# Trying to Find Coords of a Local Maximum and Minimum

1. Jun 5, 2015

### cmkluza

1. The problem statement, all variables and given/known data
(c) Show that $tan(x) + cot(x) \equiv 2csc(2x)$
(d) Hence or otherwise, find the coordinates of the local maximum and local minimum points of the graph of $y = tan(2x) + cot(2x), 0≤x≤\frac{π}{2}$

2. Relevant equations
Most likely a lot of different trigonometric formulas to help with this, but I'm not sure which specifically.

3. The attempt at a solution
I've proven that $tan(x) + cot(x) \equiv 2csc(2x)$, but I don't know what to do next. I can see that my current equation is a compression of the previous one, but I don't know how that plays into finding the local maximum and minimum.

Any suggestions will be greatly appreciated. Thanks!

2. Jun 5, 2015

3. Jun 5, 2015

### Staff: Mentor

Since you have shown that tan(x) + cot(x) = 2csc(2x) is an identity, what do you suppose that tan(2x) + cot(2x) is equal to?

4. Jun 7, 2015

### cmkluza

Hello, sorry to reply to this so much later, but I've been busy. Off the top of my head, since there's a compression by a factor of 2x occurring in the equation, I would assume that $tan(2x) + cot(2x) = \frac{2csc(2x)}{2x}$, but that seems off to me. Would it also be equivalent to $2csc(4x)$? But I still am unsure how to answer this, unless I'm completely off in my idea of an inverse graph. If I transform this into a csc function/graph, the inverse of a sin function/graph, wouldn't there be asymptotes, making any maxima and minima between 0 and π undefined?

5. Jun 7, 2015

### Staff: Mentor

And it is off. You shouldn't be thinking about compressions or any other transformation -- only substitution.
You have tan(x) + cot(x) = 2csc(2x). If you replace x in this identity by some other expression, the identity will still be true, barring only values that make the individual terms undefined.
You're on the right track, but what is the whole equation?

6. Jun 7, 2015

### cmkluza

So the entire equation would be $tan(2x) + cot(2x) = 2csc(4x)$? I'd imagine after this, the easiest way to find local maxima and minima would be to look at the graph of $y = 2csc(4x)$, but now I'm confused as to what counts as a "local" maximum or minimum. I just read that the local minimum for $y=csc(x)$ would be the point corresponding to the local maximum of $y=sin(x)$, and vice versa for the local maximum. I can see where this is coming from, since the point for local minimum would be at the bottom of the curve it's a part of, but it would be higher than the local maximum in this case. This is kind of counter-intuitive to me, since I thought maxima and minima, local or not, were supposed to be at the top and bottom of the graph as a whole. Am I on the right track with the $y = 2csc(4x)$ part, and is there a reason behind the maxima and minima placement on a cosecant graph?

Thanks for all your help so far!

7. Jun 7, 2015

### Staff: Mentor

Yes.
Not necessarily. The graph of y = csc(x) will have a local minimum at $(\pi/2, 1)$ and will have a local maximum at $(3\pi/2, -1)$. "Local" in these terms means largest or smallest in a relatively small interval, so it is possible for a local minimum to actually be larger than a local maximum. Keep in mind that the graph of y = csc(x) is a series of disconnected curves. "Local" pertains to each of these disjoint curves.

Last edited: Jun 7, 2015
8. Jun 7, 2015

### cmkluza

Thank you very much for your continuous help with this problem! I think I have a much better grasp on this now.

Last edited by a moderator: Jun 7, 2015