##### Like what you saw?

##### Create FREE Account and:

- Watch all FREE content in 21 subjects(388 videos for 23 hours)
- FREE advice on how to get better grades at school from an expert
- FREE study tips and eBooks on various topics

# Graphing the Reciprocal Trigonometric Functions - 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.

We're ready to graph the cotangent function, but first I want to show you how some identities can reveal that the cotangent functions is really just a transformation of tangent.

First of all, cotangent of theta equals tangent of pi over 2 minus theta. That's one of the co-function identities and you can actually factor a -1 out of this and get a negative theta minus pi over 2 right? That's just Algebra and then the negative angle identity allows me to pull this minus sing outside the tangents because tangents is an odd function so I get minus tangent of theta minus pi over 2.

So cotangent theta is really minus tangent theta minus 2. It's a reflection of tangent across the x axis so we're going to flip it down and it's a shift to the right so I'm going to remember that as I'm making a table of data for tangent.

Now to make a table of data, I need to refer to the graph of tangent and hopefully you know that the tangent of pi over 4 is 1, tangent of 0 is 0, tangent has an asymptote at p over 2 and negative pi over 2 and because tangent is odd, at -5 pi over 4 is equal -1, so let me fill that in on the table, alright undefined for negative pi over 2, negative pi over 2 and -1 0, 0 pi over 4, 1 and pi over 2 undefined.

Now again while reflecting across the x axis that means I have to multiply all the y values by -1 so undefined becomes undefined, these just switch signs I get 1, 0 -1 let me fill this up here this is theta and cotangent theta and then these values are going to get shifted pi over 2 to the right so I have to add pi over 2 to all these values so minus pi over 2 becomes 0, 3 pi sorry pi over 4, pi over 2, now 3 pi over 4 and pi.

This is one period of tangent, so this is going to end up being one period of cotangent, let's graph this one period. I'll first graph the asymptotes and I've got an asymptote and I've got an asymptote at 0 and at pi, so I draw x equals 0 and x equals pi and because cotangent is going to inherit the period from tangent of pi, it's going to have an asymptote every pi. All the reciprocal functions have asymptotes pi units apart do you notice this? Anyway so I fill in negative pi and with that I can fill in this one too -2 pi, so I've got five asymptotes, that's all I need. This will give me a bunch of periods.

So let me plot these points; between 0 and pi I need a quarter of pi, half and 3 quarters, here is pi over 2 where I expect the value to be 0. At pi over 4, the value is 1 and at 3 pi over 4, the value is -1 so this is what one period of cotangent looks like. Remember it's a reflection of the tangent function.

The tangent function has a tendency to increase, it's got these increasing pieces, cotangent decreases and this is a full period do I can just duplicate this here, here we go and the nice thing about the key points of cotangent as well as tangent is that they line up, so you can easily copy them and here is the third period nice shape, alright one more, and that's four periods of y equal cotangent theta.

Notice you will recall that cotangent is cosine theta over sine theta that's one of our identities. Notice that the cotangent has asymptotes where sine has zeros at the integer multiples of pi and that cotangent has zeros where cosine does at the odd multiples of pi over 2. Remember this identity if you forget where the asymptotes and the zeros go.

Please enter your name.

Are you sure you want to delete this comment?

###### 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!”

##### Sample Problems (3)

Need help with a problem?

Watch expert teachers solve similar problems.

## Comments (0)

Please Sign in or Sign up to add your comment.

## ·

Delete