Example 4 The domain of the function $f(x)$ is $\mathbf{R}$, for any $a, b \in \mathbf{R}$, there is $f(a+b)+f(a-b)=$ $2 f(a) f(b)$, and there exists $c>0$, such that $f\left(\frac{c}{2}\right)=0$, try to determine: Is $f(x)$ a periodic function? If so, find one of its periods; if not, please explain the reason.
Analyzing the structure of $f(a+b)+f(a-b)=2 f(a) f(b)$, we can associate it with the sum-to-product formula $\cos (\alpha+\beta) + \cos (\alpha-\beta)=2 \cos \alpha \cos \beta$, and $\cos \frac{\pi}{2}=0$. This allows us to hypothesize that $y=\cos x$ is a prototype of $f(x)$, and since the smallest positive period of $y=\cos x$ is $2 \pi$, we conjecture that $f(x)$ is a periodic function with a period of $2 c$.

To prove that $2 c$ is a period of $f(x)$, we need to show that $f(x+2 c)=f(x)$. Given the condition $f\left(\frac{c}{2}\right) = 0$ and $f(a+b)+f(a-b)=2 f(a) \cdot f(b)$, we need to choose appropriate values of $a$ and $b$ to make $\int\left(\frac{c}{2}\right)$ and $f(x)$ appear in the condition.

Solution: Let $a=x+\frac{c}{2}, b=\frac{c}{2}$, and substitute into $f(a+b)+f(a-b)=2 f(a) f(b)$, we get $f(x+c) = -f(x)$. Therefore, $f(x+2 c) = f[(x+c)+c] = -f(x+c) = f(x)$.
So, $f(x+2 c) = f(x)$.
Thus, $f(x)$ is a periodic function with a period of $2 c$.
Explanation: For some propositions, when the solution approach is not clear, we can explore through observation, association, special cases, and seeking general structures. Finding a prototype is often an effective method, which can quickly lead us to the solution for certain functional equations.

Common prototypes of abstract functions include: a function $f(x)$ that satisfies $f(x y)=f(x)+f(y)$ and $f\left(\frac{x}{y}\right)=f(x)-f(y)$ has a prototype $\log _{0} x$; a function $f(x)$ that satisfies $f(x+y)=f(x)+f(y)$ and $f(x-y)=f(x)-f(y)$ has a prototype $k x$; a function $f(x)$ that satisfies $f(x+y)=f(x) \cdot f(y)$ and $f(x-y)=\frac{f(x)}{f(y)}$ has a prototype $a^{\prime}$; and so on.