已知函f(x)=cosxcos(x-π/3)-cos2x-1/4,x∈R.求f(x)的单调递增区间.求f(x)在［-π.

f(x)=cosxcos(x-π/3)-cos2x-1/4
=cosx[(1/2)cosx+(√3/2)sinx]-cos2x-1/4
=(1/2)cos²x+(√3/2)sinxcosx-cos2x-1/4
=(1/4)(cos2x+1)+(√3/4)sin2x-cos2x-1/4
=(√3/4)sin2x-(3/4)cos2x
=(√3/2)sin(2x-π/3)
单增：
2x-π/3∈[2kπ-π/2,2kπ+π/2]
x∈[kπ-π/12,kπ+5π/12]
所以单调增区间为
[kπ-π/12,kπ+5π/12] k∈z
［-π/6,π/4］
2x-π/3∈[-2π/3,π/6]
所以
f(x)的值域为 [-√3/2,√3/4]
最大值为 √3/4 最小值为 -√3/2

（1）f(x)=cosxcos(x-π/3)-cos2x-1/4
=cosx[cosxcos(π/3)+sinxsin(π/3)]-cos2x-1/4
=cosx[(1/2)cosx+(√3/2)sinx]-cos2x-1/4
=(1/2)cos²x+(√3/2)sinxcosx-cos2x-1/4
=(1/4)(c...