Verified answer

[tex]\displaystyle\bf\\1)\\\\Cos(-5^\circ)\cdot Cos35^\circ=Cos5^\circ\cdot Cos35^\circ=\frac{1}{2}\Big[Cos(5^\circ+35^\circ)+Cos(5^\circ-35^\circ)\Big]=\\\\\\=\frac{1}{2} \Big(Cos40^\circ+Cos30^\circ\Big)=\frac{1}{2} Cos40^\circ+\frac{1}{2} \cdot\frac{\sqrt{3} }{2} =\frac{1}{2} Cos40^\circ+\frac{\sqrt{3} }{4} \\\\\\2)\\\\Sin32^\circ\cdot Sin28^\circ=\frac{1}{2}\Big[Cos(32^\circ-28^\circ)-Cos(32^\circ+28^\circ)\Big]=[/tex]

[tex]\displaystyle\bf\\=\frac{1}{2} \Big(Cos4^\circ-Cos60^\circ\Big)=\frac{1}{2} Cos4^\circ-\frac{1}{2} \cdot\frac{1}{2} =\frac{1}{2} Cos4^\circ-\frac{1}{4} \\\\\\3)\\\\2Cos18^\circ\cdot Cos(-28^\circ)=2Cos18^\circ\cdot Cos28^\circ=\\\\\\=2\cdot\frac{1}{2} \Big[Cos(18^\circ+28^\circ)+Cos(18^\circ-28^\circ)\Big]=Cos46^\circ+Cos10^\circ[/tex]

[tex]\displaystyle\bf\\4)\\\\2Sin(\alpha +2\beta )\cdot Cos (\alpha -2\beta )=2\cdot\frac{1}{2} \Big[Sin(\alpha +2\beta +\alpha -2\beta )+Sin(\alpha +2\beta -\alpha +2\beta )\Big]=\\\\\\=Sin2\alpha +Sin4\beta \\\\\\5)\\\\2Cos(3\alpha +\beta )\cdot Sin (3\alpha -\beta )=2\cdot\frac{1}{2} \Big[Sin(3\alpha -\beta +3\alpha +\beta )+Sin(3\alpha -\beta -3\alpha -\beta )\Big]=\\\\\\=Sin6\alpha -Sin2\beta[/tex]

[tex]\displaystyle\bf\\6)\\\\2Sin(4\alpha -3\beta )\cdot Sin (4\alpha +3\beta )=\\\\2\cdot\frac{1}{2} \Big[Cos(4\alpha -3\beta -4\alpha -3\beta )-Cos(4\alpha -3\beta +4\alpha +3\beta )\Big]=\\\\\\=Cos6\beta 2\alpha -Cos8\alpha[/tex]