Welcome back, everyone. Here, we're going to take our first look at what's called the unit circle, which is just a circle of radius 1 where all of the angles going around our circle correspond to specific xy coordinates on our graph. Now, the unit circle is going to be massively important in this course because you're going to need it to solve a bunch of different trigonometric problems. And the first time you see the unit circle, it may be a bit overwhelming because there is a ton of information here, and you might not know any of it yet. But that's totally okay because here I'm going to completely break down the unit circle for you throughout this chapter, and soon you'll be able to look at a completely blank unit circle and fill in all of that missing information completely on your own. So let's go ahead and get started with what the unit circle even is. Now, like I mentioned, the unit circle is a circle of radius 1 and, like any circle, it goes around from 0 to 360 degrees. Now this might be really familiar to you, but we want to look specifically in the context of this unit circle, so we're going to go around and fill in all of these angle measures. Now remember, the convention when looking at angle measures is to start measuring from the x-axis from 0 degrees here. And going a quarter of the way around my circle, I reach 90 degrees, and then halfway around my circle I am at 180 degrees, and then 270 degrees at the bottom of my circle here until I reach a full rotation of 360 degrees.
Now we don't just want to know these angle measures in degrees because it's also important to know our angle measures in radians, especially when working with the unit circle. So remember, a full rotation around 360 degrees is also equal to 2π radians. So looking at our unit circle here, 0 degrees is 0 radians, and then going up to 90 degrees, this is π/2 radians. Then halfway around our circle, we reach π radians. And then at the bottom here, we have 3π/2 radians until we reach a full rotation, of course, of 2π radians at the end here. Now the angle measures are not the only important thing here, and this might not have been any new information. But remember, all of these angle measures on our unit circle specifically correspond to an xy coordinate. So looking at our graph here at 0 degrees and noticing the scale of my graph, I see that 0 degrees is located at this point (1,0). Then up to 90 degrees, this is the point (0,1). Over at 180 degrees is the point (-1,0) and then down here at the bottom of my circle this is the point (0,-1). So all of these angle measures correspond to these ordered pairs as well.
Now knowing that the radius of our unit circle is 1 and seeing all of these points on our graph, you may have also noticed that our center is right here at the origin, so our unit circle is centered at the point (0,0). Now with this knowledge in mind, we can also represent our unit circle using an equation. Now here we have the general equation of a circle, but knowing that our center is at (0,0), this h and this k go away. And we also know that our radius is 1. So this gives us our equation for the unit circle as \(x^2 + y^2 = 1\). So that means any xy point on my unit circle, I can plug back into this equation, and it would give me a true statement. So all of these points that I have on here, I could plug into that equation and get a true statement because I know that they're all on the unit circle. But you might actually be given a completely random point and be asked to verify whether it's on the unit circle or not because these are not the only 4 points on my graph, right? I have all of these points in between them.
So let's go ahead and take a look at our first example here. In this first example, we're asked to identify which points are on the unit circle and then label them on the graph. So the first point that we're given here is this point (1,1), and we want to verify if this is on the unit circle. Now this is actually a point that's going to be really easy to just go ahead and plot, so let's go ahead and just do that first. So looking at my graph, going up to (1,1), here is the location of that point. And I can clearly see that this is not on my unit circle. It's a little bit outside of it, so I can already say that this is not on the unit circle. But if we want to perform an extra check here, we can also go ahead and just plug this point into our unit circle equation. So doing that, I get \(1^2 + 1^2\). And I want to know, is that equal to 1 because if it is, then it's on the unit circle. Now one squared is just 1, so this is 1+1. And is this equal to 1? Well, 1+1 is equal to 2, so this is definitely not a true statement, and 2 is not equal to 1, so this verifies what we already saw that this point is not on our unit circle. Let's look at one other point here. We have the point \((\frac{1}{2}, \frac{\sqrt{3}}{2})\). Now, this seems like a rather random point, and it's definitely not one that I can just plot on my graph easily like (1,1) because I don't know exactly where \(\frac{\sqrt{3}}{2}\) is. So let's just go ahead and plug this right into our equation to verify whether it's on the unit circle or not. So plugging in our x and y values here, I have \((\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2\). And we're verifying, is this equal to 1? So squaring these values, remember, with fractions, we want to square both the numerator and the denominator. So \(\frac{1}{2}^2\) is going to be \(\frac{1}{4}\). And then squaring \(\sqrt{3} / 2\), I know that it will cancel out that square root in the numerator giving me a 3. And then squaring that denominator, I get a 4 again. And we're verifying, is this equal to 1? Now adding these two fractions together, this gives me 4 over 4. And is this equal to 1? Well, 4 over 4 is indeed 1. So one equals one. This is definitely a true statement, which tells me that this point, \(\frac{1}{2}, \frac{\sqrt{3}}{2}\), is actually on our unit circle. And we can go ahead and plot it there as well. So looking at our unit circle on our coordinate system over here, if I go over to the point 0.5 on that x-axis, I know that wherever it's touching the unit circle that is this point \(\frac{1}{2}, \frac{\sqrt{3}}{2}\). So my point is right here, \(\frac{1}{2}, \frac{\sqrt{3}}{2}\). But remember, every single point on our unit circle also corresponds to an angle. And it just so happens that this point corresponds to the angle 60 degrees. Now we'll continue to see this angle and this point throughout this chapter. But for now, since we know what the unit.circle is and how to verify if a point is on it, let's get a bit more practice. Thanks for watching, and let me know if you have questions.
