设函数f（x）=2√3sin^2x-(sinx-cosx)^2且x∈R

更新时间2019-01-12 01:24:48

 

1.求f（x）单调区间（增和减的都要）

2.把f（x）图象横坐标伸长到原来2倍，纵坐标不变，再向左平移π/3个单位得到y=g（x）的图象

求g（-π/3）

f（x）=2√3sin²x-(sinx-cosx)²

=√3（1-cos2x）-1+sin2x

=-√3cos2x+sin2x+√3-1

=2[sin2xcos(π/3)-cos2xsin(π/3)]+√3-1

=2sin(2x-π/3)+√3-1

①当2kπ-π/2≤2x-π/3≤2kπ+π/2，即：x∈[kπ-π/12,kπ+5π/12],k∈Z时，f（x）=2√3sin²x-(sinx-cosx)²单调递增

②当2kπ+π/2≤2x-π/3≤2kπ+3π/2，即：x∈[kπ+5π/12,kπ+11π/12],k∈Z时，f（x）=2√3sin²x-(sinx-cosx)²单调递减

y=g（x）=2sin(x-π/6+π/3)+√3-1=2sin(x+π/6)+√3-1

g（-π/3）=2sin(-π/3+π/6)+√3-1=√3-2