 Hello, and welcome to this first screencast for Math 123 Trigonometry, where we will discuss the definition of the circular functions, the cosine and sine functions. These will be the two primary functions that we will be working with throughout the course. The main thing we will use to make this definition is the so-called unit circle. We will place that circle with a center at the origin, and it has a radius of 1. There's a picture of the unit circle, and up at top you see the equation for that unit circle. The key idea is to measure the length of an arc along the circle. To do this, we always start our measurement at the point 1, 0. Now, if the arc has a positive direction, then we measure it, starting at the point 1, 0, and moving in a counterclockwise direction such as this. The arc will come and at a certain point there that will be called the terminal point of the arc, and generally we're going to use x, y to represent the coordinates of the terminal point of the arc. On the other hand, if the arc has a negative direction, we still start at the point 1, 0, but then we measure clockwise along the circle, so we might measure to place something like that. Now, a very important thing to remember is the circumference of the entire circle. If you recall, the circumference from the circle is equal to 2 pi times the radius. Since this circle has a radius of 1, it is a unit circle, we get a circumference of 2 pi. So if we go all the way around the circle one time, we will have an arc of length 2 pi. Other important points to keep in mind are kind of the points where the circle intersects the coordinate axes, which would be these four points that I'm drawing in there right now. And if we go one half of the way around the circle, then we have an arc of length pi, the one half of 2 pi. So the terminal point for an arc of length pi would be at the point minus 1, 0. And again, we can now divide this into quarters. And so an arc of length pi over 2 would finish up at that point right there. And an arc of length 3 pi over 2, which would be 3 quarters of the way around the circle, would end up at this point. So keep that in mind. In the next slide, we will show some arcs of special length, which are shown right here. These arcs all have our multiples of pi over 6 and pi over 4. So an arc of length pi over 6 would look something like that. And now if you can see if I can take certain multiples of pi over 6, such as 4 times pi over 6, we will get 4 sixth times pi. And if we reduce that fraction, 4 sixth is equal to 2 thirds, we get 2 thirds pi. So that's why you see something like 2 thirds pi there. That is a multiple of pi over 6. Some of the other ones, it's easier to see such as 7 pi over 6. And the pi over 3 one such as 5 pi over 3 is actually 10 times the multiple pi over 6. And again, if you work with that fraction, you get 10 sixth pi. And now we divide both numerator and denominator by 2 and we get 5 thirds pi. So that would correspond to that point there. It would go all the way around the circle to that point there. We also, as I said, we measure arcs in the negative direction as well. And those arcs are shown in the next slide. And you can see the measurements are pretty much the same except we're going in the negative direction. So again, you see minus 2 pi over 3 and you'll see minus 5 pi over 3 as we saw before. And again, this time the minus pi over 3 arc goes all the way around and has a terminal point in the first quadrant. So here's our definitions of the special functions that we will be working with throughout the course. And it's always good to make sure you understand these definitions as we will rely upon them from time to time throughout the course. And the ideas of t represents the length and direction of an arc. Notice again the word direction in there for our conventions for positive direction and negative direction. The initial point is at 1, 0. And the terminal point we're labeling as x, y. And for example, if a picture might look something like this and we would see a terminal point here of x, y, if that arc is length t, then we say the cosine of t is equal to x. Our shorthand notation for this, which is what we will be using throughout the course, is right there. And again, notice the function notation cosine of t equals x. The textbook will quite frequently leave off the parentheses and write cosine of t equals x. I prefer to include the parentheses in this and we'll try to do so throughout the course as it really emphasizes the function nature that we're talking about. This is a function of the real number t. Again, we measure that a length along the unit circle with our direction conventions and we come up with cosine of t equal to the x coordinate of the terminal point. And in the same way, the sine of t is equal to the y coordinate of the terminal point. And you see that right there. Another thing to really keep in mind always is the equation for the unit circle. That's a very, very important fundamental fact about the circle. And in fact, in the next slide, we translate that equation into one involving the cosine of t and the sine of t. Okay, as you can see, we have the equation for the unit circle in the first line. And what we have done there is simply made the substitution x equal to cosine of t and y equal to sine of t into this equation. That has given us the second equation in the line there. And you can now see we get this equation here. Now, there's kind of a funny thing about notation for trigonometric functions or in this case, the circular functions. The second line is actually the correct notation that we would use here but it has become convention over a long period of time to shorten this notation. And so instead of writing cosine of t, that whole thing squared, we shorten it to this. And unfortunately, maybe we sometimes read that as cosine squared of t. Actually, it is cosine of t quantity squared. And the way we will usually see this identity is right there. And that is called the Pythagorean identity, as you can see by the title of the slide. Now, this next problem is a typical type of problem that we will encounter in our trigonometry course. And it basically gives the value of one of the trigonometric functions and asks you to find the value or possible values of the other trigonometric function. And the primary tool we use is this identity right here, the Pythagorean identity. So you have to be careful with squares and square roots here. And the idea now is simply if sine of t equals one-third, to substitute that into the equation and you get cosine of t squared plus one-third squared equals one. And with a little bit of algebra and arithmetic, we get this equation of one-third squared becomes one-ninth. And now we subtract one-ninth from both sides of the equation. And we have to be a little careful with fractions, but we get one minus one-ninth, which is equal to eight-ninths. Now, here's where we start being careful with squares and square roots. We have cosine of t squared equals eight-ninths. The idea is to take the square root of both sides of the equation. But remember, there are two square roots, and so we will write cosine of t equals plus or minus the square root of eight-ninths. And we can do a little bit of simplification of that by realizing that the square root of nine is equal to three, and so we would finally be able to write cosine of t equals plus or minus the square root of eight over three. Now, it's interesting to know what has happened here. By giving the value for sine of t at one-third, what we have done is basically gone on the unit of y-axis up to about, let's say, that's one-third, and cut across that. And you can see what happens there. That intersects the circle at two points. And although my pencil width here doesn't show much difference here, the coordinate of that point now will have y-coordinate of one-third and the x-coordinate will be the square root of eight over three. So the coordinates of that point are listed as eight-thirds, one-third. The point on the left over here, create a little room up here, will have coordinates. The x-coordinate now is negative, so we get minus the square root of eight over three one-third as the coordinates of that point. So that's our first type of problem in the trigonometry course, and I hope you will find the screencast to be useful for you to give you examples to work with and so forth as you study the course. Thank you, and see you next time.