求(1-sin^6x-cos^6x)/(1-sin^4x-cos^4x)的值
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求(1-sin^6x-cos^6x)/(1-sin^4x-cos^4x)的值
求(1-sin^6x-cos^6x)/(1-sin^4x-cos^4x)的值
求(1-sin^6x-cos^6x)/(1-sin^4x-cos^4x)的值
sin^6x+cos^6x
=(sin^2x+cos^2x)(sin^4x-sin^2xcos^2x+cos^4x)
=sin^4x-sin^2xcos^2x+cos^4x
=(sin^2x+cos^2x)^2-3sin^2xcos^2x
=1-3sin^2xcos^2x
所以分子=3sin^2xcos^2x
sin^4x+cos^4x
=(sin^2x+cos^2x)^2-2sin^2xcos^2x
=1-2sin^2xcos^2x
所以分母=2sin^2xcos^2x
所以原式=3/2
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