Question

已知函数f （x）= cosxsin（ x+π/3）-√3cos²x+√3/4,x属于R

. (1)求f(x)的最小正周期；(2)求f(x)在闭区间[-π/4,π/4]上的最大值和最小值.

Answer 1

解：
（1）f（x）=cosxsin(x+∏/3)-根号3cos^x+根号3/4
=cosx(sinxcosπ/3+cosxsinπ/3)-√3/2(1+cos2x)+√3/4
=1/2sinxcosx+√3/2cos²x-√3/2(1+cos2x)+√3/4
=1/4sin2x+√3/4(1+cos2x)-√3/2(1+cos2x)+√3/4
=1/4sin2x-√3/4cos2x-√3/2
=1/2(1/2sin2x-√3/2cos2x)-√3/2
=1/2sin(2x-π/3)-√3/2
f(x)最小正周期T=2π/2=π

(2)∵x∈[-π/4,π/4]
∴2x-π/3∈[-5π/6,π/6]
当2x-π/3=-π/2时，f(x)min=-1/2-√3/2
当2x-π/3=π/6时，f(x)min=1/4-√3/2

追问

我不知道错的这步

我算出来了，你的答案后面不应该加✔3/2

Answer 2

（1）f(x)=cosxsin(x+π/3)-√3cos²x+√3/4
=cosx(1/2sinx+√3/2cosx)-√3cos²x+√3/4
=1/4sin2x-√3/2cos²x+√3/4
=1/4sin2x-√3/4(cos2x+1)+√3/4
=1/4sin2x-√3/4cos2x
=1/2sin(2x-π/3)
最小正周期T=π
-π/2<2X-π/3<π/2，递增
-π/12<X<5π/12
-7π/12<X<-π/12， 递减
在【-π/4,π/4】上，-7π/12<-π/4<-π/12,最小值f(-π/12)=-1/2
-π/12<π/4<5π/12, 最大值f(π/4) =1/4