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# The Sine Addition Formulas - 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.

We're talking about the sine addition formulas, here is a particularly challenging example. It says if cosine of alpha equals 7 over 25 and alpha is between 3 pi over 2 and 2 pi and sine of beta is 3/5 with beta between pi over 5 and pi, find the sine of alpha minus beta and the sine of alpha plus beta.

In order to solve this problem, I'm going to need to use the unit circle because I need more than just the cosine of alpha and the sine of beta.

So let's start with alpha. Angle alpha is between 3 pi over 2 and 2 pi, so that puts alpha in the fourth quadrant 3 pi over 2, 2 pi right in here and the cosine of alpha is 7 over 25, so that means that the first coordinate of my point on the unit circle is 7 over 25 so I'll put that right here and I don't know the y coordinate but I'll find it the second, so this is alpha.

Now remember this is the unit circle and the unit circle has the equation x² plus y² equals 1 so every point on the unit circle satisfies this equation including this one and that means that 7 over 25² plus y² this y² is 1. Let's square this we get 49 over 625 plus y² equals 625 over 625 and we subtract y² equals well something over 625. 625 minus 49, 625 minus 50 would be 575 so minus 49 would be 576 and it turns out that these are both perfect squares, 576 is 24² and 625 is 25², so this is plus or minus 24 over 25. So we have to decide whether it's plus or minus, but the fourth quadrant the y coordinate is going to have to be negative so it's got to be negative, -24 over 25. All right that takes care of alpha.

Now we know the cosine of alpha and the sine of alpha what about beta? Well beta is between pi over 2 and pi and it's sine is 3/5 so between pi over 2 and pi, put beta in this quadrant second quadrant and if sine is 3/5 that means the y value of the point on this circle is going to be 3/5 well it's about here so something 3/5.

So this could be my angle beta here. To find the x coordinate I did the same trick I did here I used the equation x² plus y² equals 1, so I get x² plus 3/5 squared equals 1 and that means x² plus 9 over 25 is equals 25 over 25 so I subtract x² equals 16 over 25 and then extract the roots to get plus or minus 4/5.

Now the x coordinate is going to have to be negative because we are in the second quadrant, so the answer is -4/5, 3/5 and that gives me the cosine and the sine of my angle beta and now this is all I need to find the sine of alpha minus beta and the sine of alpha plus beta.

Now remember the formula for the sine and the difference, it's sine, cosine, cosine, sine alpha beta, alpha beta, alpha beta and then when you have a minus sign with a sine formula it stays the same, so I just fill in values. The sine of alpha -24 over 25 and the cosine 7 over 25, so -24 and 7 over 25 and what about beta? Cosine of beta is -4/5, sine of beta is 3/5, -4/5, 3/5.

So we've got -24 over 25 times -4 over 5, the two negatives will cancel, our denominator is going to be 125 25 times 5 and the numerator 24 times 4, 24 times 2 is 48 times another 2 is 96 and then minus denominator is still 125, I get 21 on top and so it's mainly 96 minus 21, that's 75. So 75 over 125 which reduces really nicely to 3/5 and that's my answer for the sine of alpha minus beta.

Now let's find the sine of alpha plus beta. We still have a formula sine, cosine, cosine, sine, but with the sine of a sum formula there's going to be a plus in between, it's alpha, beta, alpha beta. Once again sine and cosine of alpha right here cosine is 7 over 25, sine is -24 over 25 so -24 over 25 7 over 25 and then beta cosine is -4/5, sine is 3/5 -4/5, 3/5 we get 24 times 4 again 96 over 125 plus 21, 7 times 3 over 125. 96 plus 21 is 117 over 125 and that is the sine of alpha plus beta.

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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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