3.283. $\cos \left(270^{\circ}-2 \alpha\right) \operatorname{ctg}\left(30^{\circ}-2 \alpha\right) \operatorname{tg}\left(240^{\circ}-2 \alpha\right)(2 \cos 4 \alpha-1)$.
## Solution.

$$
\begin{aligned}
& \cos \left(270^{\circ}-2 \alpha\right) \operatorname{ctg}\left(30^{\circ}-2 \alpha\right) \operatorname{tg}\left(240^{\circ}-2 \alpha\right)(2 \cos 4 \alpha-1)= \\
& =\cos \left(270^{\circ}-2 \alpha\right) \operatorname{ctg}\left(30^{\circ}-2 \alpha\right) \operatorname{tg}\left(270^{\circ}-\left(30^{\circ}+2 \alpha\right)\right)(2 \cos 4 \alpha-1)= \\
& =-\sin 2 \alpha \operatorname{ctg}\left(30^{\circ}-2 \alpha\right) \operatorname{ctg}\left(30^{\circ}+2 \alpha\right)(2 \cos 4 \alpha-1)= \\
& =-\frac{\sin 2 \alpha \cos \left(30^{\circ}-2 \alpha\right) \cos \left(30^{\circ}+2 \alpha\right)(2 \cos 4 \alpha-1)}{\sin \left(30^{\circ}-2 \alpha\right) \sin \left(30^{\circ}+2 \alpha\right)}= \\
& =\left[\cos x \cos y=\frac{1}{2}(\cos (x-y)+\cos (x+y))\right. \\
& \left.\sin x \sin y=\frac{1}{2}(\cos (x-y)-\cos (x+y))\right]=
\end{aligned}
$$

$$
=-\frac{\sin 2 \alpha \cdot \frac{1}{2}\left(\cos 4 \alpha+\cos 60^{\circ}\right)(2 \cos 4 \alpha-1)}{\frac{1}{2}\left(\cos 4 \alpha-\cos 60^{\circ}\right)}=
$$

$$
=-\frac{\sin 2 \alpha\left(\cos 4 \alpha+\frac{1}{2}\right)(2 \cos 4 \alpha-1)}{\cos 4 \alpha-\frac{1}{2}}=
$$

$=-\frac{\sin 2 \alpha(2 \cos 4 \alpha+1)(2 \cos 4 \alpha-1)}{2 \cos 4 \alpha-1}=-\sin 2 \alpha(2 \cos 4 \alpha+1)=$

$=\left[\cos 2 x=1-2 \sin ^{2} x, \sin 3 x=3 \sin x-4 \sin ^{3} x\right]=$

$=-\sin 2 \alpha\left(2\left(1-2 \sin ^{2} 2 \alpha\right)+1\right)=-\sin 2 \alpha\left(2-4 \sin ^{2} 2 \alpha+1\right)=$

$=-\sin 2 \alpha\left(3-4 \sin ^{2} 2 \alpha\right)=-\left(3 \sin 2 \alpha-4 \sin ^{3} 2 \alpha\right)=-\sin 6 \alpha$.

Answer: $-\sin 6 \alpha$.