7. Given that $F_{1}$ and $F_{2}$ are the left and right foci of a hyperbola $C$ centered at the origin with symmetric axes as the coordinate axes, a line $l$ passing through $F_{2}$ intersects the right branch of the hyperbola at points $A$ and $B$. $I_{1}$ and $I_{2}$ are the incenter of $\triangle A F_{1} F_{2}$ and $\triangle B F_{1} F_{2}$, respectively. If the eccentricity of the hyperbola $C$ is $2$, $\left|I_{1} I_{2}\right|=\frac{9}{2}$, and the sine of the inclination angle of line $l$ is $\frac{8}{9}$, then the equation of the hyperbola $C$ is $\qquad$
7. $\frac{x^{2}}{4}-\frac{y^{2}}{12}=1$.

Let the equation of the hyperbola $C$ be
$$
\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>0, b>0) \text {, }
$$

with the semi-focal distance $c=\sqrt{a^{2}+b^{2}}$.
Let the points where the incircle of $\triangle A F_{1} F_{2}$ touches the $x$-axis, $A F_{1}$, and $A F_{2}$ be $E$, $P$, and $Q$, respectively. Then
$A P=A Q, F_{1} E=F_{1} P, F_{2} E=F_{2} Q$.
Thus, $A F_{1}-A F_{2}=2 a, F_{1} F_{2}=2 c$
$\Rightarrow F_{1} E=c+a, F_{2} E=c-a$.
Therefore, $E$ is the right vertex of the hyperbola.
Let the points where the incircle of $\triangle B F_{1} F_{2}$ touches the $x$-axis, $B F_{1}$, and $B F_{2}$ be $E^{\prime}$, $R$, and $S$, respectively. Similarly, $E^{\prime}$ is the right vertex of the hyperbola. Hence, points $E$ and $E^{\prime}$ coincide.
Thus, $I_{1} I_{2} \perp x$-axis.
Given the eccentricity of the hyperbola $C$ is $\frac{c}{a}=2$, we have
$$
c=2 a \Rightarrow F_{2} Q=F_{2} S=F_{2} E=c-a=a \text {. }
$$

Connect $I_{1} Q$ and $I_{2} S$.
In the right trapezoid $I_{1} I_{2} S Q$, $\angle Q I_{1} I_{2}$ is the inclination angle of line $l$, i.e.,
$$
\sin \angle Q I_{1} I_{2}=\frac{8}{9} .
$$

Then $2 a=Q S=I_{1} I_{2} \sin \angle Q I_{1} I_{2}=\frac{9}{2} \times \frac{8}{9}=4$
$$
\Rightarrow a=2, c=4, b=\sqrt{c^{2}-a^{2}}=2 \sqrt{3} \text {. }
$$

Therefore, the equation of the hyperbola $C$ is $\frac{x^{2}}{4}-\frac{y^{2}}{12}=1$.