函数f(x)=2sinxcos(x-π/3)-√3cos^2x+sian(x+π/2)sinx将函数f(x)按向量a平移后得到函数g(x),且当x=π/3时,g(x)取最大值3,求向量a和g(x)

Question

函数f(x)=2sinxcos(x-π/3)-√3cos^2x+sian(x+π/2)sinx
将函数f(x)按向量a平移后得到函数g(x),且当x=π/3时,g(x)取最大值3,求向量a和g(x)

Answer 1

f(x)=2sinxcos(x-π/3) - √3cos²x + sin(x+π/2)sinx
f(x)=2sinx(cosxcos(π/3) + sinxsin(π/3)) - √3cos²x + (sinxcos(π/2) + sin(π/2)cosx)sinx
f(x)=2sinx(1/2cosx + √3/2sinx) - √3cos²x + sinxcosx
f(x)=sinxcosx + √3sin²x - √3cos²x + sinxcosx
f(x)=2sinxcosx + √3(sin²x-cos²x) //2倍角公式：sin2A=2sinA·cosA cos2A=cos²A-sin²A
f(x)=sin2x -√3cos2x
f(x)=2(cos(π/3)sin2x - sin(π/3)cos2x)
f(x)=2sin(2x-π/3)

∵f(x)=sinx,当x=2kπ+π/2时f(x)取得最大值1
∴f(x)=2sin(2x-π/3),当x=kπ+5/12π时f(x)取得最大值2

∵f(x)是周期为π的周期函数,g(x)是由f(x)平移后得到的,∴g(x)也是周期为π的周期函数
∵当x=π/3时,g(x)取最大值3,且g(x)是周期函数
∴5π/12-π/3=π/12 既π/3=5π/12 - π/12
∴a=（-π/12+kπ,1） k∈Z

g(x)=2sin(2(x-(-π/12-kπ))-π/3)+1 k∈Z
g(x)=2sin(2x-π/6+2kπ)+1 k∈Z
g(x)=2sin(2x-π/6)+1