f(x)=2sin(2x+π/6),求g(x)=f(x-π/12)-f(x+π/12)的单调递增区间
问题描述：
f(x)=2sin(2x+π/6),求g(x)=f(x-π/12)-f(x+π/12)的单调递增区间
答
f(x)=2sin(2x+π/6)
g(x)=f(x-π/12)-f(x+π/12)
=2sin(2x-π/6+π/6)-2sin(2x+π/6+π/6)
=2sin2x-2sin(2x+π/3)
=2sin2x-2[sin2x*(1/2)+cos2x*(√3/2)]
=sin2x-√3cos2x
=2sin(2x-π/3)
求单调递增区间：
令2kπ-π/2＜2x-π/3＜2kπ+π/2,k∈Z
kπ-π/12＜x＜kπ+5π/12,k∈Z所以单调递增区间是(kπ-π/12,kπ+5π/12),k∈Z
答
g(x)=f(x-π/12)-f(x+π/12)
=2sin(2x)-2sin(2x+π/3)
=2[sin(2x)-sin(2x+π/3)]
=2sin(2x-π/3)
单调递增区间由
2kπ-π/2≤2x-π/3≤2kπ+π/2
得kπ-π/12≤x≤kπ+5π/12