函数y=[cos(x-π/12)]^2+[sin(x+π/12)]^2-1是周期为___的____(奇/偶)函数.
问题描述：
函数y=[cos(x-π/12)]^2+[sin(x+π/12)]^2-1是周期为___的____(奇/偶)函数.
答
y=[cos(x-π/12)]^2+[sin(x+π/12)]^2-1
=[1+cos(2x-π/6)]/2+[1-cos(2x+π/6)]/2-1
=[cos(2x-π/6)-cos(2x+π/6)]/2
=(1/2)*(-2)sin2xsin(-π/6)
=(1/2)sin2x
所以填入：π,奇