函数y=[cos(x-π/12)]^2+[sin(x+π/12)]^2-1是周期为___的____(奇/偶)函数._数学解答

函数y=[cos(x-π/12)]^2+[sin(x+π/12)]^2-1是周期为___的____(奇/偶)函数.

人气：369 ℃　时间：2019-09-23 09:21:29

解答

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