Question

$\frac{1}{2} - 2 \sin ^{2} ( \alpha ) \cos ^{2} ( \alpha )$
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Answer 1

Ответ:

$\dfrac{1}{2}-2\, sin^2a\cdot cos^2a=\dfrac{1}{2}-2\cdot \Big(sina\cdot cosa\Big)^2=\dfrac{1}{2}-2\cdot \Big(\dfrac{1}{2}\, sin2a\Big)^2=\\\\\\=\dfrac{1}{2}-\dfrac{1}{2}\, sin^22a=\dfrac{1}{2}\cdot \Big(1-sin^22a\Big)=\dfrac{1}{2}\cdot cos^22a\\\\\\\\P.S.\ \ \ 2\, sinx\cdot cosx=sin2x\\\\sin^2x+cos^2x=1\ \ \ \to \ \ \ cos^2x=1-sin^2x$