Question

已知函数f(x)=5sinxcosx-5√3cos^2x+[(5√3)/2] (x∈R)求：函数最小正周期和单调区间

Answer 1

解；依题意得y=(5/2)sin2x-(5√3/2)cos2x
=(5/2)sin(π-2x)+(5√3/2)cos(π-2x)
=5sin(π-2x+a)
tana=(5√3/2)/(5/2)=√3
a=π/3
y=5sin(π-2x+a)
=5sin(π-2x+π/3)
=5sin(2x-π/3)

y=sinx的单调递增区间是
2kπ-π/2<x<2kπ+π/2
单调递减区间是
2kπ+π/2<x<2kπ+3π/2

所以y=5sin(2x-π/3)
的单调递增区间是
2kπ-π/2<2x-π/3<2kπ+π/2
2kπ-π/6<2x<2kπ+5π/6
kπ-π/12<x<kπ+5π/12

单调递减区间是
2kπ+π/2<2x-π/3<2kπ+3π/2
2kπ+5π/6<2x<2kπ+11π/6
kπ+5π/12<x<kπ+11π/12