6. As shown in the figure, in the rectangular prism $A B C D-A_{1} B_{1} C_{1} D_{1}$, $A B=4, B C=C C_{1}=2 \sqrt{2}$, $M$ is the midpoint of $B C_{1}$, and $N$ is the midpoint of $M C_{1}$. If the angle between the skew lines $A N$ and $C M$ is $\theta$, and the distance is $d$, then $d \sin \theta=$ $\qquad$
$B C C_{1} B_{1}$ is a square $\Rightarrow C M \perp B C_{1} \Rightarrow C M \perp$ plane $A B C_{1} \Rightarrow \theta=90^{\circ}$.
$E$ is the trisection point of $A B$ $\Rightarrow E M / / A N \Rightarrow A N / /$ plane $M C E$,
and the distance $d$ from $A$ to plane $M C E$ is $\frac{1}{2}$ of the distance from $B$ to plane $M C E$.
$$
\begin{array}{l}
E M^{2}=E C^{2}-C M^{2}=E B^{2}+B C^{2}-C M^{2}=\left(\frac{8}{3}\right)^{2}+4=\frac{100}{9} \\
\Rightarrow E M=\frac{10}{3} \Rightarrow S_{\triangle M C E}=\frac{1}{2} C M \cdot E M=\frac{10}{3} .
\end{array}
$$

Thus $V_{E-B C M}=V_{B-M C E} \Rightarrow S_{\triangle B C M} \cdot E B=S_{\triangle M C E} \cdot 2 d$
$$
\Rightarrow 2 \cdot \frac{8}{3}=\frac{10}{3} \cdot 2 d \Rightarrow d=\frac{4}{5} \Rightarrow d \sin \theta=\frac{4}{5} \text {. }
$$