• 2022-06-06
    设tanα=1,且cosα<0,则sinα=()
    A: -√2/2
    B: -1/2
    C: 1/2
    D: √2/2
  • A

    内容

    • 0

      cos(2α)=cos^2(α)-sin^2(α)=2cos^2(α)-1=1-2sin^2(α)

    • 1

      设\(z = \int_ { { x^2}}^y { { e^t}\sin t} dt\),则\({z_{xx}=}\) A: \(2{e^ { { x^2}}}\left[ {\left( {1 + 2{x^2}} \right)\sin {x^2} + 2{x^2}\cos {x^2}} \right]\) B: \( - 2{e^ { { x^2}}}\left[ {\left( {1 + 2{x^2}} \right)\sin {x^2} - 2{x^2}\cos {x^2}} \right]\) C: \( - 2{e^ { { x^2}}}\left[ {\left( {1 + 2{x^2}} \right)\sin {x^2} + 2{x^2}\cos {x^2}} \right]\) D: \( - 2{e^ { { x^2}}}\left[ {\left( {1 + 2{x^2}} \right)\cos {x^2} + 2{x^2}\sin {x^2}} \right]\)

    • 2

      17e0b849d3a4a3b.jpg,计算[img=19x34]17e0ab14a855463.jpg[/img]的实验命令为( ). A: syms x; f=diff((1+sin(x)^2)/cos(x),1)f=2*sin(x) + (sin(x)*(sin(x)^2 + 1))/cos(x)^2 B: f=diff((1+sinx^2)/cosx,1)f=1/2/x^(1/2)/(1-x)^(1/2) C: syms x;f=diff((1+sinx^2)/cosx,1)f=2*sin(x) + (sin(x)*(sin(x)^2 + 1))/cos(x)^2

    • 3

      已知α、β属于(0,π),且tan(α-β)=1/2,tanβ=-1/7,且2α-β的值

    • 4

      若tan a=2且sin a<0,则cos a的值是