4. E. $\operatorname{E}\{\mathrm{A} a=(\cos \alpha, \sin \alpha), b=(\cos \beta, \sin \beta)$, $|a-b|=\frac{2 \cdot 5}{5}$. Given $0<\alpha<\frac{\pi}{2},-\frac{\pi}{2}<\beta<0, \mathrm{H}$. si: $\beta=-\frac{5}{12}$. Then $\sin \alpha=(\quad$.
(A) $\frac{31}{65}$
(B) $\frac{32}{65}$
(C) $\frac{33}{65}$
(D) $\frac{34}{65}$
Prompt: As shown in the figure
13, take points $A(\cos \alpha, \sin \alpha), B(\cos \beta, \sin \beta)$.

Given $|A B| = \frac{2 \sqrt{5}}{5}$, then $\angle A O B = \alpha - \beta$, and
$$
\begin{array}{l}
\cos (\alpha - \beta) = \frac{1^{2} + 1^{2} - \left(\frac{2 \sqrt{5}}{5}\right)^{2}}{2 \times 1 \times 1} = \frac{3}{5} \\
\Rightarrow \sin (\alpha - \beta) = \frac{4}{5} \\
\Rightarrow \sin \alpha = -\frac{5}{13} \cdot \frac{3}{5} + \frac{12}{13} \cdot \frac{4}{5} = \frac{33}{65}
\end{array}
$$