函数y=sin²(x+π/12)+cos²(x-π/12)-1的周期T=?,奇偶性为

答案:T=π 奇函数 解答:y=1-(cos(x+π/12))^2+(cos(x-π/12))^2-1 =(cos(x-π/12))^2-(cos(x+π/12))^2 =(cos(x-π/12)-cos(x+π/12))×(cos(x-π/12)+cos(x+π/12)) =2cosxcosπ/12×2sinxsinπ/12 =(2sinxcosx)×(2sinπ/12cosπ/12) =s