6.下列三角函数:①sin(nπ+4/3π) ②cos(2nπ+π/6) ③sin(2nπ+π/3) ④cos[(2n+1)π-π/6]
问题描述：
6.下列三角函数:①sin(nπ+4/3π) ②cos(2nπ+π/6) ③sin(2nπ+π/3) ④cos[(2n+1)π-π/6]
⑤sin[（2n+1）π-π/3]（n∈Z)其中函数值与sinπ/3相同的是235,为什么?
答
sin(nπ+4/3π)=±sin(4/3π)=±sinπ/3cos(2nπ+π/6)=cosπ/6=sinπ/3sin(2nπ+π/3)=sinπ/3cos[(2n+1)π-π/6]=cos[2nπ+π-π/6]=cos(π-π/6)=-cosπ/6=-sinπ/3sin[（2n+1）π-π/3]=sin[2nπ+π-π/3]=sin(...n∈Z是指所有选项？第1个为什么不等于±sin π比3或±cosπ/3呢，sin（nπ+4/3π）=sin[π+（nπ+π/3)]=sin（nπ+π/3)，到这里n是奇数或者偶数未知啊?Z 表示整数当 n 为偶数sin（nπ+4/3π）=sin[π+π/3)]=-sinπ/3当n为奇数, n+1为偶数sin（nπ+4/3π）=sin[(n+1)π+π/3)]=sinπ/3所以用±