分类：教育/科学-学习帮助
问题：三角函数的高手进！①函数y=sin(2x/3)+cos(2x/3 
①y=sin(2x/3)+cos(2x/3+π/6）的图象中相邻的两条对称轴的距离是？？②（4x^2+-6x+5)(1+1/x)^5的展开式中，常数项是？？
回答：解:一:
原式=sin(2x/3) + (√3/2)cos(2x/3)-(1/2)sin(2x/30

    =(1/2) * sin(2x/3) + (√3/2)cos(2x/3)

    =cos(π/3) * sin(2x/3) + sin(π/3)cos(2x/3)

    =sin(2x/3+π/3)

最小正周期T=2π/ (2/3) =3π

所以 相邻的两条对称轴的距离=3π/2


二:
(1+1/x)^5=1 + 5/x + 10/x^2 + 10/x^3 + 5/x^4 + 1/x^5

所以 常数项=4x^2 * 10/x^2 + 5/x * 6x + 5*1 

           =40+30+5

           =75