51问答网 > 已知函数f(x)=sin^2(x+π⼀12), g(x)=1+1⼀2sin2x x属于[-π⼀12,11π⼀12]

已知函数f(x)=sin^2(x+π⼀12), g(x)=1+1⼀2sin2x x属于[-π⼀12,11π⼀12]

求函数h(x)=f(x)-g(x)的单调递减区间

2025-01-06 11:13:54

推荐回答（2个）

回答1：

h(x)'=f(x)'-g(x)'=sin(2x+π/6)+cos2x=sin(2x+π/6)+sin(2x+π/2)=2sin(2x+π/3)*sin(-π/3)
所以 2kπ-π/2<=2x+π/3<=2kπ+π/2 (k是整数)
kπ-5π/12<=x<=kπ+π/12(k是整数)
求[kπ-5π/12,kπ+π/12]与[-π/12,11π/12]交集得[-π/12,π/12]∪[7π/12,11π/12].

回答2：

f(x)=sin²(x+π/12)
=[1-cos(2x+π/6)]/2
=1/2-1/2(cos2xcosπ/6-sin2xsinπ/6)
=1/2-√3/4cos2x+1/4sin2x
h(x)=f(x)-g(x)
=1/2-√3/4cos2x+1/4sin2x-1-1/2sin2x
=-1/2(√3/2cos2x+1/2sin2x)-1/2
=-1/2sin(2x+π/3)-1/2
令kπ+π/2≤2x+π/3≤kπ+3π/2(k∈N)
得kπ/2+π/12≤x≤kπ/2+7π/12
即单调递减区间为[kπ/2+π/12,kπ/2+7π/12](k∈N)