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# The Inverse Tangent Function - Concept

###### 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.

Since tangent is not a one-to-one function, the domain must be limited to -pi/2 to pi/2, which is called the restricted tangent function. The graph of the **inverse tangent function** is a reflection of the restricted tangent function over y = x. Note that the vertical asymptotes become horizontal, at y = pi/2 and y = -pi/2, and the domain and ranges swap for the inverse function.

I want to introduce the inverse tangent function, we start with the tangent function y equals tan x and you'll notice that the tangent function is very much not one to one so in order to be able to invert this function we first have to restrict its domain to a portion that is one to one. So we pick the center period from negative pi over 2 to pi over 2 and that's how we define the restricted tangent function y=tan x for x between negative pi over 2 and pi over 2. And that's just this piece of it here let me just trace over, this is the restricted tangent function. Okay and of course the restricted tangent function has 5 key points if you include the 2 asymptotes that bound it you've got negative pi over 2 where tangents are defined, negative pi over 4 negative 1, 0, 0, pi over 4, 1 and again another vertical asymptote that pi over 2 which is because tangents are undefined.

Now the inverse of the restricted tangent functions called y equals inverse tangent x and it's also written y equals arc tan. And I want to graph it, in order to graph it I take the points from here and I switch the x and y coordinates. So for example for this point negative pi over 4 to negative 1 you get negative 1, negative pi over 4. So this point goes right about here, it's kind of hard to tell. This is my negative 1 and negative pi over 4 is half way between 0 and negative pi over 2, so right about there. And then 0, 0 is a point and pi over 4, 1 becomes 1 pi over 4 so 1 is about here. pi over 4 is half way between 0 and pi over 2 so there's my 1 pi over 4 then we draw a smooth curve and keeping in mind that any inverse function is going to be a reflection of the original function around the line y=x. So something like this and this.

Now I haven't yet drawn the asymptotes but what's interesting is in this graph I've basically interchanged all the x and y coordinates of this graph. So I also interchanged the x and y coordinates for the asymptotes, the asymptotes are originally x equals negative pi over 2, x equals pi over 2. The new asymptotes are y equals negative pi over 2 and y equals pi over 2. So let me draw those, here's negative pi over 2 and here's positive pi over 2. So this is sort of an interesting function because it's probably the first function we've looked at where there were actually 2 horizontal asymptotes y equals pi over 2 inverse tangent approaches that as x approaches infinity and as x approaches negative infinity the graph goes to y equals negative pi over 2. So this is the graph of y equals inverse tangent x or arc tan x.

And one more thing, the domain and range of this function you can probably tell by looking at the graph of the domain is going to be all real numbers. So inverse tangent can take all real numbers and the range you restrict it to between negative pi over 2 and pi over 2. So it's kind of like the range of the inverse sine function only the n points are not included so you can never get pi over 2 out of the inverse tangent function or negative pi over 2. So that's the inverse tangent function.

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###### Norm Prokup

PhD. in Mathematics, University of Rhode Island

B.S. in Mechanical Engineering, Cornell University

He uses really creative examples for explaining tough concepts and illustrates them perfectly on the whiteboard. It's impossible to get lost during his lessons.

Thiswas EXCELLENT! I am a math teacher and have been looking for an easy/logical way to explain the lateral area of a cone to my students and this was incredibly helpful, thank you very much!”

I just learned more In 3 minutes of polygons here than I do in 3 weeks in my math class”

Hahaha, his examples are the same problems of my math HW!”

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