.已知 $\sin \left(\alpha+\frac{\pi}{6}\right)-\cos \alpha=\frac{1}{2}$, 则 $\sin \left(2 \alpha+\frac{\pi}{6}\right)=$
$ \text{A.} $ $-\frac{1}{2}$ $ \text{B.} $ $\frac{1}{2}$ $ \text{C.} $ $-\frac{3}{4}$ $ \text{D.} $ $\frac{3}{4}$
【答案】 B

【解析】 $\because \sin \left(\alpha-\frac{\pi}{6}\right)-\cos \alpha=\frac{\sqrt{3}}{2} \sin \alpha-\frac{1}{2} \cos \alpha$ $=\sin \left(\alpha-\frac{\pi}{6}\right)=\frac{1}{2}$,
$\therefore \sin \left(2 \alpha+\frac{\pi}{6}\right)=\sin \left[2\left(a-\frac{\pi}{6}\right)+\frac{\pi}{2}\right]=$ $\cos 2\left(\alpha-\frac{\pi}{6}\right)=1-2 \sin ^2\left(\alpha-\frac{\pi}{6}\right)=\frac{1}{2}$, 故选 B.
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