f(x)=3/2根号3sinx*cosx+cos^2(x-π/6)-3/2sin^2x+1/4 若f(X)=4/5(0<

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  • f(x)=3/2√3sinx*cosx+cos^2(x-π/6)-3/2sin^2x+1/4

    =3√3/4sin2x+1/2(1+cos(2x-π/3)-3/4(1-cos2x)+1/4

    =3√3/4sin2x+1/2cos(2x-π/3)+3/4cos2x

    =3√3/4sin2x+1/4cos2x+√3/4sin2x+3/4cos2x

    =√3sin2x+cos2x

    =2(√3/2sin2x+1/2cos2x)

    =2sin(2x+π/6)

    f(θ)=2sin(2θ+π/6)=4/5

    sin(2θ+π/6)=2/5

    0<θ<π/2,π/6<2θ+π/6<7π/6

    cos(2θ+π/6)=√21/5或cos(2θ+π/6)=-√21/5

    sin2θ=sin[(2θ+π/6)-π/6]=√3/2sin(2θ+π/6)-1/2cos(2θ+π/6)

    =√3/2*2/5-1/2*√21/5=(2√3-√21)/10或

    sin2θ=√3/2*2/5+1/2*√21/5=(2√3+√21)/10

    即sin2θ=(2√3-√21)/10或sin2θ=(2√3+√21)/10