y=sin^4x+2√3sinxcosx-cos^4x

题目

y=sin^4x+2√3sinxcosx-cos^4x
1,求最小正周期和最小值
2,写出该函数在〔0,π〕上的单调递增区间

答案

1、
y=sin^4x+2√3sinxcosx-cos^4x
=(sin^2x+cos^2x)(sin^2x-cos^2x)+2√3sinxcosx
=-cos2x+√3sin2x
=2(cos2π/3cos2x+sin2π/3sin2x)
=2cos(2x-2π/3)
可得：最小正周期为T=2π/2=π
当cos(2x-2π/3)=-1时有最小值为-2
2、 因余弦函数的递增区间为：[2kπ-π,2kπ]
所以有：2kπ-π≤2x-2π/3≤2kπ
解得：kπ-π/6≤x≤kπ+π/3
因此可得该函数在〔0,π〕上的单调递增区间为：(0,π/3]∪[5π/6,π)