若α∈(0,π/2),证明sin^3α+cos^3α
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若α∈(0,π/2),证明sin^3α+cos^3α
α∈(0,π/2)
sin^3α+cos^3α
=sin^3α+cosα*cos^2α
=sin^3α+cosα*(1-sin^2α)
=sin^3α+cosα-sin^2αcosα
=sin^3α-sin^2αcosα+cosα
=sin^2α(1-cosα)+cosα
=(1-cos^2α)(1-cosα)+cosα
=1-cos^2α-cosα+cos^3α+cosα
=1-cos^2α+cos^3α
=1-cos^2α(1-cosα)
∵cos^2α>0,1-cosα>0
∴cos^2α(1-cosα)>0
∴1-cos^2α(1-cosα)<1
sin^3α+cos^3α
=sin^3α+cosα*cos^2α
=sin^3α+cosα*(1-sin^2α)
=sin^3α+cosα-sin^2αcosα
=sin^3α-sin^2αcosα+cosα
=sin^2α(1-cosα)+cosα
=(1-cos^2α)(1-cosα)+cosα
=1-cos^2α-cosα+cos^3α+cosα
=1-cos^2α+cos^3α
=1-cos^2α(1-cosα)
∵cos^2α>0,1-cosα>0
∴cos^2α(1-cosα)>0
∴1-cos^2α(1-cosα)<1
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