Example 3 Given $x, y \in\left[-\frac{\pi}{4}, \frac{\pi}{4}\right], a \in \mathbf{R}$, and
$$
\left\{\begin{array}{l}
x^{3}+\sin x-2 a=0, \\
4 y^{3}+\sin y \cdot \cos y+a=0 .
\end{array}\right.
$$

Then $\cos (x+2 y)=$ $\qquad$
(1994, National High School Mathematics Joint Competition)
【Analysis】This problem appears similar to Example 2, but directly comparing the left sides of the two equations does not yield a suitable function. We can perform a simple transformation on the system of equations:
$$
\left\{\begin{array}{l}
x^{3}+\sin x=2 a, \\
(2 y)^{3}+\sin 2 y=-2 a .
\end{array}\right.
$$

We can construct the function
$$
f(x)=x^{3}+\sin x,
$$

which is an odd function.
Thus, $f(x)=2 a=-f(2 y)=f(-2 y)$.
Given $x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, then
$$
f^{\prime}(x)=3 x^{2}+\cos x>0 \text {. }
$$

Therefore, $f(x)$ is a monotonically increasing function on the interval $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.
Hence, $x=-2 y$, which means $x+2 y=0$.
So, $\cos (x+2 y)=1$.