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Finding the maximum and minimum points of a cosine function... 
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#1
Jan1208, 07:18 PM

P: 492

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]\left4\,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! 


#2
Jan1208, 07:39 PM

HW Helper
P: 1,662

I'd make the suggestion of factoring the "4" out and then applying the sumtoproduct trig identity cos A  cos B = 2 · sin[ (A+B)/2 ] · sin[ (AB) / 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... 


#3
Jan1208, 07:59 PM

P: 492

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]\left8\,sin\left(\frac{2\pi}{5}\right)\,sin\left(30\,X\right)\right[/tex] But how do I solve for extrema now? 


#4
Jan1208, 08:17 PM

HW Helper
P: 1,662

Finding the maximum and minimum points of a cosine function...
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? 


#5
Jan1208, 09:11 PM

P: 492

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] 


#6
Jan1208, 09:59 PM

HW Helper
P: 1,662

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 halfperiods further along. Where will the first maximum to the right of x = 0 be? 


#7
Jan1208, 10:33 PM

P: 492

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! 


#8
Jan1208, 10:52 PM

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P: 1,662




#9
Jan1208, 11:23 PM

P: 492

[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. 


#10
Jan1208, 11:33 PM

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P: 1,662




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