已知函数f(x)=(1+1⼀tanx)sin^2x+msin(x+π⼀4)sin(x-π4),当m=0时，求f(x)在区间[π⼀8,3π⼀4]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

当m=0时，求f(x)=(1+1/tanx)sin^2x=(1+cosx/sinx)sin^2x=sin^2x+sinxcosx=(1-cos2x)/2+(sin2x)/2
=(sin2x-cos2x)/2+1/2=[(√2)/2]{[(√2)/2]sin2x-[(√2)/2]cos2x}+1/2=[(√2)/2]sin(2x-π/4)+1/2
当π/8=f(x)在区间[π/8,3π/4]上的取值范围为[0,(1+√2)/2]

f(x)=(sinx)^2 + sinx*cosx = (1-cos2x)/2 + (1/2)sin2x =(1/2)(sin2x - cos2x) + 1/2
=(√2/2)sin(2x - π/4) + 1/2
由已知：π/8≤ x ≤3π/4
则：π/4≤ 2x ≤3π/2
0≤ 2x-π/4 ≤5π/4
sin(2x - π/4)∈[-√2/2，1]
f(x)∈[0，√2/2+1/2]

f(x)=1/2+根号2*sin(2x-π/4)的取值范围是[0,(1+根号2)/2]

[0,(1+√2)/2]

谢拉