Verified answer

[tex]\displaystyle\bf\\Sin4\alpha Cos\alpha +Sin2a Cos5\alpha =Sin6\alpha Cos\alpha\\\\\\Sin6\alpha Cos\alpha -Sin4a Cos\alpha =Sin2\alpha Cos5\alpha\\\\\\Sin6\alpha Cos\alpha -Sin4a Cos\alpha =Cos\alpha \cdot(Sin6\alpha -Sin4\alpha )=\\\\\\=Cos\alpha \cdot 2Sin\frac{6\alpha -4\alpha }{2} Cos\frac{6\alpha +4\alpha }{2} =\underbrace{2Cos\alpha Sin\alpha}_{Sin2\alpha } Cos5\alpha =\\\\\\=Sin2\alpha Cos5\alpha \\\\\\Sin2\alpha Cos5\alpha=Sin2\alpha Cos5\alpha[/tex]

Тождество доказано