y=sin2/3x+cos(2x/3)+cos(2x/3+π/6)的图像中相邻两对称轴的距离是

问题描述:

y=sin2/3x+cos(2x/3)+cos(2x/3+π/6)的图像中相邻两对称轴的距离是

由题可将其化简为y=3/2cos(2/3x) (1-厂3/2)sin(2/3x) 将其降次收缩得Y=ASin(2/3x k)则得周期T=2兀/(2/3)=3兀;所以相邻对称距离为T/2=3/2兀 答题不易求采纳

y=sin2/3x+cos2x/3+cos2x/3*根号3/2-sin2x/3*1/2 =1/2sin2x/3+(根号3/2+1)cos2x/3=根号[1/4+(根号3/2+1)^2]sin(2x/3+@)其中最小正周期T=2Pai/(2/3)=3Pai故图像中相邻二对对称轴的距离是半个周期,即有T/2=3Pai/2...