涵数y= √ 3cos^2x+sincosx- √ 3/2的最小正周期是?

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涵数y= √ 3cos^2x+sincosx- √ 3/2的最小正周期是?
涵数y= √ 3cos^2x+sincosx- √ 3/2的最小正周期是?

涵数y= √ 3cos^2x+sincosx- √ 3/2的最小正周期是?
y=√3cos^2x+sinxcosx-√3/2
=√3(1+cos2x)/2+1/2*sin2x-√3/2
=√3/2+√3/2*cos2x+1/2*sin2x-√3/2
=√3/2*cos2x+1/2*sin2x
=sin(2x+π/3)
所以:f(x)=sin(2x+π/3)
T=2π/2=π
最小正周期:π