##### 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 1

###### 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 some of the reciprocal trigonometric functions. I'm going to start with the secant function. Recall that secant theta is 1 over cosine theta. Because of that I'm going to make a table of cosine values from negative pi over 3 to 2 pi just so that I can make sure I get one full period of secant, then I'll graph that period of secant and extend it.

Now you may not have all the values of cosine memorized, but even if you don't, there are some tricks you can use. To start with just a few values and fill out your table. I've got 0, 1 pi over 3, Â½, pi over 2, 0 and you should know these values, try to memorize them. Now if you remember that cosine is an even function, that means that when you put in opposite inputs, you get the same output.

I can plug in minus pi over 3 and know that I'm going to get the same output that I got for pi over 3 Â½. You also have this identity cosine of theta plus pi equals negative cosine theta, this kind of identity is actually true for all 6 of the trig functions, you can add pi to the input and you just get the opposite output. Let me use that to extend this table to the right.

If I add pi to this input, I get 2 pi over 3 and all I have to do is take the opposite output -Â½. Add pi opposite output, add pi, opposite output and keep going like that. 3 pi over 2, 0, 5 pi over 3, Â½, 2 pi, 1 so that could guarantee to be a full period of secant when I take the reciprocal and that's what I have to do next because remember secant is 1 over cosine so let me start with minus pi over 3, 0 pi over 3 and so on and I'll fill them in as I go.

Just going back to the table here I started at Â½, the reciprocal of that is 2, reciprocal is 1, reciprocal is 2 2, 1, 2 pi over 2. Pi over 2 the cosine was 0, so the reciprocal of that is going to be undefined I'm going to just write u because I don't want to take too much space here. 2 pi over 3, -2, 3 pi over 2, -1 4 pi over 3, -2 and I'll fill in the rest of the points in a minute let me just get started graphing this thing.

Pi over 2 on my grid is right here. Now because the function is undefined we're going to have a vertical asymptote there let me draw that in really quickly. I'll draw my asymptotes in blue. Now because the cosine function has period 2 pi, the secant function will inherit that period, it will also have period 2 pi, so you'll see a copy of this every 2 pi.

Now that means you're going to have another asymptote here at -3 pi over 2 because that's 2 pi to the left, so let me draw that in and I'm going to fill in some points. 2, 1, 2, 0, 1 is here if this is pi over 2, then pi over 3 is about here and I get a value of 2, negative pi over 3 is here, I get a value of 2 and I'm going to have a graph that looks like this.

Now let me remind you cosine is an even function and secant is going to inherit that property, it's also going to be an even function so in addition to the two asymptotes I have shown, there's going to be a mere image of this asymptote here. Let me draw that in and again using periodicity there will be another asymptote 2 pi to the right which is at 3 pi over 2 here and so you can see that the secant function just like the tangent function has asymptotes every pi units.

So let's plot some points in here we almost have our full period. We have 2 pi over 3, 3 pi over 2 and I'm sorry this is wrong, this should be pi I'm sorry about that 3 pi over 2 s the asymptote we just plotted so we put pi there. 2 pi over 3 is -2, -1, 2. So at pi we have -1, at 2 pi over 3 which is about here we have -2 and at 4 pi over 3 which is about here, we have -2 and so we get a copy of this but upside down. It's exactly the same shape.

This is actually a full period of the secant function so if you want to continue this you can just copy for example this little piece over here if you wanted to go all the way to 2 pi so this point goes here, this point goes here and you have maybe just half of that u-shape and then let's copy this piece over here, you get another downward u. Maybe I'll do on more half over here, the right half goes here right this is 2 pi to the left and then there you go, I've got exactly two periods here, that counts as a period, that counts as a period.

So this is the function of y equals secant theta and if you recall cosine looks like this, so it's really easy to remember where the turning points of the secant function are, they're precisely at the turning point of cosine. The turning points happen at the integer multiples of pi 0, pi, 2 pi and so forth. That's the secant function.

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