1. Given $x, y \in\left[-\frac{\pi}{4}, \frac{\pi}{4}\right], a \in \mathbf{R}$, and $x^{3}+$ $\sin x-2 a=0,4 y^{3}+\sin y \cos y+a=0$, then $\cos (x$ $+2 y)=$ $\qquad$ .
1. 1.

Solution: From the given, we have $x^{3}+\sin x=2 a=(-2 y)^{3}+\sin (-2 y)$. Let $f(t)=t^{3}+\sin t$, then $f(x)=f(-2 y)$. Since $f(t)=t^{3}+\sin t$ is an increasing function for $t \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, it follows that $x=-2 y$, i.e., $x+2 y=0$. Therefore, $\cos (x+2 y)=1$.