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# Transforming the Graphs of Sine and Cosine - 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 transformations of sine and cosine right now I want to focus on transformations of cosine of this type y equals a times cosine bx. But first I need to recall the graph of cosine. Y equals cosine theta.

The graph looks like, this cosine has a period of 2 pi, I’ve got two periods down here. Whenever I’m graphing the transformations of cosine I like to use the key points, the points where the values is (1,0) or -1. So I’m going to write the coordinates of these points down, make a little table. So theta cosine theta, this point is 0,1 this point is pi over 2,0, pi -1, 3 pi over 2, 0 and finally 2 pi 1. I’ll use this anytime I transform the graph of a cosine function. Let’s take a look at our problem.

The problem says graph y equals -½ cosine 2x. Identify the amplitude and period. Let’s do the amplitude and period at the end. First if I want to graph y equals -½ cosine 2x. First thing I have to do Is identify what kinds of transformations these numbers are going to give me. The -½ out in front is going to give me a vertical compression if I factor 1/2. The negative means I’m going to flip my graph across the x axis so there is a reflection. Now the way you get that reflection and the compression is you just take the y values here and you multiply them by -½. So it would be really easy. I’ll just multiply this by -½ and get -½, 0, ½,0 and -½.

This coefficient gives me a horizontal compression. Remember it’s kind of intuitive when you have horizontal transformations. The two actually means a horizontal compression, if you want a horizontal stretch you need to have a coefficient smaller than one. So this is going to be a horizontal coefficient by a half that means I have to take these values and multiply them by a half. So I get 0 times a half 0, pi over 2 times a half pi over 4. Pi times a half pi over 2, 3 pi over 2 times a half, 3 pi over 4, 2 pi times a half is pi.

When you are doing this you can kind of see this represents one period of cosine you can kind of see what the new period is it's pi. But II'll refer to the formula later on. Let’s plot these points this will give me one period of my new graph. So I’ve got 0,-½ so I’ll make this -½. Let’s make this pi over 2 and this pi. I’ve got pi over 4, 0 pi over 4 is half way to pi over 2 so this point, pi over 2, ½ and this is ½ over here. Pi over 2 ½ gives me this point, 3 pi over 4 is 0 and pi -½.

So I draw my cosine curve, I get something like this. Now if your teacher wants you to graph 2 or even 3 periods of this function, all you got to do is use this as your template. This is one cycle of your graph. So just extend it to the left, So a point here and point here back down and at the beginning again. So this will work exactly like my first period. Now let’s just connect the points, we get a nice two periods of y equals -½ cosine 2x.

Amplitude, remember the formula equals absolute value of a where a is the coefficient in front of the cosine or sine and the absolute value of -½ is ½. The amplitude tells you ½ the distance between the maximum value and the minimum value and the period comes from then formula 2 pi over b, where b is the coefficient in front of the x here, so 2 pi over 2. This is just pi as we solved before and that’s it.

Graphing sine or cosine make sure you use your key points, do the transformation on a table first then plot your data and you’ll get one really nice period of sine or cosine and you can extend that backwards or forwards amplitude and period just get from the formula.

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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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