# Homework Help: Finding the maximum and minimum points of a cosine function

1. Jan 12, 2008

### VinnyCee

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):

$$\left|4\,cos\left(\frac{2\pi}{5}\,-\,30\,X\right)\,-\,4\,cos\left(\frac{2\pi}{5}\,+\,30\,X\right)\right|$$

2. Relevant equations

3. The attempt at a solution

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

$$120\,sin\left(\frac{2\pi}{5}\,-\,30\,X\right)\,+\,120\,sin\left(\frac{2\pi}{5}\,+\,30\,X\right)$$

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

$$-sin\left(\frac{2\pi}{5}\,-\,30\,X\right)\,=\,sin\left(\frac{2\pi}{5}\,+\,30\,X\right)$$

I can't solve that!!!

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

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

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...

$$X\,=\,\frac{\pi}{60}\,+\,\frac{2\,n\,\pi}{30}$$

and for the minima...

$$X\,=\,\frac{n\pi}{30}$$

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
2. Jan 12, 2008

### dynamicsolo

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...

3. Jan 12, 2008

### VinnyCee

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:

$$\left|-8\,sin\left(\frac{2\pi}{5}\right)\,sin\left(-30\,X\right)\right|$$

But how do I solve for extrema now?

4. Jan 12, 2008

### dynamicsolo

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

$$8\,sin\left(\frac{2\pi}{5}\right)\left|\,sin\left(-30\,X\right)\right|$$.

The function inside the bars is equal to

$$-\,sin\left(30\,X\right)$$ ,

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

$$8\,sin\left(\frac{2\pi}{5}\right)\left|\,sin\left(30\,X\right)\right|$$ ,

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. Jan 12, 2008

### VinnyCee

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...

$$0\,=\,sin(30\,X)$$

take inverse sine? So...

$$X\,=\,\frac{sin^{-1}(0)}{30}$$

6. Jan 12, 2008

### dynamicsolo

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?

7. Jan 12, 2008

### VinnyCee

The first maxima would be half way between the first and second minima, right?

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

$$X\,=\,\frac{\pi}{60}\,+\,\frac{2\,n\,\pi}{30}$$

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

Thanks!

8. Jan 12, 2008

### dynamicsolo

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 $$\frac{\,n\,\pi}{30}$$. 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 .

9. Jan 12, 2008

### VinnyCee

$\frac{n\pi}{30}$ is the text answer for the minima.

The function value at the minima and maxima is 0 and ~7.6085, respectively.

10. Jan 12, 2008

### dynamicsolo

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: $$8\,sin\left(\frac{2\pi}{5}\right)$$ is indeed ~7.6085.