函数y=sinx/[sinx+2sin(x/2)]的最小正周期是多少?
问题描述：
函数y=sinx/[sinx+2sin(x/2)]的最小正周期是多少?
答
4π
答
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.