问题描述: 函数y=sin²(x+π/12)+cos²(x-π/12)-1的周期T=?,奇偶性为 1个回答 分类:数学 2014-09-30 问题解答: 我来补答 答案:T=π 奇函数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)=sin2x×sinπ/6=(1/2)sin2x所以最小正周期T=π奇函数 展开全文阅读