toán lượng giác 10

T

trantien.hocmai

$sin^6x-cos^6x=(sin^2x-cos^2x)(sin^4x+sin^2xcos^2x+cos^4x)$
$=-cos2x(1-2sin^2xcos^2x+sin^2xcos^2x)$
$=-cos2x(1-sin^2xcos^2x)$
$=-cos2x(1-\frac{1}{4}sin^22x)$
 
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