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In need of serious help!

  1. Jan 18, 2007 #1
    For what values of y are the maxima of the functions cos(x-y) and -cos(x) located at the same x values?
    (n/2), where n is an integer
    (1+n/2), where n is an integer
    (2n+1/2), where n is an integer
    n , where n is an integer
    (2n+3/2), where n is an integer
    2n, where n is an integer
    (2n+1), where n is an integer
    (n+1/2), where n is an integer

    I'm no physicist, but in order to graduate I need to take physics. I skipped Trigonometry and went straight to calculus, learning whatever Trig I needed to pass the class. I feel I made a horrible mistake because I can't do this problem or any problems like it. If anyone is willing to help me I would be much appreciative and even offer a cash bonus to anyone that is willing to do so. I have several problems similar to the one above. Thanks.
     
  2. jcsd
  3. Jan 18, 2007 #2

    Hootenanny

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    Note that;

    [tex]\cos(A-B) = \cos(A)\cos(B)+\sin(A)\sin(B)[/tex]

    P.S. This is probably better placed in PreCalc math
     
  4. Jan 18, 2007 #3
    I know that, but how does that relate to this problem?
     
  5. Jan 18, 2007 #4

    Hootenanny

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    Well, where are the maximas of -cos(x) located?
     
  6. Jan 18, 2007 #5
    I have no idea! 1?
     
  7. Jan 18, 2007 #6

    Hootenanny

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    Okay, what are the x values for when -cos(x)=1 (I'll give you a clue, there's only one)
     
  8. Jan 18, 2007 #7
    2pie or zero?

    Are you saying, where does the -cos of x = 1? Like on a unit circle?
     
  9. Jan 18, 2007 #8

    Hootenanny

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    Close, that would be the cases where cos(x)=1
    Yes, or in other words, where does cos(x)=-1
     
  10. Jan 18, 2007 #9
    In that case, it would be Pie
     
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