sin^4x-sin^2xcos^2x+cos^4x =(sin^2x+cos^2x)^2-3sin^2xcos^2x
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/11/08 15:03:57
sin^4x-sin^2xcos^2x+cos^4x =(sin^2x+cos^2x)^2-3sin^2xcos^2x 那么这个依据呢?老师
针对三角函数的证明,一般都是采取化简的思路来解答.
对等式右边进行化简,
=(sin^2x+cos^2x)^2-3sin^2xcos^2x =(sin^4x+2sin^2xcos^2x+cos^4x)-3sin^2xcos^2x
=sin^4x-sin^2xcos^2x+cos^4x
=等式左边
所以sin^4x-sin^2xcos^2x+cos^4x =(sin^2x+cos^2x)^2-3sin^2xcos^2x 成立.
对等式右边进行化简,
=(sin^2x+cos^2x)^2-3sin^2xcos^2x =(sin^4x+2sin^2xcos^2x+cos^4x)-3sin^2xcos^2x
=sin^4x-sin^2xcos^2x+cos^4x
=等式左边
所以sin^4x-sin^2xcos^2x+cos^4x =(sin^2x+cos^2x)^2-3sin^2xcos^2x 成立.
sin^2x+cos^2x)(sin^4x-sin^2xcos^2x+cos^4x) =sin^4x-sin^2xcos
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