6.下列三角函数:①sin(nπ+4/3π) ②cos(2nπ+π/6) ③sin(2nπ+π/3) ④cos[(2n+
来源:学生作业帮 编辑:百度作业网作业帮 分类:数学作业 时间:2024/08/19 10:25:20
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?
⑤sin[(2n+1)π-π/3](n∈Z)其中函数值与sinπ/3相同的是235,为什么?
![6.下列三角函数:①sin(nπ+4/3π) ②cos(2nπ+π/6) ③sin(2nπ+π/3) ④cos[(2n+](/uploads/image/z/18663989-5-9.jpg?t=6.%E4%B8%8B%E5%88%97%E4%B8%89%E8%A7%92%E5%87%BD%E6%95%B0%3A%E2%91%A0sin%28n%CF%80%2B4%2F3%CF%80%29+%E2%91%A1cos%282n%CF%80%2B%CF%80%2F6%29+%E2%91%A2sin%282n%CF%80%2B%CF%80%2F3%29+%E2%91%A3cos%5B%282n%2B)
sin(nπ+4/3π)=±sin(4/3π)=±sinπ/3
cos(2nπ+π/6)=cosπ/6=sinπ/3
sin(2nπ+π/3)=sinπ/3
cos[(2n+1)π-π/6]=cos[2nπ+π-π/6]=cos(π-π/6)=-cosπ/6=-sinπ/3
sin[(2n+1)π-π/3]=sin[2nπ+π-π/3]=sin(π-π/3)=sinπ/3
所以选 2,3,5
再问: 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?
再答: 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±
cos(2nπ+π/6)=cosπ/6=sinπ/3
sin(2nπ+π/3)=sinπ/3
cos[(2n+1)π-π/6]=cos[2nπ+π-π/6]=cos(π-π/6)=-cosπ/6=-sinπ/3
sin[(2n+1)π-π/3]=sin[2nπ+π-π/3]=sin(π-π/3)=sinπ/3
所以选 2,3,5
再问: 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?
再答: 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±
数列an=n^2((cos(nπ/3))^2-(sin(nπ/3))^2)
计算sin(2nπ/3+π/6)+cos(2nπ/3+π/6)
求极限(sin(2/n)+cos(3/n))^(-n)
设f(x)=cos^(nπ+x).sin^(nπ-x)/cos^[(2n+1)π-x](n∈z)求f(π/6)的值
已知向量m=(cosθ,-sinθ),n=(根号2+sinθ,cosθ),θ∈(π,3π/2),且cos(θ/2+π/8
已知向量M=(2sinπ/4,cosπ/2)n=(cosπ/4,根号3)函数f(X)=m·n
sin(n兀+4兀/3),cos(2n兀+兀/6),sin(2n兀+兀/3),cos((2n+1)兀-兀/6),sin(
已知f(x)=cos²(nπ+x)sin²(nπ-x)/cos²[(2n+1)π-x](n
已知f(n)=sin[(n+1/2)π+π/4]+cos[(n-1/2)π+π/4]+tan[(n+1)π+π/4].求
已知f(n)=sin[(n+1\2)π+π\4]+cos[(n-1\2)+π\4]+tan[(n+1)π+π\4],求f
已知cos(π+α)=1/2,计算sin(2π-α) sin[(2n+1)π+α]+sin[α-(2n+1)π]/sin
求证:(cosπ/6+isinπ/6)^n=cos(nπ/6)+i *sin( nπ/6)谢谢