Question

已知函数f(x)=cosxsin(x+π/3)求其最大值最小值

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