3. $\triangle A B C$ is inscribed in a unit circle, and the angle bisectors of the three interior angles $A$, $B$, and $C$ are extended to intersect the circle at points $A_{1}$, $B_{1}$, and $C_{1}$, respectively. Then
$$
\frac{A A_{1} \cos \frac{A}{2}+B B_{1} \cos \frac{B}{2}+C C_{1} \cos \frac{C}{2}}{\sin A+\sin B+\sin C}
$$

is ( ).
(A) 2
(B) 4
(C) 6
(D) 8
3.A.

As shown in Figure 4, connect $B A_{1}$, then
$$
\begin{array}{l}
A A_{1}=2 \sin \left(B+\frac{A}{2}\right) \\
=2 \cos \left(\frac{B}{2}-\frac{C}{2}\right) .
\end{array}
$$

Therefore, $A A_{1} \cos \frac{A}{2}$
$$
\begin{array}{l}
=2 \cos \left(\frac{B}{2}-\frac{C}{2}\right) \cdot \cos \frac{A}{2} \\
=\cos \left(\frac{\pi}{2}-C\right)+\cos \left(\frac{\pi}{2}-B\right) \\
=\sin C+\sin B .
\end{array}
$$

Similarly, $B B_{1} \cos \frac{B}{2}=\sin A+\sin C$,
$$
C C_{1} \cos \frac{C}{2}=\sin A+\sin B .
$$

Thus, $A A_{1} \cos \frac{A}{2}+B B_{1} \cos \frac{B}{2}+C C_{1} \cos \frac{C}{2}$
$$
=2(\sin A+\sin B+\sin C) \text {. }
$$