Question

$\sin {}^{2} 2x - \sin{}^{2} 3x = \\ = ( \sin 2x - \sin 3x)( \sin2x + \sin3x) = ...$

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Answer 1

Răspuns:

$sin^2x-sin^23x=(sin2x-sin3x)(sin2x+sin3x) =\\=(2sin\frac{2x-3x}{2}*cos\frac{2x+3x}{2})(2sin\frac{2x+3x}{2}*cos\frac{2x-3x}{2} )\\=(2sin\frac{-x}{2}*cos\frac{5x}{2})(2sin\frac{5x}{2}*cos\frac{-x}{2})\\=-4sin\frac{x}{2}*cos\frac{5x}{2}*sin\frac{5x}{2}*cos\frac{x}{2}$