已知函数f(x)=(1+1⼀tanx)sin2x+msin(x+π⼀4)sin(x-π⼀4)

f(x)=(1+1/tanx)sin^2x+msin(x+π/4)sin(x-π/4),
（1）
m=0,
f(x)=(1+cosx/sinx)*sin²x
=sin²x+sinxcosx
=1/2(1-cos2x)+1/2sin2x
=1/2sin2x-1/2cos2x+1/2
=√2/2(√2/2sin2x-√2/2cos2x)+1/2
=√2/2sin(2x-π/4)+1/2
∵x∈(0,π/2)
∴2x-π/4∈(-π/4,3π/4)
f(x)max=√2/2+1/2
2x-π/4=-π/4时，f(x)=0
f(x)值域为(0,√2/2+1/2)
(2)
∵tana=2,即sina/cosa=2
∴ sina=2cosa代入
sin²a+cos²a=1得：
cos²a=1/5,sin²a=4/5

sin(a+π/4)sin(a-π/4)
=sin(a+π/4)sin[-π/2+(a+π/4)]
=-sin(a+π/4)cos(a+π/4)
=-1/2sin(2a+π/2)
=-1/2cos2a
=-1/2(2cos²a-1)
=-1/2(2*1/5-1)
=3/10
∵f(a)=3/5,
∴f(a)=(1+1/tana)sin²a+msin(a+π/4)sin(a-π/4)
=(1+1/2)*4/5+3m/10=3/5
∴3m/10=-3/5
∴m=-2

1、m=0，f(x)=(1+1/tanx)sin^2x=(1+cosx/sinx)sin^2x=sin^2x+sinxcosx
=(1-cos2x)/2+sin2x/2=(1+sin2x-cos2x)/2
=(1+sin2x-sin(π/2-2x))/2=(1+2cosπ/4sin(2x-π/4))/2
=(1+√2sin(2x-π/4))/2
因 x∈(0,π/2)，2x∈(0,π)，2x-π/4∈(-π/4,3π/4)，-√2/2 所以 02、tana=2，cos2a=(1-tan^2a)/(1+tan^2a)=(1-4)/(1+4)=-3/5
f(a)=(1+1/tana)sin^2a+msin(a+π/4)sin(a-π/4)
=(1+1/2)sin^2a+m*[-1/2*(cos2a-cosπ/2)]
=3/2*sin^2a-m/2*cos2a
=3/2*(1-cos2a)/2-m/2*cos2a
=3/4*(1-(-3/5))-m/2*(-3/5)
=3/4*8/5+m/2*3/5
=6/5+3m/10
=3/5
3m/10=6/5-3/5=3/5
m=2