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# Graphs of the Sine and Cosine 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 want to graph the cosine function. Now I start with the unit circle. The unit circle is how we define the sine and the cosine functions, so I’ve plotted out the points (1,0), (0,1), (-1,0) and (0,-1). These corresponds to angles 0 pi over 2, pi 3 pi over 2 and 2 pi, and those are the values that I’m going to calculate first.

Now remember in the unit circle definitions of sine and cosine, the first coordinate gives me the cosine of the angle and the second coordinate gives me the sine of the angle. So I’m going to be looking at the first coordinate here.

When the angle is 0 the first coordinate of the point on the terminal side is 1, so the cosine of 0 is 1. When the angle is pi over 2, the first coordinate is, 0 so cosine of pi over 2 is 0. When the angle is pi the first coordinate is -1, cosine is -1. And here we get 0, that’s 3 pi over 2 and when the angle is 2 pi, we’ve gone through a complete revolution we are back where we started the first coordinate is 1, cosine is 1 again. Cosine repeats itself every 2 pi.

Now let’s fill in the in-between values. I want to work with the multiples of pi over 6 and pi over 3. Pi over 6 is here, pi over 6 thirds is exactly a third of pi over 2. Now we’ve calculated the values of sine and cosine here before, we got root 3 over 2 and ½ the cosine of pi over 6 was root 3 over 2 and the sine was ½ so let me fill that in. Pi over 2, pi over 6 rather root 3 over 2.

Now where else will the cosine be root 3 over 2? Well it's the x values, so the only other point on the circle with the x value would be root 3 over 2 is down here. This point is a reflection across the x axis of this point and this happens to be the point on the circle corresponding to the angle 11 pi over 6. So with 11 pi over 6 is also going to give me a cosine of root 3 over 3. I don’t need to write down the sine value because I’m just graphing cosine here.

Now what about over here? How is the cosine of this angle related to the cosine of pi over 6? Well the x coordinate is going to be the opposite of the x coordinate here because this is the reflection across the y axis of this. So we’ll get minus root 2 over 3 here. What’s this angle that’s 5 pi over 6 because it's pi minus pi over 6. So from my angle 5 pi over 6, I get negative root 3 over 2. Where else will the cosine be negative root 3 over 2? Down here. We get exactly the same x coordinate down here because this point is a reflection of this point across the x axis.

So what will this angle be? It will be pi plus pi over 6 seven pi over 6, so I get negative root 3 over 2. So what remains to fill in the multiples of pi over 3 and the cosine of pi over 3 is gained by the first coordinate of this point. This point is a reflection of this point about the line y equals x. And whenever you reflect the number line y equals x, x and y values switch. So this point has coordinates of ½, root 3 over 2. So the cosine is ½ cosine pi over 3 is ½.

And again using the symmetry across the x axis, the cosine will be ½ down here ½ something. What will this angle be? Well I have to go all the way around the circle to here and incidentally every two of these little sectors measures an angle of pi over 3, so that’s pi over 3, 2 pi over 3, 3 pi over 3, 4 pi over 3, 5 pi over 3 that will get me to this point. So the angle will be 5 pi over 3 which is here. It will be ½.

Now let me use symmetry across the y axis. This point and this point are going to have x coordinates of -½. So what are the angels that correspond to those points? This point corresponds to pi over 3 plus pi over 3 will be 2 pi over 3.

So that’s up here and we will get -½. This point down here is another two wedges that’s I’m sorry it’s a 2 pi over 3 plus another pi over 3 plus another pi over 3 gives me this point 4 pi over 3. That’s plenty of points to draw a really nice graph of cosine, so let’s start with that.

So I’m going to fill this in. I’m going to start with these values these the 1’s 0’s and -1’s, they are the easiest to plot and they give you the idea of a rough shape. Now cosines starts out at 1. With an input of 0 when the angle is pi over 2 we get 0 at pi we get -1, 3 pi back to 0, 2 pi back up to 1. Now we are going to fill in the ½’s and -½’s. The ½’s and -½’s happen in multiples of pi over 3.

So when I look over here remember that pi over 2 divide by pi over 3 is pi over 6 this is pi over 6 this is pi over 3 which is 2 pi over 6. So that’s pi over 3 I get ½. Another pi over 3 I get -½ another pi over 3 gets to be pi and then 4 pi over 3, -½. 5 pi over 3 positive ½ so those are the ½’s and the -½’s. And then root 3 over 2 this is now this nicer number is ½ root 3 over 2 is approximately .87. .87 is roughly 7/8. So when I plot it this is ½, halfway between a half and one is three quarters and halfway between half and that is 7/8. So this point 7/8 good enough for our by-hand graph so that’s root 3 over 2 approximately.

And then the opposite of that down here, right there and here and then we get back up to root 3 over 2 over here. So we’ve plotted our points we have a really good idea of what the shape of this graph is so let me fill in the curve.

Draw a nice smooth curve. This is y equals cosine theta. In the future when you are graphing cosine you'll probably just use the points where the values are 1, 0 or -1 those are the easiest to plot. And now that you know the shape you don’t need all these other points. But it’s important to remember that the cosine has this nice smooth shape, it flattens at the bottom and at the top. And when it crosses the x axis it crosses at an angle of -45 degrees or 45 degrees.

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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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