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:神马作文网作业帮 分类:数学作业 时间:2024/11/11 06:08:49
化简 [Sin(π+α)\tan(π+α)] ×[ cot(2π-α)\cos(π+α)]×[sec(2π-α)\csc(π-α)]
[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
已知tanα=2,求cos(α-7π/2)+2sin(π-3α)/csc(π+α)+sec(π/2+α)的值
化简:[cos(α-5π)+2sin(3π-α)]/[csc(3π+α)+sec(5π+α)]
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
一道三角函数证明题~求证:(sinα+cscα)^2+(cosα+secα)^2=tan^2α+cot^2α+7
化简tanα(cosα-sinα)+(sinα+tanα)/(cotα+cscα)