[Example 1.5.1] Let the domain of the function $f(x)$ be the set of real numbers $\mathbf{R}$. For any real numbers $x$ and $y$, we have
$$
f(x+y)+f(x-y)=2 f(x) f(y),
$$

and there exists a positive number $c$ such that
$$
f\left(\frac{c}{2}\right)=0 \text{. }
$$

Is $f(x)$ a periodic function? If so, find one of its periods. If not, explain why.
This problem is to determine whether the function $f(x)$ is a periodic function, that is, whether there exists a non-zero constant $T$ such that for all real numbers $x$,
$$
f(x+T)=f(x).
$$

Observing the functional equation given in the problem,
$$
f(x+y)+f(x-y)=2 f(x) f(y).
$$

We can think of a special case that satisfies this equation, which is a sum-to-product formula for trigonometric functions:
$$
\cos (x+y)+\cos (x-y)=2 \cos x \cos y.
$$

Since
$$
\cos \frac{\pi}{2}=0,
$$

there exists a constant $c=\pi$ such that
$$
f\left(\frac{c}{2}\right)=f\left(\frac{\pi}{2}\right)=\cos \frac{\pi}{2}=0
$$

holds.
Therefore, the cosine function $f(x)=\cos x$ is a special case that meets the given conditions.
Since the cosine function $f(x)=\cos x$ is a periodic function with a period of $2 \pi (2 \pi=2 c)$, we can infer that $f(x)$ is a periodic function, and its period can be analogously determined to be $2 c$.
Guided by this special case, our solution goal becomes clear, that is, we need to prove
$$
f(x+2 c)=f(x).
$$

From the given condition
$$
f(x+y)+f(x-y)=2 f(x) f(y)
$$

and
$$
f\left(\frac{c}{2}\right)=0,
$$

we have
$$
\begin{aligned}
f(x+c)+f(x) & =f\left(x+\frac{c}{2}+\frac{c}{2}\right)+f\left(x+\frac{c}{2}-\frac{c}{2}\right) \\
& =2 f\left(x+\frac{c}{2}\right) f\left(\frac{c}{2}\right)=0,
\end{aligned}
$$

so we have
$$
\begin{array}{l}
f(x+c)=-f(x), \\
f(x+2 c)=-f(x+c)=f(x).
\end{array}
$$

Therefore, $f(x)$ is a periodic function with a period of $2 c$.
Here, the cosine function $f(x)=\cos x$ is a special case of the function under study. Starting from this special case, we can guess that the period of $f(x)$ is $2 c$, which clarifies the direction of the proof.