1. Given $\sin \alpha \cdot \cos \beta=-\frac{1}{2}$. Then the range of $\cos \alpha \cdot \sin \beta$ is ( ).
(A) $\left[-1, \frac{1}{2}\right]$
(B) $\left[-\frac{1}{2}, 1\right]$
(C) $\left[-\frac{3}{4}, \frac{3}{4}\right]$
(D) $\left[-\frac{1}{2}, \frac{1}{2}\right]$
$-1 .(D)$
$$
\text { Method i: } \begin{aligned}
& \cos ^{2} \alpha \cdot \sin ^{2} \beta=\left(1-\sin ^{2} \alpha\right)\left(1-\cos ^{2} \beta\right) \\
& =1-\left(\sin ^{2} \alpha+\cos ^{2} \beta\right)+\sin ^{2} \alpha \cdot \cos ^{2} \beta \\
& =\frac{5}{4}-\left(\sin ^{2} \alpha+\cos ^{2} \beta\right) \\
& =\frac{5}{4}-(\sin \alpha+\cos \beta)^{2}+2 \sin \alpha \cdot \cos \beta \\
& =\frac{1}{4}-(\sin \alpha+\cos \beta)^{2} \\
& \leqslant \frac{1}{4} . \\
\therefore-\frac{1}{2} \leqslant & \cos \alpha \cdot \sin \beta \leqslant \frac{1}{2} .
\end{aligned}
$$

When $\alpha=-\frac{\pi}{4} ， \beta=\frac{\pi}{4}$, $\cos \alpha \cdot \sin \beta=\frac{1}{2}$; when $\alpha=-\frac{\pi}{4}, \beta=-\frac{\pi}{4}$, $\cos \alpha \cdot \sin \beta=-\frac{1}{2}$. Therefore, the range of $\cos \alpha \cdot \sin \beta$ is $\left[-\frac{1}{2}, \frac{1}{2}\right]$.
Method 2: From the given information,
$$
\begin{array}{l}
\sin (\alpha+\beta)=\sin \alpha \cdot \cos \beta+\cos \alpha \cdot \sin \beta \\
=-\frac{1}{2}+\cos \alpha \cdot \sin \beta \in[-1,1] .
\end{array}
$$

Similarly,
$$
\begin{array}{l}
\sin (\alpha-\beta)=-\frac{1}{2}-\cos \alpha \cdot \sin \beta \in[-1,1] \\
\therefore\left\{\begin{array}{l}
-\frac{1}{2} \leqslant \cos \alpha \cdot \sin \beta \leqslant \frac{3}{2} \\
-\frac{3}{2} \leqslant \cos \alpha \cdot \sin \beta \leqslant \frac{1}{2}
\end{array}\right. \\
-\frac{1}{2} \leqslant \cos \alpha \cdot \sin \beta \leqslant \frac{1}{2} .
\end{array}
$$

When $\alpha=-\frac{\pi}{4}, \beta=\frac{\pi}{4}$, $\cos \alpha \cdot \sin \beta$ can take the maximum value $\frac{1}{2}$; when $\alpha=-\frac{\pi}{4}, \beta=-\frac{\pi}{4}$, $\cos \alpha \cdot \sin \beta$ can take the minimum value $-\frac{1}{2}$.