6. As shown in Figure 2, 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 $d \sin \theta=$ _
Solution: $\frac{4}{5}$.
[Method 1] As shown in Figure 2, take the midpoint $K$ of $C C_{1}$, then $K N / / C M$, so $\angle A N K$ is the angle formed by the skew lines $A N$ and $C M$, i.e., $\angle A N K=\theta$.
Since $A B \perp$ plane $B C C_{1} B_{1}, C M \perp B C_{1}$, it follows that $C M \perp A N$ (by the theorem of three perpendiculars).
Also, $K N / / C M$, so $K N \perp A N$, i.e., $\theta=90^{\circ}$.
Since $C M / / K N$, $C M / /$ plane $A N K$, hence the distance between the skew lines $A N$ and $C M$ is equal to the distance from line $C M$ to plane $A N K$, which is the distance from point $M$ to plane $A N K$.
Noting that plane $A B N \perp$ plane $B C C_{1}$, the distance from point $M$ to plane $A N K$ is equal to the distance from point $M$ to line $A N$.
In the right triangle $\triangle A B N$, $A B=4, B N=3$, so the distance from point $B$ to line $A N$ is $\frac{12}{5}$. Therefore, the distance from point $M$ to line $A N$ is
$$
d=\frac{12}{5} \times \frac{1}{3}=\frac{4}{5} \text {. }
$$

Thus, $d \sin \theta=\frac{4}{5} \sin 90^{\circ}=\frac{4}{5}$.
[Method 2] Establish a spatial rectangular coordinate system $D-x y z$ as shown in Figure 3, then we have $A(2 \sqrt{2}, 0,0) 、 C(0,4,0) 、 M(\sqrt{2}, 4, \sqrt{2}) 、 N\left(\frac{\sqrt{2}}{2}, 4\right.$, $\left.\frac{3 \sqrt{2}}{2}\right)$. Thus, $\overrightarrow{A N}=\left(-\frac{3 \sqrt{2}}{2}, 4, \frac{3 \sqrt{2}}{2}\right), \overrightarrow{C M}=(\sqrt{2}, 0, \sqrt{2})$.
So $\overrightarrow{A N} \cdot \overrightarrow{C M}=0$, then $A N \perp C M$, i.e., $\theta=90^{\circ}$.
Let a vector perpendicular to both $\overrightarrow{A N}$ and $\overrightarrow{C M}$ be $\mathbf{n}=(x, y, 1)$, then
$$
\left\{\begin{array}{l}
- \frac { 3 \sqrt { 2 } } { 2 } x + 4 y + \frac { 3 \sqrt { 2 } } { 2 } = 0 , \\
\sqrt { 2 } x + \sqrt { 2 } = 0 .
\end{array} \text { Solving, we get } \left\{\begin{array}{l}
x=-1, \\
y=-\frac{3 \sqrt{2}}{4} .
\end{array}\right.\right.
$$

So $\mathbf{n}=\left(-1,-\frac{3 \sqrt{2}}{4}, 1\right)$.
Also, $\overrightarrow{A C}=(-2 \sqrt{2}, 4,0)$,
So $d=\frac{|\mathbf{n} \cdot \overrightarrow{A C}|}{|\mathbf{n}|}=\frac{\sqrt{2}}{\frac{5 \sqrt{2}}{4}}=\frac{4}{5}$.
Thus, $d \sin \theta=\frac{4}{5} \sin 90^{\circ}=\frac{4}{5}$.