3.325. $\cos ^{2} 2 \alpha-3 \sin ^{2} 2 \alpha$.

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3.325. $\cos ^{2} 2 \alpha-3 \sin ^{2} 2 \alpha$.
Solution.

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

$=2\left(\cos 4 \alpha-\cos 60^{\circ}\right)=\left[\cos x-\cos y=-2 \sin \frac{x+y}{2} \sin \frac{x-y}{2}\right]=$

$=2\left(-2 \sin \left(2 \alpha+30^{\circ}\right) \sin \left(2 \alpha-30^{\circ}\right)\right)=4 \sin \left(30^{\circ}+2 \alpha\right) \sin \left(30^{\circ}-2 \alpha\right)$.

Answer: $4 \sin \left(30^{\circ}+2 \alpha\right) \sin \left(30^{\circ}-2 \alpha\right)$.