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Finding the maximum and minimum points of a cosine function

  1. Jan 12, 2008 #1
    1. The problem statement, all variables and given/known data

    This is part of a larger engineering problem, I have reduced it to this mathematical equation, which should be simple, right?

    I need to find when the following EQ has Maximum's (in terms of x):

    [tex]\left|4\,cos\left(\frac{2\pi}{5}\,-\,30\,X\right)\,-\,4\,cos\left(\frac{2\pi}{5}\,+\,30\,X\right)\right|[/tex]



    2. Relevant equations





    3. The attempt at a solution

    I guess take a derivative to find when the maximum and minimums are...

    [tex]120\,sin\left(\frac{2\pi}{5}\,-\,30\,X\right)\,+\,120\,sin\left(\frac{2\pi}{5}\,+\,30\,X\right)[/tex]

    set that equal to zero and solve for X, so I get the following equation...

    [tex]-sin\left(\frac{2\pi}{5}\,-\,30\,X\right)\,=\,sin\left(\frac{2\pi}{5}\,+\,30\,X\right)[/tex]

    I can't solve that!!!

    I put it into maple, this is what it gave...

    [tex]-\frac{\pi}{300}\,-\frac{1}{30}\,arctan\left[4\,cos\left(\frac{3\pi}{10}\right)\,sin\left(\frac{3\pi}{10}\right)\,+\,2\,cos\left(\frac{3\pi}{10}\right)\right]\,\,=\,\,-0.0524[/tex]


    The answer is given in the text, but I can't seem to arrive at the same conclusion!

    Supposedly, the answer (for the maxima) is...

    [tex]X\,=\,\frac{\pi}{60}\,+\,\frac{2\,n\,\pi}{30}[/tex]

    and for the minima...

    [tex]X\,=\,\frac{n\pi}{30}[/tex]

    but when I solve the original EQ (before taking the derivative) by graphing to find the max's - it's 1.1 - 3.3 - 5.5 - etc. which is a factor of 5 off from the book answer just above! not sure how to proceed from here!
     
    Last edited: Jan 12, 2008
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  3. Jan 12, 2008 #2

    dynamicsolo

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    Question: are there supposed to be absolute value signs on this function?

    I'd make the suggestion of factoring the "4" out and then applying the sum-to-product trig identity

    cos A - cos B = -2 · sin[ (A+B)/2 ] · sin[ (A-B) / 2 ] .

    I believe this will leave you with a product where one of the terms now becomes a constant and you just have one trig function to deal with. That may make it easier to look for the extrema...
     
  4. Jan 12, 2008 #3
    Yes, the absolute value is part of the function. I'm not too worried about that though.

    I converted the function to the simpler one:

    [tex]\left|-8\,sin\left(\frac{2\pi}{5}\right)\,sin\left(-30\,X\right)\right|[/tex]

    But how do I solve for extrema now?
     
  5. Jan 12, 2008 #4

    dynamicsolo

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    Since [tex]\,sin\left(\frac{2\pi}{5}\right)[/tex] is just a number, you can turn this into

    [tex]8\,sin\left(\frac{2\pi}{5}\right)\left|\,sin\left(-30\,X\right)\right|[/tex].

    The function inside the bars is equal to

    [tex]-\,sin\left(30\,X\right)[/tex] ,

    since sine is an odd function. The absolute value will eliminate the minus sign, leaving you with

    [tex]8\,sin\left(\frac{2\pi}{5}\right)\left|\,sin\left(30\,X\right)\right|[/tex] ,

    which is the absolute value of a sine function with period 2(pi)/30 . All the "valleys" going down to -1 now become "peaks", doubling the frequency of the maxima.

    Now, where does sin(30x) have maxima and minima?
     
  6. Jan 12, 2008 #5
    I've no idea how to get an expression for the maxima and minima of an EQ. I forgot!

    We know the minima will be zero...

    [tex]0\,=\,sin(30\,X)[/tex]

    take inverse sine? So...

    [tex]X\,=\,\frac{sin^{-1}(0)}{30}[/tex]
     
  7. Jan 12, 2008 #6

    dynamicsolo

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    Ah, but we won't be using the function arcsin(x); we just want to know where sin(30x) is either zero or has its maximum value of 1. There will be an infinite number of solutions.

    There is a minimum at x = 0, as you say. The next one (to the right) will be half a period away. Since the period is 2(pi)/30, that would put it at x = (1/2)·2(pi)/30 = (pi)/30. The rest will be successive half-periods further along.

    Where will the first maximum to the right of x = 0 be?
     
  8. Jan 12, 2008 #7
    The first maxima would be half way between the first and second minima, right?

    So the first maxima occurs at [itex]\frac{\pi}{60}[/itex]? And then at every integer multiple of the period afterwards?

    [tex]X\,=\,\frac{\pi}{60}\,+\,\frac{2\,n\,\pi}{30}[/tex]

    It's right! So how did you get that the period of the sin(30x) was [itex]\frac{2\pi}{30}[/itex]? I suppose it's a fundamental EQ where the period is just 2[itex]\pi[/itex] divided by the coefficient of the x term within the sin expression, right?


    Thanks!
     
  9. Jan 12, 2008 #8

    dynamicsolo

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    This is the one part of the given solution I was puzzled by. If you taking the absolute value, the "valleys" half a period away should become "peaks", so there would be maxima every half-period along, which would be [tex]\frac{\,n\,\pi}{30}[/tex]. That's partly why I was asking if the absolute value signs were supposed to be there; I'm wondering if the person who wrote the answer forgot about them.

    The function sin(kx) can be seen as a transformation of sin(x) which involves "horizontal compression" about the y-axis by a factor of k (for k > 1). Since sin(x) has a period of 2(pi) , the function sin(kx) will have period 2(pi)/k .
     
  10. Jan 12, 2008 #9
    [itex]\frac{n\pi}{30}[/itex] is the text answer for the minima.

    The function value at the minima and maxima is 0 and ~7.6085, respectively.
     
  11. Jan 12, 2008 #10

    dynamicsolo

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    Yes, there will be two minima per period of 2(pi)/30, which correspond to the zeroes of sin(30x). For |sin(30x)|, there should also be two maxima per period, corresponding to the location of +/-1 for sin(30x), which is why I think something was overlooked in the printed answer.

    That checks: [tex]8\,sin\left(\frac{2\pi}{5}\right)[/tex] is indeed ~7.6085.
     
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