433. $2 \sin \alpha \cos \alpha\left(\cos ^{2} \alpha-\sin ^{2} \alpha\right)$.
Solution. According to formulas (9) and (10), we have $2 \sin \alpha \cos \alpha\left(\cos ^{2} \alpha-\sin ^{2} \alpha\right)=\sin 2 \alpha \cos 2 \alpha$.

Multiplying and dividing the product by 2, we get

$$
\sin 2 \alpha \cos 2 \alpha=\frac{2 \sin 2 \alpha \cos 2 \alpha}{2}=\frac{\sin 4 \alpha}{2} \text {. }
$$