f(x)=sin²ωx+√3sinωxsin(ωx+二分之∏)+2cos²ωx(ω>0)的图象的相邻

1个回答

  • (1)

    f(x)=sin²ωx+√3sinωxsin(ωx+二分之∏)+2cos²ωx

    =1/2(1-cos2wx)+√3/2sin2wx+1+cos2wx

    =√3/2sin2wx+1/2cos2wx+1/2

    =sin(2wx+π/6)+1/2

    ∵f(x)图象的相邻两对称轴间的距离为∏

    ∴f(x)的半周期T/2=π,T=2π

    由2π/(2w)=2π得w=1/2

    ∴f(x)=sin(x+π/6)+1/2

    (2)

    ∵bcosC=(3a-c)cosB

    本剧正弦定理:

    a=2RsinA,b=2RsinB,c=2RsinC

    ∴sinBcosC=(3sinA-sinC)cosB

    ∴3sinAcosB=sinBcosC+cosBsinC=sin(B+C)

    ∵sin(B+C)=sin(180º-A)=sinA>0

    ∴3sinAcosB=sinA

    ∴cosB=1/3 ,sinB=2√2/3

    ∴f(B)=sin(B+π/6)+1/2

    =sinBcosπ/6+cosBsinπ/6+1/2

    =2√2/3*√3/2+1/3*1/2+1/2

    =(2+√6)/3