求证:[1-(sinα)^6-(cosα)^6]/[1-(sinα)^4-(cosα)^4]=3/2

1-(sinα)^6-(cosα)^6=1-sin^6α-cos^6α= sin^2α(1-sin^4α) + cos^2α(1-cos^4α)= sin^2α[sin^2α(1-sin^2α) + cos^2α] + cos^2α[cos^2α(1-cos^2α)+sin^2α]=sin^2α*cos^2α*(1+sin^2α) + sin^2α*cos^2α*(1+cos^2α)=sin^2α*cos^2α*(1+1+sin^2α+cos^2α)=3sin^2α*cos^2α1-(sinα)^4-(cosα)^4=sin^4α-sin^4α=2sin^2α*(1-sin^2α)=2sin^2α*cos^2α 两个相除即为3/2 收起