4. Given $\tan \alpha=-\frac{1}{2}$. Then the value of $\frac{\sin 2 \alpha+2 \cos 2 \alpha}{4 \cos 2 \alpha-4 \sin 2 \alpha}$ is ( ).
(A) $\frac{5}{2}$
(B) $-\frac{5}{2}$
(C) $\frac{1}{14}$
(D) $-\frac{1}{14}$
4. C.

Given $\tan \alpha=-\frac{1}{2}$, then $\tan 2 \alpha=-\frac{4}{3}$.
Therefore, the original expression $=\frac{\tan 2 \alpha+2}{4-4 \tan 2 \alpha}=\frac{1}{14}$.