Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. OK, we're going to look at graphing these on a number line. So when angle was 0°, our cosine value was 1. So at 0° we had one. At 60°, we had a half 90°, we were at 0, 45, we were at Route 2 / 2, which is approximately .7 O 7130 degrees. We were at root 3 / 2 which is oh about there 8660. Now we know that we have symmetry going on. So between 90° and 180° we have the same values going negative. So we're going to have a curve that then does a reflection and comes back up. If we were thinking about going in negative angle direction, we would have the curve going like. So if we think about the domain and the range, then the domain or what angles can we use is going to be any angle because we have every single value that we could have for a Theta. So the domain is negative Infinity to Infinity. If we look at the range for the cosine Theta, how high and how low can it go? The highest the cosine ever goes is the furthest to the right it went on a unit circle, or the highest it could ever go is 1 here. If you see the height of one on the unit circle, what's the furthest to the left it could ever go would be -1 or on this graph here of Theta and cosine, Theta -1's the lowest it's going to go. So the range would be -1 to one the domain and range for sine. Remember sine Theta is like our Y value. So at 0° we didn't go up or down any. So we have a sine value of 0. At 30° we had a sine value of 1/2. At 45° we had a sine value of root 2 / 260°. Our height was root 3 / 290°. We had a height of one. We have symmetry so in the 1st and 2nd quadrants the Y values are positive. So our signs are positive. In first and 2nd and 3rd and 4th quadrant our sign is negative because we were going below. Now we want to plot it for Theta as our input and sine Theta as our output. So at angle zero, we have 0 as our sine value. At 30° we have our sine value coming out as 1/2. At 45° we have our sine value coming out at root 2 / 2. At 60° it's root 3 / 2. At 90° it was 1 hole the symmetry. So we're going to be positive for our Y values in the second quadrant coming down. So we'd have 90 here, we'd have 120, we'd have 135, we'd have 150 and 180 at 0 again. So if we graph these points and connect them with a smooth continuous curve, we get a sine graph. Looking at this graph down here, the domain is going to be all all real numbers because we can put in any value we want for the Theta and we get AY out the range. What's the highest and lowest it can ever go? Well, the highest and lowest it's ever going to go is -1 to one -1 being the lowest one being the highest domain and range for secant. By definition, secant Theta is just one divided by cosine Theta. So if secant Theta is 1 divided by cosine Theta, then we're going to look at all of our cosine values, which were one 1/2 root 2 / 2, root 3 / 2 and 0 in that first quadrant. And then we just have symmetry. So depending on which direction we're going, it changes it from positive to negative. So 1 / 1 is one, 1 / 1/2 is 2/1 divided by root 2 / 2 is just two over root 2, and we rationalize by multiplying by root 2 over root 2. The twos then cancelled, leaving us sqrt 2 one divided by root 3 / 2. Once again we need to rationalize, but this time we multiply by root 3 over root 3 to get 2 root 3 / 3 1 / 0 is undefined. So if we looked at the cosine graph and we thought about the fact that we need one over cosine to be our secant, what happens every time the graph crosses the Theta axis? Well, the Theta axis, if I cross it and I want 1 divided by, well 1 / 0 is undefined. So each of those locations is going to have an asymptote. It can't be defined there. So every time we cross that axis with Theta on it, we're going to have it undefined. So now 1 / 1 was just one. 1 / 1/2 made it bigger, right? It made it 21 divided by root. 2 made it even bigger. So what's going to happen is we're going to get closer and closer and closer to that asymptote over here. If we were talking about zero to -90°, those would still be positive. If we were looking at here, would share the highest or lowest point and still get close to the asymptotes down here. It will look like this and like this. So if the purple is our secant Theta graph, when we look here, we're going to have the domain being all all real numbers, except it can't ever be 90° or 270°. And usually we write these in terms of π it's a little easier. So Pi halves, 3 Pi halves 5 Pi halves. So the domain's going to be all reals where Theta doesn't equal π halves plus K Pi where K is some integer. So Π halves plus a whole π would give us me give us to three Pi halves. 3 Pi halves plus a whole π would take us to five Pi halves. Pi halves minus a full π would take us to negative π halves. So those are all the lines that are undefined. Now the range is going to be the Y values, IE what comes out from the secant Theta. So negative Infinity to -1 and we're going to include -1 union one to Infinity. So it's just the opposite, the inverse of cosine. Cosine was defined from -1 to one. Well, secant's defined from negative Infinity to -1 union one to Infinity for its range. Its domain was everything except the values that gave an undefined or an asymptote. Looking at the domain and range for cosecant, cosecant is just one divided by sine. So if sine is the height the Y of our unit circle, we have 0 at angle 01, half at 30 root, 2 / 2 at 45, root, 3 / 2 at 60 and one at 90. Then by symmetry we know that over here between 90 and 180 we still have positive Y values or positive signs and then from 180 to 360 they all should be negative. So if we look at our graph for sign, we know that zero to 90 was positive. So here's my sine value. Zero to 90 positive. 90 to 180 are also +180 to 360 are negative, with our highest and our lowest points occurring at 90 and 270 respectively. Now Co secant because it's 1 divided by sine 1 / 0 is undefined. 1 / 1/2 is 2/1 divided by root 2 / 2 is root 2, 1 divided by root 3 / 2 is 2, root 3 / 3, once we rationalize 1 / 1 is 1. So on this graph, when we have our Theta being along the horizontal and our cosecant Theta being along the vertical, we can't ever have 1 / 0. So at the sine graph, we're going to have an asymptote every time the sine graph crosses the Theta line because we can't have 1 / 0. So those are our asymptotes. 1 / 1 is one, 1 / 1/2 is 2. So what's going to happen is we're going to be close to these asymptotes. We're going to share the highest and or lowest point with our sine. Highest and lowest points are going to be shared, so when we look at this one, our domain is going to be all real values except when we have an asymptote. So Theta can never ever equal 180 or 360. If we think about this in terms of π's that's π and 2π and 0 and negative π. So Theta can never be K * π where K is an integer. The range is going to be where the purple is occurring. So from 1:00 to Infinity and also from negative Infinity to 1. So negative Infinity to -1 union one to Infinity. The domain and range for tangent. Tangent by definition is sine divided by cosine or the Y values divided by the X values. If we look at our unit circle at 0°, our cosine is one when our sine is 0. So 0 / 1 is 0. If we look at our 30°, our cosine is root 3 / 2 when our sine is 1/2. So 1/2 divided by root 3 / 2 is one over root 3, which rationalizes to root 3 / 3 at 45. Root 2 / 2 divided by root 2 / 2 is one at 60. The sine is root 3 / 2 when the cosine's 1/2, so that simplifies to root 3 At 91 / 0 is undefined. So if we thought about plotting this with Theta being along the horizontal and tangent Theta being the vertical at Pi halves or 90°, we'd have undefined. Every Pi halves we would have undefined not whole integers of π, but at every fractional integer of π. At 0° our tangent was 0. Between 0° and π halves, all of my tangent values were positive. If we thought about zero to -π halves, IE these down here on the unit circle, the tangent values would have all been negative because the cosine would have been positive and the sine would have been negative. This is a periodic function. It means it repeats, and it's going to repeat every π. So the domain here is going to be all reals that capital R except when Theta can't equal π halves plus K Pi where K is an integer. If we look at the range, what are the values on this graph that it can have the output be? It's going to be negative Infinity to Infinity and the period here, how often does it repeat? It repeats every π domain and range for cotangent. By definition, cotangent is 1 divided by tangent or cosine over sine or X / y. In our unit circle we have our tangent values being 0, root 3 / 3, one root 3, and undefined for our zero Pi 6, Pi fours π, thirds, Pi halves or zero thirty degrees, 40-5 degrees, 60° and 90°. So when we take the inverse of that, 1 / 0 is now undefined 1 divided by root 3 / 3, which would be 3 over root 3. When we rationalize it and simplify, we get root three. 1 / 1 is 1 1 divided by root 3. Once again, when we rationalize it, we get root 3 / 3 and then 1 divided by undefined. We can't really do. But if we thought about that undefined was really our 1 / 0, so now we're just switching our X and our Y values. So now we have 0 / 1 which really is 0. So if we thought about graphing this where we have Theta along the horizontal line and our cotangent Theta on the vertical at zero, we had undefined. So our asymptote now is going to occur at zero. It's also going to occur at π and 2π and negative π and -2π because it's a periodic function. From zero to π halves, our cotangent values, not including zero, are going to be positive. So we're going to be close to the asymptote at Pi halves, we get 0. From π halves to Pi, we're going to have negative values, but we have the symmetry. It is a periodic function, so that's going to repeat over and over again. The period here is going to be π. It repeats every π. Our domain is going to be all real numbers where Theta can't equal K * π where K is an integer. Our range is going to be all real numbers or negative Infinity to Infinity. If we look at our cosine graph and we think about how often does it repeat. If we look at the highest value here and we look to see when does the next highest value occur, that tells me that that is a cycle and the cycle repeats every 2π. This is an even function because it has symmetry around the Y axis. So here, this point here is the same as this point over here. So cosine of negative Theta equal cosine Theta for a sine graph. If we look at this point here at the origin where we're starting to go up in our graph, and we look for the next time that we cross the origin when we start to go up in a graph, that tells us our period here is 2π again. This time it's an odd function because our Y value here is the opposite of our Y value over here. So sine of negative Theta equal negative sine Theta. Now our secant and cosecant are very similar. Our secant graph, if you recall, we just put in asymptotes every time our cosine graph went through the axis because we can't divide by zero and then they share the highest and the lowest points. So the black graph is the secant where the red graph is the cosine. For the secant, it's going to repeat. If we look here, the next time we have that point at 1 occurs at 2π. So the period for secant and cosine are going to be the same. They're also both going to be even function. So secant of negative Theta equal secant Theta. For our cosecant graph, we put in our asymptotes every time our sine graph crosses the axis. The sine graph here is being the blue, the cosecant graph is being the purple. The period from this point to this point occurs every 2π, so it's an odd function. Actually, asymptote to asymptote, or if we go from here to here, it'd be π halves to three Pi halves, which is a distance of 2π, so the period's 2π. It's an odd function. Cosecant and negative Theta is going to equal the negative cosecant Theta for our tangent and cotangent graphs. Our tangent graph, remember was sine divided by cosine. It repeats if we look at where it crosses the axis here to where it crosses the axis again, its. Is π, not 2π. It is an odd function. So our tangent of negative Theta equal negative tangent Theta. If we think about our cotangent graph, our cotangent graph was cosine divided by sine. So it looks like this. If we thought about the asymptote being at 0, the next asymptote occurs at π, so its. Is π. It's an odd function, so cotangent of negative Theta equal negative cotangent Theta. Pythagorean identities In our unit circle, we knew that we could put any right triangle in, and that because it's a right triangle, X ^2 + y ^2 equal 1 ^2. But if it's a unit circle or X really represents our cosine value and our Y really represents our sine value since our cosine was the adjacent side of the right triangle over the hypotenuse or X / 1, and our sine was really the opposite value over the hypotenuse or y / 1. So if X ^2 + y ^2 equal 1 ^2, then cosine squared Theta plus sine squared Theta has to equal 1 ^2, 1 ^2 being one. We have two other identities that are also considered Pythagorean identities. If we took this cosine squared Theta plus sine squared Theta equal 1 formula and we divided each of the terms by cosine squared Theta, we would get cosine squared Theta divided by cosine squared Theta which is 1 sine squared Theta divided by cosine squared Theta which is tangent squared Theta equaling 1 divided by cosine squared Theta which is secant squared Theta. If we took the original cosine squared Theta plus sine squared Theta equal 1 and divided each term by sine squared Theta, we would get another Pythagorean identity, which would say that cosine squared over sine squared is cotangent squared Theta, sine squared divided by sine squared is 1 and 1 divided by sine squared is cosecant squared. Sometimes it's easy to remember that in one equation the S and the T go together, so the secant and tangent. In another equation the C's go together, the cotangent and cosecant. So we actually have 3 Pythagorean identities. It's cosine squared Theta plus sine squared Theta equal 11 plus tangent squared Theta equals secant squared Theta and cotangent squared Theta plus one equal cosecant squared Theta. Thank you and have a wonderful.