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Trying to Find Coords of a Local Maximum and Minimum

  • Thread starter cmkluza
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  • #1
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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!
 

Answers and Replies

  • #3
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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?
 
  • #4
112
1
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?
 
  • #5
33,631
5,288
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.
cmkluza said:
Would it also be equivalent to ##2csc(4x)##?
You're on the right track, but what is the whole equation?
cmkluza said:
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?
 
  • #6
112
1
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!
 
  • #7
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5,288
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.
cmkluza said:
, 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.
cmkluza said:
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!
 
Last edited:
  • #8
112
1
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.
 
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