9.已知函数f(x)=sinx*2+2√3sinxcos+3cos*2x (2)已知f(a)=3,且a∈(0,π),求a

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  • 9.f(x)=sin^2x+2√3sinxcosx+3cos^2x

    =2cos^2x+√3sin2x+1

    =cos2x+√3sin2x+2

    =sin(2x+π/6)+2

    所以f(a)=sin(2a+π/6)+2=3

    即sin(2a+π/6)=1

    则有2a+π/6=2kπ+π/2(k∈N,a∈(0,π))

    可得a=π/6

    10.f(x)=sin(π/2+x)cosx-sinxcos(π-x)

    =cosxcosx-sinx(-cosx)

    =(1+cos2x)/2+(sin2x)/2

    =√2/2sin(2x+π/4)+1/2

    因f(A)=1且A为锐角,易知A=π/4

    由正弦定理可求得AC长

    11.|OP|=√(1/4+cos⁴θ),|OQ|=√(sin⁴θ+1)

    sin(α+β)=sinαcosβ+cosαsinβ

    =[(sin²θ*cos²θ)+1/2]/√(1/4+cos⁴θ)√(1+sin⁴θ)

    =1/4sin²2θ+1/2/√[1/4+1/4(1+cos2θ)²]√[1+1/4(1-cos2θ)²]

    =1/4+1/2/√(1/2)√(5/4)

    =(3/4 )/ √5/8

    =3√10/10