##### 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
- Attend and watch FREE live webinar on useful topics

# Transforming Secant and Cosecant - 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.

Let’s graph a transformation of the secant function; y equals -0.5 secant. X over 3 plus pi over 6. Let’s start by recalling the key points of secant. Secant has two vertical asymptotes at negative pi over 2 and pi over 2.

So at these points, secant is undefined. And then half way between a 0, secant is 1. A third away from negative pi over 2 to 0 negative pi over 3, secant is 2. And it’s the same at pi over 3 because secant is an even function.

So what’s our transformed graph going to look like? X over 3 plus pi over 6. Well let’s make the substitution u equals x over 3 plus pi over 6. If we do that, we can solve this for x. I subtract pi over 6 and then multiply everything by 3. I get 3u minus pi over 2 equals x. That means to get my x values all I have to do, is multiply my u values by 3 and subtract pi over 2. So multiply this by 3 and I get -3pi over 2, minus pi over 2 is -4 pi over 2, -2pi. Negative pi over 3 times 3 is negative pi, minus pi over 2 is -3pi over 2.

0 times 3 minus pi over 2, negative pi over 2. Pi over 3 times 3, pi, minus pi over 2, pi over 2. And then pi over 2 times 3 is 3pi over 2, minus pi over 2 is pi.

Now how do I get the y values? Well this is just secant u. It’s exactly the same as this. Only I have to multiply these values by -0.5. So undefined times -0.5, is still undefined here, and here and then 2 times -0.5 is -1. -1 and then 1 times -0.5 is -0.5. So this means I’m going to have vertical asymptotes so s equals -2pi and an x equals pi.

Let me plot those first. So here’s x equals -2pi, here’s x equals pi. And once you have 2 asymptotes graphed, you know that the asymptotes are going to be separated by 3pi, just like these are. And so if I go 3pi to the right I’ll get another asymptote, here at 4pi. If I go 3pi to the left, I’ll get another so I have 1 -5pi. And so on. Let me plot some points.

So we have these 3 points in the middle negative pi over 2, -0.5. Negative pi over 2 is right here between these two vertical asymptotes and I get -0.5 that’s right here. Then add -3pi over 2, I get -1. -3pi over 2 is 1 third the way from -2pi, to here at pi over 2 right here. I get -1 and then at pi over 2, I get -1. This is pi so that’s pi over 2 -1 again.

You will notice my first half period is upside down and that’s because of the vertical reflection. This -0.5 gives me a reflection across the x axis. Now once I have a half period graphed, I can get another half period by taking this flipping it across the x axis and shifting right half period.

Now in this case, it looks like the period is 6pi, so half a period is 3pi. So I take each of these points flip and shift 3pi. So -1/2 becomes positive a half then I shift 3pi to the right. Right there, this point -1 because +1. Shift to the right and I get this point here. And this point flipped up and shifted right here and so I get my second half period.

And once you have two half periods, you can extend this in both directions really easily using periodicity. So this u shape, if I shift it 2 up here to the left I get a point here, a point here another half period of my secant graph. That I can do it over here too. So I get this point shifted 3pi 6 pi to the right becomes this, this one becomes, this and I’ll just do half of the u shape. That looks like 1 and 3/4 periods of my secant graph. So just remember when you got a secant graph, drop graph 1 half period then flip it, shift it to the right, get another half period then use periodicity to fill out your graph to the left and right.

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!”

##### Concept (1)

##### 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