Question

己知函数（fx）=（sinx-cosx）sin2x/sinx求函数fx的定义域最小正周期单调递增区间

Answer 1

f（x）=（sinx-cosx）sin2x/sinx的定义域
x不等于kπ.
f(x）=（sinx-cosx）sin2x/sinx
=2（sinx-cosx）sinxcosx/sinx
=2sinxcosx-2cos^2x
=sin2x-cos2x-1
=√2sin(2x-π/4)-1
最小正周期T＝π
递增区间：2kπ-π/2<=2x-π/4<=2kπ+π/2
kπ-π/8<=x<=kπ+3π/8,k是整数

Answer 2

f(x)=（sinx-cosx）sin2x/sinx=（sinx-cosx）2sinxcosx/sinx=2(sinx-cosx)cosx=2sinxcosx-2cosx^2
=sin2x-(1+cos2x)=sin2x-cos2x+1=根2sin(2x-π/4) , 定义域满足sinx不等于0,即x不等于kπ.
最小正周期 T=π,
单调递增区间 令2kπ-π/2《2x-π/4《2kπ+π/2得[kπ-π/8,kπ+3π/8]

Answer 3

[0,3pi/8]

 

用matlab作图如下

ezplot(@(x)(sin(x)-cos(x))*sin(2*x)/sin(x),[0,2*pi])

grid on