Hi everyone, and welcome back. I thought we could just make a table first to summarize what we have found so far. So we're not scrambling to put it all together. Let's do that. I'm going to use radians only. Right now, what do we have? I have Theta, then we get a bunch of values and I would really love to know what sine of Theta is and cosine of Theta. So what are some values that we know? We're going to have mostly in quadrant 1, and then using symmetry of the circle, we know quadrant 1 really well. You can always flip things and use symmetry to get the other values. So I really want zero, and zero is the same radian, so zero degrees, zero radians. I'd love to know 30 or Pi over 6 radians, 30 degrees. We have this number, we have 45 degrees or Pi over 4, we have 60 degrees or Pi over 3, and then we have 90 degrees or Pi over 2. Obviously, we can throw more of these on, keep going. Maybe we want Pi over there, sure. We could put it together if you want, find it in the last one. It will do that one as the last one as well. So here we go. So what is sine? Sine of zero we said was zero, sine of Pi over 6, this was one-half, Pi over 4 for sine was root 2 over 2, on our axis that was half the square. We'll just fill in cosine right now, they are in fact equal. Then Pi over 3 for sine, that was your root 3 over 2, and then Pi over 2 is 1, and then sine of Pi is zero. This is all coming from the unit circle. We're realizing that the triangles formed are either half a square or half an equilateral triangle. For cosine of zero. Well, our cosine turned out to be as one, then we had switched on us. Remember root 3 over 2, and then half, and then cosine Pi over 2, that's the top of the unit circle was the x-value right there at zero, and then Pi is negative 1. So I wanted to include Pi on this because some people, what they do is they try to memorize this table and they say, "Aha, there's a pattern like, yes, the circle is symmetric." So sure, reading this line across is the same as reading this backwards, but if you just rely on these patterns and not intuition, you'll miss things where it changes, that only gets you so far and usually people don't understand stuff and once you start forgetting things, there's no recovery and then things go poorly. So this is our foundational table, feel this energy, go along as you find some more pieces and now let's use it to solve some problems. So for our first example, I'm going to give you just a bunch of things that are not on the table. I'm going to Show you how to use the table to find them. So for our first example, I want to find sine of 135 degrees. We're going to bounce back and forth. Again, all of these questions could have been asked with radians, but just so we gain some intuition, the key here is to draw a picture. So I have my unit circle, like so, what is 135 degrees? Where does it live? Let's see, it's obviously bigger than 90, but how much bigger? So it's somewhere in quadrant 2, where you have quadrant 1, top right, quadrant 2, top left. So somewhere between quadrant 1 or quadrant 2, it's larger than 90, but it's not as big as 180 degrees. So we can see the difference here. So we have a 180 degrees minus 135, play around with that, and that's 45 degrees. So if I were to draw this angle, it's going to split quadrant 2 into half. The angle that I'm talking about is all the quadrant 1 and then half of quadrant 2. Again, this is all inscribed on the unit circle, and I want sine of this number. This is where you say, "Wait a minute, this isn't on my table. How am I supposed to know this?" Draw the unit circle here on my graph so that we have it. This is circle of radius 1. What do I want? I want sine of 135 degrees. What is the sine value, x or y? You remember, sine is the y value, so I want the y value. So I want whatever y value this is, remember it's height of the x-axis. It is when I form the triangle that's inscribed by unit circle. But here's where we're going to use our knowledge of quadrant 1. The table is mostly quadrant 1. This angle, the y value by symmetry, I can also think of this triangle and flipping, going across the quadrant 1 and flipping the picture over. Understanding that if I have 45 degrees off of the negative x-axis, I'm going to have 45 degrees off of the positive x-axis. There's so much symmetry. My pictures are doing justice because it's hard to draw a perfect picture, but maybe you're an artist that mean you're getting the idea. Try this carefully. The y-value of sine of 135 is exactly the same as the y-value on 45 degrees. Take any angle you can and bring it back into quadrant one, and realize that all I want is the y-value. The y-value, the height off of the x-axis, is the same for a 135 degrees versus 45 degrees. It does not matter. So in that regard, sine of 45, that one we know. This is, of course, just root 2 over 2. Our answer here to sine of 135 is root 2 over 2. Let's do another one. Let's try sine of 5 Pi over 4. Now once again, I want the y-value here. Now this is obviously radians. We get the Pi is going on here. So remember I want the y-value. In your head, draw the unit circle and try to get a handle on what is 5 Pi over 4. 5 pi over 4 is one and a quarter Pi, like 5 over 4 is a mixed number. So what's that? I'm going to go all the way to Pi, Pi is 180 degrees, so I go four-half rotation, and then a quarter more. I come down. Now, I'm in quadrant three, so one, two, and three. Now I want sine of this value. You can imagine a little triangle is sitting down here. I want to know what the y-value is at this point, at x,y, where this ray hits the unit circle. One thing to notice here is that whatever my answer is, so if I want to see this, it has to be negative. When I'm in quadrant three, my y-value and below the x-axis, it has to be negative. Now I don't know a lot about quadrant three nor do I really want to. So I'm going to reflect this angle across the origin. I'm going to really bring it back to quadrant one. I always wanted to do this, and do it in such a way where you have this beautiful symmetric picture. My angle here, 5 pi over 4. So I realize I went like Pi and then Pi over 4. So Pi over 4, once again, is the angle that's cut out on the short side. So by finding the y-value of 5 Pi over 4 is exactly the same as finding the y-value of Pi over 4. The triangles are exactly the same. So whenever you get a question not about quadrant one, always bring it back to quadrant one. The key here though is that so Pi over 4, once again, the key is negative. If I just give this back, this is a little incorrect. If I just say sine of Pi over 4, which we know is 45 degrees, this is root 2 over 2, that's a little wrong. The reason why is because whenever you want the y-value, you went from below the x-axis to above the x-axis, and that's why I wrote a little reminder to myself to negate the answer here. So really I should have sine of Pi over 4 is the negation of sine of Pi over 4. Bring it all back into the first quadrant. Let's do another one. Now, I keep doing examples here of Pi over 4. Obviously, that doesn't always have to be the case. We can change it up, and so here we go. So let's do another. What is 2 times sine of negative 9 Pi over 4 times cosine of negative 9 Pi over 4? 9 Pi over 4, that is definitely not on my table. Negative 9 Pi over 4 is also not on my table. So what do I do? I draw the unit circle and I try to get a handle on what this is. Negative, for our sake, it means clockwise. We're going to start on the x-axis and we're going to move downwards for that. So 9 Pi over 4, let's just like go off on the side, use graph paper, try to figure out what this is. The negative, that just means clockwise and then 9 over 4 Pi, what is this? This is really, think of it like a mixed number. This is like two and a quarter Pi. So I'm going to do two Pi and then I'm going to do a quarter more, but I'm going clockwise. Imagine with your pen, I go around, two Pi is around the full lap. So it doesn't even matter. Then I just do a quarter Pi more, and now I'm down in quadrant four, for Pi over 4. So you do a full lap and then down a quarter, and you just land. Well, basically quarter over Pi in quadrant four. So now you're in Q4. What do we want here? I need cosine and sine of this value. A couple things about the sine here. Sine is your y-value at this point, wherever it meets the unit circle x,y. The y-value or below, so this one's going to be negative. I always write that down just to remember things. The cosine, that's my x-value, positive, whatever my x-value is at that exponent there, so this is positive. So it's 2 times a negative times a positive number. I should get a negative number back at the end of the day. So we'll find out. So let's see. What do I want to do here? I'm in quadrant 4. I don't like quadrant 4. I don't know much about quadrant 4, let's reflect this thing up and hang out in quadrant 1. So it gets us in Pi over 4 and that I know. This is the same as Pi over 4. Here we are again, back to where we started. So what does this become? The number 2 comes along for the ride. Sine is the y value. Sine of negative Pi over 4 is the same as negative sine of Pi over 4. So 9 Pi over 4 is negative sine of Pi over 4, the y value has to change signs, because it went from quadrant 4 to quadrant 1. Cosine x value, let's just go to our cosine of Pi over 4. They have the exact same. You should draw the triangles and see that they touch the exact same x values. Put it all together, sine of Pi over 4, cosine Pi over 4, these are on my table. So I get 2 times negative root 2 over 2 and then positive root 2 over 2. That's nice because I got 2 that cancels and then root 2 times root 2 is negative 2 over 2 and has a nice little answer here of negative one. So that expression is negative one. We like that. Now, let's do one more. Let's do another one, go back to degrees. Let's give you a really really large one. So like sine of 690 degrees. So what is this thing? So 690, that's huge. One lap around the circle is 360. That's too big. We're out of range, so let's subtract a lap off. We're in degrees. We can stay degrees. So we'll do 690 minus 360. So what does that work out to be? Three hundred and thirty. Okay. So 330 degrees, that one I can handle. Let's see, so I draw the inner circle and I start working my way around. I'm close to 360. How close am I? I'm 30 degrees off. So I go all the way around and I land somewhere in quadrant 4. But I'm 30 degrees off. So the small angles are called reference angles. The small angle is 30 degrees, and I want to know the sine is the y value. If I form my little triangle, what is the y-value at the point where this ray meets the inner circle. So obviously 330 degrees. I'd rather study with 30-degree piece because that's the same. This one if I come down with 30 degrees short. So why don't I reflect just like I did before up into quadrant 1 and I'll have a nice little 30-degree angle there. That's nice assigned. Now we got to keep in mind though, I was in quarter 4, I brought it up to quadrant 1. I really want to see sine of this thing is the y-value. So whatever my answer would be, I'm going to cheating a little bit by moving this thing, but it comes at a cost and that's just like watch out, whatever it is. This is going to be negative sine of 30. So I'm really using the symmetry here to get this, just watch out. Don't lose your minus signs. Sine of 30 degrees is on my table. That's good old negative one half. That's nice too and just because I have the space to do it. If I asked you for cosine of 690 degrees, well for all the same reasons that's the x value. Moving from quadrant 4 to quadrant 1 doesn't change the sign of the x value. It's all still very positive. This also be cosine of 30 degrees and that of course is just from the table root 3 over 2. So this is where I say, if you have a strong working knowledge of the table, understand how to move things inside of Q1 wherever they give you an angle, there's some people who are trying to memorize every single value, don't do that. Just understand quadrant 1, move things into quadrant 1, and just be mindful that the cost of that might be to put back a negative sign or just keep track where things are going. Okay. All right, so you do lots more examples on these and practice makes perfect. So keep working on these, go backwards to the video again and just keep doing them, always putting things back inside of quadrant one. Great job on this one. I'll see you next time.