已知函数f(x)＝2cos^2(ωx/2)+cos(ωx+π/3)(ω＞0）

题目

已知函数f(x)＝2cos^2(ωx/2)+cos(ωx+π/3)(ω＞0）
就是这样

答案

f(x)＝2cos^2(ωx/2)+cos(ωx+π/3)
=1+cos(ωx)+cos(ωx+π/3)
=1+cos(ωx)+cos(ωx)cosπ/3-sin(ωx)sinπ/3
=1+3cos(ωx)/2-√3sin(ωx)/2
=1+√3*[√3/2cos(ωx)-1/2sin(ωx)]
=1+√3*[sinπ/3cos(ωx)-cosπ/3sin(ωx)]
=1+√3*sin(π/3-ωx)
或者直接用和差化积
f(x)＝2cos^2(ωx/2)+cos(ωx+π/3)
=1+cos(ωx)+cos(ωx+π/3)
=1+2cos[(ωx+ωx+π/3)/2]cos(ωx-ωx-π/3)/2]
=1+2cos(ωx+π/6)cos(π/6)
=1+√3cos(ωx+π/6)