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(1-(sin^4x-sin^2xcos^2x+cos^4x)/sin^2x +3sin^2x
人气:201 ℃ 时间:2020-03-13 04:43:31
解答
sin^4x-sin^2xcos^2x+cos^4x
=sin^4x+2sin^2xcos^2x+cos^4x-3sin^2xcos^2x
=(sin^2x+cos^2x)^2-3sin^2xcos^2x
=1-3sin^2xcos^2x
所以分子=3sin^2xcos^2x
所以原式=3sin^2xcos^2x/sin^2x+3sin^2x
=3cos^2x+3sin^2x
=3(sin^2x+cos^2x)
=3
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