15. Given the plane point set $A=\{(x, y) \mid x=2 \sin \alpha+2 \sin \beta, y=2 \cos \alpha+2 \cos \beta\}$
$$
B=\{(x, y) \mid \sin (x+y) \cos (x+y) \geqslant 0\}
$$

Find the area of the region represented by $A \cap B$.
15. First, find the specific regions of $A$ and $B$:
$A$ is represented by the points determined by $\left\{\begin{array}{l}\frac{x}{2}-\sin \alpha=\sin \beta \\ \frac{y}{2}-\cos \alpha=\cos \beta\end{array}\right.$. Squaring and adding the two equations: $(x-2 \sin \alpha)^{2}+$ $(y-2 \cos \alpha)^{2}=4$ represents a circle with center $C(2 \sin \alpha, 2 \cos \alpha)$ and radius 2, as shown by the solid line. When $C$ moves along the dashed line $(x^{2}+y^{2}=4)$ for one complete cycle, the region reached by the solid line is $x^{2}+$ $y^{2} \leqslant 16$, which is a circular area with center $O$ and radius 4. Therefore, $A$ represents a circular area with center $O$ and radius 4. $B$ represents $\sin 2(x+y) \geqslant 0$, which means $2(x+y) \in [2k\pi, 2k\pi+\pi]$, or $x+y \in \left[k\pi, k\pi+\frac{\pi}{2}\right]$, where $k \in \mathbb{Z}$. The region is shown by the shaded areas between the parallel lines in the figure below. Consider the line $l: x+y=0$. Observing: If the shaded and blank parts below $l$ are swapped, the entire figure is symmetric about $l$.

Therefore: Regardless of the radius of the circle represented by $A$, the shaded part in the lower left semicircle can exactly fill all the blank parts in the upper semicircle.
So the area of the region $A \cap B$ is:
$$
\frac{1}{2} \times \pi \times 4^{2}=8 \pi .
$$