8. Given $x, y \in\left[-\frac{\pi}{4}, \frac{\pi}{4}\right], a \in \mathbf{R}$, and
$$
\left\{\begin{array}{l}
x^{3}+\sin x-2 a=0, \\
4 y^{3}+\frac{1}{2} \sin 2 y+a=0 .
\end{array}\right.
$$

then the value of $\cos (x+2 y)$ is
8.1.

Let $f(t)=t^{3}+\sin t$. Then $f(t)$ is monotonically increasing on $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

From the original system of equations, we get $f(x)=f(-2 y)=2 a$, and since $x$ and $-2 y \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, it follows that $x=-2 y$, hence $x+2 y=0$. Therefore, $\cos (x+2 y)=1$.