f(x)=3sin^2(wx+θ)+根号3sin(wx+θ)cos(wx+θ)-3/2,w>0,T=π,且x=π/12,

1个回答

  • (1)、f(x)=3sin^2(wx+θ)+根号3sin(wx+θ)cos(wx+θ)-3/2

    =根号3/2sin(2wx+2θ)+3/2[2sin^2(wx+θ)-1]

    =根号3/2sin(2wx+2θ)+3/2cos(2wx+2θ)

    =根号3[1/2sin(2wx+2θ)+根号3/2cos(2wx+2θ)]

    =根号3[sin(2wx+2θ)cosπ/3+cos(2wx+2θ)sinπ/3]

    =根号3sin(2wx+2θ+π/3)

    因为f(x)的T=π,所以T=2π/2w=π,即:w=1,

    所以f(x)=根号3sin(2x+2θ+π/3)

    又x=π/12是其对称轴

    所以2*(π/12)+2θ+π/3=kπ+π/2,k属于整数,即:θ=kπ/2,k属于整数

    所以θ=π/2,故f(x)=根号3sin(2x+π+π/3)=-根号3sin(2x+π/3).

    (2)、因为0=