2.1.15 ** In $\triangle A B C$, it is known that $\sin A \cdot \cos ^{2} \frac{C}{2}+\sin C \cdot \cos ^{2} \frac{A}{2}=\frac{3}{2} \sin B$. Find the value of $\cos \frac{A-C}{2}-2 \sin \frac{B}{2}$.
From the given, we have
$$
\begin{array}{c}
\sin A \cdot \frac{1+\cos C}{2}+\sin C \cdot \frac{1+\cos A}{2}=\frac{3}{2} \sin B \\
\sin A+\sin C+\sin A \cdot \cos C+\cos A \cdot \sin C=3 \sin B \\
\sin A+\sin C+\sin (A+C)=3 \sin B \\
\sin A+\sin C=2 \sin B \\
2 \sin \frac{A+C}{2} \cdot \cos \frac{A-C}{2}=4 \sin \frac{B}{2} \cdot \cos \frac{B}{2}
\end{array}
$$

Therefore, $\cos \frac{A-C}{2}=2 \sin \frac{B}{2}$, which means $\cos \frac{A-C}{2}-2 \sin \frac{B}{2}=0$.