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# Transforming Secant and Cosecant - 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.

To graph **secant and cosecant**, find values of the reciprocal functions and plot them on the coordinate plane. Unlike the graphs of sine and cosine, secant and cosecant have vertical asymptotes whenever the cosine and sine equal zero, respectively. Graphing transformations is made easier by substituting theta for the quantity in parenthesis and solving for x. Also, notice that neither graph has x-intercepts.

I want to graph transformations of the secant and cosecant functions but first let's take another look at the graph of y equals secant theta. Let's recall that secant is 1 over cosine theta and so it inherits a lot of properties from cosine theta. For example cosine theta is even so is secant and it's easy to show that. Secant of negative theta is 1 over cosine of negative theta and that's of course cosine theta. So 1 over cosine theta is secant theta, so that proves that secant is even, now we'll use that right away.

Let's plot some points for secant theta, now first let's start with 0 secant theta of 0. What's cosine of 0 it's 1 right? So secant of 0 will be 1 over 1 which is 1. Let's try pi over 3 I like pi over 3 because cosine of pi over 3 is a half. So secant will be 2, and then something interesting happens at pi over 2 cosine is 0 so secant is going to be undefined. Now let's keep in mind that secant is an even function, so secant of negative pi over 3 is going to be the same as pi over 3 will be 2. And at negative pi over 2 it'll be undefined just as it is here. And so this gives us enough to plot half a period of the secant function, let's do that right now. We have vertical asymptotes at negative pi over 2 and pi over 2, and then we've got 3 points to use 0, 1 pi over 3, 2 and negative pi over 3, 2. So let's plot this, pi over 3 is two thirds the way from 0 at pi over 2 alright that's pi over 6, pi over 3 and then negative pi over 6, negative pi over 3.

And so half a period of secant looks like this, it's kind of a u-shaped graph with asymptotes on either side. Now what happens elsewhere, let's make the observation that secant inherits another property from cosine, the add pi property. What's secant of theta plus pi? It's 1 over cosine of theta plus pi, and when we add pi to the argument of cosine we get negative cosine. So this is 1 over minus cosine theta and that means this will be minus secant theta. What this tells us is if I add pi to an angle I take the opposite value, so for example, let me start here if I add pi to negative pi over 3 I get 2 pi over 3. I take the opposite value negative 2, this makes it really easy to extend this table downward. So if I add pi to 0 I get pi I take the opposite value negative 1 add pi to this 4 pi over 3 take the opposite value negative 2 I add pi to this and I get 3 pi over 2 still going to be undefined.

And so I have another half period, negative 2, negative 1, negative 2 at 2 pi over 3 pi and 4 pi over 3. Add another vertical asymptote at 3 pi over 2. Let me plot that first so this is the second half period notice all the values are negative. We have negative 1 pi at 2 pi over 3 which is a third the way from pi over 2 to pi we've got negative 2. And then here we have negative 2 again, and so we've got this side down in u-shape. How are these 2 shapes related? They are exactly the same shape, take this flip it across the x axis and shift it pi units to the right and you'll get this piece here. And once you've got these 2 pieces you've got a complete period of the secant function and you can get more periods by taking this graph and shifting it to the right or left.

And so for example if you wanted to extend this to the right, we take this piece right, this value at 2 it's negative pi over 3, 2 we'd have a value that corresponds here. This value at 1, 0 1 we'd have a value that corresponds right here and then another pi over 3 to the right we'd be back up at 2. So we'd have another upward u-shape here and we could extend backwards too. We'd have another negative 1 here, another negative 2 here so you can extend it as far as you need to. But just remember that first half period to get from that first one to the second one you flip across the x axis and shift to the right pi.

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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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