（2008江西文数）函数f(x)=sinx/(sinx+2sin(x/2))为什么周期是4pi,偶函数?
网友回答
f(x)=sinx/[sinx+2sin(x/2)]f(-x)=sin(-x)/[sin(-x)+2sin(-x/2)]=-sinx/[-(sinx+2sin(x/2))]=sinx/[sinx+2sin(x/2)]所以是偶函数f(x)=sinx/(sinx+2sin(x/2))=(2sin(x/2)cos(x/2))/[2sin(x/2)cos(x/2)+2sin(x/2)]=cos...
======以下答案可供参考======
供参考答案1:
f(x)=sinx/(sinx+2sin(x/2))=SIN(X+2PI)/(SIN(X+2PI)+2SIN(X/2+2PI)=
SIN(X+4PI)/(SIN(X+4PI)+2SIN((X+4PI)/2)
所以T=4PI
因为SIN(-X)=-SINX
F(-X)=-sinx/(-sinx-2sin(x/2))=sinx/(sinx+2sin(x/2))=F(X)
所以是偶函数
供参考答案2:
4π 由f(x)=sinx/(sinx+2sinx/2)，又sinx=2sin(x/2)*cos(x/2)得
f(x) =[2sin(x/2)*cos(x/2)]/[2sin(x/2)*cos(x/2)+2sin(x/2)]
=cos(x/2)/[cos(x/2)+1]
所以 1/f(x)=1+1/cos(x/2)
即 [1/f(x)]-1=1/cos(x/2)
因为函数y=1/cos(x/2)的周期为4π。所以
[1/f(x+4π)]-1=[1/f(x)]-1 故原函数的周期为4π。