Hi, everyone. Welcome back. We're going to learn our last bit of identities here, and this will be our foundational identities that we can go use to do anything else we want. The next identities that we're going to do, they're not as easy as to derive as the other one of the Pythagorean identities where we just look at the triangle and say, "Ha-ha, Pythagorean theorem in disguise." These identities will involve sine and cosine of the sum and difference of two angles. These should be kept handy. You're not to memorize them, but they should be really kept handy because these are going to be frequently needed as we go through and work with our things. Let me just give these to you and you can copy them down. This is called the sum and difference identities, and here we go. We're given that sine of any two values, x_1, and we'll do plus or minus, it doesn't matter. X_1 plus or minus x_2 is equal to sine of x_1, cosine of x_2 plus or minus cosine of x_1, sine of x_2. Let me give you the other one and then we'll talk about it. If I have cosine, I can't remember this, [inaudible] sine and cosine, and the rest you can always manipulate from these. X_1, we'll do plus or minus x_2, we get a cosine of x_1, cosine of x_2 minus plus, I'll talk about that in a second, sine of x_1, sine of x_2. Here we are. In this sum and difference formulas, so they're called sum and difference because they're plus or minus. When the top or bottom sine is chosen, then you choose the top or bottom sine on the formula on the right. What does that mean? It means if it's a plus coming in, then on the side over here, it's a plus. If it's a minus going in, then it's a minus going out. Whatever the sign is, you just match the top or bottom. For the cosine, sum turns into a minus, so this switches. We'll do examples of this to drive this home, but if it's a plus coming in, it's a minus over here. On the right side, if it's a negative coming in on the left, then it's a positive on the right. So sines, same, this is how I remember the sine, s for sine, s for same, same sine, sine same, cosine, you got switch. To prove this, we'll put a link somewhere, but it's not as obvious why this is true. We are looking at that and saying why it's a very nice, clever geometric argument that could take up a whole page that we're not going to do. I'll put a link to it, a video where they explain it quite nicely. But for now, just use these, keep this handy as we go through our calculations. I'm going to put it all on the same slide, so you have it. The Pythagorean identity, this is your core identity. The Pythagorean, these three make up everything. I'll write it one more time just so we have it all on the same slide. So sine squared of x plus cosine squared of x is 1. If you know these three identities, everything else follows. Wherever you saw that you can get the other two identities for free and your even or odd identities, maybe we'll throw these on there as well, you can see you have it all. Stare at this slide, let this sink in. This should be what your review sheet looks like. Every single time you have to use something else, come back to this. Here they are, sum and difference, Pythagorean, even/odd. Know these, the rest, you get for free. How are they used? Let's do an example. I'm going to keep those handy, so pause the video [inaudible] and keep those handy. Let's find the exact value here. Let's do the sine of Pi over 12. So Pi over 12, this is not on your sheet, right? I want to know the exact value, I don't want it in decimal, I want it in closed form. What's nice about Pi over 12, is that you can write Pi over 12 either way, and check this here. But you can write Pi over 12 as a sum or difference of our known values. So Pi over 3 minus Pi over 4 is in fact Pi over 12. Pause the video, convince yourself that that is true. Once you write an angle as a sum or difference of two values, well, then you can use the formula for the sum and difference of values. So sine of Pi over 12, I want to think of Pi over 12 as now sine of the difference of Pi over 3 minus Pi over 4. We have a formula for this. Remember, sine keeps the same sign. It says take the sine of the first, so sine of Pi over 3, cosine of the second, so cosine Pi over 4. Keep the same sign, so minus, and then you switch it. Cosine of the first, cosine pi over 3, sine of the second. I always say sine of the first and second whenever I have a formula. I don't want to think of formulas in terms of the variables that are written with. The reason for that is because you can also see this in terms of Theta or maybe Alpha or Beta or some other crazy variable. Just a little advice to you, whenever you have a formula, say the first or a second, the sine gets the same sign here, minus is a minus, so that doesn't change. Sine of the first, cosine of the second. Keep the same sign. Cosine of the second, sine of the first. Again, if you're just trying to, just as you say it, and then different parts of the brain are activated and maybe one day you don't have to look at up as much, but for now, keep it handy, check my work, make sure you agree. Then I claim all these values are known. Now we go to the table and find what these are. Sine of pi over 3, that is root 3 over 2, cosine pi over 4. This is root 2 over 2, we've done all these. Cosine of pi over 3 is a 0.5, and then sine of pi over 3 is root 2 over 2. You can clean this up a little bit. What do they have here? If you clean this up, you get root 3. I'm going to keep the root separate and just combine over a common denominator of 4. The reason for doing that is that again, all these answers are fine if we're just trying to clean it up as much as possible, there's a root 2 and a 4 in both expressions. I'll do root 2 over 4, we'll factor that out and we get root 3 minus 1. So this would be our closed form, this is our answer in terms of square root. Again, a calculator will never be able to tell you that this is the expression. True, it'll give you a five decimal, six decimal, seven decimals, but it won't be able to tell you, and I doubt that you would be able to look at that decimal and tell me that it's root 2 over 4 times a square root 3 minus 1. So this is just an example here of using the sine formula. Let's do one other example. We'll switch it up and we'll do the cosine formula just to get practice using these things. Let's do another one about cosine of 7 pi over 12. Again, the tricky part sometimes, if you get something off the table, you're looking for how do I write this in terms of something that I know. So 12 is usually a common denominator for these practice problems because they're usually somehow written as plus or minus of pi over 3 and pi over 4. Seven pi over 12, you can come off on the side and you can check this, 7 pi over 12 is pi over 3 plus pi over 4. Again, pause the video, convince yourself that, that is true. When I want to compute cosine of 7 pi over 12, I really want to think of this as the sum of pi over 3 plus pi over 4. In that regard, I have a formula for the cosine of the sum of two values. Here we go. It's cosine of the first, say with me. Cosine of the first, cosine of the second, say first and second, don't say x_1 and x_2. Now cosine, we have to change the plus to a minus, that's why it's minus plus in the equation. Sine, same, sign, keep the same SSS, cosine you got to switch. Cosine the first, times cosine the second minus sine of the first, so sine of pi over 3, times sine of the second, pi over 4. All these values are known, so you can look them up if you need to, but you get 1.5 times root 2 over 2 minus root 3 over two times root 2 over 2. Clean that up again, combine things and you get a very similar root 2 over 4 parentheses 1 minus square root of 3. This is just working with these two formulas. All the other identities that we're going to need are obtained from these core values. For example, if you wanted to know what the formula for the tangent of two things were, so maybe we can work that out. I would never want you to know this or memorize this, but just to show you, because you'll see it on a big scary sheet. If I gave you tangent of two things, well, I'm not going to look that up, I'm not going to do other work, I'll just use my formulas for sine and I'll use sine over cosine, so I'll just do this. Yeah, it's going to be a little longer and sure if I do a lot of algebra, I can work it out and clean it up, and if you want to go look up the clean up formula, I'll leave it as an exercise if you want, but you can check. This will turn out to be tangent, and this is what you see sometimes, tangent of x_1 plus or minus tangent of x_2 over 1 minus plus, from the cosine, tangent of x_1 tangent of x_2. So this is where, I'm going to say it is 1,000 times, yes, this exists, but I don't think about it this way. I don't work with this. I don't have this memorized. I wouldn't even care about this. I would just do this. The cost of doing this, I'm not memorizing these requests is that I have two things to calculate, sine of the enumerator and cosine of the denominator, and I got to use a formula twice, but I'm happy to do that. I don't mind doing that because as long as I keep going back to sine and cosine, I can answer every single question they throw at me here. In particular, if you want to just go through this, we can look at tangent of Pi over 12. Pi over 12, we've seen this before. Pi over 12, you can write as Pi over 3 minus Pi over 4, and you say, well, "Do I need to know tangent to any snow this angle?" No just write it as sine over cosine. We have the values for sine and cosine. So let's write sine of Pi over 12 divided by cosine of Pi over 12. We worked out sine of Pi over 12 in the last slide. Remember that was root 2 over 4 parenthesis, root 3 minus 1. So the worst thing is to come in and finding the denominator, so maybe I'll skip ahead here and do that. So root 2 over 4, root 3 minus 1. We did that one already. So the work's going to come into finding cosine of Pi over 12. So lets go off on the side. We'll do those scrap work here. Let's go find cosine of Pi over 12. Cosine Pi over 12, we're going to write this as cosine of Pi over 3 minus Pi over 4. This is our difference formula here. So here we go, remember it's cosine of the first times cosine of the second. Now, cosine you have to change the sign. So that's plus sine of the first, sine of the second. All these values are known. So cosine of Pi over 3 is a half, Pi over 4 is root 2 over 2 plus sine of Pi over 3, that is root 3 over 2, and then cosine Pi over 4 is root 2 over 2, put all together, and factor you get root 2 over 4, 1 plus root 3. So I'll take that scrap work, and I'll go back to my fraction, and I get root 2 over 4, 1 plus square root of 3. What's nice about this is that the square root of 2 over 4 and the square root of 2 over 4, they will cancel. So same number of numerator and denominator cancel, and you're left with a cleaner version of root 3 minus 1 over root 3 plus 1. So notice I never needed this fancy formula. Yes, they're nice to have, but you don't need them. You absolutely don't need them. Final answer, root 3 minus 1 over root 3 plus 1. If you want, check it on a calculator, make sure it's in radians, and you can compare that the decimals are absolutely the same. There's another set of formulas identities that come with these sum and difference formulas, they are called double angle formulas. Let me write down the first one here. Let's write this down as x plus x. It happen when the two values are the same. So what if I wanted to add these two things together? So in that case, from our Identity, remember it's sine of the first, cosine to the second. Now sine keeps the same sine, so plus and then sine of the second cosine and fairly they're the same, they're exactly the same. So when you put it all together, you get two sine x, cosine x. This guy comes up a lot. I'm going to write an x plus x, you tend to write that as sine of 2x. But notice how like if I know the first one, I get the second one for free. So this one comes up enough that I'm going to put it down. But it's an immediate consequence of the sum formula for sine. So it's immediate, which is nice. So if we do the same thing for cosine, like what happens if I take the same thing for cosine and I want to add x plus x, which is also cosine of 2x. You get cosine of the first, cosine of the second, I'm going to put a minus sign there now, and then sine of the first, sine of the second. Cosine times cosine is cosine squared minus sine squared of x. So this is another way to see the identity cosine of 2x. What's interesting about this one is that you can write it a bunch of different ways. It's very versatile. So in short cosine 2x is cosine squared x minus sine squared x. Because it's a minus sign, be careful, order matters. You have to have cosine squared first. But if you combine this, because now you're starting to see sine squared and cosine squared. Remember you have your Pythagorean identity that says cosine squared x plus sine squared x is 1. So you can rearrange some things. This means you can write this as cosine squared x is 1 minus sine squared x. You can certainly rearrange this as sine squared x is 1 minus cosine squared. So these are all equivalent. You can plug these in and substitute, and sometimes you see it that way as well. Let's substitute the first one in, for cosine, you can write this as 1 minus sine squared x and then minus sine squared x. When you put that together, you have 1 minus sine squared minus sine squared, this becomes 1 minus 2sine squared x, and that's another equivalent way to see cosine of 2x. I'll summarize that in a second, but I want you to see these both ways, so you can substitute for cosine squared or you can substitute for sine squared and if you do that, you get cosine squared of x minus and watch out for parentheses here, 1 minus cosine squared of x. Clean that up and one with the minus sign is going to distribute to both pieces, so get cosine squared plus cosine squared. So we'll put that up front., that's 2 cosine squared of x minus 1. There's three ways to see this just depending on what flavor you want it maybe I'll summarize it down here. Cosine of 2x is using the sum formula for x plus x cosine squared x minus sine squared x, perfectly good way to do it. It's also absolutely equivalent to 1 minus 2sine squared x and also equivalent to 2 cosine squared x minus 1. There's a couple of ways, couple of flavors that this formula can come in, pick your poison, whatever one you need, whatever all of sluggishness. Sine of two x, though two sine x, cosine x, this one comes up a bunch, we're going to use this one later, so keep these in mind as you go through. Let's do a couple of examples with these and then we'll wrap things up. So let's solve for x and let's put x, just keep it one lap around the unit circle. So like 0-2 Pi and let's do as our first example, cosine of x is sine of 2x. Now our strategy for here is going to be to write things similar, but it's already in terms of sine and cosine. It's already in terms of sine and cosine. What's the next strategy to look at? Well, sine is with the 2x, so I have a double and cosine is with a single. Watch everything turns the sine and cosine, you want to get things out of double. You want to make the insides match so that you can cancel stuff or do whatever you want to do. Sine of 2x is our double-angle formula. We just saw this, so we really want to take of sine as when we rewrite this as cosine of x is 2sine x, cosine of x, very good. So be careful here, some folks want to divide by cosine of x, it's a little dangerous. The reason is because if you divide by a fraction, it could be zero. So I would caution against that, you're going to lose some solutions here. Instead, why don't we write this as cosine of x minus 2sine x, cosine x. So don't cancel functions, move it over to zero, factor out cosine 1 minus 2sine of x is 0, and then you have two things that multiply together that give you zero. So that means cosine of x is 0 or 1 minus 2sine of x is 0. Cosine of x is 0, where does that happen? That occurs at x equals, remember, cosine of the value zero, when x is zero, top of the circle, Pi over 2 or bottom of the circle 3 Pi over 2, which is still one lap around, so just those two values. The other one, if you do some manipulation, and you get sine of x equals to half, you're going to get two values there as well. You've seen these before, Pi over 6 or 5 Pi over 6, so corresponds to quadrants one and two. We actually have four solutions total, so this you can go back and plug them in and check they all work. How do you divide it out and canceled the cosine x with the lost two of the four. Be careful, don't do that. Same thing, let's do another example. Let's practice our cosines. So again, solve for x. Let's go 0-2 Pi, and I want sine. Find all solutions, sine of x is cosine of 2x. Now, cosine 2x, there's many ways to do, we have three versions of this double formula here. Same idea, I don't want it as a double one, I want the left side is the single. So let's convert this, I think to put it in terms of sine if possible. So why don't we pick any of the ones with sine and wouldn't be nice if we had only sines. Again, there's probably a few ways to do this, but I'm going to think of this as sine of x, and I'm going to pick the identity where I have absolutely no cosine. It's can I get it in terms of just sine would now be better. The one I'm going to use here, of course, is 2sine squared of x. Moving things around, same idea. Let's write this as sine of x minus 1 plus 2. Sine squared of x, and we'll set everything equal to 0. It's a little weird to see this is in out of order. Let's rearrange. So 2 sine squared x, we'll put the higher degree upfront, plus sine of x second minus 1 is 0. Now, this may not be obvious, but stare at this for a minute, these factors. These factors are sine of x plus 1 and 2 sine of x minus 1, and this is all equal to 0. So when two things multiply together to equal 0, we're going to get to that sine of x plus 1 is 0 or 2 sine of x minus 1 is 0, which means that sine of x is minus 1 and 2 sine of x is positive 1, or sine of x is 1.5. Sine of x is negative 1. That of course occurs, think of the y-value that's bottom of the circle, that's at 3 Pi over 2, and sine of x equals a half. Well, we just saw that, that's x equals Pi over 6 or 5 Pi over 6. In this particular example, there are three solutions that make or solve this expression. Let's do one more just to practice our formulas. Let's solve for x, same thing. Solve for x between 0 and 2 Pi. So one lap around the circle, 1 equals sine of x plus cosine of x. Now, this equation, this is little tougher. This one doesn't look like we're going to use some formulas we know. There's nothing squared in this example. Sometimes in math when you don't have the thing you want, you force it there. So I would really love for stuff to have square. I can use either Pythagorean identities or someway other formulas that have sine of cosine squared. What's a way to make squares appear when they don't naturally were never there? Well, you can square both sides. When you square both sides, this is perfectly bigger thing to do, you get 1 squared is equal to sine of x plus cosine of x squared. This is tricky because most students don't think about this or don't try this first. But you can certainly square both sides. When you do that, again, whatever you do on one side, you must do the other side. You get 1, 1 squared is 1. We're going to foil this out. So it's like first sine squared of x outside inside, plus 2 sine of x, cosine of x, plus cosine squared of x. Now you can see why this probably was a good idea, because now I have the Pythagorean identity applying to the first and last term. So sine squared plus cosine squared is 1. Then all of a sudden, I have my double angle. This is nice. So you get both things up here. 2 sine and cosine x is my double angle, so plus sine of 2x. So a very clever move, very cheeky move. It's cool. But once you see it, you keep this little strategy in your back pocket. The 1's cancel, so subtract 1 from both sides. Basically this is a question that looks tough to begin with, simplifies to 0 equals sine of 2x. So to have sine of 2x equal to 0, 2x then has to be, so when I assign 0, it's either add 0 or Pi. We'll say Pi or 2Pi, since we included 2Pi in our solution set. So just setup all the equations, 2x equals 0, 2x equals Pi, 2x equals 2Pi. That gives you that x is 0, x is Pi over 2, and x is Pi. This will turn out to be all of our solutions. It's never bad idea to go back and check that they actually work, and I leave that as an exercise. So check in two ways if you want. Just because it's such a neat little example. Graph this thing and see the graph and get a handle of it, and then of course, go back and plug in or substitute. I'll leave that as an exercise. If you didn't see this strategy to solve square things, again, I wouldn't expect you to until someone shows it to you, but this little tricky. Square it, keep it in mind, use your identities, combine identities as needed and solve for your solutions. Great job on this one. Try some more and just keep at them. Remember, if you ever stuck, think of some things outside the box. Check a graph, plug-in, have fun, no matter it isn't brainstorm. Try stuff. Good job on this one. I will see you next time.
