[Music] hello AP pre-cal kids this is Mr Bean I'm going to fill in for Mr bruss for this very first lesson of the last half of unit 3 on the tangent function today Mr breast will cover the rest of unit 3 I'm just doing this one lesson here a little Cameo here so uh I want to remind you when we get into tangent tangent if you remember the the SOA TOA stuff remember that tangent is the TOA so that is where we have opposite over adjacent and if I were to draw a quick little triangle right triangle where we have Theta there then it would be the opposite divided by the adjacent now I'm just reminding you that really fast so you can see how this fits in on a circle on a unit circle in particular so if we take array so draw yourself a quick little aray here and it doesn't really matter it can be exactly like mine if you want uh in that angle and if we just put this array anywhere somewhere in here and then we're going to label the point of intersection on the circle itself as P so we'll call that P then we have the ability to figure out for this angle here so this is the standard position angle Theta then tangent is the opposite in this case this y value here divided by the adjacent which in this case is the x value so if we have y / X when we're doing this on any line that y value divided by that x value from the origin that gives us slope so tangent is the slope of the terminal Ray on a circle so if you draw any angle in standard position then the tangent of that angle is actually the slope of that terminal rray okay that's an important thing because that's the way some of the questions will be be phrased is they say find the slope of the terminal rray you have to know that just means you're finding tangent okay so now another thing is the slope of the terminal array can also be described by the ratio of this change in y over change in X you can also think of tangent as the ratio of s and cosine therefore we have tangent theta equals sin Theta / cine Theta so that is a quick way of being able to figure out what is tangent it's if you know s and you know cosine you just do s divided cosine as long as cosine does not equal zero because if cosine equal zero denominator would be zero and then the universe blows up and we can't do that so tangent would be undefined okay so moving on let's put it into practice so here we have a problem that says in XY plane you've got an angle Theta in standard position and the angle has a me measure of pi over 3 what is the slope of the terminal Ray of the angle so when you see that what's the slope of the terminal Ray it just means what is tangent so I drew myself a quick little circle put it on a coordinate grid and then just think where is pi over 3 this is just to help remind us which of these three dots is pi over 3 it's the top one there it's this third one so what is the coordinate point of that pi over 3 that coordinate point is2 comma < TK 3 /2 so if we know the coordinate points of these angles on a unit circle that will help us quite a bit because then all we have to do is say that this is going to be the S value divided by the cosine value or in this case it would be the < TK of 3 /2 /2 and whenever you have these you know how you're going to have these twos on bottom the one Hales those one Hales will cancel and so all you're going to have is the square < TK of 3 over 1 or in other words square otk of three so the slope of that terminal array right there that terminal Ray slope is the square otk of three okay uh next up we're doing exactly the same thing so finding the tangent values is the same as finding the slope of the terminal rray these two things mean exactly the same thing it's just worded differently so if you see evaluate tangent of an angle it's the same thing as saying what is the slope of the terminal array all right so let's do this one together real quick what is the coordinate point of pi over 4 so that is going to be square < TK of 2/ 2 comma Square < TK of 2 over 2 so if you can write down the coordinate point that'll probably help you quite a bit and then you're just saying oh well that's just the Y divided by the X2 over two divided by2 over two that equals one super easy okay 5 pi over 6 where's that one 5 pi over 6 is Right close to there somewhere around there something like that it's almost at Pi so that is < TK 3 over2 comma and then what's the Y value 12 then we're doing y / X so we get 12 / < TK 3/2 and then that's going to equal the halves just cancel so all we have on top is a one /un 3 now this is correct answer 1 ID3 or just 1 over 3 but if it's a multiple choice exam often they may just rationalize this so you they would multiply top and bottom by square Ro of three to then get a negative the negative is from right there Square < TK of 3 over three that's how you rationalize these fractions again this answer right here is correct but you want to practice rationalizing just in case you you don't see if this might be the answer on the multiple choice and you think this is the answer and you don't see that these two things are equivalent then that's a problem you just want to know that these two things are equivalent okay next is uh tangent of 3 Pi / 2 where is 3 pi over 2 on the unit circle it is right there on bottom 3 pi over 2 and so what's that coordinate point that is an x value of zero a y value of1 and so this then we're going to have --1 over 0 uhoh that is undefined we cannot divide by 0o so tangent at 3 pi/ 2 is undefined nothing there all right that's how we find tangent values that's how we find the slope of the terminal array so let's now move on to some other stuff all right here we're going to get ready for what in the world is tangent doing like what's the graph of it doing let's start with a unit circle I'm going to take this ter Ray and I'm going to rotate it around the unit circle just to get a feel for what tangent is doing so remember tangent is the slope of this thing so what do the slope start off with when we're in standard position right at the very beginning the slope starts at zero slope of a horizontal line is zero so now as we start moving around the unit circle what's happening the slope is positive and the slope is positive and more positive and more positive it's getting really really steep what's happening to it gets larger until it approaches Infinity it's approaching Infinity as it gets bigger and bigger and bigger and bigger until we get right to Pi / 2 now all of a sudden you have this vertical line what's the slope of a vertical line undefined so we now have undefined there then what happens next as we keep going the slope right there as we pass that point the slope is now negative and that's getting less negative less negative so the slope's actually getting larger so it was super negative and it's getting larger as we get towards zero once we reach right here what is that angle right there that angle is pi so once we get to Pi we are now back to a slope of zero again and then we're going to keep going and so as soon as we pass that what is happening we have a positive slope the slope will be the same as it was in quadrant 1 this is now the same slope now you can see here if I extend this line you can see that would be the same exact slope as when I was way over here in quadrant one and it's just going to now repeat itself so if I keep going around here and I get down it's the same as when it was way up here in quadrant one so it's a repeat so once I made all the way here to Pi it's going to start over and do the same exact tangent values again and then it's going to start over and do the same tangent values so how often does that happen every one half Revolution every half of a circle that's when the T the tangent function repeats so the period of tangent how far did we have to go for the period the period is only Pi not like s and cosine s and cosine was 2 pi you had to go a full circle to get a repeat values tangent you only have to go half a circle and then it starts repeating again so period of tangent is pi now one thing is remember how we had this undefined thing so where did the undefined slopes happen it happens when cosine is zero so that's my next part here cosine theta equals 0 when excuse me when does that happen it happens at these angles Pi / 2 and then it happens again at 3 pi over 2 and then it happens again at 5 pi over 2 and it just keeps happening in that you just add Pi add Pi add Pi so we're going to have a way to say that here in just a minute we'll talk about that but that is going to be where our vertical asmp tootes are so what you're going to do now is you're going to sketch some vertical ASM tootes real quick that will look like this there's one vertical ASM toote right at pi over two yours won't look as cool cuz it won't be red uh and then well maybe some of you will and then we're going to do another one where's the other one at what 3 Pi / 2 and so notice that we have a repeat every Pi values I could go the other way to negative Pi / 2 so these are all the places where cosine is equal to zero and cosine being zero causes tangent to be undefined so there's our vertical ASM tootes so that's an important thing for you to see where these vertical ASM tootes are when we start sketching things you're always going to start sketching the vertical ASM tootes first if you can't do the vertical ASM tootes it's going to be really hard to sketch the graph of these all right so then next up where are these vertical ASM tootes well this is how we say that they appear every pi over 2 plus K Pi for integer values of K you might sometimes teachers or books will say n Pi it doesn't really matter what that variable is that that that value is as long as they say the same thing I think Mr breust in his lessons is probably going to change it to an N don't freak out about that it's the same thing okay so K Pi 2 plus K Pi for integer values of K so let me go back there so that just says we're at pi over 2 and we're adding a pi we add one period to get all the other vertical ASM tootes and you just keep adding periods over and over again but what if it's not tangent Theta what if it's tangent B Theta like there's a little constant in front of of little coefficient in front of theta then it changes the period the period is then pi over B that's the period of tangent where you have a coefficient here please don't get this confused remember when we did s and cosine s and cosine was 2 pi over B we've done that a whole bunch for sinusoids do not get that confused with Tangent because with Tangent it's not 2 pi over B it is pi over B you've got to be able to Interchange between the two of them knowing when you're dealing with Tangent when you're dealing with s and cosine periods are different Okay so if you have this tangent of B Theta then that's going to change how often you have your vertical ASM tootes uh I'm going to explain this a little bit more here in just a second so just get this written down and then when we practice it it'll make more sense so here's these two things here the vertical ASM toote is going to appear at every pi over 2B so it's going to actually be half of the period is where the first vertical ASM toote appears and then it's going to repeat every period so it's just K times a period pi over B is the period of tangent so it's integer values of K you just do multiples of the period all right that might seem a little confusing let's just do a problem real quick here on here and I think this will makes sense okay so the period what's the period for this thing the period is going to be pi over B which is two so if I have I'm going sketch this real quick if I have a my vertical ASM tootes surrounding the origin here the tangent is going to be right in the middle of this in between these where are these vertical toes well if it's if the entire period is pi/ 2 that means this distance right here right there that distance is half of Pi / 2 which is pi over 4 so we have vertical ASM tootes I'm going to just abbreviate this by saying vertical asmp tootes at x equals a positive and a negative pi over4 that's where the vertical ASM tootes are so we're going to then say here's our formula we say the vertical ASM toote is at pi over 4 it's our first vertical ASM toote and then we're going to add every period is another one so we say k pi/ 2 for integer values of K all right that right here is the whole answer that whole thing oh squeezing it in oh I almost crossed that off that whole thing is the answer to this there's our little formula pi over 4 plus K * the period so we start with the first vertical ASM toote and we add just add periods to it and that gives us all of the vertical asmp tootes that for that function all right let's do another one so here what's this one the period is going to be pi over B in this case b is 2/3 well that makes this multiply by the reciprocal we get 3 Hales Pi so again what are we doing here we have vertical ASM toote there a vertical ASM toote there that whole thing is 3 Pi / 2 so half of that where's this first that distance from here to here here that distance is half of a period which would be 3 pi over 4 okay that's where that comes from so then I can say all right so I've got a vertical ASM toote vertical ASM toote at x equals there's a positive 3i over 4 I can just say 3 pi over 4 sometimes you might say plus or minus because that that one represents the negative one but I've got one of them at 3 pi over 4 so then I can say my answer to this is that I have a vertical ASM toote at 3 pi over 4 plus the a period which is 3 pi/ 2 so plus every K 3 Pi / 2 every period where I'm multiplying it by K for integer I can't write integer values of K there we go and then box that kind of a box that's awful all right there's our answers for how to do those hopefully that makes sense to you because you will have some problems on that and you'll see some of those in AP classroom as well okay now let's talk about some characteristics and get to the graph of these things cuz I know you love graphing all right characteristics of the tangent function this is what it looks like this is the original Parent graph of a tangent function so you see here I've got my tangent down here and what is happening each time between these set of ASM tootes after we figure out where the ASM tootes are the function is always increasing right it's never going down here it's just always going up and then its graph is going to change from concave down to concave up so notice this this is concave down and then right here it changes and it becomes concave up and that happens over and over again it repeats itself that point where it changes concavity that is remember that's called a point of inflection that point I'm going to fill in as a DOT is really important because we first find the vertical asmp tootes and then we find the point of inflection when we're trying to graph these things vertical ASM tootes point of inflection and from there we can sketch the graph of it so those are the important things things being able to look at these graphs all right with that we don't just have a whole bunch of really easy graphs though right we're not going to graph tangent Theta over and over again we're going to do things to it we're going to multiply it by a number in front multiply the Theta on the inside add or subtract something to it and then add something at the very end we're going to have all these different transformations of the original function so what do all these things do I think a lot of this you're going to already be able to recognize because it's the same thing we've been doing with all the different types of functions so what's a do in front it creates a vertical d it stretches it up and down or it shrinks it up and down so it stretches or shrinks vertically by a factor of the absolute value of a now if a is negative if a is negative what's it do it flips it right so there's a reflection over the x axis okay so those things nothing new there uh I will talk about a again when I get to the graph I'll talk about how a is also helpful on the graph uh the constant B what's that do the B on the inside that gets a little bit trickier that creates a horizontal dilation right and it's always the opposite so it changes the period by a factor of 1 over B so if B is large the period shrinks if B is smaller than one like between zero and one if it's a small number the period actually grows so it does the opposite of what you'd expect um and if B is negative oh yeah if B is negative there's a reflection over the y- axis it flips left and right uh for uh the reflection process all right constant C remember this so we have a plus c on the inside that creates a horizontal translation we're now shifting it left or right and it's always the opposite so in this case it would it says plus C there it's a minus C so nothing new here we've been doing this stuff over and over again and then the constant D at the very end this plus d it's going to create a vertical translation it's shifting it up or down by D units okay so now we're going to take all this information put it to use and graph here we go what should we start with there's there's all these things going on we need to know where the vertical ASM tootes are identify the vertical ASM tootes shift them according to any shifts that we have for space shifts and then we can get points of inflection so vertical ASM tootes we've already practiced this let's take the period figure out what that is so the period is going to be oh wait wait wait sorry if you already wrote that let's do something else first let's take uh this two and Factor it out so that we can see what the phase shift is so 2 * Theta minus this just leaves us with pi right no that's wrong so Theta minus what is it oh pi over 4 oh that was tricky yeah because now if I distribute the two I would then get back up here all right so now that helps us with the phase shift so now let's do the period the period is pi over B not 2 pi over B just pi over B in this case a two so if we have that that means I have a vertical ASM toote at half of that right so I have a vertical ASM toote at x equal half of that pi over 4 so don't graph it I'm just going to show you where that would be that would be right there again do not sketch this yet because I want you to see what we're doing um so then from that what else do I have going on I have a few things going on I need to say that it is a uh a shift right uh pi over 4 so just one line and then I also have a uh it's going to go up one okay so here's how we do that let's take our vertical ASM toote and I'm going to now you want to sketch this we're going to move it to the right a pi over4 which is just one line right so there I have that now where was my other vertical ASM tootes uh let me show you let me pull this back the vertical ASM tootes would be here and here maybe that's how you should be able to see it cuz that's a period of pi/ 2 that whole period positive pi over 4 negative pi over 4 vertical ASM tootes now we're going to take these and shift them right pi over 4 to there almost right there okay so now that's where I going to take and draw more vertical ASM tootes so it looks like we're going to have vertical ASM tootes every two lines which is every uh pi over 2 so sketch all of those okay that's a lot of vertical ASM tootes okay now what we've got uh let's find the point of inflection so the point of inflection will be right in the middle of these things but um we had to shift it up one right so this up one comes into play so there's my up one that's a point of inflections go ahead and Dot all the points of inflection real quick okay your graph should look something like this so far that's what our sketch is now we're going to remember that tangent is going to go up it's going to go concave down and then concave up what I'm telling you here is that the a value helps us with a quick little sketch the a value is 1/2 and this is kind of a quick a nice little trick to help with the sketch if you go halfway from the point of inflection to the vertical ASM toote halfway in between and then you go up a units which in this case is a half that is actually the value right there so halfway between the point of inflection and the vertical ASM toote you go up a units so you can do that real quick and then you do the same thing the other direction so you go the other direction but You' go down a units and so then it would be like that go ahead and put those dots real quick and then from here it's sketching now be careful here this is not a straight line right we're not just connecting these we want to go on the right side here it where it's going up it's going to be concave up so just kind of try to make it uh if you're any good at it concave up for each of those and then on the left side it's concave down and you just try to hit that thing and then the ver the vertical ASM toote will will push your graph down so again it's just a sketch uh the vertical ASM tootes are very important important important and the uh and the point of inflection you want to make sure those are accurate for sure and then from there it's just kind of a little bit of a sketch for understanding the graphs all right number eight this would be a really good one for you to pause this video try this one on your own I will then have the answer with all the explanation to see if you get the same thing as I do okay how' you do this is what it should look like if you got that you're crushing it you're ready to move on if you didn't get that just look over what I have I've got the period here is 2 pi uh vertical ASM toote at plus or minus Pi but then I had to instead of the vertical ASM toote being here at pi and negative pi I had to shift it left Pi so that's why everything shifted over the left and then another trick was that there was a vertical uh this negative here in front caused it to be a vertical reflection so it's decreasing instead of increasing for this tangent function uh the three again also that three there in front helped me with these other coordinate points halfway to the vertical ASM toote helped me to identify where those other points might be to help with the sketch okay that's everything this is Mr Bean signing off rock that Mastery check I will not see you back until unit 4 for those of you who are doing unit 4 uh so uh good luck with Mr breast and I'll see you later