Biến đổi cho mk sin⁶x + cos⁶x Thành 1- ¾ sin²2x 07/08/2021 Bởi Lyla Biến đổi cho mk sin⁶x + cos⁶x Thành 1- ¾ sin²2x
$\sin^6x+\cos^6x$ $=(\sin^2x+\cos^2x)(\sin^4x+\cos^4x-\sin^2x\cos^2x)$ $=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x-\sin^2x\cos^2x$ $=1-3\sin^2x\cos^2x$ $=1-\dfrac{3}{4}.4\sin^2x\cos^2x$ $=1-\dfrac{3}{4}\sin^22x$ Bình luận
$sin^6x+cos^6x$ $=(sin^2x)^3+(cos^2x)^3$ $=(sin^2x+cos^2x)(sin^4x-sin^2xcos^2x+cos^4x)$ $=(sin^2x+cos^2x)^2-3sin^2xcos^2x$ $=1-3.(\dfrac{1}{2}sin2x)^2$ $=1-\dfrac{3}{4}sin^22x$ Bình luận
$\sin^6x+\cos^6x$
$=(\sin^2x+\cos^2x)(\sin^4x+\cos^4x-\sin^2x\cos^2x)$
$=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x-\sin^2x\cos^2x$
$=1-3\sin^2x\cos^2x$
$=1-\dfrac{3}{4}.4\sin^2x\cos^2x$
$=1-\dfrac{3}{4}\sin^22x$
$sin^6x+cos^6x$
$=(sin^2x)^3+(cos^2x)^3$
$=(sin^2x+cos^2x)(sin^4x-sin^2xcos^2x+cos^4x)$
$=(sin^2x+cos^2x)^2-3sin^2xcos^2x$
$=1-3.(\dfrac{1}{2}sin2x)^2$
$=1-\dfrac{3}{4}sin^22x$