求解一三角题
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(1-cos^6x-sin^6x)/(1-cos^4x-sin^4x)
={1-(cos^2x+sin^2x)(cos^2x-sinxcosx+sin^2x)}/{1-(cos^2x+sin^2x)^2+2sinxcosx}
=(1-1+sinxcosx)/2sinxcosx
=1/2
={1-(cos^2x+sin^2x)(cos^2x-sinxcosx+sin^2x)}/{1-(cos^2x+sin^2x)^2+2sinxcosx}
=(1-1+sinxcosx)/2sinxcosx
=1/2
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1-cos^6 x-sin^6 x=1-(cos^2x+sin^2x)(cos^4x-cos^2xsin^2x+sin^4x)=1-cos^4x-cos^4x+cos^2xsin^2x与下式相除=1+cos^2xsin^2/1-cos^4x-cos^4x=1+1=2
注:1-cos^4x-cos^4x=1-(cos^2x+sin^2x)^2+cos^2xsin^2x=cos^2xsin^2x
注:1-cos^4x-cos^4x=1-(cos^2x+sin^2x)^2+cos^2xsin^2x=cos^2xsin^2x
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