Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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The Inverse Tangent Function - Problem 3

Norm Prokup
Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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I want to graph a transformation of the inverse of the tangent function. Let’s graph y equals negative inverse tangent x plus pi over 2. Now whenever you’re doing a transformation graph, you want to have some key points that you can use in your transformation and if you’re not 100% sure of what points to use, remember that the inverse tangent function is the inverse of the restricted tangent function.

Let’s take a look at some points here. These three points of the tangent function are the ones we would use if we were graphing a transformation of tangent. They are negative pi over 4, -1, here 0,0 here and here we have pi over 4,1. For inverse tangent I would use the interchange of these. You switch the x and the y coordinates, so I’d get negative 1, negative pi over 4, again 0, 0 and 1, pi over 4.

The other thing I have to keep track of because I’m basically transforming this blue graph are its two horizontal asymptotes. It’s the asymptote at y is equals pi over 2 and y equals negative pi over 2. I can apply the transformations to those, those asymptotes as well. So we’ve got negative inverse tangent x plus pi over 2.

Now what kinds of transformations do these numbers represent? The -1 represents a reflection across the x axis. And to get that I take these y values and I multiply them by -1 and the plus pi over 2 represents a shift upwards by pi over 2 units and I can get that by adding pi over 2 to the result. I take each of these numbers, multiply by -1 and then add pi over 2. Let’s do that. So negative pi over 4 times negative one is pi over 4 plus pi over 2 is 3 pi over 4. Times -1 plus pi over 2 is pi over 2. Times -1 plus pi over 2 is pi over 4. And when you look at this function, there’s no actual horizontal transformation going on, so the x values will be unaffected.

So I just translate these over, -1, zero and 1. This gives me three points that I can plot for my new transformed function. Negative one, 3/4pi, zero pi over 2 and 1 pi over 4. Let’s plot those.

Negative 1,3 pi over 4 would go about here, 0,pi over 2 and 1 pi over 4. Now what happens to the asymptotes? Well I can apply all the transformations I applied to these y values to the asymptotes as well because they’re y values as well. So I multiply these by -1 which basically reverses them and then I add pi over 2. So this first one becomes times -1 gives me negative pi over 2 plus pi over 2 gives me y equals zero. Times -1 gives me pi over 2 plus pi over 2 gives me y equals pi. These are my two new asymptotes. Let me plot those. Y equals pi and y equals 0. And now I’m ready to finish the graph. Don’t forget to show your asymptotic behavior.

Now this is the graph of y equals negative inverse tangent x plus pi over 2 and the reason this graph is so interesting is that it's actually identical to inverse cotangent. Which suggests the identity; inverse cotangent x equals negative inverse tangent x plus pi over 2. This identity is actually related to the co-function identity for the tangent and cotangent.

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