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Trigonometri Question (HELP)

  1. Aug 31, 2005 #1
    Hi

    I have this here Trigonometri function:

    F(x) = Ax cos(x) + Bx sin(x) + C cos(x) + D sin(x)


    where F'(x) = x cos(x)

    I'm suppose to find A, B, C , D any hints or idears on how I do that?

    Sincerely
    Bob
     
  2. jcsd
  3. Aug 31, 2005 #2

    LeonhardEuler

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    Gold Member

    Take the derivative of F(x). You know that F'(x)=x cos(x). This means the coefficient of (x cos(x)) should be 1. What can you say about the coefficients of all other terms?
     
  4. Aug 31, 2005 #3
    Since its trigonometry then A,B,C,D can't higher than one?

    F'(x) = A (cos(x) - x *sin(x)) + B(x *cos(x) + sin(x)) + C ( -sin(x)) + D(cos(x) = x cos(x)

    But how I proced from here ?

    /Bob
     
  5. Aug 31, 2005 #4

    TD

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    Homework Helper

    No, these are just coefficients so they can be higher than 1.
    It's cos(a) (and sin(a))) which are bounded between -1 and 1.

    Your derivative seems correct, now just check what the coefficients have to be to get xcos(x).
     
  6. Aug 31, 2005 #5
    The best way of doing this is that to make the left side of the expression shorter ?

    /Bob

     
  7. Aug 31, 2005 #6

    TD

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    Homework Helper

    Well, we have:

    [tex]\begin{array}{l}
    F\left( x \right) = Ax\cos x + Bx\sin x + C\cos x + D\sin x \\
    F'\left( x \right) = A\left( {\cos x - x\sin x} \right) + B\left( {\sin x + x\cos x} \right) - C\sin x + D\cos x \\
    \end{array}[/tex]

    Rearranging gives:

    [tex]F'\left( x \right) = \left( {A + D} \right)\cos x + \left( {B - C} \right)\sin x - Ax\sin x + Bx\cos x[/tex]

    We want this to equal [itex]F'\left( x \right) = x\cos x[/itex]

    So:

    [tex]\left( {A + D} \right)\cos x + \left( {B - C} \right)\sin x - Ax\sin x + Bx\cos x = x\cos x[/tex]

    That should give a very simple system for the unknown coefficients A -> D
     
  8. Aug 31, 2005 #7
    Thanks then A = 0 , D = 0, B = C = 1

    Right ?

    /Bob
     
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