3. Given $\sin (\alpha+2 \beta)=\frac{3}{5}, \sin (2 \alpha-3 \beta)=$ $\frac{4}{5}, \alpha \in\left[\frac{\pi}{12}, \frac{\pi}{4}\right], \beta \in\left[0, \frac{\pi}{12}\right]$. Then $\sin (8 \alpha-5 \beta)$ $=(\quad)$.
(A) $\frac{4}{25}$
(B) $-\frac{4}{125}$
(C) $-\frac{4}{5}$
(D) $\frac{4}{5}$
3.C.

Notice
$$
\begin{array}{l}
\sin (8 \alpha-5 \beta) \\
=\sin [2(\alpha+2 \beta)+3(2 \alpha-3 \beta)] \\
=\sin 2(\alpha+2 \beta) \cdot \cos 3(2 \alpha-3 \beta)+ \\
\quad \cos 2(\alpha+2 \beta) \cdot \sin 3(2 \alpha-3 \beta) .
\end{array}
$$

From the problem, we know that $\alpha+2 \beta \in\left[\frac{\pi}{12}, \frac{5 \pi}{12}\right]$,
$$
\begin{array}{l}
2 \alpha-3 \beta \in\left[-\frac{\pi}{12}, \frac{\pi}{2}\right] . \\
\text { Hence } \cos (\alpha+2 \beta)=\frac{4}{5}, \cos (2 \alpha-3 \beta)=\frac{3}{5}, \\
\sin 2(\alpha+2 \beta)=2 \times \frac{3}{5} \times \frac{4}{5}=\frac{24}{25}, \\
\cos 2(\alpha+2 \beta)=2\left(\frac{4}{5}\right)^{2}-1=\frac{7}{25}, \\
\cos 3(2 \alpha-3 \beta)=4\left(\frac{3}{5}\right)^{3}-3 \times \frac{3}{5}=-\frac{117}{125}, \\
\sin 3(2 \alpha-3 \beta)=3 \times \frac{4}{5}-4\left(\frac{4}{5}\right)^{3}=\frac{44}{125} .
\end{array}
$$

Then $\sin (8 \alpha-5 \beta)$
$$
\begin{array}{l}
=\frac{24}{25} \times\left(-\frac{117}{125}\right)+\frac{44}{125} \times \frac{7}{25} \\
=-\frac{4}{5}
\end{array}
$$