Question

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Answer 1

$\it{ \sin {}^{2} {}^{} 2x - \sin {}^{2} 3x = \frac{1 - \cos4x }{2} - \frac{1 - \cos6x }{2} = \frac{1}{2} \times (1 - \cos4x - 1 + \cos6x) = }$

$\it{ \frac{1}{2} \times ( - \cos4x + \cos6x) }$

Answer 2

Răspuns:

$sin^2x-sin^23x=\frac{1-cos4x}{2}-\frac{1-cos6x}{2} = \frac{1-cos4x-1-cos6x}{2} =\\=\frac{cos4x-cos6x}{2}$

Voi face separat -cos4x-cos6x

$-1(cos4x+cos6x)= -1(2cos\frac{4x+6x}{2} *cos\frac{4x-6x}{2})=-2cos\frac{10x}{2} *cos\frac{-2x}{2} \\=-2cos5x*cos(-x)=-2cos5x*cosx$

Acum inlocuim in ecuatie

$\frac{-2cos5x*cosx}{2}=-cos5x*cosx$