Example 5 As shown in Figure 9, in the cube $A B C D-$ $A_{1} B_{1} C_{1} D_{1}$, $E$ and $F$ are the midpoints of $A B$ and $A A_{1}$, respectively. The sine value of the plane angle formed by the dihedral angle between plane $C E B_{1}$ and plane $D_{1} F B_{1}$ is ( ).
(A) $\frac{1}{2}$
(B) $\frac{\sqrt{2}}{2}$
(C) $\frac{\sqrt{3}}{2}$
(D) 1
Solution: Let the edge length of the cube be 1. Then
$$
E C=E B_{1}=\frac{\sqrt{5}}{2} \text {. }
$$

Since $E$ and $F$ are the midpoints of $A B$ and $A A_{1}$ respectively, it follows that $D_{1} F$, $D A$, and $C E$ must intersect at a point $G$.

Connecting $B_{1} G$. By the theorem of parallel lines cutting intersecting lines into proportional segments, we have
$$
\frac{E G+\frac{\sqrt{5}}{2}}{E G}=2 \text {. }
$$

Therefore, $E G=\frac{\sqrt{5}}{2}$.
Since $E G=E B_{1}=E C=\frac{\sqrt{5}}{2}$, it follows that $\angle G B_{1} C=90^{\circ}$.
Similarly, $\angle G B_{1} D_{1}=90^{\circ}$.
In the equilateral $\triangle B_{1} D_{1} C$, we have
$$
\angle C B_{1} D_{1}=60^{\circ} \text {. }
$$

By the trihedral angle formula, we get
$$
\cos \theta=\frac{\cos 60^{\circ}-\cos 90^{\circ} \cdot \cos 90^{\circ}}{\sin 90^{\circ} \cdot \sin 90^{\circ}}=\frac{1}{2} .
$$

Thus, $\theta=60^{\circ}$.
Therefore, $\sin \theta=\frac{\sqrt{3}}{2}$.