###### 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.

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# Graphs of the Sine and Cosine Functions - Problem 1

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.

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I want to graph the sine function. Now I’ve drawn the unit circle to help me out and remember with unit circles definitions of sine and cosine the first coordinate gives me the cosine value, the second coordinate gives me the sine value. So as I go around when theta is 0, I have a sine value of 0. When theta is pi over 2, I have a sine value of 1. When it's pi a sine value of 0. And when it's 3 pi over 2 a sine value of -1. These are the quadrilateral angles and I want to fill in those values first.

Sine of 0 is 0, sine of pi over 2 is 1, sine of pi is 0, 3 pi over 2 -1 and 2 pi brings me all the way back to where I started and so the sine is 0. And let me also plot these points as I go, so I’m going to call this -1, this would be pi over 2, this would be pi, 3 pi over 2. Alright they started (0,0) right this point up here, pi over 2, 1 pi 0, 3 pi over 2 -1 and 2 pi 0.

Now if you’ve never seen a graph the sine function before you can want to plot more points than this. I’m not going to plot all the points I know I’m just going to plot the multiples of pi over 6 and that’s what I’ve shown here on the unit circle, so this for example is pi over 6.

Now the sine of pi over 6 is a half, it comes from the coordinates of this point root 3 over 2, ½. So the sine of pi over 6 is ½, I’m going to write that in. Now where else is the sine going to be y is equals to ½ remember this is the y coordinate. If I flip over to 5 pi over 6 I get a point that is the reflection of this point around the y axis, so that also has a y coordinate of ½. It's got an x coordinate of negative root 3 over 2 that doesn’t matter right now.

So again at 5 pi over 6 I get ½, at 7 pi over 6 which brings me down to this point it’s a reflection across the x axis, so this point it’s going to have a y coordinate of -½ I will right the x coordinate I don’t really need it. Let’s plot that down 7 pi over 6 is pi over 6 more than pi -½ and then at 11 pi over 6 I’ve got a reflection of this point around the y axis it will also have a y coordinate of -½. 11 pi over 6 is pi over 6 short of 2 pi, -½.

So let me plot the ½’s and the -½ , the ½’s happen at pi over 6 which is third the way from 0 to pi over 2 here. 5 pi over 6 which is 2/3 the way from pi over 2 to pi, now we have the -½'s they happen at 7 pi over 6 and the 11 pi over 6. And you can already draw a pretty good graph, but let me just go over for the sake of review go over the values at pi over 3, 2 pi over 3, 4 pi over 3 and 5 pi over 3 the multiples of pi over 3.

Pi over 3 terminates at this point and this point is symmetric with this point around the line y equals x. So its coordinates are the interchange of these two, we get ½ 3 over 2. So the sine of pi over 3 is root 3 over 2. Now 2 pi over 3 is the reflection, the point that we get of 2 pi over 3 is the reflection of this point around the y axis, so it will have the opposite first coordinate in the same second coordinate.

Root 3 over 2, so the sine of 2 pi over 3 is also root 3 over 2. And then 4 pi over 3 is down here it’s the reflection of this point across the x axis. This point will have the opposite y coordinate and so will this point 5 pi over 3.

So let me put these guys on my table, negative root 3 over 2, negative root 3 over 2. Negative root 3 over 2 oh sorry, root 3 over 2 point (8,7) point 7 is around 7/8 so when I’m plotting at half way between a half and 1 is these quarters, half way between three quarters and 1 is 7/8 so this is about where root 3 over 2 is. So pi over 3, I have root 3 over 2 and I also have it at root 2 pi over 3 right about here. I have negative root 3 over 2 here and here, so I’ll just estimate it and I’ve got a really nice set of points to draw 1 wave of the sine graph.

There you go, that’s y equals sine of theta. Now in the future when you are drawing the sine graph you may need to use the quadrant values, 0 pi over 2 pi 3 pi over 2 and 2 pi. But having graphed it this way you see what the shape is, you see that it's round at the top, you see that it crosses through the x axis at about a 45 degree angle and crosses down here about a -45 degree angle.

That’s the graph of the sine function.