Example 1 Given that $f(x)$ is an even function defined on $\mathbf{R}$, if $g(x)$ is an odd function, and $g(x)=f(x-1)$, find the value of $f(2007)$.
Analysis: It seems that solving this problem directly is infeasible; one must explore patterns from the given conditions, i.e., utilize the periodicity of the function to find a solution.

Solution: Given the conditions \( f(-x) = f(x) \), \( g(-x) = -g(x) \), and \( g(x) = f(x-1) \), we have \( g(-x) = f(-x-1) = f(x+1) = -g(x) \), which implies \( g(x) = -f(x+1) \). Therefore, \( f(x-1) = -f(x+1) \), i.e., \( f(x) = -f(x+2) = f(x+4) \).
Thus, \( f(x) \) is a periodic function with a period of 4.
Since \( g(x) \) is an odd function on \( \mathbf{R} \), we have \( g(0) = 0 \), so \( f(-1) = g(0) = 0 \). Therefore,
\[
f(2007) = f(4 \times 502 - 1) = f(-1) = 0.
\]

Explanation: The key to solving this problem is to discover the periodicity of the function and to uncover the hidden condition \( g(0) = 0 \).