化简:[cos(α-5π)+2sin(3π-α)]/[csc(3π+α)+sec(5π+α)]
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:2024/09/30 16:36:28
化简:[cos(α-5π)+2sin(3π-α)]/[csc(3π+α)+sec(5π+α)]
[cos(α-5π)+2sin(3π-α)]/[csc(3π+α)+sec(5π+α)]
=[cosacos5π+sinasin5π+2(sin3πcosa-cos3πsina)]/[1/sin(3π+a)+1/cos(5π+a)]
=[-cosa+2sina]/[-1/sina-1/cosa]
=(-cosa+2sina)*sinacosa/(-cosa-sina)
=(-1+2tana)tana/(-1-tana)
=(tana-2tan²a)/(1+tana)
=[cosacos5π+sinasin5π+2(sin3πcosa-cos3πsina)]/[1/sin(3π+a)+1/cos(5π+a)]
=[-cosa+2sina]/[-1/sina-1/cosa]
=(-cosa+2sina)*sinacosa/(-cosa-sina)
=(-1+2tana)tana/(-1-tana)
=(tana-2tan²a)/(1+tana)
化简:[cos(α-5π)+2sin(3π-α)]/[csc(3π+α)+sec(5π+α)]
化简 [Sin(π+α)\tan(π+α)] ×[ cot(2π-α)\cos(π+α)]×[sec(2π-α)\csc
已知tanα=2,求cos(α-7π/2)+2sin(π-3α)/csc(π+α)+sec(π/2+α)的值
化简[(secα-cosα)*(cscα-sinα)]/2sinα*cosα
求证:(cosαcscα-sinαsecα)/(cosα+sinα)=cscα-secα
求证(cosαcscα-sinαsecα)/(cosα+sinα)=cscα-secα
π/2的6个三角函数值 sin cos tan sec csc cot
求证:(tanα -cotα )/(secα -cscα )=sinα +cosα
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
化简:1/(1+sin²α)+1/(1+cos²α)+1/(1+sec²α)+1/(csc
证明 sin^2αtanα+cos^2αcotα+2sinαcosα=secαcscα