f(x)=(1+cotx)(sinx^2)-2sin(x+π/4)sin(x-π/4) x∈[π/12,π/2]

f(x)=(1+cotx)(sinx^2)-2sin(x+π/4)sin(x-π/4) =（1+cosx/sinx)(sinx^2)+2sin(x+π/4)sin(π/4-x)
==（1+cosx/sinx)(sinx^2)+2cos(π/4-x)sin(π/4-x)
=sinx^2+sinxcosx+sin(π/2-2x)=sinx^2+1/2sin2x+cos2x=(1-cos2x)/2+1/2sin2x+cos2x
=1/2(sin2x+cos2x)+1/2=(根号2倍）sin(2x+π/4)/2+1/2
因为x∈[π/12,π/2],求得2x+π/4∈[5π/12,5π/4】,sinx∈[-二分之根号2,1】,从而原式取值范围【0,1】.