• 2022-06-06
    cos(x)*cos(x/2)*cos(x/4)*cos(x/8).cos(x/(2^(n-1))
  • cos(x)*cos(x/2)*cos(x/4)*cos(x/8).cos(x/(2^(n-1))=cos(x)*cos(x/2)*cos(x/4)*cos(x/8).cos(x/(2^(n-1))sin(x/(2^(n-1))/sin(x/2^(n-1))=cos(x)*cos(x/2)*cos(x/4)*cos(x/8).sin(x/(2^(n-2))/2sin(x/2^(n-1))=...=cos(x)*cos(x/2)*cos(x/4)*cos(x/8)sin(x/8)/2^(n-4)*sin(x/2^(n-1))=cos(x)*cos(x/2)*cos(x/4)sin(x/4)/2^(n-3)*sin(x/2^(n-1))=cos(x)*cos(x/2)*sin(x/2)/2^(n-2)*sin(x/2^(n-1))=cos(x)*sin(x)/2^(n-1)sin(x/2^(n-1))=sin(2x)/[2^nsin(x/2^(n-1))]

    内容

    • 0

      【单选题】化简 sin( x + y )sin( x - y ) + cos( x + y )cos( x - y ) 的结果是 A. sin 2 x B. cos 2 y C. - cos 2 x D. -cos 2 y

    • 1

      求微分方程[img=143x21]17da5f14490e50e.png[/img]的通解,实验命令为(). A: dsolve(D2y-2*Dy+5*y=sin(2*x),x)ans =exp(x)*sin(2*x)*C2+exp(x)*cos(2*x)*C1+1/17*sin(2*x)+4/17*cos(2*x) B: dsolve('D2y-2*Dy+5*y=sin(2*x)','x')ans =cos(2*x)*(sin(4*x)/17 - cos(4*x)/68 + 1/4) - sin(2*x)*(cos(4*x)/17 + sin(4*x)/68) + C1*cos(2*x)*exp(x) - C2*sin(2*x)*exp(x) C: dsolve(D2y-2*Dy+5*y=sin(2*x),'x','y')ans =exp(x)*sin(2*x)*C2+exp(x)*cos(2*x)*C1+1/17*sin(2*x)+4/17*cos(2*x)

    • 2

      已知\( y = {x^3}\cos 2x \),则\( y'' \)为( ). A: 0 B: \( 6x\cos 2x{\rm{ + }}12{x^2}\sin 2x - 4{x^3}\cos 2x \) C: \( 6x\cos 2x - 12{x^2}\sin 2x{\rm{ + }}4{x^3}\cos 2x \) D: \( 6x\cos 2x - 12{x^2}\sin 2x - 4{x^3}\cos 2x \)

    • 3

      \( \int {\cos \ln xdx} = \)( ) A: \( {x \over 2}(\cos \ln x + \sin \ln x) + C \) B: \( {x \over 2}(\cos \ln x - \sin \ln x) + C \) C: \(- {x \over 2}(\cos \ln x + \sin \ln x) + C \) D: \(- {x \over 2}(\cos \ln x - \sin \ln x) + C \)

    • 4

      求微分方程[img=634x60]17da653955cf9e7.png[/img]的特解。 ( ) A: sin(2*x)/3 - cos(x) - cos(x)/3 B: sin(2*x)/3 - cos(x) - sin(x)/3 C: cos(2*x)/3 - cos(x) - sin(x)/3 D: sin(2*x)/3 - sin(x) - sin(x)/3