cos^2 \frac{x}{2} - 1 \\
= \sin x + \cos x \\
= \sqrt{2} \sin \left(x + \frac{\pi}{4}\right).
\end{array}
$$

Therefore, the maximum value of $f(x)$ is $\sqrt{2}$, the smallest positive period is $2\pi$, and $f(x)$ is monotonically increasing on $\left[0, \frac{\pi}{4}\right]$ and monotonically decreasing on $\left[\frac{\pi}{4}, \frac{\pi}{2}\right]$.

Explanation: After learning trigonometric functions, many people believe that the sine-type function $f(x) = A \sin (\omega x + \varphi) (x \in M)$ is always a periodic function, but this is not always the case.
(1) When $M$ is a finite interval, such as $[0, 3\pi]$, $f(x)$ is clearly not a periodic function because the domain of a periodic function must be unbounded on at least one end.
(2) When $M$ is an infinite interval, $f(x)$ is not necessarily a periodic function, for example:
$f(x) = \sin x, x \in M, M = \left\{x \mid x = 2^n, n \in \mathbf{N}\right\}$, because $M$ is not a periodic domain.
(3) When $M$ is a periodic domain, $f(x)$ is not necessarily a periodic function, for example:
$$
f(x) = \sin x, x \in M (=\mathbf{Z})
$$
The inherent period of $f(x) = \sin x$ is $T_1 = 2\pi$, and the period of $M = \mathbf{Z}$ is $T_2 = 1$. Since $\frac{T_1}{T_2} = 2\pi$ (an irrational number), they do not match, so $f(x)$ is definitely not a periodic function.
(4) The smallest positive period of $f(x) = A \sin (\omega x + \varphi) (x \in \mathbf{R})$ is $T_1 = \frac{2\pi}{|\omega|}$, and the smallest positive period of $M$ is $T_2$. When $\frac{T_1}{T_2} = \frac{2\pi}{|\omega T_2|}$ is a rational number, $f(x) = A \sin (\omega x + \varphi) (x \in M)$ is a periodic function, and the least common multiple of $T_1$ and $T_2$ is a period of $f(x)$.
(III) Graphical Method
For some functions, the period is not easily discernible and does not meet the conditions for using formulas. If using the definition to find the period is also cumbersome, consider using a graph (not necessarily drawn) to determine the period.