3.107. $\frac{\sin (2 \alpha+2 \pi)+2 \sin (4 \alpha-\pi)+\sin (6 \alpha+4 \pi)}{\cos (6 \pi-2 \alpha)+2 \cos (4 \alpha-\pi)+\cos (6 \alpha-4 \pi)}$.

Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.

3.107. $\frac{\sin (2 \alpha+2 \pi)+2 \sin (4 \alpha-\pi)+\sin (6 \alpha+4 \pi)}{\cos (6 \pi-2 \alpha)+2 \cos (4 \alpha-\pi)+\cos (6 \alpha-4 \pi)}$.
Solution.

$$
\begin{aligned}
& \frac{\sin (2 \alpha+2 \pi)+2 \sin (4 \alpha-\pi)+\sin (6 \alpha+4 \pi)}{\cos (6 \pi-2 \alpha)+2 \cos (4 \alpha-\pi)+\cos (6 \alpha-4 \pi)}= \\
& =\frac{\sin (2 \pi+2 \alpha)-2 \sin (\pi-4 \alpha)+\sin (4 \pi+6 \alpha)}{\cos (6 \pi-2 \alpha)+2 \cos (\pi-4 \alpha)+\cos (4 \pi-6 \alpha)}=\frac{\sin 2 \alpha-2 \sin 4 \alpha+\sin 6 \alpha}{\cos 2 \alpha-2 \cos 4 \alpha+\cos 6 \alpha}= \\
& =\frac{2 \sin 4 \alpha \cos 2 \alpha-2 \sin 4 \alpha}{2 \cos 4 \alpha \cos 2 \alpha-2 \cos 4 \alpha}=\frac{2 \sin 4 \alpha(\cos 2 \alpha-1)}{2 \cos 4 \alpha(\cos 2 \alpha-1)}=\frac{\sin 4 \alpha}{\cos 4 \alpha}=\tan 4 \alpha .
\end{aligned}
$$

Answer: $\tan 4 \alpha$.