$4.59 \frac{\cos 2 \alpha-\cos 6 \alpha+\cos 10 \alpha-\cos 14 \alpha}{\sin 2 \alpha+\sin 6 \alpha+\sin 10 \alpha+\sin 14 \alpha}$.
### 4.59 Finding

\[
\begin{aligned}
& \frac{\cos 2 \alpha - \cos 6 \alpha + \cos 10 \alpha - \cos 14 \alpha}{\sin 2 \alpha + \sin 6 \alpha + \sin 10 \alpha + \sin 14 \alpha} = \\
& = \frac{2 \sin 8 \alpha \sin 6 \alpha - 2 \sin 8 \alpha \sin 2 \alpha}{2 \sin 8 \alpha \cos 6 \alpha + 2 \sin 8 \alpha \cos 2 \alpha} = \frac{\sin 6 \alpha - \sin 2 \alpha}{\cos 6 \alpha + \cos 2 \alpha} = \\
& = \frac{2 \sin 2 \alpha \cos 4 \alpha}{2 \cos 2 \alpha \cos 4 \alpha} = \tan 2 \alpha
\end{aligned}
\]