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

by VinnyCee
Tags: cosine, function, maximum, minimum, points
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VinnyCee
#1
Jan12-08, 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]\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!
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dynamicsolo
#2
Jan12-08, 07:39 PM
HW Helper
P: 1,662
Quote Quote by VinnyCee View Post

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]
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...
VinnyCee
#3
Jan12-08, 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]\left|-8\,sin\left(\frac{2\pi}{5}\right)\,sin\left(-30\,X\right)\right|[/tex]

But how do I solve for extrema now?

dynamicsolo
#4
Jan12-08, 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?
VinnyCee
#5
Jan12-08, 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]
dynamicsolo
#6
Jan12-08, 09:59 PM
HW Helper
P: 1,662
Quote Quote by VinnyCee View Post
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]
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?
VinnyCee
#7
Jan12-08, 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!
dynamicsolo
#8
Jan12-08, 10:52 PM
HW Helper
P: 1,662
Quote Quote by VinnyCee View Post
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]
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.

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?
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 .
VinnyCee
#9
Jan12-08, 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.
dynamicsolo
#10
Jan12-08, 11:33 PM
HW Helper
P: 1,662
Quote Quote by VinnyCee View Post
[itex]\frac{n\pi}{30}[/itex] is the text answer for the minima.
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.

The function value at the minima and maxima is 0 and ~7.6085, respectively.
That checks: [tex]8\,sin\left(\frac{2\pi}{5}\right)[/tex] is indeed ~7.6085.


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