Question

已知f(x)=sinx(sinx-cosx).(1)求f(x)的最小正周期;(2)求F(x)取最大值时x的取值集合;(3)求f(x)的单调增区间

要过程。（之前学的过太久 都忘记了。。）

Answer 1

1
f(x)=sinx(sinx-cosx)
=sin²x-sinxcosx
=1/2(1-cos2x)-(1/2)sin2x
=-1/2(sin2x+cos2x)+1/2
=-√2/2sin(2x+π/4)+1/2
∴周期为2π/2=π
2
F(x)取最大值时
sin(2x+π/4)=-1
2x+π/4=2kπ-π/2
x=kπ-3π/8(k∈Z)
3
当2kπ+π/2≤2x+π/4≤2kπ+3π/2时，-√2/2sin(2x+π/4)单调递增
∴kπ+π/8≤x≤kπ+5π/8，f(x)递增
即求f(x)的单调增区间为[kπ+π/8,kπ+5π/8]