Question

设函数f(x)=sin(π/3x-π/6)-2cos²πx/6, 求(1)f(x)最小正周期及单调递增区间.

Answer 1

∵f(x)=sinxsin(x
π/2)-√3cos²(3π
x)
√3/2
=sinxcosx-√3cos²x
√3/2
=(1/2)sin2x-(√3/2)cos2x
=sin(2x-π/3)
(1)∴T=2π/2=π
(2)令2x-π/3=2kπ-π/2，解得x=kπ-π/12，k∈Z
令2x-π/3=2kπ
π/2，解得x=kπ
5π/12，k∈Z
∴f(x)的单调递增区间为x∈[kπ-π/12，kπ
5π/12]，k∈Z
(3)令2x-π/3=kπ
π/2，解得x=kπ/2
5π/12
∴f(x)的对称轴方程为x=kπ/2
5π/12，k∈Z
令2x-π/3=kπ，解得x=kπ/2
π/6
∴f(x)的对称中心坐标为(kπ/2
π/6，0)，k∈Z