3.309. $1-\cos (\pi-8 \alpha)-\cos (\pi+4 \alpha)$.
## Solution.

$$
\begin{aligned}
& 1-\cos (\pi-8 \alpha)-\cos (\pi+4 \alpha)=1+\cos 8 \alpha+\cos 4 \alpha= \\
& =1+\cos 2(4 \alpha)+\cos 4 \alpha=1+2 \cos ^{2} 4 \alpha-1+\cos 4 \alpha=2 \cos ^{2} 4 \alpha+\cos 4 \alpha= \\
& =2 \cos 4 \alpha\left(\cos 4 \alpha+\frac{1}{2}\right)=2 \cos 4 \alpha\left(\cos 4 \alpha+\cos \frac{\pi}{3}\right)=
\end{aligned}
$$

$=\left[\cos x+\cos y=2 \cos \frac{x+y}{2} \cos \frac{x-y}{2}\right]=$

$=2 \cos 4 \alpha \cdot 2 \cos \left(2 \alpha+\frac{\pi}{6}\right) \cos \left(2 \alpha-\frac{\pi}{6}\right)=$

$=4 \cos 4 \alpha \cos \left(2 \alpha+\frac{\pi}{6}\right) \cos \left(2 \alpha-\frac{\pi}{6}\right)$.

Answer: $4 \cos 4 \alpha \cos \left(2 \alpha+\frac{\pi}{6}\right) \cos \left(2 \alpha-\frac{\pi}{6}\right)$.