ts**051435adsl2011-07-01TA获得超过449个赞关注由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。