若 $\sin \left(\frac{\pi}{6}-a\right)=\frac{1}{3}$, 则 $\cos \left(\frac{2 \pi}{3}+2 a\right)=($ )
由题意, 可得 $\cos \left(\frac{2 \pi}{3}+2 a\right)=-\cos \left[\pi-\left(\frac{2 \pi}{3}+2 a\right)\right]=-\cos \left(\frac{\pi}{3}-2 a\right)=-\cos \left[2\left(\frac{\pi}{6}-a\right)\right]$ $=-\left[1-2 \sin ^{2}\left(\frac{\pi}{6}-a\right)\right]=-\frac{7}{9}$, 故选 $\mathrm{A}$.