##### 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

# Graphing the Reciprocal Trigonometric Functions - Problem 2

###### 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.

I'm graphing this reciprocal trigonometric functions and right now I'd like to graph y equals co-secant theta. Remember that co-secant theta is 1 over sine theta, it's the reciprocal of sine so I need to have a good table of sine values before I get started.

Now let's suppose I only remember these three sine values, maybe a couple more, but mostly the sine values in the first quadrant. I can easily extend those using two identities, sine of negative theta equals negative sine theta and sine of theta plus pi equals negative sine theta.

First I can extend into the negative direction by using this identity, this means that sine is an odd function, so if I have opposite inputs I get opposite outputs. For example if I want to plug in minus pi over 6, I should get the opposite of the output I got with pi over 6, so I should get minus Â½.

I can use this identity to extend this table forward. This identity says that if I add pi to my input, I get the opposite output. For example if I add pi to this, I get 5 pi over 6 and I should get the opposite output Â½. Add pi opposite output, add pi, add pi to get 3 pi over 2, -1, I get 11 pi over 6 and negative Â½, 2 pi, 0, so that's a pretty complete table of sine values and I can take the reciprocal of all these I've got a nice table of co-secant values.

Remember I know this is confusing, but remember the reciprocal of co-secant is sine and the reciprocal of secant is cosine, I know that that's kind of weird and confusing, but it's a mistake that allow people make reversing those. So I'm going to put these values down here. 5 pi over 6, pi I'll continue this in a second.

So I need to take the reciprocals of these, I started with negative a half, so I'm going to get -2, the reciprocal of 0 is undefined and 2. Now that undefined I'll write it u that' going to give me an asymptote I'll come back to that in a moment.

The reciprocal of 1 is 1, the reciprocal of Â½ is 2 and then sine of pi is 0, so reciprocal is undefined. Let me graph this part of the co-secant graph this part of the co-secant graph right now because I have two asymptotes, one at x equals 0 and one at x equals pi.

The x equals 0 asymptote is right here on the y axis and then x equals pi. Now one of the things that the co-secant function inherits from the sine function is the period 2 pi and that means that whatever cycle I come up with from over at intervals length 2 pi, that's going to repeat itself infinitely many times, so for example this asymptote will have a repetition here at 2 pi and this asymptote at pi will repeat down here negative pi and you can see that you're going to have asymptotes every integer multiple of pi and so let me just draw those five and I'm done with asymptotes.

I'm going to plot let's say I plot these three points right now, pi over 3 will plot first that's right here pi over 2, 1 pi over 6 the way from 0 to pi over 2, so here I'm going to get 2 and 5 pi over 6 is 2/3 the way from pi over 2 to pi, so right here, and I fill it out. And you'll notice if you've graphed secant before, this shape is very similar to one of the u shapes of secant, in fact they have identical shapes they're shifts of one another.

So let me plot let's see I'll plot this point at negative pi over 6, I get -2 where I'm going to assign and then I need to take a few more values. I've got 7 pi over 6, 3 pi over 2 and 11 pi over 6. The reciprocals are -2, -1, -2, 7 pi over 6, 3 pi over 2, 11 pi over 6, -2, -1, -2. I already have the asymptotes I'm not going to fill in the2 pi value. But at 3 pi over 2, I've got -1, so down here, 7 pi over 6 is just pi over 6 past pi, so about here I get -2 and 1 pi over 6 is pi over 6 than 2 pi so right here and that's a complete period of the co-secant function, so all I have to do is repeat this here.

So I'm going to repeat this u-shape here and that's one and a half periods and we'll repeat this u-shape over here and that's two periods. So this is two periods of y equals co-secant theta and if you recall the graph of sine theta, you can see where sine and co-secant are kind of connected to one another. This will help you draw the co-secant graph in the future. Remember that the turning points of co-secant like this point here and this point here and this point and this point, they are exactly the same as the turning points of sine.

So the odd multiples of pi over 2 and also the asymptotes for co-secant are exactly the same as the zeros of sine, that's something to remember whenever you're graphing co-secant in the future.

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!”

###### Get Peer Support on User Forum

Peer helping is a great way to learn. Join your peers to ask & answer questions and share ideas.

##### 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