sin^8X+cos^8X+4sin^2Xcos^2x-2sin^4xcos^4X(化简)
来源:学生作业帮 编辑:神马作文网作业帮 分类:数学作业 时间:2024/10/05 17:17:52
sin^8X+cos^8X+4sin^2Xcos^2x-2sin^4xcos^4X(化简)
原式=sin^8X-2sin^4xcos^4X+cos^8X+4sin^2Xcos^2x
=(sin^4x-cos^4x)^2+4sin^2Xcos^2x
=[(sin^2x-cos^2x)(sin^2x+cos^2x)]^2+4sin^2Xcos^2x
=(sin^2x-cos^2x)^2+4sin^2Xcos^2x
=sin^4x-2sin^2Xcos^2x+cos^4x+4sin^2Xcos^2x
=sin^4x+2sin^2Xcos^2x+cos^4x
=(sin^2x+cos^2x)^2
=1
=(sin^4x-cos^4x)^2+4sin^2Xcos^2x
=[(sin^2x-cos^2x)(sin^2x+cos^2x)]^2+4sin^2Xcos^2x
=(sin^2x-cos^2x)^2+4sin^2Xcos^2x
=sin^4x-2sin^2Xcos^2x+cos^4x+4sin^2Xcos^2x
=sin^4x+2sin^2Xcos^2x+cos^4x
=(sin^2x+cos^2x)^2
=1
sin^2x+cos^2x)(sin^4x-sin^2xcos^2x+cos^4x) =sin^4x-sin^2xcos
化简cos^4x+sin^2xcos^2x+sin^2x
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