Question

Помогите, пожалуйста, с решением

Answer 1

$sin2alpha=2sinalpha cosalpha /()^2$

$sin^22alpha=4sin^2alpha cos^2alpha /cdot(-2)$

$-2sin^22alpha=-8sin^2alpha cos^2alpha /+1$

$1-2sin^22alpha=1-8sin^2alpha cos^2alpha$

$cos4alpha=1-8sin^2alpha cos^2alpha$

$8sin^2alpha cos^2alpha=1-cos4alpha /:8$

$sin^2alpha cos^2alpha= frac{1}{8} - frac{1}{8} cos4alpha$

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$sin^6alpha+cos^6alpha=$

$(sin^2alpha)^3+(cos^2alpha)^3=$

$(sin^2alpha+cos^2alpha)(sin^4alpha-sin^2alpha cos^2alpha+cos^4alpha)=$

$sin^4alpha-sin^2alpha cos^2alpha+cos^4alpha=$

$(sin^4alpha+cos^4alpha)-sin^2alpha cos^2alpha=$

$(sin^2alpha+cos^2alpha)^2-2sin^2alpha cos^2alpha-sin^2alpha cos^2alpha=$

$1-3sin^2alpha cos^2alpha=$

$1-3 cdot left( frac{1}{8} - frac{1}{8} cos4alpha right) =$

$1- frac{3}{8}+ frac{3}{8} cos4alpha =$

$frac{5}{8}+ frac{3}{8} cos4alpha$