求解一道三角函数题 谢谢

sinx=2sin(x/2)cos(x/2)
所以f(x)=2sin(x/2)cos(x/2)/[2sin(x/2)cos(x/2)+2sin(x/2)]
=cos(x/2)/[cos(x/2)+1]
=[cos(x/2)+1-1]/[cos(x/2)+1]
=[cos(x/2)+1]/[cos(x/2)+1]-1/[cos(x/2)+1]
=1-1/[cos(x/2)+1]

cos(x/2)周期是4π
所以整个函数的周期是4π

f(-x)=sin(-x)/[sin(-x)+2sin(-x/2)]
=-sinx/[-sinx-2sin(x/2)]
=sinx/[sinx+2sin(x/2)]
=f(x)

定义域
sinx+2sin(x/2)不等于0
若sinx+2sin(x/2)=0
则2sin(x/2)[cos(x/2)+1]=0
sin(x/2)=0,cos(x/2)=-1
x/2=kπ
x=2kπ
所以定义域是x不等于2kπ，关于原点对称
所以f(x)是偶函数，不是奇函数