Verified answer

[tex]\displaystyle\bf\\2)\\\\-2Sin\Big(3-\alpha \Big) Cos\Big(\alpha +3\Big)=-2\cdot\frac{1}{2} \Big[Sin\Big(3-\alpha -\alpha -3\Big)+Sin\Big(3-\alpha +\alpha +3\Big)\Big]=\\\\\\=-\Big[Sin\Big(-2\alpha \Big)+Sin6\Big]=Sin2\alpha-Sin6[/tex]