已知函数f(X)=(1+1/tanx)sin^2x+msin(x+π/4)sin(x-π/4) 当tana=2时,f(a

2个回答

  • 将函数f(X)=(1+1/tanx)sin^2x+msin(x+π/4)sin(x-π/4) 化简得:

    =(1+cosx/sinx)*2sinxcosx+m(sinxcosπ/4+cosxsinπ/4)(sinxcosπ/4-cosxsinπ/4)

    =2sinxcosx+2(cosx)^2+1/2[(sinx)^2-(cosx)^2]m

    由tana=2得

    sina=2cosa

    所以:f(a)=(sina)^2+2(cosa)^2+1/2*3m(cosa)^2

    =1+(cosa)^2+3m/2*(cosa)^2

    =3/5

    所以得:3m/2*(cosa)^2+2/5[(cosa)^2+(sina)^2]+(cosa)^2=0

    3m/2*(cosa)^2+7/5*(cosa)^2+2/5*(sina)^2=0

    因为tana=2

    将上式同除于(cosa)^2得((cosa)^2由tana=2知不会为零):

    3m/2+7/5+2/5*4=0

    所以:m=-2