f(x)=sin(x+π/6)cosx+cosxcos(x-π/3)-1/2
cos(x-π/3)=cos(π/3-x)
=cos[π/2-(x+π/6)]
=sin(x+π/6)
所以,f(x)=sin(x+π/6)cosx+cosxsin(x+π/6)-1/2
=sin(x+π/6+x)-1/2
=sin(2x+π/6)-1/2
所以,最小正周期T=2π/2=π

祝你開心!希望能幫到你,如果不懂,請追問,祝學(xué)習(xí)進(jìn)步!O(∩_∩)O“=sin(x+π/6+x)-1/2=sin(2x+π/6)-1/2”化簡后為什么不是sin(2x+π/3)-1/2？x+π/6+x=2x+π/6啊，為啥會有π/3出來？弱弱的問一句 這里用的是那個公式？- =我不知道這里是怎么化簡出來的sin(x+π/6)cosx+cosxsin(x+π/6)=sin[(x+π/6)+x]用的是：sin(α+β)=sinαcosβ+cosαsinβ那如果按sin[(x+π/6)+x]化簡 不就是sin(x+π/6)cosx+cos(x+π/6)sinx么？囧了~~看錯了，不好意思了正f(x)=sin(x+π/6)cosx+cosxcos(x-π/3)-1/2cos(x-π/3)=cos(π/3-x)=cos[π/2-(x+π/6)]=sin(x+π/6)所以，f(x)=sin(x+π/6)cosx+cosxsin(x+π/6)-1/2=2sin(x+π/6)cosx-1/2=2[(√3/2)sinx+(1/2)cosx]cosx-1/2=√3sinxcosx+cos²x-1/2由倍角公式：sin2x=2sinxcosx，cos2x=2cos²x-1得：sinxcosx=(√3/2)sin2x，cos²x=(1+cos2x)/2所以，f(x)=(√3/2)sin2x+(1/2)cos2x=sin(2x+π/6)所以，最小正周期T=2π/2=π