##### 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
- FREE study tips and eBooks on various topics

# More Transformations of Sine and Cosine - Problem 3

###### 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 the sine and cosine functions. I've got a problem here that asks me to graph y equals 10 minus 8 times the sine of pi over 6 times the quantity x minus 3.

This has every kind of transformation we have. It's got vertical stretching and shrinking, horizontal stretching and shrinking, reflection, translation up, translation sideways. So the first thing I want to do is take the translations out of the picture and just graph the untranslated graph y equals -8 sine pi over 6 times x. And the way I do that is I start with my key points for the sine curve 0,0, pi over 2, 1, pi 0, 3 pi over 2, -1 and 2 pi 0.

And then I want to figure out what kinds of stretches or shrinks these numbers represent. This -8 is a vertical stretch of the sine curve by a factor of 8 and it's a reflection across the x axis. And the way I figure out where the points are going to go is I multiply these guys by -8 because this affects the y values, the values of sine, so multiplying these by -8 I get 0, -8, 0, 8, 0 and this is -8 sine pi over 6x.

And then I look at this guy, this is my b value and determines the horizontal stretching or compression with pi over 6 that's about one-half so this is a horizontal stretch of some kind. If you want to know the stretch factor you take the reciprocal, 6 over pi and you multiply that by the x values. 6 over pi times 0 is 0, 6 over pi times pi over 2 is going to give me 3, times pi will give me 6, 6 over pi times 3 pi over 2 gives me 9 and 6 over pi times 2 pi gives me 12.

So I'm going to plot these points and graph this function now this is the un-shifted function, I'll graph both functions and I'll label them so you can see which is which, but first I want to plot these points and I need to label my axis here. So since I have what looks like a period of 12 and amplitude of about 8, I'm going to make this 8, this -8 and I will make this 12 and that 6.

So I start with 0,0 I'll put that point here and then 3,-8 down here 6,0 here 9,8 here and finally 12,0 it's this point here and so I connect these and I get my sine curve. It's a reflected sine curve so it should look upside down and I'm going to extend this one period to the right. Teachers often ask you to graph two periods of a sine or cosine curve, so it's just extended one period to the right. And all I'm doing is taking these five points and shifting the whole thing one period to the right.

So I've got two periods of my un-shifted curve, now what I need to do is figure out what kinds of shifts I'm dealing with here. It looks like I've got some kind of horizontal shift and a vertical shift. Now x minus 3 that's a horizontal shift to the right of three units. This number gives you the horizontal shift which is also called the phase shift. This is your x minus h and h is the phase shift which is 3. So I'm going to shift to the right 3.

Now this 10 is being added to my y values, so that's a shift upward by 10 so each of these points needs to be shifted to the right 3 and up 10 and I need to figure out what 10 is on this axis. If this is 16, this is 4, that's 2 that distance is 10 right there. So 0,0 is going to go to the right 3 and up 10. This point is going to go to the right 3 and up 10, so up to 2. This point will go to the right 3 and up 10. This point goes to the right 3 and up 10, and up 10 will be up to 18. This point goes to the right 3 and up 10. This point goes to the right 3 and up 10 up to 2, -8 plus 10 is 2, to the right 3 and up 10, to the right 3 and up 10 and this is at 8, so up 10 means 18 and finally to the right 3 and up 10 and this is my final curve, down, up. Okay and this final curve is y equals 10 minus 8times the sine of pi over 6 times the quantity x minus 3.

The other curve y equals -8 sine pi over 6 times x. Okay well we figured out what the phase shift is, what's the amplitude and period? The amplitude comes from this term, remember the amplitude is the absolute value of a, the coefficient in front of the sine or cosine. The coefficient in front of the sine is -8, so the absolute value of that is 8 and then period, that comes from the formula 2 pi over b where b is this coefficient the pi over 6, that's going to be 2 pi over pi over 6, which is the same as 2 pi times 6 over pi. I cancel the pi's and I'm left with 12 and that verifies what I saw in my table of data that this thing is a period of 12 and they both do.

So this was a particularly difficult example, we had every kind of transformation, we had vertical stretching, reflection, horizontal stretching, a shift to the right and a shift up, but remember you can always do this in two steps. You can graph the un-shifted function first, graph that and then shift it in your final step.

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!”

##### 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