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

1个回答

  • 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