已知函数f(x)=(1+ 1 tanx )sin 2 x+msin(x+ π 4 )sin(x- π 4 )

1个回答

  • (1)当m=0时,函数f(x)=(1+

    1

    tanx )sin 2x=

    sinx+cosx

    sinx •sin 2x=sin 2x+sinxcosx=

    1-cos2x

    2 +

    1

    2 sin2x=

    1

    2 +

    2

    2 sin(2x-

    π

    4 ).

    π

    8 ≤x≤

    4 ,∴0≤2x-

    π

    4 ≤

    4 ,∴-

    2

    2 ≤sin(2x-

    π

    4 )≤1,0≤f(x)≤

    1+

    2

    2 ,

    故f(x)在区间[

    π

    8 ,

    4 ]上的取值范围为[0

    1+

    2

    2 ,].

    (2)∵当tana=2时,f(a)=

    3

    5 ,∴sin 2a=

    4

    5 ,cos 2a=

    1

    5 .

    再由f(a)=(1+

    1

    tana )sin 2a+msin(a+

    π

    4 )sin(a-

    π

    4 )=

    3

    2 sin 2a+m(

    1

    2 sin 2a-

    1

    2 cos 2a )=

    12+3m

    10 ,

    可得

    12+3m

    10 =

    3

    5 ,解得m=-2.