Question

已知函数f（x ）=(sinx-cosx)sin2x ／sinx (1)求f(x )的定义域及最小正周期 (2)求f(x)的单调增区间

Answer 1

解答：
（1）sinx≠0， x≠kπ，k∈Z
f(x)=(sinx-cosx)sin2x/sinx
=(sinx-cosx)*2cosx
=2sinxcosx-2cos²x
=sin2x-cos2x-1
=√2sin(2x-π/4)-1
T=2π/2=π
（2）增区间为
2kπ-π/2≤2x-π/4<2kπ或 2kπ<2x-π/4≤2kπ+π/2
2kπ-π/4≤2x<2kπ+π/4或 2kπ+π/4<2x≤2kπ+3π/4
kπ-π/8≤x<kπ+π/8或 kπ+π/8<x≤kπ+3π/8
增区间【 kπ-π/8，kπ+π/8）和（ kπ+π/8，kπ+3π/8】，k∈Z

Answer 2

函数f(x)=(sinx一cosx)sin2x/sinx

sinx≠0，所以x≠kπ，k∈Z.
函数定义域是{x|x≠kπ，k∈Z}.

f(x)=(sinx一cosx)sin2x/sinx
=(sinx一cosx)*2sinxcosx/sinx
=(sinx一cosx)*2cosx
=2sinxcosx-2cos²x
=sin2x-(1+cos2x)
=sin2x-cos2x-1
=√2(√2/2*sin2x-√2/2*cos2x)-1
=√2 sin(2x-π/4)-1
所以函数的最小正周期是π.

2kπ-π/2≤2x-π/4≤2kπ+π/2，k∈Z.
所以kπ-π/8≤x≤kπ+3π/8，k∈Z.
注意到函数的定义域是{x|x≠kπ，k∈Z}，
所以函数的单调递增区间是[kπ-π/8, kπ），（kπ，kπ+3π/8].

Answer 3

(1)由sinx≠0得：x≠kπ,所以f(x )的定义域为{x|x≠kπ,k∈z}
f（x ）=(sinx-cosx)sin2x ／sinx= (sinx-cosx)2sinx cosx／sinx =2cosx(sinx-cosx)=2sinxcosx-2cos^2x=sin2x-cos2x-1=√2sin(2x-π/4)-1，所以最小正周期为2π/2=π
(2)由-π/2+2kπ≤2x-π/4≤π/2+2kπ得：-π/8+kπ≤x≤3π/8+kπ,所以f(x)的单调增区间为
[-π/8+kπ,3π/8+kπ](k∈z)

Answer 4

这是正确答案

Answer 5

(1)原式=（sinx-cosx）2sinx*cosx/sinx=2(sinx-cosx)cosx=2sinx*cosx-2cosx*cosx=sin2x-(2cos2x+1)=sin2x-2cos2x-2,(x不等于kπ)
最小正周期T=2π/2=π
(2)