利用公式求下列三角函数値:(l).cos(-420°);(2).sin(-7/6π);(3).sin(-1300°);(4).cos(-79/6π)化简(1)sin(α+180°)cos(-α)sin(-α-180°)(2)sin^3(-α)cos(2π+α)tan(-α-π)要过程啊!
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/30 11:56:27
![利用公式求下列三角函数値:(l).cos(-420°);(2).sin(-7/6π);(3).sin(-1300°);(4).cos(-79/6π)化简(1)sin(α+180°)cos(-α)sin(-α-180°)(2)sin^3(-α)cos(2π+α)tan(-α-π)要过程啊!](/uploads/image/z/1596234-66-4.jpg?t=%E5%88%A9%E7%94%A8%E5%85%AC%E5%BC%8F%E6%B1%82%E4%B8%8B%E5%88%97%E4%B8%89%E8%A7%92%E5%87%BD%E6%95%B0%E5%80%A4%3A%28l%29.cos%28-420%C2%B0%29%3B%282%29.sin%28-7%2F6%CF%80%29%3B%283%29.sin%28-1300%C2%B0%29%3B%284%29.cos%28-79%2F6%CF%80%EF%BC%89%E5%8C%96%E7%AE%80%EF%BC%881%EF%BC%89sin%EF%BC%88%CE%B1%2B180%C2%B0%EF%BC%89cos%EF%BC%88-%CE%B1%EF%BC%89sin%EF%BC%88-%CE%B1-180%C2%B0%EF%BC%89%EF%BC%882%EF%BC%89sin%5E3%28-%CE%B1%EF%BC%89cos%EF%BC%882%CF%80%2B%CE%B1%EF%BC%89tan%EF%BC%88-%CE%B1-%CF%80%EF%BC%89%E8%A6%81%E8%BF%87%E7%A8%8B%E5%95%8A%21)
利用公式求下列三角函数値:(l).cos(-420°);(2).sin(-7/6π);(3).sin(-1300°);(4).cos(-79/6π)化简(1)sin(α+180°)cos(-α)sin(-α-180°)(2)sin^3(-α)cos(2π+α)tan(-α-π)要过程啊!
利用公式求下列三角函数値:(l).cos(-420°);(2).sin(-7/6π);(3).sin(-1300°);(4).cos(-79/6π)
化简(1)sin(α+180°)cos(-α)sin(-α-180°)
(2)sin^3(-α)cos(2π+α)tan(-α-π)要过程啊!
利用公式求下列三角函数値:(l).cos(-420°);(2).sin(-7/6π);(3).sin(-1300°);(4).cos(-79/6π)化简(1)sin(α+180°)cos(-α)sin(-α-180°)(2)sin^3(-α)cos(2π+α)tan(-α-π)要过程啊!
(1).cos(-420°)=cos60°=1/2
(2).sin(-7/6π)=sin(π/6)=1/2
(3).sin(-1300°)=sin140°=sin40°
(4).cos(-79/6π)=-cos(π/6)=-√3/2
(1)sin(α+180°)=-sinα
cos(-α)=cosα
sin(-α-180°)=sinα
(2)sin^3(-α)=-sin^3(α)
cos(2π+α)=cosα
tan(-α-π)=-tanα
cos(-420)=cos(-360-60)=cos(-60)=cos(60)=1/2;
Sin(-7/6pai)=sin(-1/6pai)=-sin(1/6pai)=-1/2;
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