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. $-\frac{1}{14}$
Solve $\frac{2 \sin \alpha \cos \alpha+2\left(\cos ^{2} \alpha-\sin ^{2} \alpha\right)}{4\left(\cos ^{2} \alpha-\sin ^{2} \alpha\right)-8 \sin \alpha \cos \alpha}=\frac{-\tan ^{2} \alpha+\tan \alpha+1}{-2 \tan ^{2} \alpha-4 \tan \alpha+2}=\frac{1}{14}$, so the answer is $C$.