Question

函数y=sinx/[sinx+2sin(x/2)]的最小正周期是多少?

Answer 1

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.