Example 1 (1994 National High School Mathematics League Question) Given $x, y \in\left[-\frac{\pi}{4}, \frac{\pi}{4}\right], a \in \mathbf{R}$,
and $\left\{\begin{array}{l}x^{3}+\sin x-2 a=0, \\ 4 y^{3}+\sin y \cos y+a=0 .\end{array}\right.$ Find the value of $\cos (x+2 y)$.
Analysis: The feature of this problem is a very small population. The required solution is not closely related to the given conditions. Pay attention to comparing the structure of the two expressions, think boldly, and by constructing a function, a successful path can be found.

Solution: The conditions can be transformed into $\left\{\begin{array}{l}x^{3}+\sin x-2 a=0 \\ 8 y^{3}+\sin 2 y+2 a=0\end{array}\right.$, thus $x^{3}+\sin x=-8 y^{3}-\sin 2 y$. Clearly, $x,-2 y \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. Construct the function $f(x)=t^{3}+\sin t, t \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. It is known that this function is monotonic, so $x=-2 y$, and thus $\cos (x+2 y)=1$