21.12 ** 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 $\cos (x+2 y)-$
Analyzing the system of equations in the question, it can be transformed into
$$
\left\{\begin{array}{l}
x^{3}+\sin x-2 a=0, \\
(2 y)^{3}+\sin 2 y+2 a=0 .
\end{array}\right.
$$

Construct the function \( f(t)=t^{3}+\sin t \). Clearly, when \( -\frac{\pi}{2} \leqslant t \leqslant \frac{\pi}{2} \), \( f(t) \) is monotonically increasing and is an odd function. From (1) and (2), we get \( f(x)=2 a=f(-2 y) \), so \( x=-2 y \), which means \( x+2 y=0 \). Therefore, \( \cos (x+2 y)=1 \).