Example 9 $\triangle A B C$ is inscribed in a unit circle, the angle bisectors of the three interior angles $A, B, C$ are extended to intersect this circle at points $A_{1}, B_{1}, C_{1}$, respectively. Find the value of $\frac{A A_{1} \cos \frac{A}{2}+B B_{1} \cos \frac{B}{2}+C C_{1} \cos \frac{C}{2}}{\sin A+\sin B+\sin C}$.
Using the Law of Sines to convert sides into angles and handle them as trigonometric expressions.

Solution: As shown in Figure 12-7, connect $B A_{1}$. Then, $A A_{1}=2 \sin \left(B+\frac{A}{2}\right)=2 \cos \frac{B-C}{2}$, so $A A_{1} \cos \frac{A}{2}=2 \cos \frac{B-C}{2} \cos \frac{A}{2}=\sin C+\sin B$. Similarly, $B B_{1} \cos \frac{B}{2}=\sin A+\sin C$, and $C C_{1} \cos \frac{C}{2}=\sin A+\sin B$. Substituting these into the original expression, we get $\frac{A A_{1} \cos \frac{A}{2}+B B_{1} \cos \frac{B}{2}+C C_{1} \cos \frac{C}{2}}{\sin A+\sin B+\sin C}=$ $\frac{2(\sin A+\sin B+\sin C)}{\sin A+\sin B+\sin C}=2$.