Răspuns:
[tex]sin^2x-sin^23x=(sin2x-sin3x)(sin2x+sin3x) =\\=(2sin\frac{2x-3x}{2}*cos\frac{2x+3x}{2})(2sin\frac{2x+3x}{2}*cos\frac{2x-3x}{2} )\\=(2sin\frac{-x}{2}*cos\frac{5x}{2})(2sin\frac{5x}{2}*cos\frac{-x}{2})\\=-4sin\frac{x}{2}*cos\frac{5x}{2}*sin\frac{5x}{2}*cos\frac{x}{2}[/tex]
