7. Let the dihedral angle between a lateral face and the base of a regular quadrilateral pyramid $P-ABCD$ be $\alpha$, and the dihedral angle between two adjacent lateral faces be $\beta$. Then $2 \cos \beta + \cos 2 \alpha =$ $\qquad$ .
7. -1 As shown in the figure, draw $PO \perp$ plane $ABCD$, then $O$ is the center of the square $ABCD$. Let the side length of the base $ABCD$ be $a$, and the side length of the lateral edge be $b$.
Since the projection of the lateral face $\triangle PCD$ on the base $ABCD$ is $\triangle OCD$,
by the projection area theorem, we have $\cos \alpha=\frac{S_{\triangle OCD}}{S_{\triangle PCD}}=\frac{a^{2}}{4 S_{\triangle PCD}}=\frac{a^{2}}{2 b \cdot CF}$.
Draw $CF \perp DP$ from point $C$, and connect $AF$. It is easy to see that $AF \perp DP$, so the plane angle of the dihedral angle formed by the adjacent lateral faces is $\angle AFC=\beta$.
Thus, $\cos \beta=\frac{AF^{2}+CF^{2}-AC^{2}}{2 AF \cdot CF}=1-\frac{2 CO^{2}}{CF^{2}}=1-\frac{a^{2}}{CF^{2}}$.
In $\triangle PCD$, by the area relationship, we have $\frac{1}{2} b \cdot CF=\frac{1}{2} a \cdot \sqrt{b^{2}-\left(\frac{a}{2}\right)^{2}}$, so $\frac{a^{2}}{CF^{2}}=\frac{4 b^{2}}{4 b^{2}-a^{2}}$. Therefore, $\cos \beta=\frac{-a^{2}}{4 b^{2}-a^{2}}$, and $\cos^{2} \alpha=\left(\frac{a}{2 b}\right)^{2} \cdot \frac{a^{2}}{CF^{2}}=\left(\frac{a}{2 b}\right)^{2} \cdot \frac{4 b^{2}}{4 b^{2}-a^{2}}=\frac{a^{2}}{4 b^{2}-a^{2}}$, hence $\cos^{2} \alpha+\cos \beta=0$, which means $\cos 2 \alpha+2 \cos \beta=-1$.