Question

函数的f(x)=[(sinx)^4+(cosx)^4+(sinx)^2*(cosx)^2]/[2-sin(2x)]最小正

Answer 1

a=sinx,b=cosx,则a^2+b^2=1

分子=a^4+b^4+a^2b^2=a^4+2a^2b^2+b^4-a^2b^2

=(a^2+b^2)^2-a^2b^2

=1-a^2b^2

=1-(sinxcosx)^2

=1-(1/4)(sin2x)^2

=[1+(1/2)sin2x][1-(1/2)sin2x]

分母=2[[1-(1/2)sin2x]

所以f(x)=(1/2)+(1/4)sin2x

所以T=2π/2=π

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