Question

已知函数f(x)=2sinxcosx+sin(2x+π/2)+1 1.求他的最小正周期和最大值 2.求y=f(x)的单调

Answer 1

f(x)=2sinxcosx+sin(2x+π/2)+1

=sin2x+cos2x+2

=√2sin（2x+π/4）+1

1.最小正周周期T=2π/2=π

当sin（2x+π/4）=1时取最大值

即当2x+π/4=2kπ+π/2,x=kπ+π/8时,y的最大值为√2+1

2.、单调增区间：

2kπ-π/2 ≤2x+π/4≤2kπ

则单调增区间为[kπ-3π/8,kπ-π/8]

单调减区间：2kπ ≤2x+π/4≤2kπ +π/2

则单调增区间为[kπ-π/8,kπ+3π/8]