\(B=3\left(sin^4x+cos^4x\right)\left(sin^4x-cos^4x\right)+4cos^6x-8sin^6x+6sin^4x\)
\(=3\left(sin^4x+cos^4x\right)\left(sin^2x+cos^2x\right)\left(sin^2x-cos^2x\right)+4cos^6x-2sin^6x+6\left(sin^4x-sin^6x\right)\)
\(=3\left(sin^4x+cos^4x\right)\left(sin^2x-cos^2x\right)+4cos^6x-2sin^6x+6sin^4x\left(1-sin^2x\right)\)
\(=3sin^6x-3cos^6x-3sin^4x.cos^2x+3cos^4x.sin^2x+4cos^6x-2sin^6x+6sin^4x.cos^2x\)
\(=sin^6x+cos^6x+3sin^4x.cos^2x+3cos^4x.sin^2x\)
\(=\left(sin^2x+cos^2x\right)^3=1^3=1\)
Đúng 2
Bình luận (0)
