已知函数f(x)=(1+1/tanx)sin^2x+msin(x+π/4)sin(x-π4),当m=0时,求f(x)在区间[π/8,3π/4]上的取值范围

已知函数f(x)=(1+1/tanx)sin^2x+msin(x+π/4)sin(x-π4),当m=0时,求f(x)在区间[π/8,3π/4]上的取值范围
已知函数f(x)=(1+1/tanx)sin^2x+msin(x+π/4)sin(x-π4),当m=0时,求f(x)在区间[π/8,3π/4]上的取值范围

已知函数f(x)=(1+1/tanx)sin^2x+msin(x+π/4)sin(x-π4),当m=0时,求f(x)在区间[π/8,3π/4]上的取值范围
m=0,所以f(x)=)=(1+1/tanx)sin^2x=sin^2x+sinxcosx=1-cos^2x+1/2sin2x=1-(cos2x+1)/2+1/2sin2x=-1/2cos2x+1/2sin2x+1/2=二分之根号二sin(2x-π/4)+1/2,下面的应该就好做了吧