3. As shown in the figure, from the right-angle vertex $C$ of the right triangle $\triangle A B C$, lines $C E$ and $C F$ are drawn to the trisection points of the hypotenuse $A B$. It is known that $C E=\sin \alpha, C F=\cos \alpha$ (where $\alpha$ is an acute angle). Then $A B=$
3. $\frac{3 \sqrt{5}}{5}$.

As shown in the figure, let $A C=b$, $B C=a$. Draw $E M \perp A C, F N \perp A C$, with the feet of the perpendiculars being $M, N$ respectively.
Since $B E=E F=F A$,

then
$$
\begin{array}{l}
E M=\frac{2}{3} a, F N=\frac{1}{3} a, \\
C M=\frac{1}{3} b, C N=\frac{2}{3} b .
\end{array}
$$

In the right triangle $\triangle C E M$,
$$
\left(\frac{2}{3} a\right)^{2}+\left(\frac{1}{3} b\right)^{2}=\sin ^{2} \alpha \text {, }
$$

In the right triangle $\triangle C F N$,
$$
\begin{array}{l}
\left(\frac{1}{3} a\right)^{2}+\left(\frac{2}{3} b\right)^{2}=\cos ^{2} \alpha, \\
\therefore \frac{5}{9} a^{2}+\frac{5}{9} b^{2}=1, a^{2}+b^{2}=\frac{9}{5} .
\end{array}
$$

Therefore, $A B=\sqrt{a^{2}+b^{2}}=\frac{3 \sqrt{5}}{5}$.