4. Given that the base of the regular tetrahedron $S-ABC$ is an equilateral triangle with side length 1, and the side edges are 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 ( ).
(A) $\frac{\sqrt{15}}{10}$
(B) $\frac{4 \sqrt{15}}{15}$
(C) $\frac{\sqrt{15}}{15}$
(D) $\frac{2 \sqrt{15}}{15}$
4. D.

Let the section intersect edge $SC$ at point $D$. It is easy to see that $D$ is the midpoint of edge $SC$.

Take the midpoint $E$ of $AB$, and connect $EC$, $DE$, and $SE$. Then $\angle DEC$ is the plane angle of the dihedral angle formed by the section and the base, denoted as $\theta$.
In $\triangle SEC$,
$$
SE=\frac{\sqrt{15}}{2}, \quad EC=\frac{\sqrt{3}}{2}, \quad SC=2 \text{.}
$$

Then the median $DE=\frac{\sqrt{5}}{2}$.
In $\triangle DEC$, applying the cosine rule yields
$$
\cos \theta=\frac{2 \sqrt{15}}{15} .
$$