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^2α
1-(sinα)^4-(cosα)^4
=sin^4α-sin^4α
=2sin^2α*(1-sin^2α)
=2sin^2α*cos^2α
我都算了相反数的
对结果没影响的
两个相除为2/3
采纳下哈
谢谢
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