Algebra (all content)
Sal evaluates the cosine of the sum of 60° and another angle whose right triangle is given. To do this, he must use the cosine angle addition formula. Created by Sal Khan.
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- At3:10, when finding the cosine of 60 degrees, and at3:54. when finding the sine of 60 degrees, why do you use the example triangle that you drew? Why not just use the triangle given with the problem?(12 votes)
- Why can you not just say that the cosine of B is 15/17, take the inverse to get the angle, add 60 and take the cosine of all of that? You get the same answer. Will this not always work?(18 votes)
- Why is this so important to learn for life?(0 votes)
- The field of robotics requires knowlage of kinkamatics in order for a programmer and/or mechanical enginear to effectively program machinery. It takes a great deal of math to tell an artificial hand exactly how to help a disabled veteran tie his shoe.
The artificial hand has to know where it's parts are in virtual space, as compared to real space. This requires a working knowlage on how to mathematically describe angles and their position in third dimentional space.(36 votes)
- At4:45how did you determine that 8/17 multiplied by square root of 3 equaled 4 square root of 3 over 17?(9 votes)
- In multiplication you can divide the factors with each other. I can't really show you since I'm limited by a keyboard, but I'll try:
8/17* sqrt 3/2 <=>
8/2*sqrt 3/17 and 8/2 is 4, which he multiplied with sqrt 3/17. He's not doing anything fancy, just writing a fraction directly in simplest form. Excuse my bad English, I'm doing my best to explain.(7 votes)
- I've always wondered, why can't we just use the distributive property and simplify cos( x + y) to cos( x) + cos(y)?(4 votes)
- Couldn't we just use the law of sines here? We have all the sides and one angle, so can't we find the other angles, add 60, and call it good that way?(6 votes)
- You can use the law of sines as well.
Here are the other ways to solve this problem I can think of:
- Using the law of cosines
- Using the inverse trig functions
One good thing of the way used in the video is that you don't have to use inverse functions, hence you don't get repeating decimal.(1 vote)
- if that's assumed to be a 30 60 90 then the cos(<abc + 60) = cos(90) then will the last simplified terms at4:50simplify to 0?(2 votes)
- Yes, that would be true if the triangle was a 30 60 90 triangle and <abc was the 30 degree angle. However, you can't assume that that's the case in this problem. The 8, 15, 17 side lengths don't align with the x, xsqrt(3), and 2x side lengths that every 30 60 90 triangle has, meaning that <abc is not 30 degrees in the problem.(5 votes)
- Isn't it easier to just take arccos(15/17)≃28.07°, and then cos(28.07°+60°)?(3 votes)
- In what video is this angle addition formula derived?(3 votes)
- Got a question: While proving the angle addition formulas, I somehow got a triangle who shares a line that goes outside, and that angle was y, now from what I could see, the other angle was pi-y. And it says that side is -cosy while the hypotenuse was 1. The sine and cosine have a period of pi, which means any angle plus or minus pi is going to be same negative result of each which means the -cosy is true. Is there a video where Sal explains this a bit further? I could prove this also with angle addition formula, but I can't prove an angle addition formula with angle addition formula that just doesn't make sense now does it. So if anyone could give me the link of Sal showing angle and angles that have the same negative cosine/ sine values of that angle I'd be mighty grateful.(3 votes)
We've got triangle ABC here. It looks like a right triangle. We can verify that it is because it satisfies the Pythagorean theorem. 8 squared is 64 plus 15 squared is 225. 64 plus 225 is 289, which is 17 squared. So it satisfies the Pythagorean theorem. So we know that this right over here is a right triangle. But what they're asking us is-- what is the cosine of angle ABC, which is this angle right over here plus 60 degrees. Now, just like this, I don't have any clear way of evaluating it. But we do have some trig identities in our toolkit that we could use to express this in a way that we might be able to evaluate it. In particular, we know that the cosine of A plus B is equal to the cosine of A times the cosine of B minus the sine of A times the sine of B. So we could do the exact same thing here with the cosine of angle ABC plus 60 degrees. This is going to be equal to-- and let me just write out the whole thing-- the cosine of angle ABC times the cosine of 60 degrees minus the sine of angle ABC times the sine of 60 degrees. So this right over here is angle ABC. This is angle ABC. And this is 60 degrees, and this is 60 degrees. So let's evaluate each of these things. What is the cosine of angle ABC going to be equal to? I'll do that over here. The cosine of angle ABC. Well, we could go back to Sohcahtoa. Let me write it down. Cosine is defined as the adjacent-- for this angle-- the adjacent over the hypotenuse. So cosine of angle ABC is equal to 15 over 17. So this right over here is 15 over 17. What is the cosine of 60 degrees? Well, for there, we have to turn back to our knowledge of 30, 60, 90 triangles. So if I have a triangle like this-- so let me do my best here-- to construct a 30, 60, 90 right triangle. So that's a 60-degree side. This is a 30-degree side. We know-- and we've seen this multiple times-- if the hypotenuse has length 1, the side opposite the 30-degree side is one half. And then, the side opposite the 60-degree side is square root of 3 times this. So it's square root of 3 over 2. So the cosine of 60 degrees-- if you're looking at this side right over here. And let me write it in a color I haven't used yet. So I care about the cosine of 60 degrees. The cosine of 60 degrees is going to be equal to, once again, adjacent over hypotenuse. One half over 1. it's going to be equal to one half over 1, which is equal to one half. Now, let's think about the sine. Now, let's think about the sine of angle ABC. So that's this one right over here. So we have our triangle right here. Sine is opposite over hypotenuse. So opposite has length 8. Over hypotenuse, is 17. So it's equal to 8 over 17. And then, finally, we need to figure out what the sign of 60 degrees is. Well, for the 60-degree side of this right triangle, opposite over hypotenuse. So square root of 3 over 2 over 1, which is just the square root of 3 over 2. So we have all of the information we need to evaluate it. So this was the sine of 60 degrees. This whole thing is going to evaluate to cosine of angle ABC is 15 over 17 times cosine of 60 degrees is one half. So times one half. And then, we're going to subtract sine of ABC, which is 8 over 17. And then, times sine of 60, which is square root of 3 over 2. So times the square root of 3 over 2. And now, we just have to simplify it a little bit. So this is going to be equal to-- if I multiply one half times this, let's see. It's going to be 15 over 34 minus-- and let's see. We're dividing by 2. So it's 4/17ths. I'll write this 4 square roots of 3 over 17. And that's about as simple as I could do it. If I wanted, I could have a common denominator here of 34. And then, I could add the 2, so it could be 8 squared. So 3 over 34, but that still won't simplify it that much. So this is a reasonable good answer for what they are asking for. 15 over 34 minus 4 square roots of 3 over 17.