题目

求函数最小正周期{要有过程}
FX=sin{2x+PI/6}+sin{2x-pi/6}+cos2x

提问时间：2021-03-14

答案

f(x)=[sin(2x+π/6)+sin(2x-π/6)]+cos(2x)=2sin{[(2x+π/6)+(2x-π/6)]/2}cos{[(2x+π/6)-(2x-π/6)]/2}+cos(2x) =2sin(2x)cos(π/6)+2cos(2x)sin(π/6)=2sin(2x+π/6) 式中振幅A=2,角频率ω=2,初相φ0=π/6,又sinx...