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

f(x)=cosxsin(x+π/3)-√3cos^2x+(√3)/4
=cosx(1/2sinx+√3/2cosx)-√3cos^2x+(√3)/4
=1/4sin2x-√3/2cos^2x+(√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