Trying to Find Coords of a Local Maximum and Minimum

[cmkluza]

Homework Statement

(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 $y = tan(2x) + cot(2x), 0≤x≤\frac{π}{2}$

Homework Equations

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

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!

 

[theodoros.mihos]

See http://cims.nyu.edu/~kiryl/Precalculus/Section_5.4-More%20Trigonometric%20Graphs/More%20Trigonometric%20Graphs.pdf

 

[Mark44]

Homework Statement

(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 $y = tan(2x) + cot(2x), 0≤x≤\frac{π}{2}$

Homework Equations

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

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.

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?

 

[cmkluza]

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?

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?

 

[Mark44]

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.

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.

Would it also be equivalent to $2csc(4x)$?

You're on the right track, but what is the whole equation?

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?

 

[cmkluza]

You're on the right track, but what is the whole equation?

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!

 

[Mark44]

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 $y = 2csc(4x)$

Yes.

, 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.

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.

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!

 

[cmkluza]

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.

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