6. In $\triangle A B C$, $A$ is a moving point, $B\left(-\frac{a}{2}, 0\right)$, $C\left(\frac{a}{2}, 0\right)(a>0)$ are fixed points, and the trajectory equation of the moving point $A$ is $\frac{16 x^{2}}{a^{2}}-\frac{16 y^{2}}{3 a^{2}}=1(y \neq 0)$ of the right branch. Then the three interior angles of $\triangle A B C$ satisfy ( ).
(A) $\sin C-\sin B=\frac{1}{2} \sin A$
(B) $\sin C-\sin B=-\frac{1}{2} \sin A$
(C) $\sin C+\sin B=\frac{1}{2} \sin A$
(D) $\sin C-\sin B=\sin A$
6. A.

Write the trajectory equation as $\frac{x^{2}}{\left(\frac{a}{4}\right)^{2}}-\frac{y^{2}}{\left(\frac{\sqrt{3} a}{4}\right)^{2}}=1$.
From the definition of a hyperbola, we have
$$
|A B|-|A C|=2 \times \frac{a}{4}=\frac{a}{2}=\frac{1}{2}|B C| \text {. }
$$

By the Law of Sines, we know
$$
\begin{array}{l}
|A B|=2 R \sin C,|A C|=2 R \sin B, \\
|B C|=2 R \sin A .
\end{array}
$$

Substituting these into the above equation and simplifying, we get
$$
\sin C-\sin B=\frac{1}{2} \sin A .
$$