7. Let $O$ be the circumcenter of $\triangle A B C$. If
$$
\overrightarrow{A O}=\overrightarrow{A B}+2 \overrightarrow{A C} \text {, }
$$

then the value of $\sin \angle B A C$ is $\qquad$
7. $\frac{\sqrt{10}}{4}$.

Assume the circumradius $R=2$ of $\triangle ABC$.
From the given condition, we have
$$
2 \overrightarrow{A C}=\overrightarrow{A O}-\overrightarrow{A B}=\overrightarrow{B O} \text {. }
$$

Thus, $A C=\frac{1}{2} B O=1$.
Let $M$ be the midpoint of $A C$, then $O M \perp A C$.
Combining with equation (1), we know $O M \perp B O$, and points $B$ and $A$ are on the same side of line $O M$.
Therefore, $\cos \angle B O C=\cos \left(90^{\circ}+\angle M O C\right)$
$=-\sin \angle M O C=-\frac{M C}{O C}=-\frac{1}{4}$.
In $\triangle B O C$, by the cosine rule, we get
$$
\begin{array}{l}
B C=\sqrt{O B^{2}+O C^{2}-2 O B \cdot O C \cos \angle B O C} \\
=\sqrt{10} .
\end{array}
$$

In $\triangle A B C$, by the sine rule, we get
$$
\sin \angle B A C=\frac{B C}{2 R}=\frac{\sqrt{10}}{4} \text {. }
$$