3.287. $3-4 \cos 4 \alpha+\cos 8 \alpha-8 \cos ^{4} 2 \alpha$.

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3.287. $3-4 \cos 4 \alpha+\cos 8 \alpha-8 \cos ^{4} 2 \alpha$.
## Solution.

$$
\begin{aligned}
& 3-4 \cos 4 \alpha+\cos 8 \alpha-8 \cos ^{4} 2 \alpha= \\
& =\left[\cos 2 x=2 \cos ^{2} x-1 ; \cos ^{2} \frac{x}{2}=\frac{1+\cos x}{2}\right]= \\
& =3-4 \cos 4 \alpha+2 \cos ^{2} 4 \alpha-1-8\left(\frac{1+\cos 4 \alpha}{2}\right)^{2}= \\
& =2-4 \cos 4 \alpha+2 \cos ^{2} 4 \alpha-2\left(1+2 \cos 4 \alpha+\cos ^{2} 4 \alpha\right)= \\
& =2-4 \cos 4 \alpha+2 \cos ^{2} 4 \alpha-2-4 \cos 4 \alpha-2 \cos ^{2} 4 \alpha=-8 \cos 4 \alpha \\
& \text { Answer: }-8 \cos 4 \alpha
\end{aligned}
$$