化简 [Sin(π+α)\tan(π+α)] ×[ cot(2π-α)\cos(π+α)]×[sec(2π-α)\csc
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:2024/07/22 16:59:33
化简 [Sin(π+α)\tan(π+α)] ×[ cot(2π-α)\cos(π+α)]×[sec(2π-α)\csc(π-α)]
![化简 [Sin(π+α)\tan(π+α)] ×[ cot(2π-α)\cos(π+α)]×[sec(2π-α)\csc](/uploads/image/z/3798211-67-1.jpg?t=%E5%8C%96%E7%AE%80+%5BSin%28%CF%80%2B%CE%B1%29%5Ctan%28%CF%80%2B%CE%B1%29%5D+%C3%97%5B+cot%282%CF%80-%CE%B1%29%5Ccos%28%CF%80%2B%CE%B1%29%5D%C3%97%5Bsec%282%CF%80-%CE%B1%29%5Ccsc)
[Sin(π+α)\tan(π+α)] ×[ cot(2π-α)\cos(π+α)]×[sec(2π-α)\csc(π-α)]
= [(-sinα)\tanα] ×[(-cotα)\(-cosα)]×[secα\cscα]
= [(-sinα)×cosα\sinα] ×[(cosα\sinα)\cosα)]×(sinα\cosα)
=-cosα×(1\sinα)×(sinα\cosα)
=-1
= [(-sinα)\tanα] ×[(-cotα)\(-cosα)]×[secα\cscα]
= [(-sinα)×cosα\sinα] ×[(cosα\sinα)\cosα)]×(sinα\cosα)
=-cosα×(1\sinα)×(sinα\cosα)
=-1
化简 [Sin(π+α)\tan(π+α)] ×[ cot(2π-α)\cos(π+α)]×[sec(2π-α)\csc
π/2的6个三角函数值 sin cos tan sec csc cot
证明tan^2α-cot^2α/sin^2α-cos^2α=sec^2α+csc^2α
已知tan^2α+cot^2α+sec^2α+csc^2α=7,则sinαcosα
求证:(tanα -cotα )/(secα -cscα )=sinα +cosα
证明 sin^2αtanα+cos^2αcotα+2sinαcosα=secαcscα
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sin²π/2=?还有类似的,比如cos²π/2,tan、cot、sec、csc,还有sinπ、ta
(tan^2α-cot^2α)/(sin^2α-cos^2α)=sec^2α•csc^2α
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化简tanα(cosα-sinα)+(sinα+tanα)/(cotα+cscα)