Example 1 In $\triangle A B C$, $a, b, c$ are the sides opposite to angles $A, B, C$ respectively. Given $a+c=2 b, A-C=\frac{\pi}{3}$, find the value of $\sin B$.
Analysis We use the Law of Sines to transform the relationship between sides and angles.
Solution By the Law of Sines $a=2 R \sin A, b=2 R \sin B, c=2 R \sin C$, substituting into $a+c=2 b$ yields $\sin A+\sin C$ $=2 \sin B$, so $2 \sin \frac{A+C}{2} \cos \frac{A-C}{2}=4 \sin \frac{B}{2} \cos \frac{B}{2}$. Also, $A-C=\frac{\pi}{3}$, we can get: $\sin \frac{B}{2}=\frac{\sqrt{3}}{4}$, then $\cos \frac{B}{2}=\frac{\sqrt{13}}{4}$, so $\sin B=2 \sin \frac{B}{2} \cos \frac{B}{2}=\frac{\sqrt{39}}{8}$