y=cosπx*cos(πx-π/3)+sin^2πx的周期、单调区间、值域
问题描述：
y=cosπx*cos(πx-π/3)+sin^2πx的周期、单调区间、值域
答
y=cosπx*cos(πx-π/3)+sin^2πx=[cos(πx+πx-π/3)+cos(πx-πx+π/3)]/2+sin^2πx=cos(2πx)/4+sin(2πx)*√3/4+1/4+sin^2πx=cos(2πx)/4+sin(2πx)*√3/4+1/4+(1-cos2πx)/2=[cos(2πx+2π/3)]/2+5/4下面的会...