Example 2 Given that the base of the regular tetrahedron $S-ABC$ is an equilateral triangle with a side length of 1, and the side edge length is 2. If a section through the line $AB$ divides the volume of the regular tetrahedron into two equal parts, then the cosine value of the plane angle of the dihedral angle formed by the section and the base is $\qquad$ . [2]
【Analysis】As shown in Figure 5, establish a spatial rectangular coordinate system. Let $O$ be the projection of point $S$ on the base plane, $O S=h$. Then
$$
\begin{array}{l}
A(0,0,0), \\
B(1,0,0), \\
C\left(\frac{1}{2}, \frac{\sqrt{3}}{2}, 0\right), \\
O\left(\frac{1}{2}, \frac{\sqrt{3}}{6}, 0\right), \\
S\left(\frac{1}{2}, \frac{\sqrt{3}}{6}, h\right) .
\end{array}
$$

From $S A=2$, we get
$$
\begin{array}{l}
\sqrt{\left(\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{6}\right)^{2}+h^{2}}=2 \\
\Rightarrow h=\frac{\sqrt{33}}{3} \Rightarrow S\left(\frac{1}{2}, \frac{\sqrt{3}}{6}, \frac{\sqrt{33}}{3}\right) \\
\Rightarrow D\left(\frac{1}{2}, \frac{\sqrt{3}}{3}, \frac{\sqrt{33}}{6}\right) .
\end{array}
$$

By property (7), the equation of plane $D A B$ is
$$
\sqrt{33} y-2 \sqrt{3} z=0 \text {. }
$$

The equation of plane $A B C$ is $z=0$, so the cosine of the angle between the normal vectors of the two planes is
$$
\cos \theta=\frac{2 \sqrt{15}}{15} \text {. }
$$

Therefore, the cosine of the dihedral angle formed by the two planes is $\frac{2 \sqrt{15}}{15}$.