4. In a regular tetrahedron $P-ABC$, it is known that the angle between the lateral edge and the base is $45^{\circ}$. Then the cosine value of the dihedral angle between two adjacent lateral faces is $\qquad$
4. $\frac{1}{5}$.

As shown in Figure 2, let the base edge length of the regular tetrahedron $P-ABC$ be $a$, and $E$ be the midpoint of $AB$.

Then $\angle PCE$ is the angle between the lateral edge $PC$ and the base $ABC$, i.e., $\angle PCE=45^{\circ}$.
Draw $AF \perp PC$ at point $F$ from point $A$.
By symmetry,
$BF \perp PC$.
Thus, $\angle AFB$ is the dihedral angle between the faces $PAC$ and $PBC$.
In the isosceles right $\triangle EFC$,
$$
EF=\frac{\sqrt{2}}{2} CE=\frac{\sqrt{6}}{4} a \text {. }
$$

Therefore, $AF=BF=\sqrt{BE^{2}+EF^{2}}=\frac{\sqrt{10}}{4} a$.
In $\triangle AFB$,
$$
\cos \angle AFB=\frac{AF^{2}+BF^{2}-AB^{2}}{2 AF \cdot BF}=\frac{1}{5} .
$$