1. In square $A B C D$, $M$ is the midpoint of side $B C$, and $N$ is the midpoint of side $C D$. Then the value of $\sin \angle M A N$ is $\qquad$
$$
\text { II. } 1 . \frac{3}{5} \text {. }
$$

As shown in Figure 4, let the side length of the square be $S_{\triangle A R M}=S_{\triangle N N N}=1, S_{\triangle C U N}=$ 0.5. Therefore, $S_{\angle M M N}=4-1-1-$ $0.5=1.5$. Also, $A M=A N=\sqrt{5}$, $S_{\triangle M M N}=\frac{1}{2} A M \cdot A N \sin \angle M A N$, so $\sin \angle M A N=\frac{2 \times 1.5}{\sqrt{5} \times \sqrt{5}}=\frac{3}{5}$.