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Y = p + q cos x

  1. Jun 3, 2007 #1
    1. The problem statement, all variables and given/known data

    I have a graph of 5 = cos 0 and 1 = cos 180. The question is find the values for integers p and q
    y = p + q cos x

    p and q are integers

    How do I go about doing this

    Thx
     
  2. jcsd
  3. Jun 3, 2007 #2

    Gib Z

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    Excuse me? I don't think anyone will understand that...5= cos 0? Not only is that not true..how you graph something like that is out of the question, similarly to the 1=cos 180. State the question more clearly please.
     
  4. Jun 3, 2007 #3
    sorry. I mean the graph cos x starts at y = 1 when x = 0 and goes into a bucket like shape. The graph I'm given is the same as y = cos x EXCEPT that it's when x = 0 y = 5 and when x = 180 y = 1... so the lowest value of y is 1 and it's highest is 5, if you see what i mean?

    Thx
     
  5. Jun 3, 2007 #4

    VietDao29

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    Ok, so when x = 0, then y = 5, it means that:
    5 = p + q cos(0o) = p + q

    When x = 180o, y = 1, that means:
    1 = p + q cos(180o) = p - q

    So you have 2 equations:
    [tex]\left\{ \begin{array}{ccc} 5 & = & p + q \\ 1 & = & p - q \end{array} \right.[/tex]
    From the system of equations above, can you solve for 2 unknowns, namely, p, and q?
    Can you go from here? :)
     
    Last edited: Jun 3, 2007
  6. Jun 3, 2007 #5
    similtaneous equations
    p = 1 + q

    so 5 = 1 + q + q
    so q = 2
    therefore p = 3

    right?
     
  7. Jun 3, 2007 #6

    VietDao29

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    Perfectly correct. Congratulations. :)
     
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