Example 9 (1994 National High School League Question) Given $x, y \in\left[-\frac{\pi}{4}, \frac{\pi}{4}\right]$, 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 $\cos (x+2 y)=$ $\qquad$
Solution: Fill in 1. Reason: Consider the function $f(t)=t^{3}+\sin t$, which is monotonically increasing on $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. Let $f(x)=2a$, i.e., $x^{3}+\sin x-2a=0$, and $f(2y)=-2a$, i.e., $4y^{3}+\sin y \cdot \cos y+a=0$. From $f(x)=2a=f(-2y)$, we get $x=-2y$.
Since $2y \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, we know $y \in\left[-\frac{\pi}{4}, \frac{\pi}{4}\right]$, hence $\cos (x+2y)=1$.