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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More Transformations of Sine and Cosine - 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'm graphing transformations of sine. Here is a problem that asks me to graph y equals 4 sine ½ x minus pi over 4. Now this is almost in the form that I like. I need to make one alteration. This equals 4 times the sine of ½ and I'm factoring the ½ out of both of these terms. So I'm left with x minus pi over 2. I have to do that because I want to be able to recognize what the horizontal shift is and when I don't have the ½ factored out, it's hard to recognize that. So the horizontal shift is going to end up being pi over 2.

Now when I graph a stretched and shifted sine or cosine graph, I usually do it in two steps. First I make a table of data for the un-shifted so without the x minus pi over 2 and then I graph it and then I shift the graph over, so I usually end up with two graphs.

Now we start off with the key points of the sine function 0,0 pi over 2, 1, pi 0, 3 pi over 2, -1, and 2 pi 0 and I'm just trying to graph the stretched or shrunk graph which is y equals 4 sine ½ x. So I'm not including the shift yet.

Now the first thing I have to figure out is what kind of transformations do these numbers give me? The 4 is a vertical stretch by a factor of 4 and that's going to affect the y values, so all these y values will be multiplied by 4. 0, 4, 0, -4 and 0. Well the ½ gives me a horizontal stretch. Remember that it's counter intuitive when you have horizontal transformations so you're going to have to multiply the values of x by 2, there's the horizontal stretch by a factor of 2, that's the reciprocal of the b value.

So multiplying these by 2, I get 0, pi, 2 pi, 3 pi and 4 pi. So let me graph this which is my un-shifted graph 0, 0, pi 4 I want to make this pi and this 4. 2 pi 0, 3 pi -4 and 4 pi 0. So this is one period of my un-shifted graph and I can extend this to the left really easily.

Well now that I've got a nice graph of y equals 4 sine ½ x, I can throw in the horizontal shift, so let me get another color, I use black for the final graph. Remember the horizontal shift is a shift to the right pi over 2 remember that the phase shift is exactly equal to the h value. This is the phase shift or the horizontal shift. So I'm going to shift this graph to the right pi over 2. So half of this distance here so this point goes to here and this point goes half the way to here so here, this point goes to here, this one to here and this guy goes here, so let me just connect those and I can extend this backwards too.

This guys goes to here, this guy goes to here, one here and here. I'm connecting those and you've got yourself a really nice graph of both y equals 4 sine ½ times the quantity x minus pi over 2 and you've also got the un-shifted graph y equals 4 times sine of ½ x.

I think whenever you have two graphs shown in your final answer, you should label them so that you tell the teacher that you know which one is which, but basically the steps are key points, make a table of the un-shifted graphs values, graph the un-shifted sine curve and then graph the shifted sine curve.

One more thing we've got to do amplitude and period. The amplitude remember is the absolute value of a, this coefficient. The absolute value of 4 is 4 and the period is 2 pi over b, b is the ½, 2 pi over ½ is 4 pi. So the amplitude of this is 4, the period is 4 pi and the phase shift is pi over 2.

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