Question

己知函数f(x)=2c0sxsin(x-兀/3)+根号3sinx^2+sinxc0sx.(l)f

x)的最小正周期

Answer 1

解：f(x)=2cosxsin(x-兀/3)+√3sinx²+sinxcosx =2cosx.(sinxcos兀/3-cosxsin兀/3)+√3sin^2 x+sinxcosx
=2cosx(1/2•sinx-√3/2•cosx)+√3sin^2 x+sinxcosx
=cosxsinx-√3cos²x+√3sin^2 x+sinxcosx
=sin2x-√3cos2x
=2sin(2x-兀/3)
∴Ymax=2
Ymin=-2
T=2兀/2=兀

Answer 2

f(x)=2cosx(1/2•sinx-√3/2•cosx)+√3sin^2 x+sinxcosx
=sin2x-√3cos2x
=2sin(2x-π/3)
T=2π/2=π.