3. $\triangle A B C$ is inscribed in a unit circle, and the angle bisectors of the three interior angles $A, B, C$ are extended to intersect the circle at points $A_{1}, B_{1}, C_{1}$. Then the value of $\frac{A A_{1} \cos \frac{A}{2}+B B_{1} \cos \frac{B}{2}+C C_{1} \cos \frac{C}{2}}{\sin A+\sin B+\sin C}$ is
A. 2
B. 4
C. 6
D. 8
Solve:
As shown in the figure, $\cos \frac{A}{2}=\frac{B D}{A_{1} I}=\frac{A E}{A I}=\frac{B D+A E}{A A_{1}}$
$$
\begin{array}{l}
=\frac{\frac{a}{2}+\frac{b+c-a}{2}}{A A_{1}}=\frac{\frac{b+c}{2}}{A A_{1}} \Rightarrow A A_{1} \cos \frac{A}{2}=\frac{b+c}{2} . \\
\text { Therefore } \frac{A A_{1} \cos \frac{A}{2}+B B_{1} \cos \frac{B}{2}+C C_{1} \cos \frac{C}{2}}{\sin A+\sin B+\sin C} \\
=\frac{a+b+c}{\sin A+\sin B+\sin C}=2 R=2 . \text { So the answer is } A .
\end{array}
$$