Question

已知函数f(x)=2cosxcos(x-π/6)-√3sin∧2x+sinxcosx.求f(x)的

已知函数f(x)=2cosxcos(x-π/6)-√3sin∧2x+sinxcosx.求f(x)的最小正周期

Answer 1

首先化简
f(x)=2cosxcos(x-π/6)-√3sin^2x+sinxcosx
=2cosxcos(x-π/6)-2sinx(√3/2sinx-1/2cosx)
=2cosxcos(x-π/6)-2sinx(sinxcosπ/6-cosxsinπ/6)
=2cosxcos(x-π/6)-2sinxsin(x-π/6)
=2cos(x+x-π/6)
=2cos(2x-π/6)
f(x)的最小正周期T=2π/w=2π/2=π
看完了采纳哦~~

Answer 2

解f(x)=2cosxcos(x-π/6)-√3sin∧2x+sinxcosx.

=cos（x+x-π/6)+cos（x-x+π/6)-√3sin∧2x+sinxcosx
=cos（2x-π/6)+√3/2-√3sin∧2x+1/2sin2x
=cos（2x-π/6)+√3/2-√3/2×2sin∧2x+1/2sin2x
=cos（2x-π/6)+√3/2（1-2sin∧2x）+1/2sin2x
=cos（2x-π/6)+√3/2cos2x+1/2sin2x
=√3/2cos2x+1/2sin2x+√3/2cos2x+1/2sin2x
=√3cos2x+sin2x
=2（√3/2cos2x+1/2sin2x）
=2sin（2x+π/3)
故T=2π/2=π。