（5分）已知$\sin (\alpha -\beta )=\dfrac{1}{3}$，$\cos \alpha \sin \beta =\dfrac{1}{6}$，则$\cos (2\alpha +2\beta )=($　　$)$
A．$\dfrac{7}{9}$              B．$\dfrac{1}{9}$              C．$-\dfrac{1}{9}$              D．$-\dfrac{7}{9}$
答案：$B$
分析：由已知结合和差角公式先求出$\sin \alpha \cos \beta$，再求出$\sin (\alpha +\beta )$，然后结合二倍角公式可求．
解：因为$\sin (\alpha -\beta )=\sin \alpha \cos \beta -\sin \beta \cos \alpha =\dfrac{1}{3}$，$\cos \alpha \sin \beta =\dfrac{1}{6}$，
所以$\sin \alpha \cos \beta =\dfrac{1}{2}$，
所以$\sin (\alpha +\beta )=\sin \alpha \cos \beta +\sin \beta \cos \alpha =\dfrac{1}{2}+\dfrac{1}{6}=\dfrac{2}{3}$，
则$\cos (2\alpha +2\beta )=1-2\sin ^{2}(\alpha +\beta )=1-2\times \dfrac{4}{9}=\dfrac{1}{9}$．
故选：$B$．
点评：本题主要考查了和差角公式，二倍角公式的应用，属于中档题．