2 câu trả lời
$\sin^6x+\cos^6x$
$=(\sin^2x+\cos^2x)(\sin^4x+\cos^4x-\sin^2x\cos^2x)$
$=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x-\sin^2x\cos^2x$
$=1-3\sin^2x\cos^2x$
$=1-\dfrac{3}{4}.4\sin^2x\cos^2x$
$=1-\dfrac{3}{4}\sin^22x$
$sin^6x+cos^6x$
$=(sin^2x)^3+(cos^2x)^3$
$=(sin^2x+cos^2x)(sin^4x-sin^2xcos^2x+cos^4x)$
$=(sin^2x+cos^2x)^2-3sin^2xcos^2x$
$=1-3.(\dfrac{1}{2}sin2x)^2$
$=1-\dfrac{3}{4}sin^22x$
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