(1)求角 $A 、 B 、 C$;
(2) 若 $b=2$, 延长 $B C$ 至 $D$, 使 $\triangle A B D$ 的面积为 $\frac{3 \sqrt{3}}{2}$, 求 $\sin \angle C A D$.
【答案】 18.【解析】(1) 由 $A+B+C=\pi, \therefore A+C=\pi-B$,
\begin{aligned} & \cos B=-\cos (A+C), \\ & \therefore \cos (A-C)-\cos (A+C)=\frac{3}{2}, \\ & \therefore \sin A \cdot \sin C=\frac{3}{4}, \end{aligned}

\begin{aligned} & \therefore \sin ^2 B=\sin A \cdot \sin C=\frac{3}{4}, \\ & \therefore \sin B=\frac{\sqrt{3}}{2}, \\ & \text { 方法一: } \therefore|\cos B|=\frac{1}{2} \text {, 又 } \cos B=\frac{a^2+c^2-b^2}{2 a c}= \end{aligned}

$\frac{a^2+c^2-a c}{2 a c} \geqslant \frac{2 a c-a c}{2 a c}=\frac{1}{2}$, 当且仅当 $a=c$ 时, 等 号成立,
\begin{aligned} & \therefore \cos B=\frac{1}{2}, a=c \text {, 又 } 0 < B < \pi, \\ & \therefore A=B=C=\frac{\pi}{3} . \end{aligned}

\begin{aligned} & \text { (2) } \because b=2, \therefore|A B|=2, \\ & \therefore S_{\triangle A B D}=\frac{1}{2} \cdot|A B| \cdot|B D| \cdot \sin 60^{\circ}, \\ & \text { 即 } \frac{1}{2} \times 2 \cdot|B D| \cdot \frac{\sqrt{3}}{2}=\frac{3 \sqrt{3}}{2}, \therefore|B D|=3, \therefore|C D| \\ & =1 \text {, 由余弦定理: 在 } \triangle A C D \text { 中, }|A D|^2=|A C|^2 \\ & +|C D|^2-2|A C| \cdot|C D| \cos \angle D C A=2^2+1^2 \\ & -2 \times 2 \times 1 \times\left(-\frac{1}{2}\right)=7, \\ & \text { 又 由 正 弦 定 理: } \frac{|A D|}{\sin 120^{\circ}}=\frac{|C D|}{\sin \angle C A D}, \\ & \therefore \frac{\sqrt{7}}{\frac{\sqrt{3}}{2}}=\frac{1}{\sin \angle C A D}, \\ & \therefore \sin \angle C A D=\frac{\sqrt{3}}{\sqrt{7}}=\frac{\sqrt{3}}{2 \sqrt{7}}=\frac{\sqrt{21}}{14} . \cdots \cdots \cdots 12 \text { 分 } \end{aligned}