Transcript for:
Polar Graphs and Polar Coordinates Overview

[Music] what up what up if you are watching this you are beginning unit one of AP precalculus we're going to go through these four sections so uh let's learn a few things and let's practice a few problems and see how long this takes probably about an hour give or take this is a periodic function a periodic function is basically something that repeats cycle over a certain amount of time now when you're looking at a periodic function just from a graphical point of view you're looking at a few things you are looking at for a its period which means how long does it take to complete a cycle okay so the period is basically going to be the distance from this x value to that x value another thing that we'll care about is the amplitude the amplitude is not the very tippy top to the the very bottom but the very tippy top minus the very bottom which gives you the total distance but divide that by two so the amplitude is going to be half the distance from the top to the bottom whatever that distance is periodic functions repeat themselves so a lot of problems are going to be like hey look at this guy but give me an answer of something down the road in part of the graph that you can't see we'll deal with that when we have to deal with that that's a periodic function okay the most basic kinds of periodic functions are s and cosine now before we get into anything different let's look at the most basic version of s and cosine that you may have learned back in the day you probably remember SOA and what SOA meant back then was s of an angle meant its opposite over its hypotenuse and the C part of soaa meant the coine sign of an angle equals the adjacent side over the hypotenuse now when we look at this picture right here the sign of my angle is going to equal the opposite over R we're going to look at sign in two different ways and we're going to look at cosine in two different ways way number one is what if you're dealing with something called the unit circle what if you're dealing with the unit circle the unit circle is when r is equal to one otherwise we'll deal with everything else so the old e e everything else now when you have the unit circle and R is equal to one s of theta is going to equal y its opposite side over its hypotenuse R so sine theta equals y/ r which is y over one which is y so sin Theta is just the Y value oh how nice I wonder if cosine will give us something similar I don't know let's get a different color and find out cosine is the adjacent side over the hypotenuse So Co Theta is going to equal the adjacent side which is X over the hypotenuse which is R but R is one so X X over 1 is X so cosine Theta is equal to just the x value so good news if I give you the unit circle and you know R is equal to 1 you have an X and A Y value of cos x or cos Theta sin Theta your x value is cos Theta and your y value is sin Theta well how nice well what about these everything else problems that we were talking about over here where R is just some number that we don't know about well if R is just some number we don't know about then sinine Theta is just going to equal y over R if I want to kind of recreate this guy over here and I multiply R to both sides that gives me R sin theta equals y oh and as we probably are going to figure out if R sin theta equals y I could do the same thing with cos Theta cos Theta is going to be X over R and If I multiply both sides by R then R cos Theta is my x value so when I give you problems where the radius is not one I could use this point here so very often very often we will be given problems s that have to do with coordinates sometimes the radius is one sometimes it's not and if the radius is one we will refer to the X as our cos Theta the Y as our sin Theta and if the r is not one then we will refer to the x coordinate as R Co Theta and the y-coordinate as R sin Theta so that's basically the basic definition of what s and cosine means from a circular triangular point of view now speaking of unit circle there she is probably the most important thing that you could possibly learn in pre-calculus is the unit circle let's start off very basic the unit circle assumes that R is one thank you Jesus now if R is one that gives us the point right here of one Zer that gives us the point right here of 01 that gives us the point right here of -10 and that gives us the a point right here of 01 now with the unit circle we have measurements that we care about we know that an angle is measured starting from the right and working our way counterclockwise okay an entire circle in radians is 2 pi so you are zero radians if the whole circle is 2 pi radians that means half the circle is regular pi and half of half a circle is Pi / 2 and 34 of 2 pi is going to be 3 pi over two now we also need to come up with these guys you are one3 of my way to pi over two you are half of my way to pi over two and you are 2third of my way to pi over 2 so this angle right here uh multiply 1/3 to Pi pi over 2 gives you pi over 6 okay pi over 6 multiply 1/ 12 to pi over 2 and you have pi over4 and multiply 2/3 to pi over 2 and you have pi over 3 okay this is going to be very much the same thing okay we are at at pi over two and we're making our way to Pi so oneir of our way between pi over 2 to Pi is 2 pi over 3 halfway in between pi over 2 and Pi is going to be 3/4 PI and 2/3 of our way from pi over 2 to Pi is going to be 5 pi over 6 now we're at pi and the easy way of figure figuring these guys out is just add pi to all six of them okay add Pi so add 6 over six or add yeah add six over six to that that gets us s pi over six add pi to that that's 5 pi over 4 add one to that you get 4 Pi over3 add a whole pi to that and you get what is that 5 Pi over3 Okay add 4 pi over4 to that you get 7 pi over 4 and add 6 pi over 6 to this and you get 11 pi over six all right so there's our angle measures probably my least favorite part okay now the rest of it is coming up with the actual coordinates now when we deal with coordinates we always think of it like this this way think of it as the square root of something over two the square root of something over two starting from the x axis and working your way up you're going to start with three so < tk3 over two < tk2 over 2un one over two and that's going to be your x value so this is going to be < tk3 over two this is going to be < tk2 over 2 and this is going to beunk 1 over2 but the square root of one is just one so 1 over two how do I figure out the Y values do the same thing in Reverse < tk3 over two < tk2 over two < TK one which is 1 over two and now I make my way around keeping in mind that I'm now in different quadrants okay these guys are the basically the same exact distance and the same EX exact y value so U are going to be2 because it's a negative x value but my y value stays the same okay you line up with that so the negative x value because we're in quadrant 2 < tk2 over 2 < tk2 over two okay and now U are negative < tk3 over 2 which makes sense because you're closest to the x axis and you are one hand okay these guys match up with those guys similarly these guys match up but now I'm in the third quadrant which means both my X and my y are negative so you are going to be < tk3 over2 the x value does not change but the Y value is now -1 over two you are basically the same as that which is < tk2 2 over 2 but also netive < tk2 over2 because we're in the third quadrant which everything is negative you line up with that so negative a half < tk3 over two negative tk3 over2 don't want to forget that to complete this process you guys line up with that but now we have positive X's again so you are 12 negative root 3 over 2 you are positive < tk2 over 2ga < tk2 over2 Quadrant 4 is positive negative and you are uh < tk3 over 2 one2 that's the unit circle it's a mess but it's useful it's very useful all right s and cosine graphs we're just going to basically draw out the very basic forms of the sign and cosine graph let's start out with s again everything's periodic s starts out at 01 okay Stein starts out at 01 and it Peaks out at regular one bottoms out at negative one and has a period of two Pi so it's going to start down here it's going to peek out and then bottom out at regular Pi which means it's going to peek out at half a pi when it comes back down here it bottoms out at 3 pi over two and then comes back up and hits the x axis again at two Pi okay so attempting to draw these is never a good idea for me but I'm going to give it my best shot start out at 0 p PE out at one come back down to Pi come back down to -1 and then come up to two Pi get what you pay for when you watch my videos but that's what sine X looks like that's what sin x looks like cosine has very very very similar behaviors but this guy starts out at 01 Okay so you start out at 01 and you bottom out at negative 1 eventually but we don't bottom out yet we start out at 01 we hit the x axis at pi over two we bottom out at Pi we come back out and hit the x axis again at 3 pi over two and then we complete its period process at 2 pi okay so drawing this guy out we start out at 01 we hit the x axis at pi over two I missed it I missed it I missed it see now no one will ever know uh bottom out at Nega one at Pi hit the x axis again at 3 pi over 2 and then repeat the process at 2 pi so that's the difference between sin x and cosine X okay so we're going to need to keep the these in mind as we answer a buttload of problem s that are coming our way starting now all right oh yeah oops tangent what is tangent we might see tangent once or twice tangent is literally whatever s is divided by cosine is that is the tangent value you don't see tangent a ton until the second half of unit two but you will probably see a teensy teensy teensy bit of tangent uh every now and then so tangent is just s divided cosine tangent is just s divid cosine tangent is just s / cosine let's move my face now we're starting right off the bat with a periodic graph as you can see right there right away I ask you what the period is which means if I were to pick like a spot like right there how long will it take me to get to the same exact spot well here so this peak occurs at seven and then re occurs again at 15 so if you're wondering what is the period you do 15us 7 which is eight okay that's the period how long it takes to complete a cycle and this Cycle takes eight units to do just that what is the amplitude Okay well the amplitude is the very very very tippy top to the very very very bottom but divid div that by two okay so the very tippy top is five the very bottom is one so the length from top to bottom is 5 minus one but you're going to divide that by two and four / two last time I checked is two now we need to plug in 41 well I don't have 41 but this is what I know this is what I know I know my period is is eight okay so whatever happens at one can be found by doing this 8 divided by 41 okay uh or 41 technically divided by 8 that goes into a five times that's 40 remainder of one now why is that important your remainder is telling you that 41 behaves in the same exact way as regular one at regular one when X is regular one I'm at three so F of one is three which means F of 41 behaves the same exact way because if I were to cut this guy and move it over a bunch of times we will end up with uh that guy right there and we can kind of see that because this ends at 20 so if I were to bump this over again and then shift it over one more we can see that one right there so this kind of does work out looking at it graphically if I want to look at it graphically but that's how you find out if you can't look at it graphically I can't extend that graph anymore because I'm out of space I just can't do it it's like my NASA hat I'm out of [Laughter] space oh oh man let's move my face again all right given an interval for the measure of alpha oh what are we talking about me I'm not Alpha the measure of alpha is oh this whole angle right there they want me to find sin Alpha they want me to find COS Alpha and they want me to find tan Alpha o this is the unit circle how do I know that well because the radius is one so if I'm using the unit circle I can use the fact that COs in this case Alpha is my x s in this case Alpha is my y I have a point of 0.6 0.8 so my S Alpha is just my yalue and my y value is just 0.8 so sin Alpha is just 0.8 similarly Co Alpha is just my x value and my x value is just 0.6 now to find tangent we do s divided by cosine so tangent is going to be negative uh 0.8 divided by 0.6 so that's the same thing as basically saying 8 over 6 which is 1.2 no 1.3 repeating so my tangent is going to be -1.3 repeating or-1 and A3 okay so when I give you the unit circle and I ask you to find s and cosine in your given points all you have to do is just relate the sign and cosine to the points that you're given easy peasy lemon square squeezy oh but what if it's not the unit circle well let's see what we got this next one we don't have the unit circle the radius is three how do I know that well if this point goes over to three Z that means the radius is three don't overthink it and when the radius is three that means my x value is now radius cos Theta and my y value is radius s Theta okay that's my X that's my y I'm also given this point down here that I care about where Theta travels all the way over here and so the point that I have is -1.1 -2.7 N1 so I kind of wish I made this a calculator problem but I'll use Siri for it because the numbers aren't miserable it's not like I need to do like the sign of a number or anything wacky like that my job is to find sin Theta well I know from this information over here two things I know R cos theta equals x which means if I divide both sides by R cos Theta equal x over R just kind of like what we saw in that second or third slide when we began this video similarly and this is what matters here is that my sign Theta is going to be R sin theta equals my Y and if I'm being asked to find sin Theta I'm going to calculate whatever my Y is over my R so sin Theta is whatever my Y is which is -2.7 N1 over R which is three so -2.7 91 divid 3 the answer is approximately .933 9303 so we'll leave that as that okay so probably should have used a uh a calculator but it's too late now uh cosine is found by doing X over R so cosine Theta is my x value which is -1.1 ID 3 -1.1 / 3 the answer is approximately. 3666 I don't like that I don't like that but thankfully for rounding we don't have to be in fear okay so there's my co Theta and Tang Theta is basically the same thing as U divided U so 0.930 divided by 0.967 so 0.930 divided negative 0.367 it's going to be positive it's approximately 2534 2534 I like it I like it I like it a lot all right now we are being asked to find the ratio of each expression now what I'm going to do is I'm going to use that unit circle that we so eloquently did uh a little while back I'm gonna try to draw that holy cow that looks awful I'm keeping it I'm keeping it all right and what we're going to do is we're going to label the points based off the angle measurements that I Give You Now reminder that Co Theta is our x value sinine Theta is our yalue so when we look at these we are thinking oh y thoughts oh X thoughts blah blah blah blah blah blah blah now 3 Pi over4 if you remember here was zero here was Pi no it wasn't that's half a pi here was pi and here was 3 pi over 2 3 Pi over4 is literally that Center spot right there okay 3 Pi over4 was that Center spot right there that gave us a point of < tk2 over 2 and < tk2 over two because that's how the center spots work in this case the x value is negative because we're in quadrant 2 I'm asking you to find the S of 3 pi over 4 which which means I care about the yalue so the S of 3 Pi over4 is the yalue created by that angle which gives us the y coordinate < tk2 over two all right on to the next one cosine of 3 pi over 2 3 Pi / 2 is a little bit after pi and if you remember that 3 pi two is actually right here there it is so right there so that point right there is actually the point 01 I want the cosine of it which is the x value so the cosine of 3 pi over 2 is the x coordinate of 01 which is zero how about that catch me outside how about that remember her bad baby all right sine of 11 pi over six that is just before 2 pi and if you remember there's three dots here that's the third dot okay whenever you are coming up with your numbers when I'm thinking about this I always think when you start from the x axis your x value is going to be that < tk3 over two and then your y value is going to be the opposite number which is one in this case now since this is Quadrant 4 Y is negative I'm asking you to find the S of 11 pi over 6 which means I need the Y value which ends up being one2 okay last but not least cos 4 pi over 3 I'll use black 4 Pi over3 is in between here and here and that would be the third dot the third dot of this guy okay now the third dot of this guy if we're thinking X values that's root3 over2 < tk2 over2 that is root one over two so one over two and then the other one is < tk3 over two X and Y are both negative so cosine wants you to find the x value and the x value is - one2 in this case all right so that's how you use the unit circle to come up with values or ratios and stuff like that if you're given a bunch of ratios which I am all right the measure of angle C OD is 2 pi over 3 radians so you are 2 pi over three which brings you to the first dot of the three dots there okay so C is going to be the first dot find the coordinates of C here's issue number one the coordinates of c are not as simple as me saying oh well this is a unit circle it is a unit circle but the radius is now four so I have to treat this as R cos Theta R sin Theta now if this was the unit circle since that is the first dot of three that gives me negative a half regular < tk3 over2 negative positive because it's the second quadrant however that's not exactly what I'm doing here if this was the unit circle cos Theta would be negative a half and sin Theta would be root3 over2 my R here is four so I'm going to do 4 * - 12 R which is four again * < tk3 over 2 so that gives me -2 and 2 < tk3 over two No 2 < tk3 sorry 2 root3 manageable manageable B is halfway in between a and C all right so if B is halfway in between a and C that gives you the angle 5 pi over 6 you're my third dot find the coordinates of B all right well if you're my third dot that means you are literally the flipped version of that so < tk3 over two uh negative and then2 however I need to multiply those guys by four so basically it's very very very similar to what we see up here 4 * < tk3 over2 four * a half is going to give me2 < tk3 two yeah yeah yeah yeah let B be the coordinate 5 < tk2 and regular 5 < tk2 find the measure of an angle rotated counterclockwise from the positive xais to B all right okay so obviously we don't have a unit circle because what this is supposed to be is first things first points are supposed to be something over two so I have an idea let me take this point and multiply this by 2 over two so I have negative so if I multiply You by 2 over two and I multiply You by two over two that looks a little bit more familiar okay so that would be -10 < tk2 over 2 and you would be regular 10 < tk2 over two now since this isn't the unit circle I'm comparing this to R cos Theta and you R sin Theta okay so what basically this is telling me is this is telling me right here that my like and let's focus on the sign first because the negative might be a bit confusing and you might look at this and be like whoa whoa whoa what did you just do root2 over two I know what that is I I've I've seen root2 over2 all over the unit circle so if I just focused on the fact that sin Theta is < tk2 over two that leaves me with a radius of 10 so similarly over here since this is negative I can't have a negative radius so I'll just pretend that the radius is 10 and the cosine is < tk2 over 2 so I don't have I'm I'm not asking you to find the radius but the radius is 10 however this is now saying Co Theta is < tk2 over2 okay in its negative form and sin Theta is < tk2 over2 so when that happens that point and of course this is not drawn the scale but that point is that direct midpoint between these two guys so I have three dots B is the middle dot which makes this 3 pi over 4 so you are 3 pi over 4 what is the distance from zero to B Ah that's my radius my radius is 10 figured it out already what is the distance from A to B if the distance is vertical okay so you are literally on top of the other okay so if you are a y value right if the distance here is a yalue then you are < tk2 over two if this is the same exact guy as that then you are also a value of < tk2 over 2 and so I have < tk2 over 2 plus < tk2 over 2 which is < tk2 * 2 over 2 which is < tk2 did you catch that very exciting very exciting very exciting stuff I hate this problem oh boy I hate this problem but they I've seen this problem all over like AP classroom and stuff like that so I'm assuming it might show up somewhere I don't know who knows we're going to do it we are given an angle Theta in standard position as shown in the figure below so boom here's my Theta the function G is given by G of a equals sin a so this is a s value uh for the angle Alpha not shown Theta is smaller than Alpha and Alpha is in between Theta and 2 pi so here's 2 pi here's Theta Alpha is somewhere over here okay Alpha is that area over here I don't know exactly what it is but it's that area over there I need to figure out what happens if I plug in Alpha and Theta so let's view this from graphing the S function point of view I care about the sign function from 0 to 2 pi s starts at Z 0 Peaks out at one bottoms out at negative one uh it hits the xaxis again at regular Pi it uh repeats itself at two Pi which means it Peaks out at half a pi and it bottoms out at 3 pi over two all right so let me attempt to draw this out which I'm always bad at let me attempt to draw this out which I'm always bad at not too bad if I say so myself now Theta is basically anything from zero all the way and Beyond 3 pi over 2 so Theta could be anything from here to like here could be Theta so this new guy Alpha is going to be between here and here so whatever I plug in for Theta whatever I plug in for Theta okay I could plug in well Theta is right here so if I plug in Theta I I get some number if I plug Theta I get like negative something okay so G of theta is like negative like point8 or something whatever I plug in for Alpha it's going to be to the right of it which means it's going to be bigger than whatever I plug in for Theta no matter what so G of alpha is going to be greater than G of theta because anything to the right of theta no matter what is going going to be higher so I hope I explained it well I feel like I didn't but um I don't know it's just a weird problem but I've seen like there's three problems on AP classroom like this like when you as a teacher when you assign homeworks there's three that look just like this and the other three don't so like they clearly AP classroom likes this type of problem so I don't know will it show up on an exam who knows but don't be surprised if it don't all right f ofx equals cos x let's just draw it out f ofx equals regular old cosine X okay cosine starts at one okay hits at uh half a pi bottoms out at Pi uh comes back up hits again at 3 pi over two and repeats itself at two Pi okay so cosine looks like gross where is f ofx cos x concave up on the interval 0 to Pi concave up look like this so this happens here so this is concave down then the switch of concavity happens right at pi over two concave up concave up concave up and then the change of concavity happens at 3 pi over 2 now at pi over 2 and 3 pi over two it's neither concave up nor concave down so we will say that this is concave up on pi over two to 3 pi over 2 when does f ofx cos x equal 1 I'm not given an interval this is a periodic function which means it hits one at zero it hits one at 2 pi so it'll hit one at 4 pi and six pi and if I go backwards -2 Pi so basically I can word it out like this basically every two Pi is when it hits one okay so we will say 2K right 2 K Pi where K is an integer okay so if I plug in zero I get zero which is an answer if I plug in and by the way I should say x equals make it look better uh if I plug in one 2 * 1 is two 2 pi is another answer okay if I plug in two 2 * 2 is four four Pi is another answer this will work if if I plug in negative 1 I get -2 Pi that'll work boom is f ofx an even function cosine X is an even function and here's why two reasons it starts up here looks like this it has reason number one reflectional symmetry over the Y AIS whatever happens on the right side of the Y AIS the exact Mirror Image happens on the left side it has reflectional symmetry the other reason why is if I were to plug in F ofx or rather plug in Negative X I would get F ofx okay so yes for the two reasons that even functions are even functions reflectional symmetry over the xaxis and also because F ofx if I were to plug in Negative X I would get get the same exact value as if I were to plug in regular X like if I were to plug in 2 pi I get one if I were to plug in -2 Pi I get one okay and the same thing will always happen left and right so yes F ofx cosine X is an even function ah let's take a look at s s Theta so we kind of looked at s let's draw it again because it's good for you you even though I'm the one drawing it uh starts at0 0 Peaks out at one bottoms out at negative one uh it repeats itself every two pi times okay so it's G to go up here at half a pi it's going to peek out at regular Pi it's going to hit the x-axis at 3 pi over two it's going to bottom out at 2 pi it repeats the process okay so starts at z z Peak hit bottom hit repeat okay where is the rate of change of G of theta decreasing on 04 Pi it's making me think a little bit first off rate of change decreasing what does that mean concave down rate of change decreasing means concave down this is going to be concave down here from zero to pi over just regular Pi just regular Pi regular Pi regular Pi regular regular regular Pi now that is when it's from zero to 2 pi it's now concave up but when it repeats itself from two pi to 3 Pi it's going to be concave down again so we must include those 2 pi to 3 pi don't include them because they're neither concave up nor concave down because those are inflection points what did I just say inflection points I did just say inflection points you're an inflection point you're an inflection point you're an inflection point so it looks like inflection points happen at zero and pi and two Pi inflection points are going to be the X values that happen at every K Pi where k is an integer there you go now I can spell that makes sense uh one Pi yep two Pi yep 3 Pi yep zero Pi zero yep negative one Pi yep pumpkin piid what is the period of G Theta come on I think we know that by now all right oh this is not a calculator problem okay no calculators yet uh let h of x equal sin x and J of x equals cos x where on 0 to Pi do h of X and J of X have the same concavity so let me move my face over here because I'm going to need to draw both of these guys on the same coordinate plane this should be a lot of fun all right uh obviously we know the magic numbers we have Pi we have two Pi we also know the magic numbers of one and negative one okay uh sin x I'm going to let you be red sin x starts at 0 0 Peaks out hits the x- axis again bottoms out and then repeats I'm going to let cosine equal blue cosine starts here uh Works its way down bottoms out a negative one crosses and then repeats okay this is gonna be interesting I want the same concavity all right red is concave down between zero and pi and concave up between pi and 2 pi blue is concave down between zero and half a pi and then between half a pi and 3 pi over two is concave up and then concave down again for the rest so both of these are concave down so I'll say that I'll be specific concave down on the interval 0 and pi over 2 and both of these are concave up on the interval pi to 3 pi over two gross where is H of X red greater than J of x I can't do that without a calculator I can't do that without a calculator so is that I wonder if uh that I should have started calculator time there I mean I I created these problems I should have just started calculator time there so this is what I'm going to do uh I'm going to take a little side uh let's look at this problem this is a calculator problem you can't do this without there's no way um this is a calculator problem so uh it's basically going to happen where what do I want I want where red is greater than blue so it happens between the intersection right here and the intersection right there so what I need to do is I need to find out those two intersections and I'll answer it that way right uh y equals first off let's make sure your mode is set to radians it should be it is uh y = sin x and cosx like that uh I'm gonna go to zoom and make it Zoom trig because that it makes it all triy so there's my guys uh I wanted to find out this intersection point right here and this intersection point right there because the one that I wanted was greater at those moments so I need that intersection point so let's go to calc second Trace Click Five intersect and um oh it's G to intersect forever so I'm G to hit enter I'm G to make sure that I give it an interval otherwise it's GNA fall apart on me and by giving it an interval it gives me one intersection point of 0.705 three blah blah blah blah blah so the x value is going to be 0.7 what is that 85 or 05 that's an eight s85 okay n uh yeah yeah yeah yeah yeah yeah yeah yeah yeah all right so we need to find the other intersection second Cal intersect and let's make sure we go over here let's make sure we go over here and give it an interval because we have a bunch of intersection points otherwise it's not going to know what he wants so there's the interval okay so everything in between that and that guy gives me 3 9269 so 3927 is going to be uh my interval so let's go back and write that out all right so when we did that this guy right here ended up being Point uh 785 was what I wrote down over here that guy ended up being 3927 no 3927 so definitely you need a calculator for that one not as uh not as simple as one would hope although there's other ways that you can do that but we're not there yet all right and I gr uh C where on the interval 0 to Pi is H equal to 0.75 so I'm dealing with s and I need to graph 0.75 and see where my sign curve is going to uh cross 0.75 so I need an intersection there and an intersection there so let's see what happens all right so now we are being asked where H h of X where H of X is 0.75 so the H of X was the sign so let's pretend that the cosine is not there and we could do that by going to the equal sign and now this it's gone okay so only sign shows up I care when it's equal to 0.75 on the interval 0 to Pi so what I'm going to do is I'm going to graph 0. 75 let's see so here's zero okay here's Pi Here's 2 pi it looks like it hits twice so I need to find both of those intersection points both of those X values so I need the intersect second Cal five and I'm going to make sure I'm very clear that I choose the interval between that guy and that guy so that it gives me what I want so that actually gave me the one I didn't want that gave me the second one 2293 so that's going to be my second one don't know why it did that 2.2945 the right maybe not too much enter enter ah and I worked at that time 0.848 0.848 so that's my other guy okay that happened at two spots 0.848 and [Music] 2294 find the average rate of change of J ofx so now we're on Coast J of X on the interval uh a quarter of a pi and 3.75 Pi all right so rate of change average rate of change is f of B minus F of A over B minus a so what I'm going to need to type out in my calculator is what am I doing J which is co co of remember you're my a you're my B Co Co of 3.75 Pi minus COS of 0.25 Pi all over 3.75 uh Pi minus 0.25 Pi well you're not going to get a pleasant answer or are we all right so this is how I'm going to handle this one I'm just going to type this all in in one foul swoop that you saw me write down a minute ago cosine and parentheses of course don't forget your parentheses when you have a numerator uh 3.75 Pi is second karat thingy close the parentheses minus cosine again of 0.25 Pi close it close it now I'm going to divide parentheses 3.75 Pi so 3.75 Pi minus 0.25 Pi second thing close it and see what terrible number that I get well zero now I get zero and if you're like that's not zero uh e to the 14 is zero so let's let's write that out and let's go back to our Google side and see if we can figure out why we got zero why well here's the thing cosine look like this cosine look like this and we're going all the way out to four Pi so let's go all the way out to four Pi all right so cosine starts up here okay there's 2 pi and there's 4 Pi okay a quarter after a quarter of a pi after my initial beginning I'm like right here a quarter Pi before four Pi I'm at the same exact level the same exact level which means if I'm like whatever number that is uh which point something N I end up at 0 n after who cares of amount of time I haven't changed anything I haven't changed at all the average rate of change of a horizontal uh line is nothing ain't that fun a that weird I don't know I think it's nice it's not like I did that on purpose thank you guys for watching um couple more of these and then we'll be done AP precalculus so uh I appreciate everybody liking and and all the comments so please don't stop that anytime soon it uh it's it's fun to see my YouTube grow just a little bit so keep that going please thank you so much I love you all thanks for watching bye hey hey hey [Music] what up what up let's talk about sinusoidal functions in AP precalculus you can't spell fun without sinusoidal functions let's start out with the most basic form of a sinos soidal function I'm actually going to rewrite that I sto this picture from the internet and upon further review I don't like it anymore the most basic form of a sinusoidal function is y or f ofx equals a sign parentheses B another parentheses x minus C close it close it plus d now there's obviously four letters that matter in this case a is the amplitude Okay now let me just draw out some random sinusoidal function looks something like that that's what a sinusoidal function is a stands for the amplitude a lot of people think the amplitude is the distance from the tippy top to the very bottom that's not true it's half the distance from the tippy top to the very bottom that is called the amplitude period is where things repeat itself so if I were to go to the tippy top here and measure how long it takes to get from this tippy top to that tippy top that gets me my period now B is not my period B is my frequency if I have the period And I want to find B I'm going to do 2 pi over B to well that'll get me period but I could do two pi over the period to get me B okay so that's that C and D are unique here they're my shifts now C represents a phase shift now we know from all the other types of problems that we've done that usually things inside the parentheses with X are backwards and that is true if if I have something like blah blah blah blah blah x + 4 that does not mean you take your sinal soidal function and move it to the right four it's opposite that actually means move it left four so that is a horizontal shift also known as a phase shift last but not least D is my vertical shift but that also introduces A New Concept called the midline now a midline is a horizontal line that cuts my sinusoidal function in half okay it is a horizontal line which is y equals some random number D gives me that number so D basically takes the middle of my cidal function and shifts it up or down depending on whether this is positive or negative and very often you will be asked hey what's the midline the midline is D okay so that is the typical equation we will use this equation a million bazillion times during this slideshow this video memorize it know it love it worship it don't worship it but just you know know it all right we talked about amplitude how amplitude effects a graph can be seen in this picture here here I give you co I give you three Co I give you five Co here's regular Coast see how it Peaks out at one here's three Coast see how it Peaks out at three here's five Coast see how it Peaks out at five that's it that's it the period does not change notice how all of these hit the x-axis at the same exact spot notice how all of these repeat its cycle at two Pi that's because the amplitude has no effect on any shift no effect on any uh period any frequency it just affects the height if I were to make one of these guy is negative then you're just going to literally take your guy and reflect it over the x- axis like you normally do when you have a negative um transformation like that okay so that's amplitude I think we probably feel comfortable if we feel comfortable with the first four sections of this uh unit we feel comfortable with amplitude same thing with period but frequency is kind of the new guy okay now again when you're given a number in front of X that number is called your frequency so in a case like this the frequency would be one okay now if and so I would say b equals 1 because the number in front of X is usually represented by B whenever we have that equation now if I wanted to find the period you find the period by doing 2 pi over the frequency so 2 pi over one means that sin x has a period of 2 pi we know that sin Pi is in Pur in purple or sin x is in purple so it starts out at 0 0 and it repeats itself at 2 pi when I multiply a number to x it compresses things okay it takes my period And it shortens it why well B in this case would be Pi so the period in this case would be 2 pi over my period or with my frequency which is pi that cancels out and gives me two 2 pi is 6 point something two is two so if you look at this guy this repeats its process and we're looking at the black line This repeats its process every two so right there is two and then it repeats it again at four and repeats it again at six and so on and so forth so multiplying a numbered X compresses your function dividing a number stretches it out okay and so I would look at this and some people might be like well there's no number in front of x x is on the the number is on the bottom listen buddy stop talking like that and you can write this out as3x which means B would be 13 so my period would be 2 pi over 13 oh my goodness a fraction on the bottom well let's just multiply the top and the Bottom by its reciprocal which is 3 over one or three and these guys cancel out so the period is 6 Pi okay so looking at this red line even though we don't see it this stretches out to 3 Pi but over here at negative 3 Pi we see that it's kind of like on the x- axis and then it goes below and then it goes above and then it repeats itself from negative -3 pi to regular 3 Pi which is 6 Pi okay so the period is every time or how many units it takes for something to cycle okay and the frequency kind of has to almost do with like a rate so to speak it actually deals with the angle measurement created by like a circle and when dealing with these things that's something that we'll see in future future videos I'm sure knowing myself last but not least we're going to look at the vertical and the horizontal shifts so in orange yellow whatever you want to call it you see your regular cosine starting out at one and then bottoming out when I take cosine and I subtract pi over two from it what I'm actually doing is I'm taking my cosine function and I'm moving it right half a pi so now it's there and it just so happens that cos x minus pi over 2 is sinx isn't that something you we we'll deal with that later we'll deal with that later cos x plus 3 on the outside means I take the regular cosine function that starts at one and I go up one two three is that four I think that's four oh that's because it starts at one yeah okay so my fault one two three okay okay all right so uh let's get started uh this is basically for sin x I'm going to sketch this knowing what I know maybe what I could do is the questions first that might be helpful if I was comparing this to y equals that is an eraser yals a s uh parentheses b x minus C I'm doing a lot and I'm wasting a whole lot of time because I don't have a minus C I don't have a b and I don't have a d all this is is this is basically just saying look that's my amplitude my amplitude is four okay uh my frequency is the number in front of X so B is invisible one the way you find your period if given B is the period is 2 pi over the frequency so period is 2 pi over 1 which is 2 pi so all this is is the sign function stretched out okay so let's try to draw that out holy cow drawing lines is not that fun and I was like maybe I should include graphs no I don't need a graph I'll just make my own I do have graphs later but I don't know one two three 4 one two three four it Peaks out at four it bottoms out at negative four this is sign which starts at z0 it'll Peak out at four and that will happen at pi over two and then it'll come down at regular pi and cross the xaxis bottom out at -4 at 3 pi over 2 and then repeat the process at 2 pi so let me try to draw this out okay starts it's a sign so it starts at 0 0 Peaks out at four and that happens at pi over two then it bottoms out at Pi well not bottoms out but it hits the x-axis completely bottoms out at 3 pi over 2 and then comes back up to repeat the process at two pi and then more and then more and then more more more all I had to worry about here uh was the amplitude that's the only difference just the amplitude all right what's next F ofx equal negative Co X sketch it all right well the only thing I have to worry about here is the fact that the amplitude is now a negative number cosine regular cosine a rough draft starts out at one and goes like this when you have a negative in front of it you are literally reflecting everything over the xaxis so this flips upside down so let me draw this out okay this won't start out at one this will start out at negative one and then it'll come up hit the x axis at pi over two it'll hit one at Pi it'll come back down and hit the x-axis again at 3 pi over two and then repeat the process at negative one at 2 pi so let me draw that out start at negative one start from the bottom like Drake it's going to come up and it's going to hit pi over two and then it's going to max out at one come back down three pi over two and then repeat right there when does f ofx equal 1 on the interval Z 2 pi well here's the interval zero and 2 pi it hits one just one and that just once is right there at Pi so when X is pi is when f ofx is equal to one okay when is f ofx concave up on uh the interval 0er to 2 pi well concave up looks like this and concave up happens here at uh there to there hits an inflection point so this is concave up concave down concave down concave down inflection point concave up concave up concave up so we are concave up between zero and pi over two but not including pi over two because you don't include it it's neither concave up nor concave down on the inflection point Union 3 pi over 2 2 pi and technically I can technically I can include zero and 2 pi because it is concave down on those and they allow me to include them so I'm allowed to include them there I'm allowed to include them there list all the inflection points well the inflection points are going to happen at pi over two 3 pi over two 5 pi over two negative pi over 2 so how can I word that oo pi over 2 * K would be like 1 Pi / 2 or 2 * pi / 2 no I can't word it out like that it has to be K plus pi over two that's how it would work K in this case and by the way these are X values so maybe I should say xal k + pi/ 2 K in this case is any integer so 0 + < / 2 is < / 2 1 + < / 2 is 3 piun / 2 - 1 + pi over 2 is negative pi over two so on and so forth so those are my inflection points uh X is an integer there you go okay K is an integer there you go there you go oh great a graph i' I've smartened up all right sketch 3 Co X over 2 now I'm comparing this to the whole function here okay now I don't have the whole function so I'm not going to write out the whole function however I am going to rewrite this out as 3 Co half of X okay now let's label what I know the number in front of X or the number in front of the whole thing is my amplitude so my amplitude is three okay 1 half is my frequency not my period I find out what my period is by doing 2 pi over 12 multiply the top and the Bottom by two these cross out and I end up with a period of four Pi okay okay so I'm going to this is a cosine which starts at 1 2 3 okay the period is four Pi which means I'm going I don't have four Pi so I can't like go up here and start all over again I didn't leave myself enough space so this is how I'm going to view it if I max out here I'm going to bottom out at half of my period so I'm going to bottom out here at three at 2 pi which which means I cross the xaxis at regular pi and I'm going to come back up and hit it again at 3 Pi now cosine is an even function and since I didn't shift anything especially left or right horizontally I know what's going to happen as I move to the left I'm going to cross uh the x axis at Pi so let me put a dot there I'm going to bottom out at -3 at -2 pi and I'm going to hit -3 Pi so let me try to draw draw this out the best that I can which I'm really bad at although this is looking okay and now I'm starting to jinx myself because I'm getting confidence and I'm running out of space and I'm not good at this and I almost made it my midline is where this guy gets cut in half horizontally well it gets cut in half at the y AIS now we've mentioned before that when you add a number to this whole thing y equals your number is the midline well I have no number added which means it's zero so the midline is the equation y equals Zer the maximum value happens at three the minimum value happens atg -3 and that ladies and gentlemen is three Co half a half a pi or half an X I mean half an X not half a pi half an X three Co half a pie would live right there now you know now you know more more more more more all you want is more more more remember the Leave Britney Alone guy that's what he said look it up just you youngsters probably have no clue what I'm talking about GH all right now I'm going to really write out the whole thing now I'm going to really compare this to a uh son B parentheses xus C close parentheses close parentheses plus d now the nice thing here is B is invisible one so I don't have to panic I don't have to fear B is invisible one my amplitude is three that minus pi means I shift this to the right Pi that's a horizontal shift to the right Pi remember it's backwards land when you deal with sign uh inside the x parentheses this means I have a horizontal or a vertical shift of positive one so I go up positive one okay so this is how I'm going to draw this out okay my midline is y equals 1 so I'm going to draw my midline to begin this out this kind of acts as my new x axis okay when I draw this out since I have a horizontal shift moving to the right Pi that means I begin my sign at Pi since my frequency is one that means my period is 2 pi so at pi Pi 3 Pi it repeats netive Pi it repeats -3 Pi it repeats okay I start out at this new point my amplitude is three which means I go up three and then recross here go down three recross so going the other way I'd go up here recross down Rec cross up recross down recross all right so again this right here is my starting point I'm going to go up halfway in between my so basically a fourth of the period which is half a pi that's when I make sure I'm up three spaces because this is sign then I every three quarters of a pi that happens again so here I am at 2 pi halfway in between go down three there you have it okay so now I'm running out of space now let's go to the left I'm going to go in between one quarter of my period go down three I'll come back up a quarter of my period is up three come back down a quarter of my period is down three go back up a quarter of my period is up three go back down try to draw this out okay this is where I began okay sign goes up like that first then makes its way down then back down again that makes its way up and do it again holy cow holy cow do it again I did it again I'm just not good at drawing these things why don't I since I'm not good at drawing them right to left why don't I draw them left to right okay I have the dots hope everyone's having a good day I hope uh now would be a time to what do YouTuber say in these moments uh smash the Subscribe button drop a comment there we go all right I did it after all of that my maximum value is four my minimum value is -2 oh look at this one this one's different let's write out our formula y equals a sin B parentheses x minus C close it close it plus d now you might be like different yeah there's more numbers uh I see parentheses here but I don't see parentheses here so this is what you must do each and every every time that you don't have that set of parentheses and you're adding or subtracting inside and you have something in front of X you need to factor out the number in front of x no matter what it is Factor it out you divide it from that guy fortunately this one's friendly enough but you have to divide it out you have to factor it out so this becomes two s factor out the two that leaves us with x minus divide that by two so 2 pi parentheses uh parentheses again minus one let's analyze what we have the number in front of sign is my amplitude so my amplitude is two my frequency is the number in front of X when it's factored like that so B is two what that means though is my period is going to be 2 pi over that b so 2 pi over two which simplifies out the pi so that's my period sometimes spelled with a t T all you youngsters know what that means um what else this is minus 2 pi inside the parentheses which means I'm going to have a horizontal or a phase shift going to the right positive 2 pi distance minus one means I go down one or in other words my midline is yal-1 so now what I have to do is I have to draw out my regular sign thing and have my new starting spot so the way I write out that starting spot is I always draw my midline first which is negative 1 so I'm going to draw out y equals ne1 right there okay this shifts over or begins at right 2 pi so that guy right there is where I'm going to start my sign okay sign starts at the origin that it's supposed to start at so it's going to start up there it's not cosine where it starts here it's sign where it starts there the period is pi so you're going to repeat every Pi so you're going to repeat here repeat here repeat here repeat here repeat here repeat here okay so what would happen is this goes up a couple comes back down crosses again goes down and repeats so while we're at it let's draw the spots where it crosses this new midline axis again so every half it's going to cross every half a period it's going to cross right there now again it's going to start here go up two and then come back down so every quarter of a period And every three4 of a period it's going to Peak out and then bottom out so let's say that every quarter of a period it's going to go up two every quarter up two every quarter up two every quarter up two every quarter up two every quarter up two every three quarters down two every three quarters down two every three quarters down to Hell is repetition the book once said the book was called Storm of the Century hell is repetition well welcome to the math teacher Cod Channel all right time to draw this puppy out so I start out here let's get a different color looks so pretty by the way um goes up comes down goes down comes up and repeat so let me do this over here because I'm not good at going right to left basically if you're bad at drawing these so am I okay and so like it's kind of like my my channel is like learn how to draw signs with a guy who doesn't have any artistic talent whatsoever all right almost there that looks bad almost there did the best I could did the best I could but there you have it and again It's Tricky there that's the trick the trick is to make this look like this by factoring out the number in front of X it has to be the number in front of X and if for any reason like that's a three then you would have to do like 43s pi and it just gets grosser and it is what it is but yeah that's that's how you do it that is how you do it oh I got to stop otherwise I'm gonna get flagged for copyright I'm not Montel Jordan show why both of these are true these are fun little trigonometric identities hey did you know that if I take s and shift it to the left pi over two that I get cosine let's show that let's show that let's show that if I were to take signs so I'm going to draw a very very very basic picture of sign it's not going to look good at all okay but I'm going to try to do this in one foul swoop right I have pi over 2 pi 3 pi over two and 2 pi okay if I were to take regular sign which looks like this okay and if I were to move this to the left pi over two so let's go over here to to the left pi over two if I were to move this to the left pi over two it's coign you see it it's cosign because that guy starts there and if I drew my lines a little bit more correctly you would get cosine well look at that look at God look at God all right right well let's try this again let's try this again if I were to take a regular cosine function which tell you what let's just use this guy let's use this guy because that is cosine now if I were to take regular cosine and move that to the left so let's extend this graph just a little bit further to the left little bit further to the left and make you negative uh uh what is that Pi okay if I were to move you to the left oh the picture's not good the picture's not good here let's do this perfect if I were to do that that is negative sign write a sign equation for the graph below all right sign equation we've got y equal a uh Time s b parentheses x minus C close it close it plus d all right what do I have first things first we should always draw the midline so it Peaks out of five bottoms out of one the midline is halfway in between that which is three so the midline is the equation y = 3 which means D is is three we max out at five we bottom out at one which means the amplitude is two that's what I know this is sign and S is supposed to start at 0 0 this does there is no uh horizontal phase shift and the period for this guy is two pi which means the frequency is one so these are the only two pieces of information that I need to come up with my equation so Y = 2 sin 1 x NO phase shift so I'll simplify this I'll clean this up in a minute plus three okay so you're not necessary Y = 2 s regular X cuz 1 * X is just regular x + 3 okay I can handle that I can handle that just fine write a cosine equation for this graph okay y equals a co uh parentheses B parentheses x - c close it close it plus d all right uh cosine starts at 1 Z that doesn't happen here now let's not panic let's not panic I need a midline which this Peaks out at one bottoms out at what appears to be -3 so the midline is going to be at1 the midline is at 1 so that gives us d of1 uh if I Peak out at one and bottom out at three that means my amplitude is two which I believe it was last time so I'm not really a creative job when I come up with these examples uh that's that now cosine is supposed to start up here and it didn't so why don't I pick a new starting spot where it is supposed to start at the tippy top and that's right here so it starts here which means I moved to the right pi to make that happen Okay so I'm moving to the right Pi I'm moving I'm moving down one I have an amplitude of two now I need my period my period is two Pi again so that makes life easy which makes my frequency one okay so my amplitude is a while y = a which is 2 cosine B which is 1 * parentheses x if I move to the right Pi then it's going to be x minus Pi close it close it and then minus one so let's clean this up Y = 2 Co distribute the One X x - pi - one all right a Ferris Wheel by the way AP pre-cal exam problems love the ferris wheel problem um Prett popular you'll see it probably on a practice test or actual test here and there Ferris wheel problem the ferris wheel has a diameter of 40 meters and rotates at a rate of 0.2 revolutions per minute the bottom of the wheel is located two uh meters above the ground create a sinusoidal equation to represent the height of the passenger as a function of time assuming the passenger board is at the boards at the lowest point okay boards at the lowest point means uh not sign so I could use sign I could use sign it doesn't really matter you could use sign you could use cosine uh it's all really good with these things but you have y equals a sinine b x minus C close it close it plus d all right diameter is 40 meters which means you're 20 you're 20 okay you are at its lowest 2 meters off the ground okay which means here it has a midline of 22 the Taylor for special and then it Peaks out at 42 the Jackie Robinson special okay so I have that going for me so this is what I know I know my amplitude is going to be 20 now before I get too much into this okay this is what's going on okay uh and you know what let me lower this a lot because we're talking about real life and you can't go underground you're starting at two you're starting at two and uh at some point you peek out at 402 and you're at the bottom and then you make your way up and then you're at the top and you make your way way down and you repeat and you repeat and you repeat repeat okay now the midline happens here at 22 the Taylor Swift special so that gives me a d of 22 I still need c and I still need B my B can be found by figuring out what my period is now this guy rotates at 02 revolutions per minute 02 revolutions per minute which means it takes five minutes for a complete Revolution that is really slow but the period is going to be five now do I want my period no I want my frequency and in order to find my frequency I take 2 pi and divide it by my period so that would be my frequency so I don't have to simplify that kind of makes life a little bit easier the trickiest part about all of this is the phase shift okay because if I start at the very bottom s or cosine doesn't start at the very bottom of anything s is supposed to start here which means I have to go write a certain amount of time now this whole thing this whole thing has a period of five all right so me going from the bottom to the middle is going to be one quarter of a turn which means a quarter of five okay so this guy is going to be 5 over 4 which means my phase shift is positive 5 over4 not pleasant at all but I don't really have to make it Pleasant I could just write this out y equals 20 s parentheses B which is 2 piun over 5 parentheses x - 5 over4 that'll actually simplify out just fine close it close it and then plus 22 the Taylor Swift special let me write that out let me simplify this out and clean it up I didn't give myself near enough space okay let me write it out here y equals 20 s distribute we get 2 pi over 5x minus 2 over 4 is a half 5 over five cancels out so half a pi so pi over two it's not too bad minus pi over two close that plus 22 that's the ferris wheel problem pretty tricky stuff not gonna lie all right last problem is a calculator problem I love this problem I invented it not gonna lie NGL not gonna lie uh made me think of the office I tried to think of what are things that behave in a sinos soidal way that's kind of floating off the ground a little bit well let's take a big movie screen let's have a big movie screen in the the DVD logo going boop boop boop boop and the way that this travels the height of the DVD logo travels in a sinusoidal function so this is what's happening a DVD logo is slowly floating along a movie screen in a movie theater the height of the logo at given times is given by the table below no problem forgot I forgot the word um so you know at zero minutes the the DVD logo is like four meters off the ground and after one minute it's seven meters off the ground and after two minutes it's 10 meters off the ground boop boop boop boop boop use a calculator to perform sinusoidal regression to find an equation representing the height above the ground as a function of time so what I'm going to do is I'm going to run to a calculator I'm going to type all this information in and that's going to give me an equation that represents a assign function for this data right here all right stat let's type out these numbers zero uh one two three four five six and that's it and then we got friendly numbers thank goodness for that 47 10 uh 7414 all right so we need to do sinusal regression so you're going to go to stat you're going to go to calc and you're going to go up because it's like right there at C sign reg you're going to hit enter uh we are going to need to use this guy later also the period it goes from four to four and it goes from 0er to six to do that so the period is six we're going to use this equation later so let's go to vars y vars enter enter see what what grossness we come up with all right so y = 3.95 4 times s parentheses 1.05 7 minus 0.527 lots of rounding up then close the parenthesis plus 5.5 69 um we're rounding up I wasn't laughing at that number I swear uh let's write that out all right so that equation ended up being y equals 3.95 4 S parentheses 1.57 x minus 0.527 plus 5569 so there's your equation sreg SRE I hope we don't sreg meaning regularly almost done use that equation to predict the height of the object at T equals 8 minutes all right so I have my equation I know I stored my equation all I have to do is just find out what happens when T equals 8 all right so now that I have this equation stored in y equals why don't we go to zoom trig so we can kind of have a good idea of what it looks like not too happy with that um this is the the the floating DVD logo so I need to kind of go a little bit higher so let's change my window from it gave me what zero so let's go zero and by the way if we're talking about real life stuff here uh let's do like negative one because we're not gonna have negative time let's bump this out to 20 I mean the future problem has eight minutes and 10 minutes so let's bump this out to 12 not 20 uh the height of this thing peaked out at uh what was it 10 so let's bump it up to 11 let's bump it up to 11 right y Max not to be confused with IMAX speaking of movie theaters and let's graph so the DVD logo is floating around and it goes boop boop boop boop and in this particular problem problem B we care about when uh or not when but where we are at moment eight so I can do second table let's go to my table and go to eight and see where we are off the ground 9512 so what is that meters 9.51 two MERS off the ground so when T was equal to eight I just plugged in eight I think I used the table and I ended up getting 9512 M off the ground so that was the answer for b c is after 10 minutes when will the logo hit the five meter Mark so since this is sinos soidal and it is a sign it's going to look like this so there's going to be multiple times where this is going to go up and down and up and down and up and down and hit five meters I care about when after 10 minutes the first time it hits the 10 meter Mark so if the 10 meter Mark starts there and this is five for example I care about the first time it hits that okay so let's see what we come up with all right last but not least after 10 minutes when will the logo hit the five meter Mark so we care about what happens after 10 so let's go to my window and since we care about after 10 let's set this uh from 10 to 20 and just see it all right so it's definitely going to hit the 5 meter Mark uh at some point but when well y equals 5 gives me that 5 meter Mark if I just enter five so graph and right around here is where it's going to hit that five meter Mark so let's find the intersection between these two lines second Trace intersect is going to give me that so let's scooch to the left because I want the first time it hits that 5 meter Mark so let's scoot to the left of it hit enter let's hit right a few times to the right of it hit enter hit enter again and it's going to tell me 12.25 61 so 12.25 6 minutes so after the 10 minute Mark we ended up with 12256 minutes so 12 minutes and 256 seconds all right that covers a pretty important chunk about AP precalculus sinusal functions hope this helps thank you so much for watching like And subscribe [Music] bye what up what up we are going to start the second half of unit 3 in AP precalculus tangents and inverses and other things just jump right into it tangent what is it well by definition tangent is s over cosine so we will write out tan theta equals sin Theta over Co Theta now at this point we should know what a s graph and what a cosine graph looks like but tangent is going to look very different one of the reasons why is you now have a Theta like an x value in the denominator so you should expect to see a few vertical asmp tootes now this is what tangent looks like graphically first things first you're used to a period of two Pi not anymore the period for tangent is regular Pi okay vertical asmp tootes never had vertical ASM tootes for S and cosine well now you do because cosine Theta is going to equal zero in certain spots like at pi/ 2 or 3 pi over two or negative pi over two so on and so forth okay now let's graph what this looks like generally obviously you know those of you who watch my videos know that I'm not good at drawing graphs but I will try my best so what did I say vertical asmp tootes at pi over 2 so we'll label you as pi/ 2 vertical ASM tootes at 3 pi over 2 so we'll label you as such and then you will be pi over 2 and you will be3 pi over 2 and so on and so forth now tangent has this weird curve looking thing that goes through Z 0 okay the tangent um cosine of zero is one so s of Z is zero so that that makes sense and then it kind of looks like this it's like this weird concave down inflection point concave up okay and then it does it again here at Pi so I why don't even write that down so concave down hits concave up same thing over here at negative pi concave down hits concave up I could have done a better job but you know uh pay me more but that that's basically what tangent looks like graphically it's different for sure the thing you're going to have to remember the most is the period is a high so that changes our answers when we solve for tangent but for the most part that's what tangent does tangent is s over cosine and many of the things that we do will be based off of that fun fact inverse trig functions and graphs now an inverse trig function function is used in A Moment Like This say you have sin xal a half and your job is to solve for x you know normally when we have something in front of X you would be like divide both sides there's no such thing as dividing a sign when solving for x what you do is you inverse sign both sides inverse sign both sides and basically what you're doing is that cancels out and you're asking yourself or myself um what angle gives me the or the sign of what angle gives me a half and then you take it from there and then you have your answers and so on and so forth okay now um one of the things that you have to remember is if I give you a sign and then I ask you to give information about the inverse sign the domain and the range flip so the domain of sign is the range of inverse sign the domain or the range of sign is the inverse is the range the range of sign is the domain of inverse sign third times a CH all the rules apply to like cosine tangent fun stuff like that now one other word is uh the word invertible okay now invertible not to be confused with convertible I think I spelled that right invertible means can you take take the inverse of something something and uh what that is is you use what's called a horizontal line test now you're used to the vertical line test but vertical line test is used if I give you a picture that looks like this right and you want to be like hey is that a function you draw yourself a vertical line you see if that vertical line passes through the graph and it does more than once it's not a function well when you take something that's the inverse you are graphing it over the yals xais and so what you're doing is you're kind of making it a little sideways and backwards at the same time so if I were to give you this picture and ask you is and let me draw the same picture is this graph invertible you would make a horizontal line like so and you would pass that through the graph and notice that in this case it doesn't hit the graph more than once so you would be like yeah that one's invertible that means the inverse of this graph exists even though the function is not a function it's not a function in itself because it fails the vertical line test but the inverse exists that's what invertible means does the inverse exist and you use the horizontal line test to do it now let's take a look at our three new friends seant cosecant cotangent okay secant a bre deviated as SEC College uh is the same as 1 over cosine Theta okay cosecant abbreviated thatat is the same as 1 over sin Theta and cotangent abbreviated that is the same as one over tangent which happens to be the uh reciprocal of regular tangent which we saw a couple slides ago so why don't we just flip those as well now we will get into graphing these guys and what the pictures look like we will get into that okay so don't you worry your pre heads off uh what's going to happen is already you will notice uhoh denominator problems and so we're going to have vertical ASM tootes and things like that but we'll worry about that when we have to worry about that all right now the last thing that we're going to look at is probably the most frustrating thing that we're going to look at because it's a lot of memorization but it's not super bad okay there's three pythagorean theorem identities that you're going to see with trigonometry sin squar theta plus cos squ Theta is going to equal one now usually when you see that that might end up turning into like one like sin squal 1 - co^ squ something like that you will see that every now and then definitely memorize this I'm not sure if you need to have these next two memorized but I'm just going to put them in here just in case secant squar Theta minus tangent SAR Theta is going to equal 1 and cosecant squar Theta minus coent squ Theta equal 1 okay definitely have the red memorized the double angle formulas the double angle identities is what if instead of giving you like s something I give you sine 2 Theta well you can rewrite sin 2 Theta as 2 sin Theta cos Theta memorize that also Co 2 Theta is going to be cos squar Theta minus sin squar Theta now you might be looking at this and saying boy oh boy that's awfully close to that it sure is but it's not the same it's very close though which might mean you might do problems where you have to rewrite things and remind yourself that cos squar by the way is the same as 1 - sin squ so technically you could say that this is also 1 - 2 sin^2 Theta this is also the same as 2 co^ 2 thet minus one just rewrite it in different ways fun stuff have both of those memorized also have and there's Co there's tangent ones too but again we're only going to uh concern ourselves with the sign and cosine because if we have to deal with the tangent ones we will use what the definition of tangent is I don't want to try to memorize too much the tangent 2 Theta is it's not pretty okay uh similarly with this the Su the sum and difference formulas if I give you sine Alpha plus beta okay you can write that as s Alpha cos beta and you know what let's let's combine them plus or minus okay you're going to get plus or minus cos alpha s beta so if I do sin Alpha plus beta it'll be sin Alpha cos beta plus cos Alpha sin beta if I do sin Alpha minus beta it'll be sin Alpha cos beta minus cos Alpha sin beta okay last one that we're going to look at is going to be cosine Alpha plus or minus beta okay this is going to be cos Alpha cos beta flip these guys s Alpha sin beta okay so this would like if I gave you cos Alpha plus beta you would write out cos Alpha Co beta minus sin Alpha sin beta and vice versa and vice versa all right let's see let me move my space all right so uh we're going to start out with a very not funl looking uh tangent graph okay and then we're gonna graph it all right let's move my face like I said although I'm not sure if I need to no I don't need to I'm good what is the period Well I have to rewrite this I have to rewrite it because again I'm kind of comparing this to my regular sinusoidal graph okay so I'm gonna write this as tan a for amplitude even though we don't have amplitude uh b x minus C close it close it plus d okay a doesn't really give us amplitude because now the picture looks like this so I'm not going to worry about amplitude there's no such thing as amplitude but it has to do with like stretch and stuff like that compression we don't have to worry about that here because there's no number there uh if I rewrite this and factor out a two I have 2 * x - piun close it close it minus one now what this looks like is normally if I have this number I would say ah the way you find the period is you do 2 pi over two now it's regular pi over two instead of 2 pi over two so my period is now pi over two this is going to shift to the right pi and shift down -1 okay so what is the period Pi / 2 what is the frequency regular two what is an equation for all vertical asmm tootes oh okay uh well if my let's draw this out and then I'll do the vertical ASM tootes okay so first things first uh my new origin so to speak has me going right Pi down two so things are going to start right there my period is pi over two right so uh what that means is pi over two to the right pi over two to the left and I'm going to go pi over two every that many times so now I'm drawing out my vertical ASM tootes this fun now as I'm doing this uh I'm going to remember what I wrote out on the previous slide what tangent looks like okay and now tangent is going to go through negative one through each of these so here here here here here here here here and it's going to kind of look like the oh gosh so bad at these all right so let's try to curve things yeah there we go all right so better kind of missed the dot but that's all right you get the idea look problem solved wowow now I'm not going to worry about the ones to the left and to the right at the very end I've done enough now the equation for all the vertical ASM tootes looks like what's happening is I could pick one vertical ASM toote like that's PI over4 right there that's Pi over4 so what's happening is it's going to be a vertical line so x equals it's going to have a period of pi over two so what I can do is if I pick uh like pi over to K and just add pi over four that should do the job okay so like if this was Zero I would have 0 plus pi over4 nailed it if this was one I'd have pi over two plus pi over four which is going to be that guy nailed it and then I could choose uh any one I want but yeah those are uh all the vertical ASM tootes for tangent is that neat given an angle Theta in standard position where sin theta equals cos theta equals < tk2 over2 what is tan Theta oh this is going to be super easy tan Theta is sin Theta over cos Theta I already told you what sin Theta and cos Theta are sin Theta is < tk2 over2 cos Theta isk22 over2 something over itself is just one now this makes sense and here's why tangent represents s which is a y value cosine represents X which is an x value okay now y overx That's like up and over this is slope and if I go up < tk2 over two and over < tk2 over two that's a perfect slope of regular one oh it makes sense how this makes sense which takes us to B what is the slope of the terminal Ray well the slope of the terminal Ray is tangent and I just found out that tan theta equals one so it's the same exact answer look at that little definition for you all right wardrobe change location change I'm at my home uh and just in time to do this ugly terrible ugly problem first things first this is the original function that I'm giving you it's a mess it goes up to three goes down to neg3 it moves to the right a little bit it has a different period and I'm giving you a domain not only is that a thing I don't even care about the graph I don't care about any of that I want you to find the inverse which is going to be a mess so what I always do when finding the inverses F ofx drives me nuts so I'm going to change it to Y so y = 3 sin PX - Pi / 2 and step number one when finding the inverses flip-flop the X and the Y so x = 3 sin Pi y minus pi over 2 so many pies it's like we're at Thanksgiving now I need to solve for the new y okay so the first thing I'm going to do is I'm going to get rid of that three because that's really the only thing I can do and I'm going to divide three from both sides and let me move that over here so X over3 I'm going to need all this space I'm going to want all this space especially for graphing uh x over3 equals you're now gone and you know what I know it's next so let me leave the space s parentheses Pi y - Pi / 2 now all the other stuff lives within the sign so the next step is to get rid of the sign how do I get get rid of the sign inverse sign both sides so inverse s u and inverse s u that allows these to cross out and now I have sin inverse of x over3 = piy minus pi over 2 which is you know manageable still gross but manageable add pi over two to both sides so inverse sign of x over3 + pi/ 2 equals Pi Y and rather divide everything by pi which I could do I'm going to multiply both sides by one over Pi because it's going to help me deal with the fact that I have two things and distributed property and a pi that's going to disappear it's going to make it look prettier at the end of the day now the nice thing is y is now all by itself Y which becomes F inverse so let me write it out as F inverse of x 1 over Pi * sin of X over 3 inverse sin of x 3 is 1 over piun * inverse sin of x 3 distribute 1 over piun * pi over two the pies cancel out as a half all right great now the graph of an inverse sign okay normally is the graph of the inverse of a regular sign okay regular sign look like this okay really graph of an inverse sign you're thinking is probably going to look like this but here's the problem that's not a function that's why I gave you that domain because this is what's going to happen now since this guy this guy right here has a Al amplitude I almost said altitude amplitude of three the range of this guy is Nega 3 to three because the amplitude is the range gives the range I have no vertical Transformations I'm not moving up or down so that is my range for sure this is my domain and if you're like well what do we care we're not graphing the original function what do we care well if I'm supposed to graph the inverse I care about the domain of the original and the range of the original because the range of the original is the domain of the inverse and the domain of the original is the range of the inverse so what I get is I get this guy right here okay it's going to look like this uh just it's going to look like that okay this is my function it's going to stretch out to -3 and three and it's going to have a range of zero to one which means it's going to have we're not going to call it midline but it's kind of like like our midline of one half okay so let me try to draw that out one more time okay because it's not that's not doesn't look good at all okay not that that looks even better one two three one two 3 one okay and it's going [Music] to go here and kind of go like that just imagine more of a curve like the change of concavity should happen here let me do this let me make it bold that way no one will ever notice how bad it looks so if you kind of can see it a slight change of concavity like right there on the Y AIS not easy not fun hopefully you'll never have to do that like on a test but we did it here and you know that's that's the battle great I have to do another one do I no oh thank you Jesus all right I have 10 f ofx equals 10 I'm going to sketch regular f of x I can handle this but instead of sketching the original I'm going to do it by making a table all right so what does that mean well let's graph regular F ofx or let's graph uh yeah let's make a table not graph but make a table of regular F ofx let me adjust my pad because all my lines are crooked um f ofx tangent has a period of Pi and starts out at Z 0 remember tangent looks like this and goes through the origin which means it only stretches out to pi over 2 so I'll start attive pi over 2 I'll do piun over 4 0 pi over 4 and pi over 2 okay the inverse or the tangent of negative pi over 2 is undefined okay vertical ASM toote vertical ASM toote it's supposed to say done at a timer I forgot that at a timer at negative pi over two or negative pi over 4 it's NE 1 at 0 it's 0er at Pi over4 it's 1 and here we're undefined again okay now I'm going to sketch the F inverse by flipping everything okay so the F inverse is going to look a little something like this so x f inverse uh I am undefined attive pi over two uh at uh Pi over4 or at negative 1 and then 0 one undefined I have negative pi over 4 0 uh pi over 4 pi over 2 okay okay now here's a thing if at X I'm undefined right that means over here was a vertical ASM toote so if over here is a vertical ASM toote what happens here is going to be a horizontal ASM toote because of the way um inverses work so at negative pi over two which I'll call this negative pi over 2 so you would be Pi over4 negative pi over4 you would be positive Pi over4 you would be positive pi over 2 we have horizontal ASM tootes that looks awful here and here and here at -1 I'm attive over4 0 0 and at positive one I'm at regular Pi over4 so the graph is going to basically limit out here and look like that okay so describing this what would this look like well this is a uh let me use AP pre-calculus words this is a horizontal shift positive six units or six units to the right and this is a vertical shift positive four units or four units up so basically take the red thing and go you got to make those noises otherwise it doesn't count also you have to write out the word right but draw an arrow from up for up because I'm an idiot all right now I have to sketch another graph and it's going to be 2 cosecant Pi Theta minus Pi + 1 all right all right cosecant if you remember is the same as one over s so this follows a lot of the sign rules but sign acts as our kind of like our blueprint okay so what we're going to do is we're going to treat this like a sign the sign is going to give us our blueprint and from there we can draw the cosine cosine and secant when graphed look like these parabas that kind of you know flip-flop go up down up down up down by drawing out s or cosine depending on if this was secant as like a blueprint if we were to draw this out the blueprint lets us know which parabas go up which parabolas go down where do they start stuff like that so I look at a guy like this right nothing in front of theta which tells me the period is 2 pi right uh two is in front of cosecant which tells me the amplitude is two okay uh a horizontal phase shift of positive Pi which means right Pi units and I go up one okay so the blueprint for this guy is going to be a sign it's going to be a sign that I go right Pi up one and would normally start right here okay now the red that I'm going to draw is going to be my blueprint the blue that I'm going to draw is going to be my actual graph okay so my blueprint period of 2 pi so you repeat the process here you repeat the process here you repeat the process here okay since this is sign and I'm not flipping it upside down I'm going to uh hit my midline every pi and since the amplitude is two I'm going to go up two hit down two so going in reverse down two oh hold on made a mistake there let's move you up because that would be going down three put a dot here so you would be going down going up going down going up now again this is my blueprint this is not at all my secant graph cosecant graph okay but by doing this this we now know how our graph is going to behave okay where the midline of my sign would be would be my vertical ASM tootes okay so if you wanted to if you really wanted to go above and beyond put your vertical asmp tootes right here okay where the midline would be the mid line uh is a line that goes through the middle goes through the inflection point goes through where you cross your x axis you know stuff like that all right almost done almost done almost almost done and then I can graph my actual graph which is not going to look good singing all right so Peaks out there bottoms out there Peaks out there bottoms out there Peaks out there bottoms out there Parabola opening upwards here Parabola opening downward Wards there Parabola opening upwards here Parabola opening downwards there Parabola opening upwards here oh gosh I was doing so well I was doing so well I have a little thing in the way let me move that over uh Parabola going downwards here there you go Parabola upwards here Parabola downwards there all right fun all right solve -6 < tk3 = 9 cose 3 uh Theta I don't like cosecant I'm GNA write it out as sign so I'm going to write this out as -6 < tk3 let me do this over one you'll see why I'm doing this although it's not a necessary step equals -9 over S 3 Theta now that I wrote it out as a proportional I can treat it as a proportional so that's what I'm going to do I'm going to multiply you to each other crisscross applesauce so you are now uh ne9 equals -6 < tk3 3 pi sign 3 Pi -6 < tk3 sin 3 Pi that's an m and we'll just pretend it's not there uh let's divide both sides by 6 < tk3 let's divide both sides by -6 < tk3 cross U out you become positive 3 over2 root3 but root3 on the bottom is bad so positive3 over 2 < tk3 I have to multiply the top and the bottom byun3 okay and what that gives us let me continue the rest let me get rid of that M okay uh okay uh 2 < tk3 * < tk3 is 6 3 < tk3 on the top that simplifies further is equal to sin 3 Theta okay that becomes < tk3 over2 equals leave some space sine 3 Theta in order to unleash uh Theta I have to inverse sign both sides so inverse sign inverse sign that crosses you out so I have the inverse sign of < tk3 over 2 equals 3 Theta now inverse sign of < tk3 over2 is going to be pi over 3 so that equals 3 Theta okay and also 2 pi over 3 because s is positive in quadrants 1 and two so 2 pi over 3 equals 3 Theta divide both sides by three or multiply both sides by a third the bottom is going to become nine on both of those so you get Theta excuse me equals pi over 9 and I'm going to leave some space and you'll see why uh and 2 pi over nine nine can I get rid of that three now okay the finger's coming up again and here's why the original problem is 3 Theta normally I'd be done but since this is three Theta uh I have a period or frequency that is three which means this cycle repeats three times if we're talking about between uh three or like and first off I didn't even say uh oh okay so that's another thing I didn't even say any bounds so I have to answer this as an equation great well what's going to happen is this is going to repeat itself every three times per 2 pi which means you're going to have a period that is now 2 pi over 3 so what's going to happen here is you're going to add 2 pi over 3 K to that and 2 pi over 3 K to that so that between zero and 2 pi you're actually going to get six possible solutions and since I didn't give you an interval I have an infinite amount of solutions so I have to say that K is an integer I guess I have to say that as well what a gross awful terrible problem gross all right for this one I actually do have an interval which makes life a little bit a little bit a little bit easier but this problem is still kind of gross except for the fact that I have a bunch of Coes uh lying all over the place so let me get rid of the one that's over here so let me subtract 2 Co Theta from you seems to make sense so now I have 2 < tk3 cos Theta sin Theta minus 3 Co theta equals nothing I have a CO in common so let's pull it out let's bring out a cosine Theta that leaves me Factor it out that leaves me with 2 < tk3 sin Theta minus 3 equal 0 now fortunately unlike the last one uh no frequency issues here everything is the way it should be thank you Lord Jesus now uh what I have to do is I have to set each one equal to zero so the first one's lovely Co theta equals z means I inverse cosine both sides and Theta is going to be the inverse cosine of zero and the inverse cosine of zero happens at two spots pi over 2 and 3 pi over 2 so Theta is Pi / 2 and 3 Pi / 2 and again don't have to worry about the weird period stuff that we saw before okay everything's exactly where it needs to be Ray now the other one is going to be a little less Pleasant 2 < tk3 sin Theta - 3 = 0 add three < tk3 sin theta = 3 divide both sides by 2 < tk3 so you have sin theta = 3 over 2 < tk3 that's against the rules so multiply the top and the Bottom by root3 and root3 okay that gives you sin theta equals 3 < tk3 over 6 which simplifies to < tk3 over2 ah much nicer now you inverse sign both sides and so the in so Theta is going to be the inverse sign of < tk3 over2 so very similar to the last one uh the inverse sign of root3 over2 is pi over 3 and 2 pi over3 and that's the only two spots as long as the period is normal that you have to worry about between zero and two Pi so my final answer is going to be U and U all eight possib all four possibilities yeah yeah pretty awesome all right solve on the interval 0 and 2 pi I have sin^2 Theta = sin Theta + 1 - sin^2 Theta let's move everything to the left okay so let's subtract sin Theta let's subtract one let's add sin squar Theta that way it equals zero because I sense some factoring that's going to happen if I add s^ 2 Theta to sin 2 Theta that gets me 2 sin^2 Theta minus sin Theta minus one now if you're looking at this and a little thought bubble pops on your head and you're thinking that's a thought bubble a 2 2 A 2 minus a - 1 factors out to 2 a + 1 time a minus one you're right but instead of sign or instead of a we have S so this factors out to 2 sin Theta + one regular sin Theta minus one = Z okay and now you just set each one equal to zero and solve uh so you would be 2 sin Theta + 1 = 0 subtract 1 S 2 sin Theta = -1 / 2 sin Theta = one over two inverse sign inverse sign and the inverse sign of Nega a half is 7 piun over 6 and 11 pi over 6 so Theta is 7 pi over 6 and then the other side which is 11 pi over 6 this one's a little bit friendlier sin Theta - 1 = 0 add one uh and so sin Theta is going to equal one inverse s inverse s and that's only going to happen once at pi over two no yeah pi over two because it Peaks out and won't happen again and that's also because this is basically asking where on the unit circle is the Y value uh one and that happens at like the 90 Dee angle which is pi over two so there you go hello 100% I did it all right another one I think maybe the last solving one this one's actually not that bad cos squ Theta I've seen that before I've seen that on cos 2 theta plus sin SAR theta equals one which means if I want to make you look more like you what I can do is I can can turn co^ 2qu Theta into 1 minus sin 2 Theta which is what I'm going to [Music] do uh and then what I will do is I will move everything to the right so I will subtract one add sin squar Theta subtract one add s^ SAR Theta and it looks like I'm going to have myself a very very very similar problem what I just did okay I'm going to have zero so I'll put it over here equals sin^2 Theta uh plus 2 sin Theta + one ah a perfect square trinomial that is sin Theta + 1^ 2ar = 0 so I don't have to do both sin Theta + 1 = 0 S Theta = -1 inverse sign both sides and that gives me Theta = 3i / 2 now no interval so s of a Theta is only going to be negative one once in the entire unit circle so all we have to do is add 2 pi K to this and then we've covered ourselves so K is an integer well we have more of these I don't know why I put so many of these solving trig functions problems on this thing but you know what more time means more YouTube money in my pocket baby now I can make about $40 a month let's um oh my gosh let's oh I see it I see it right away we have ourselves a double angle formula now uh we saw these at the very beginning remember it seems like long ago but we saw these at the very beginning where we said that regular sign of 2 Theta would be 2 sin Co Theta okay so this would be -2 sin Theta Co Theta uh similarly we have of course nothing happens with root2 sin Theta that's just normal that's just normies uh minus 2 for sin Theta cos Theta because 2 * 2 is this many now what I'm going to do is I'm going to add uh for sin Theta cos Theta to both sides like so and you know while I'm at it I'm not going to have all the space in the world in fact I'm not going to have any space in the world I'm going to subtract < tk2 sin Theta from both sides as well so everything is on the left so that'll leave us with uh two right 2 sin Theta cos Theta and if I subtract that from the right to the left minus < tk2 sin Theta set it equal to zero all right let's rip out a sign let's factor out a sign sign and that's all I can factor out I factored out the sign a of Base minus < tk2 equals z so the first one's going to be the lovely one that's going to be S Theta = 0 inverse sign both sides Theta is going to equal the inverse s of0 which happens three times it happens twice now when I created this problem see how now this is where you would normally do it you would be like zero and Pi you know because of the whole unit circle thing but since I included two Pi with this I got to include that two pi as well so it's going to happen at zero pi and two pi oh oh Nicholas or goat or whatever I don't know that was me being silly uh all right so now I have 2 Co Theta minus < tk2 = 0 add < tk2 is 2 cos Theta = < tk2 divide both sides by two and you have cos Theta = < tk2 over 2 inverse cosine both sides and you have the inverse cosine of < tk2 over 2 which is going to give me pi over equals pi over4 as an option because it's the X values and it's positive so pi over 4 and 7 pi pi over 4 yes so both of all all five of my answers are going to be zero Pi 2 pi pi over 4 and 7 pi over 4 great find the exact value of cosine pi over 12 well we're finally done solving trig equations but now we're introducing a new layer of Hell called the angle and difference trigonometric identities you see I don't know what cosine pi over 12 is but you do know what you do know what if I if I do a little bit of math magic right put this over 12 okay I know what four Pi or I know what better yet let's not say four Pi let's say I know what pi over3 is cosine of that I know the cosine of pi over four so if I were to do this and make this 4 Pi minus 3 Pi this ends up being that now this is an unnecessary step this is me just going over my thinking process so I can show what I'm really doing by splitting it up like that I can now kind of split it up like this right so this becomes 4 pi over 12us 3 pi over 12 which turns it into Co pi over 3 something I know minus Pi over4 something I know okay um cosine of pi over 3 minus Co s of Pi over4 now um when I had the difference formula in that one slide that I show you like the third or fourth slide in uh let me see if I can fit it up here co uh Alpha minus beta is the same exact thing as Co Alpha cos beta with a cosine that's the opposite of what you want so plus sin Alpha sin beta so this is going to end up being cos pi over 3 Co pi over 4 plus s pi over3 s pi over4 cos piun over 3 is a half cos piun over 4 is < tk2 over2 2 plus sin < over 3 is < tk3 over 2 to think about that one and then sine Pi over4 is also root2 over2 because these guys uh also < tk2 over two so multiply these guys 1 time root2 is < tk2 2 * 2 is 4 plusun 6 over four add the fractions don't combine the roots because you're adding them and they're different and you can't simplify any of them leave it like that that's the exact value okay not plugging anything into your calculator which we will do at some point soon I'm sure now if memory serves me correctly if we're doing the subtraction one with cosine next then we're probably doing a sign one with addition which we kind of are and we are it's backwards so if you look back which I'm not going to do because it's cheating uh when it was cosine and it was minus inside the parentheses it was cos Co plus sin s this is sin cos cos sin with a plus sign which means that this is sine Alpha plus beta so I'll call you Alpha I'll call you beta so so this is the same thing as Sin piun over 9+ 2 piun over 9 pi over 9+ 2 pi over 9 is 3 pi over 9 3 pi over 9 simplifies out to pi over 3 which gives me root3 over2 those backwards I liked it whole lot easier a whole lot easier the base of a tree is 150 feet from a mark on the ground check out these drawing skills here's a tree it's red the base of a tree is 150 feet from a mark on the ground let's actually make that on the ground and let's put a mark right here oh hi Mark the room uh the angle of elevation which is wherever you are observing from from that Mark to the top of the tree is pi over3 radians so somebody is looking up there at the top of the tree and the angle is pi over three calculate the height of the tree now geometry would tell you SOA which is good good uh and then you would think adjacent OPP opposite tangent good good I'm going to use more pre-calculus language even though I already just used the geometry language and told you what to do but if I view this you know as a x value because you know it's the the horizontal distance and if I view this as a yalue which is the vertical distance and if I don't care about the hypotenuse which is the radius then yeah I am going to use tangent and tangent of my uh angle is going to equal the Y overx X and now it's just a matter of answering the question uh tangent pi over 3 is equal to the Y value which is H over the x value which is 150 okay I don't remember what tan Theta or tan pi over 3 is off the top of my head so what I'll do is I'll quick change this to s pi over3 over cosine Pi over3 because why the heck not that's going to equal a H over 150 sin piun over 3 is < tk3 over2 cosine pi over 3 is a half and that's going to equal H over 150 so if I multiply the top and the Bottom by two that makes the half go away and that crosses out so you have < tk3 equals H over 150 multiply both sides by 150 and the height is going to be 150 root three uh feet I don't think this is a calculator problem good job drawing the Box stupid um nope no calculator so we're going to leave it like that now those of you who might kind of recognize this all right if you're good at geometry even though we're pretending like we don't know it but if you remember your 30 60 90 triangles uh Pi over3 is a 60 degree angle which makes this a 30 60 90 triangle which makes this this by the way uh yeah you would just divide it by root3 um and then this would be if I cared about this would be 300 root3 but I don't care about that I don't care about that trees The Lorax he speaks for them remember that guy uhoh there's my calculator oh it's time it's time it's time a man whose eyes are 6.1 feet off the ground is flying a kite at a moment the string of the kite is taught love that word and is 143.at is that many feet off the ground what is the angle of elevation well let's draw a man look at him he's flying a kite oh the string is taught that is a kite uh there's the thing and then like ribbons or something I don't don't fly kites anymore 2024 no one flies kites uh he is 6.1 feet off the ground I'll get to that in a moment at a moment the string of the kite is taau is 143.5 feet long the kite is 36.4 feet off the ground he is 6.1 feet off the ground uh what is the angle of elevation all right so based off of this awful picture is what I have I have a triangle I have a Theta you don't change you do uh we care about this angle right here we know how high off the ground the kite is don't care about that what I care about is how long this side is and that side is going to be 96.4 minus 6.1 which is going to be 90.3 okay which I believe is a wrap station here in southeast Pennsylvania now um if I were to to try to do like SOA again I would be like that's the opposite and that's the hypotenuse sign or we can view this as R and we can view this as Y and you would have uh s equal sin theta equals uh y over R right okay so sin Theta is going to equal 90.3 over it might be a Christian radius station I don't know inverse sign both sides now I'm curious uh inverse both sides and so that's what I'm going to need my calculator for I'm going to need my calculator to find the inverse sign of 90.3 over 143.5 so let's take a little break over and figure that out all right this will not take long not at all uh let's just make sure we're set to radians we are so quit out of that inverse sign is going to be second sign and then we have to type out 90.3 uh divided by 143.5 I guess we can close the parentheses because it's proper hit enter and that's 6823 blah blah blah blah blah radians so we'll rounded to 68 2 radians all right Mr calculator you have spoken 0.682 I've heard of Blink 182 but 0.682 not the same at all radians though we're dealing with radians because that's what AP pre-calculus deals with okay all right a pendulum ride is an amusement park don't I have a picture like and my face is probably covering it no okay a pendulum ride is an amusement park attraction that rotates in complete circles at phase two of the ride the time in seconds can be modeled by the equation t equal 4.8 over Pi inverse sign remember arc sign is inverse sign H minus 74.3 over 70.7 uh in parentheses for H feet how long after phase two begins does the height reach 100 feet so what I need to do is I need to graph this guy in my calculator and I'm going to also graph 100 in my calculator and see what I get so I believe I made a stupid mistake and said before that I needed to graph this I actually don't need to graph this um I have this backwards H is my height and it tells me to find out when the time that this hits 100 so all I have to do is type in the original formula and replace that H with 100 so uh 4.8 divided by pi Pi is second carot thingy close it inverse sign of let's open up another set of parentheses because I'm always nervous and this is where I get 100 100 minus 74.3 close it and divide it by 70.7 and that'll get me my answer so no graphing needed for this one I I was wrong so first time for everything what do we get we get 0. 5684 so we'll say 0.568 seconds not long at all all right 0.568 seconds didn't take long at all didn't take long at all solve T equals 4.8 Pi AR sign of that function for H all right so I'm going to rewrite S as inverse sign because it's the same exact thing and while I'm at it why don't I multiply both sides by pi over 4.8 so let's do that first I'm going to multiply both sides by pi over 4.8 that allows those guys to go go away so Pi T over 4.8 is going to equal the inverse sign of hus 74.3 over 7.7 okay now to undo an inverse sign what you do is regular sign so if I regular sign both sides which I don't have the space to write out I get regular sign of Pi T over 4.8 that equals hus 74.3 over 70.7 the next step to get rid of the 70.7 multiply both sides to 70.7 that gives me 70.7 s pi T over 4.8 = hus 74.3 one more step add 74.3 to both sides add 74.3 to both sides okay and I'm going to move H to the left H equal 70.7 sine Pi T over 4.8 Plus plus 74.3 if Phase 2 last 26.8 seconds how many times is the height of 127.5 reach so what I'm going to do is I'm going to graph this guy now and I'm going to also graph 127.5 and see how many times my S function is going to cross 127.5 over the interval that I'm given right there 26 8 all right now I'm going to do a little bit of the graphing and what I'm going to do is I'm going to graph the function that I just came up with 70.7 s uh what do I have Pi T So Pi second that t is X divide that by 4.8 so 48 close the parentheses plus 74 .3 now that plus 74.3 represents a vertical shift so I'm not going to see anything when I hit graph oh I stand corrected but I don't see what I really want to see um which was The Other Side of the Mountain I think that's a song uh but anyway I care about what's going on for the first 26.8 seconds so and I also care what the other magic number is what uh how many times is it 127.5 so what I'm going to do is I'm going to change my window so I need to go to 26.8 so we'll make it 0 to 26.8 because I don't really need to find the intercepts was it 26.8 let me go back and check 26.8 26.8 and then we'll bump this up to 200 okay and then what I'm also going to do is I'm going to graph the fact that we're hitting the height of 127.5 so I'm going to go down to 127.5 and this will give me how that's an eight this will give me the amount of times my sign graph is going to hit that line 1 2 3 four five six so that's my answer six name an interval where the height is decreasing so let's go to my calculator and find that out all right so to find a decreasing interval I got rid of the other part that I was dealing with what I want to do is I want to find one of these okay so from here to here or from here to here or from here to here let's do the first one because why not second Cal uh maximum will allow me to find where it begins decreasing so uh let's move to the left just a little bit hit enter let's move to the right a little bit hit enter and by hitting enter again it gives me a maximum value of what is that 2.40 so we'll say 2.4 now it starts decreasing here up until we hit that minimum value so why don't we find out what that minimum minimum value minimum value is so second Cal let's find the minimum value three let's go to the right a bit let's hit enter let's go to the right some more let's hit enter let's hit enter again and that gets me 7.1 19999 which is 7.2 so one of my options there's two others is we're decreasing between 2.4 seconds and 7.2 seconds that was a doozy by the way fun fact I have the flu I did this whole video while having the flu that's how much I care about you guys and your education I love you thanks for watching all my videos please continue to do so um one more of these and then that'll do it for unit three so fun stuff thanks for watching as always have a blessed day bye hey hey [Music] this is the end my only friend the end of unit three and technically AP pre-calculus because we only do the first three units on the test so that's exciting news today we're going to talk about polar graphs and polar coordinates and polar bears that's just a joke I'm just joshing you guys we're not going to really talk about polar bears where are we well we are going to talk about is polar coordinates versus rectangular coordinates you're used to rectangular coordinates rectangular coordinates are given to you as XY uh like for example the point 48 means you go right four you go up eight you put a dot there easy peasy lemon squeezy polar coordinates though are given to you as something completely different R and Theta R is going to be where you draw your dot compared to z0 on a polar graph now 0 is also called the pole so if R is three you're not going like right three you're going to count three units away and get ready to draw the dot where do you draw that dot well your Theta is going to tell you where you draw that dot see Theta is the same Theta that it's been all unit long if you were to go counterclockwise from the right positive Axis or in this case Zero uh if I tell you to draw the point 3 pi over 2 then you go to pi over two and you draw the dot one two three units away simple now what if I want to switch between the two if I want to turn something that is rectangular into Polar okay like into Polar what I do is in order to turn my XY into R Theta form the r is going to be found by taking the square root of x^2 + y^2 and if you're like that's kind of like Pythagorean theorem it sure is because your radius is basically you taking the X and the Y value and finding your radius which is kind of like your hypotenuse to find your Theta you're going to inverse tan the Y overx which again comes very much from SOA and stuff like that so that's how you would turn something that's rectangular into Polar how do you turn something that's polar into rectangular well it's a little bit friendlier I guess R is going to be or the Y value is going to be R Co Theta which should look familiar the Y value is going to be R sin Theta which should look familiar so if I gave you what's the point that I used I think I said 3 pi over 2 or something let's just say I wanted you to graph uh 3 Pi over4 okay this tells you how far away you're going to draw that dot from 0 and this tells you where you're going to do it Pi over4 lives there here's 0 0 one two three put a dot there now what would that look like on a regular XY plane well let's turn polar into rectangular by doing 3 cos pi over 4 and 3 sin Pi over4 okay cos pi over 4 is < tk2 over2 so this becomes 3 < tk2 over2 or 3 < tk2 over2 and sine piun over 4 is also < tk2 over2 so this also becomes 3 < tk2 over2 and if you were to run to your calculator and find out what 3 < tk2 over2 is and pretend that you're going to draw that on an XY coordinate plane you would land somewhere up here which is going to be the equivalent to that guy over there which no one said anything I put my DOT at the wrong spot should be right there because I said pi over four this whole time I could have been wrong but I wasn't let's move on to the next thing all right I moved my face because now we're talking about complex numbers and polar coordinates now a complex number involves a imaginary numbers complex numbers are usually given to us in the form a + bi I so let me give you an example of a complex number 2 + 2 I if I wanted to graph a complex number on a complex number plane it's very similar to a rectangular plane you would basically say you are the real number so let's go right to on the real axis and you the complex or imaginary number so let's go up to two on the imaginary axis and put a dot at what appears to be 22 now if I wanted to link this to what polar coordinates look like it's going to be very different a polar coordinate that is also complex is in this form R Co theta plus i r sin Theta and you might look at this and say well wait a minute R Co Theta is like the x value and R sin Theta is like the yalue so that's and when we graphed this that was basically the x value and that was basically the yalue now you're getting it by the way other teachers might write it out like this where they factor out the r it's the same exact idea but you know it all depends on the teacher that you might have uh it's the same it's all the same don't be surprised if you see either option now this was the uh complex number form form what if I want to make this polar well we're going to do what we did in the other slide but kind of opposite where I believe I gave you polar and we turned it rectangular now I'm giving you something that looks more rectangular and we're going to turn it polar okay so when I did that in the previous slide what I wrote down is in order to find the magnitude the r is we would take this square root of X2 plus y^2 well now we're viewing it as a 2 plus b^ s and if you're like is it really isn't it just the same thing it is you're my a you're my B which is basically like an X which is basically like a y so we would say that R is going to be 2^ 2ar + 2^ 2quared r is going to be 4 + 4 R is going to be the square root of 8 and you can clean that up by making that 2 < tk2 now what we saw in the last slide is a in order to get the Theta version of it we would take the inverse tangent of Y overx well I don't have y overx but it's the same exact idea if we said B over a okay and that gets us R Theta now in this case to saves some room uh B over a is 2 over two so the inverse tangent of 2 over two is the same thing as the inverse tangent of one the inverse tangent of one we technically have two answers but since this guy lives in quadrant one we only care about the quadrant one answer which is pi over4 so my Theta is pi over 4 my R is 2 < tk2 so in order to take these two pieces of information and make it look like one of those I'm going to and I'm going to make it look like the blue one because I use that much more often I'm going to say that this complex number written as a polar coordinate is now going to look like 2 < tk2 cine Theta which is pi over 4+ 2 < tk2 let's attach an i there not inside the square root because that wouldn't make sense sine pi over4 and what that would look like graphically on a polar coordinate is you would imagine what 2 < tk2 is so you know two point something three point something I actually don't know the answer and then you would go to where Pi over4 is in that direction and if this was a polar graph you would count out 2 < tk2 and put a dot there which would kind of equate to that okay all right fun all right using trig graphs or tables to make polar graphs now I you can memorize what polar graphs look like when graphing polar graphs and we'll kind of talk about what they could possibly look like as we do problems here but to me the best way to draw out a polar graph is to me either make a trigonometry graph like on a regular rectangular plane or to draw out a table on a rectangular plane uh using the information that you have Okay so let's just say that I gave you something like three let's say I want you to graph on a rectangular or on a polar graph r equal 3 cosine Theta okay and I want to graph that a very quick way of doing this and and keep in mind too AP pre-calculus is going to have a lot of multiple choices so if if this is a multiple choice problem this will be a lot faster than what you expect but if I were to go and draw the whole thing out this is how I would do it okay quickly and we will do a lot of these so don't worry but a very basic way of doing this is you know that three Co Theta is going to be up one two three down one two three cosine goes like this and stops right there regular frequency so at 2 pi so it bottoms out at regular Pi at half a pi is where we hit zero at 3 pi over two is where we hit zero again okay and then we repeat the process so maybe we draw the points that we know okay at 0 three I can go over here and say all right well when Theta is zero I have three so the magnitude is three so we go over here and where Theta is zero we have a magnitude of 1 two 3 and we put a dot there then let's put a dot at zero on pi over two okay now what this is going to look like graphically okay okay what this is going to look like graphically so far is as I go from the angle zero to pi over two my magnitude is getting smaller so what's happening is as I go from zero to pi over two I am going towards this dot so this is being created let me do a better job let me do a better job one more time as I'm going from three as I'm going from the angle 0 to pi over 2 my magnitude is getting smaller according to this picture here so I'm creating more like a circle because I'm now heading toward with zero okay as I go from pi over two to Pi I'm hitting negative3 so Pi is -3 so if Pi is -3 usually right regular three would be over one two 3 right here but if it's negative3 it's over there so I'm literally creating a circle again and then as I go from PI to 3 pi over 2 it's the negativ so I'm creating that Circle okay and as I go from 3 pi over two I'm hitting Pi again so this should be a little bit more circular so let me clean this up a little bit this should be a lot more circular but basically by using this picture I am able to draw out what this graph looks like which is just supposed to be a circle which I just simply cannot draw much better okay all right not pretty but we'll do more let's jump into some problems given the complex number graph on an imaginary axis all right if I remember what an imaginary axis looks like okay uh let me scooch this up let me scooch this up because I know exactly where this is going to live this is my real axis right here this is my imaginary axis right here the real number is five so one two three four five the imaginary number is negative five so one two three four five in the negative area and we get a DOT right there okay okay rewrite in the form R cos theta plus I sin Theta all right so very much what we did a couple slides ago I need to label u a and label u b and R is going to become the square root of a 2 + b 2 so R is going to equal big square root 5^ 2ar + 5^ SAR don't worry this just going to become 25 + 25 which is going to become 50 which is going to become factor out of five 5 < tk2 so that's the r the way we find the Theta and my lights just turned off in my room is the inverse tangent of B over a y overx in this case the inverse tangent of-5 over regular five which simplifies out to the inverse tangent of neg1 now the inverse tangent of negative 1 has two possible answers one that lives up here or one that lives down there we want the one that lives down there which means we are going to have an angle measure of 7 pi four took me a second there so putting this all together we can write out 5 < tk2 cos 7 pi over 4 plus 5 < tk2 i s 7 Pi over4 given the polar coordinate 35 pi over 6 graph the polar coordinate easy three is your magnitude okay three is how far away from the pole 0 0 you're going to put that dot so 1 two 3 or 1 two 3 whatever 5 pi over 6 is going to be the direction that you do that in so let's locate 5 pi over 6 and let's count out 1 2 three and that's how you graph that to write that in rectangular form rectangular form is found by doing R cos Theta very familiar R sin Theta R is three Theta is 5 pi over 6 so 3 cos 5 pi over 6 3 S 5 pi over 6 cosine pi over six let me see we're over here is going to be < tk3 over2 right yep 3 * < tk3 over2 and sine Pi 5 pi over 6 is going to be a half positive a half so 3times a half so simplify this out and you get -3 < tk3 over two and 3 over two so if you were graphing that on a rectangular plane you would go whatever number that is left that many up a little bit and you'll have a DOT there which should uh you know if you were to glue that on top of the other would look just like that all right all right another wardrobe change it's 6:30 a in narst toown Pennsylvania I'm about to do some polar graphing wish me luck graph this polar coordinate -4 Pi over3 all right well this is my magnitude this over here right here where I'm drawing that R and that is going to be the angle that I use usually what I do is I go to that Pi over3 which is somewhere around here and I would draw at four but this is -4 so what this means is I go to Pi over3 and I go in the opposite direction for so I go down 4 Pi over3 1 2 3 four and I put the dot there so so a negative magnitude means I go to that angle and I spin my magnitude in the opposite direction than when I'm used to so basically if I wanted to I could rewrite this guy as regular four add pi to that and it would be 4 pi over 3 and it would be the same exact thing so graphed now to rewrite that in rectangular form I'm going to use the old R Co Theta R sin Theta thing a thing and plug in what I know it doesn't matter which one of these I use because they're the same thing so let's just use the negative original version so -4 Co pi over 3 -4 sin pi over 3 cine pi over 3 is a half so -4 * a half and sin Pi over3 is < tk3 [Music] over2 so since I'm running out of space let me just write this all in one thing4 * a half is -2 no overthinking there uh -4 over 2 is -2 up top so -2 tk3 so if you were to graph this on a rectangular plane you know rectangular planes look like like this you would go left two down -2 and A3 or root3 whatever that looks like you'll have a DOT right there which correlates to that dot right there as if you were to draw an XY axis right there through the pole fun all right things get a little hectic here now I'm going to do exactly what the directions ask me to do and draw this by making a table when I'm done I will explain kind of like a shortcut to to do this where maybe it won't help with drawing but it will help with identifying like on a multiple choice test and that would make life a little bit easier okay so at the end of this problem I'm going to compare 4 sin 2 Theta to a s or cosine and Theta we'll get there right now I want to graph this by making a table okay so what I'm going to do is is I'm going to graph I'm going to make a table it's going to look like this hopefully I give myself enough space hopefully I gave myself enough space um I'm I'm graphing all the way from zero to uh Theta to 2 pi the fact that there's a two in front of theta means that things halfway through start to repeat since the frequency is two that means things start to repeat themselves halfway through okay so by putting a two in front of Pi that cuts my normal period of two Pi in half because 2 pi over two so again things repeat themselves so I probably you know if I put a Theta here I'm going to use every single number here all the way up through pi and if I have space I'll maybe even do the extra one so uh I'm going to start at zero pi over 12 12 pi over 6 pi over 4 pi over 3 5 pi over [Music] 12 pi over 2 7 pi over 12 uh 2 pi over 3 3 Pi over4 5 pi over 6 11 pi over 12 and pi and I barely just made it and at that Pi though I'm going to see the repeating process take place uh what I'm going to do in the middle here is I'm going to write out 4 sin 2 Theta and I'm going to allow this to be an area where I do a little bit of calculations in other words for S 2 Theta is going to be 4 sin 2 * 0 which is 0 and then I'll figure out what that math is over here when I get to it uh this becomes four S 2 * piun over 12 is pi over 6 4 S 2 * piun over 6 is pi over 3 for S 2 * < over 4 is < / 2 and again as we do these These are now more manageable things that we can deal with we we know that we can do all of these signs we just have to multiply four uh when we're done them uh 4 S 2 * pi over 3 is 2 pi over 3 4 S 2 * 5 pi over 12 is 5 pi over 6 4 S 2 * < / 2 is Pi 4 S 2 * 7 pi over 12 is 7 pi over 6 for S 2 * 2 piun over 3 is 4 pi over 3 4 sin 2 * 3 piun over 4 is 3 piun 2 4 sin 2 * 5 piun over 6 is 5 piun over 3 4 S 2 * 11 piun over 12 is 11 pi over 6 11 pi over 6 and 2 * pi is 2 pi which is the same as zero time to do some math time to do some math s0 is zero so 4 sin 0 is zero sine of pi over 6 is a half so four S of a half is two okay uh this becomes 2 < tk3 this becomes one so four this becomes oh no way two < tk3 okay because it's root3 over two 4 * root3 over2 is 2 < tk3 uh this becomes a half to uh sine of Pi is 0 so four time 0 is 0 this is negative a half so -2 this is < tk2 over3 no this is < tk3 over two so that would be -2 < tk3 uh this is1 yep so uh yep yeah it took me a second there so that's ne4 and some of you might be able to see the pattern forming uh this is -2 < tk3 because it's the same as this uh this becomes -2 because it's negative a half and now we're back to zero again and then it repeats the process it repeats the process it repeats the process so from here to here we get these gross numbers and then if we were to write all the way back around you would literally just cover uh the numbers that you would have so I think we're all right I think we're all right but we'll see let's let's graph what we have z0 is a dot in the middle okay uh you as we move around you go to two Okay 2 < tk3 is going to be like somewhere around here is okay and then we hit four and then 2 < tk3 again and then 5 pi over 12 is 2 okay [Music] uh then we hit zero and then for you we're at -2 so that's going to be here okay and then -2 < tk3 puts me around here and then negative uh 4 is going to put me right around here -2 < tk3 puts me around here and then -2 puts me around here and then back to zero now as we go for the rest of them we use these same EX exact numbers in this order so Pi is zero 13 pi over 12 would be 2 okay so here and then that 2un three and then that four and then that 2un three and then that two and back to zero and then here here at 19 pi over 12 we would be at -2 so that would put me here at regular two and then at 2 and a half in did I go too far no and then at 4ish and then at like three and a halfish not two and a halfish uh and then at two again and then back to zero so the the the direction and the trip that I took around this whole thing had me start here here okay loop around here okay and then continue my path around here and continue my path around here and then continue my path around here now that was the long way of doing this okay that was the long way of doing this and it took some time but what we created here was a rose and a rose is something that looks like this form now here's something that you can memorize for when you have to deal with this stuff uh um without having to draw it out like for for example if I see this I now know what this looks like and I know what all of the other a sin n thetas look like they're going to look like flowers okay when n is even when n is even what's going to happen is you're going to have two n petals in other words in other words since I have uh 2 N I created a rose with four petals if this was 3 n well hold on if if n was six I'd have 12 petals if n was four I'd have eight petals so on and so forth now if n is even or odd I would have that many petals when it's sign when it's sign okay the the flowers do not go through they don't start here when it's coine the flowers would start here because of the fact that cosine starts at like 01 or in this case it would be 04 okay so a sign would be literally the exact same flower with with the pedals going through the major axes but since this is a sign to n we can expect having the flowers or having the petals go through like the diagonal axes and have two times more petals than that number there so that's like the shortcut multiple choice way of doing a problem like this pretty intense pretty intense now knowing what know and and knowing what we just said okay when it was when it was s what was it four Theta okay we had a flower I'm sorry S 2 Theta we had four pedals and it looked like this that's awful uh now that we know that this is cosign we can expect three pedals where the one pedal is going to live right there and so probably have something that looks something like this that's what we can expect where these go out to six that's what we can expect so if this was a multiple choice question and we see a picture where there's a pedal there a pedal there and a pedal diagonally here that's going to be our guy but that's not what it's asking us to do right here so what I am G to do is I'm not gonna not going to go too crazy I I I am going to use since I know I'm going to get three pedals maybe I can just make my life a little bit easier and um uh only do well let's see how many I can do let's see how many I can do let's let's make a table again okay let's make a table again we have Theta we have 6 cos 3 Theta and we have R so so let's choose 0er let's choose pi over 12 at some point around here I can expect repeating to take place so maybe I stop at 3 Pi over4 if I have room for it let's do pi over 6 Pi over4 pi over three 5 pi over 12 uh pi over 2 7 pi over 12 2 pi over three and 3 Pi over4 is basically all the room that I have anyway and that's when we will start seeing things repeat itself okay 6 cos 3 * 0 is 0 6 Co 3 * piun over 12 is pi over 4 6 Co 3 * < / 6 is < / 2 6 cos 3 3 * < 4 is 3 piun over 4 6 cos 3 * < over 3 is < so we're halfway through 6 Co 3 * 5 pi over 12 is 5 pi over 4 I don't know why that took me so long maybe it's because it's 650 a.m 6 cos 3 pi/ 2 almost there 6 cos 7 pi over 4 6 cos 2 pi so zero so now we start repeating so this will be the same thing as 6 cos Pi over4 all right okay cosine 0 is 1 so 6 * 1 is 6 cosine pi over 4 uh is going to be Pi / 2 no uh uh ro2 over two so this becomes < tk2 over3 you're zero your negative so negative < tk2 over3 your1 so6 you're negative < tk2 over3 uh your zero your regular < tk2 over3 and then the process repeat so 6 < tk2 over three all right so 6 0 six one two three four five six < tk2 over three is really small like point something like GH like right here okay zero and then at uh pi over 4 which is U you're at negative < tk2 over3 so uh pi over 4 is like U okay and then at Pi over3 we're at -6 and then come back down to here and then Z and then 7 pi over 12 is here and then 2 pi over 3 is 6 here and then wrap back around so basically oh and I need I have one more I have um oh no I'm good all right oh I do have one more it's gon to it's going to happen right here okay all right so I start out here and then I wrap around oh gosh oh golly oh my gosh by golly it's Christmas song yeah so definitely not the prettiest thing in the world but uh it is a rose it's a three petal Rose it's supposed to be this Rose this pedal looks awful I should have moved it around a little bit more so that it looks more like this but again you know you guys aren't paying me for my art skills you're not paying me at all but I do get money I do get money as rappers like to say so here's my three petal rows just as I predicted but again like you know I would look at something like this and say like it's a it's a three petal Rose because you're odd I know it starts on the x axis because it's coine so it's not g to it's it's going to be you know rotationally symmetrical so I should expect if you're going to have a rose here you know it's kind of like a sideways Merced Mercedes symbol or Mitsubishi or whatever car you like because I don't know anything about cars uh windmill fan fan blade fan blade symbol all right all right now now that we've made a couple polar graphs by making tables maybe this is less tedious maybe maybe not but what I'm going to do is I'm going to sketch what this looks like on a regular coordinate plane and find a way to translate that on a polar graph so let me do that right now we have two + 4 cos Theta now cos Theta has a regular period of 2 pi so that's good news but this amplitude and the fact that we're adding two to it tells me we have a vertical shift with a positive two change so we're going to we're going to have a midline here at two and I'm really not going to give myself enough space so let's go one two three four five six that should do it one two three four that's going to overdo it let's add one more for funsies and we're gonna have like a midline here at regular two okay um the period is two Pi like I said before so that's going to happen Magic's going to happen at pi and then we have 3 pi over 2 and pi over two and yada yada yada the amplitude is four which means normally at cosine we start at one and in this case we'd start at four but since we're adding two to everything we're going to start way up here at one two three four five six is this six two three four five six and then we're going to hit our midline at pi over two we're going to bottom out at -2 at pi hit our midline again at 3 pi over two and start the process at two Pi so me me me me drawing a bunch of dots because it looks better than trying to draw for real because I'm sloppy now let's label what we know uh at zero I'm at six so here's 0er 1 two 3 four five six um at pi over two I'm at two so here's pi over two and here's two uh at Pi I'm at -2 so here's Pi -2 is going to live right there and at 3 pi over two which is this guy I'm at two so one two okay so what's going to happen is as I'm normally on a regular trig graph on this picture as I move towards pi over two my number goes closer to that two value so here I am my number is going to go closer to that two value so basically we have a picture that kind of looks like this okay now this is where it gets weird at some point after pi over two I hit zero so this is going to happen then I hit negative land and then I finally hit that guy right there so this pattern continues and I create this weird curl going in okay now that I'm at pi and -2 I'm going to start moving closer to zero and eventually hit that to so the opposite happens here okay and very similarly to what I saw over here now that I'm at Pi 3 pi over two at two I am moving further away and hitting that six value so this is going to kind of create like a circular looking thing here and I'm going to have this now what I have and if you are somebody who speaks French uh I apologize for what's going to happen but what I have here is a li soul that is like a weird C with a curve on it now I don't speak French I think it means heart but I don't know in America we call this a cardioid because it kind of looks like a heart it kind of does I don't know but what makes this special from the other two is the fact that I added a number to it so the number that I add the number that I could subtract is going to make this guy look like a specific type of cardioid or Li uh there's ones that's more of like what's called a dimpled version and then you have this version it all depends on what number this is and what number that is that'll tell you what you're going to get okay but these are examples of what polar graphs could look like all right now that we have a little experience graphing these things the old-fashioned way let's try to memorize what they look like at the very beginning we graphed a circle I believe it was something like 3 cosine Theta and it ended up looking like a circle over here first things first a circle is going to be something in the form R equals some number times either s or cosine okay signs open up here okay if this a was negative it would open down here now we'll explore that what does the a do well if I make the a bigger it just makes my circle bigger okay if I make the a bigger it makes my circle bigger if a was negative it shows up down there what if this was cosine all right well let's change that let's change this to cosine and now it shows up Lefty righty instead of upy Downy so a negative cosine appears on the left and a positive cosine appears on the right and I think we did something like three cosine Theta to start out the day I don't remember I don't remember now what if I were to take a number and multiply it to Theta well let's see if the number is one we have that Circle still as the number increases things explode now we're not going to worry about decimals not unless we're using a calculator when it's even as you know we double up on the amount of pedals that we get when it's odd we get the same amount of petals so something like you know 3 CO3 Theta has a row with three petals on it if I turn cosine into s it kind of shifts a little bit and instead of starting here on the positive x axis it starts up here now okay and as I make n larger so now it's even again I have eight petals if I make it five it now has five petals if I make make it negative oh it's upside down and the a just determines whe whether or not I have something that's bigger or smaller it's my magnitude now let's bring this back down to one make it a circle and let's take a look at the Limon version of it if I were to take like one and add one to this there it is it makes that Li what if I make it two it makes it bigger what if I made a and made that smaller it becomes the dimpled version where now it's more of a heart instead of the weird loopy looking thing which I believe is called a looped Limon and that's a dimpled Limon so turning into the sign would make it look I'm sorry it is sign turning it into cosine would make it the sideways version and there you have it okay so you're looking at three different versions of these things and we looked at all of them something that's a circle is going to be you know five s Theta something that's going to be a rows is going to be like three sin 5 Theta and if I add a number to it that's where we get the limit all all right now that I'm done graphing those guys for now I think uh let's do something a little different trace the part of the graph that covers just one part of the interval so I have R = 8 sin 4 Theta so 4 Theta gives me the eight petals that eight tells me that this goes out to eight which appears to make sense um you know it checks out checks out I care about tracing the part between uh pi and three pi over two so this is what I'm going to do I'm gonna approach this a little differently I'm going to graph this guy out now what this means by having the four in front of it is my period is going to be 2 pi over 4 which means my period is half a pi so whatever this does is going to repeat four times every two Pi or every half a pi now I only care about the stuff that happens between pi which is U and 3 pi over two which is U now normally I'd start you know at at Z 0 and then I'd go up to eight and down and repeat the process up to eight and then down and then repeat the process but since I only care about this I'm going to try to draw this out to the best that I can go up to eight down repeat so between pi and 3 pi over 2 I start at zero at pi so here I am at Pi okay there's my Pi I'm starting at zero as I go towards 3 pi over two what's happening is I'm going out to eight then coming back and then normally here I'd go like this but now I'm in negative land so what's G to happen is I'm starting at pi we're in positive land according to my picture I go out to eight so let me make this a little bit bigger I go out to eight let me make this a little bit bigger I go out to eight right and then I hit zero now on the picture here I am as I continue now I'm in negative territory which means not here but instead here and that is me tracing so that is me tracing from PI to 3 pi over two this little green area I traced it or for my Spanish-speaking friends I threed it that's three different languages we have going on in this one we had French we had Spanish and now we have English and it's going to stay that way uh because I'm not that talented let's get the pen back to normal plot on a complex plane well this is a complex polar point the old CPP uh what this says is I have a magnitude of two and I go to 4 Pi over3 which is U and I go positive two in that direction One Two there that's what it looks like the absolute value if given a point on a complex plane or a point in general is the distance from the pole the pole the pole to that point well if I give it to you in complex form the absolute value is just simply the magnitude if I give you something in polar form the magnitude is your absolute value absolute value just means distance so if I give you something in rectangular form the distance might mean something else like using Pythagorean theorem or the distance formula but in this case it's just the magnitude so writing this in rectangular form is me accidentally doing what I kind of said we would do all right now in rectangular form what you're going to do to find out your number is you're going to do and again it looks like this a plus bi I okay that's the rectangular form of a complex number because we have that I there the a is going to to be R cos Theta so 2 cos 4i over 3 my B is going to be R sin Theta 2 sin 4 Pi over3 okay so my a is going to be 2 * cos pi over or 4 pi over 3 which is < tk2 over 2 okay and we can kind of see that down there and we can kind of see that it's since it's in the middle of everything the numbers seem to make sense so these cross out and you end up with negative < tk2 okay let me scooch that up so a is negative < tk2 a is netive < tk2 now the B is going to basically be the same thing and since the numbers seem to check out this is going to be 2 * < tk2 over 2 so B is going to also be < tk2 now putting this all together and wrapping it up in a little bow I need to make it look like this so we are looking at and let me put it up here < tk2 minus I < tk2 okay which means you know we' go left like one point something down one point something and that seems to check out run2 is 1 point something and that seems to be exactly what we need find the distance from the pole the origin to the point on the curve when Theta is pi over4 so there's that formula not too bad I know I owed it before but it's not too bad all we're going to do is plug in pi over 4 and do a little bit of math so R of Pi over4 is going to be 2 + 3 cosine Pi over4 cosine pi over 4 is < tk2 over 2 so 2 + 3 * < tk2 over 2 2 + < tk3 over two and and I'm going to leave it like that there's zero reason why I need to combine that into one term so uh so that's that find the values of theta where there's a maximum value now I now kind of know what this picture is going to look like on a polar graph but for me I visualize these better on the regular old trig graph it's the same thing if this guy if if this guy and I kind of know what the maximum is going to be I know at some point cosine Theta is going to equal one so at some point I'm going to get 2 + 3 * 1 is five at some point I'm going to get five so this is what I would rather do I know what cosine looks like I know what cosine looks like and again now that I kind of said that out in my head I really don't need to do this but I'm going to do it anyway I know that this is going to be vertical shift of two so it's going to have a midline at two here it has an amplitude of three so it's going to go up to 1 2 3 4 five it's going to bottom out at negative 1 and since there's no horizontal shift it's going to start at 0 five it's going to go down and hit and then repeat itself at 2 pi okay so common sense would tell me that again at some point and right now close your eyes and either visualize this which means you don't have to close your eyes or visualize the polar graph at some point you know your maximum is going to be five when's that going to happen well that's going to happen when cosine of theta is one when is cosine Theta equal to one at one and only one spot zero but since I'm including 2 pi it's going to happen at 2 pi as as well so my maximum values are going to happen at zero so theta equals 0 and Theta equal 2 pi okay so whether again you view that on a polar graph and I think this picture is going to look something like this okay that's going to happen right here at the very beginning which is a very good place to start Sound of Music which I believe took place in Germany I think which I believe the German word for no is nine it's a shame we didn't have any problems that involved the number nine there's still time but nine no calculator no no calculator I move my face at everything just to find out that there's no calculator but again we really don't need one find the average rate of change on that interval right there now it's going to be ugly but we don't need a calculator for it okay the average rate of change for a polar function is the exact same process for the average rate of change that you saw back in unit one the average rate of change is if I have F of B minus F of a or R of B minus r of a you put that over B minus a so if I called you a and I called you B then what what I need to do is find out what R of 5 piun over 4 minus r of pi over 2 is over 5 Pi over4 minus pi over two so I'm G to step to the side here and just you know take my time because I'm going to need the space and show what R of 5 pi over 4 is and then I'll do R of pi over 2 directly underneath that okay so we're g to have 2+ 3 Co 5 pi over 4 5 pi over 4 is quadrant 3 so that's < tk2 over 2 so 2 + 3 * < TK 2 over 2 so 2 - 3 < tk2 over two when you multiply them together and let's leave it like that just in case something magical happens in this process right here now let's find out what R of pi over 2 is pi over2 how nice why well 3 Co Pi / 2 [Music] is going to be two why well because cos pi/ 2 is zero so now that I have this information I can replace R of 5 uh Pi over4 with 2 - 3 < tk2 over 2 in an unnecessary parentheses so let me just get rid of that parentheses now we're going to minus uh R of pi over 2 which is just two so if I had more there then I'd want the parentheses but don't need it uh why don't I change this to 2 * 2 so I can have five Pi minus 2 pi over 4 now I can combine them 2 minus 2 go away so I have 3 < tk2 over 2 all over 3 pi over four let's multiply the top and the Bottom by four over 3 pi to get rid of the 3 Pi over4 on the bottom u2s simplify out to one and two u3s go away and that leaves me with2 < tk2 over Pi for which polar function does the limit of theta approach Infinity of that polar function equals zero I have three polar functions right here okay I need to find the one where the limit equals zero well if this was chapter one and I had theta's at the top and theta's at the bottom and whatever I would only focus on the first term of each right so you would be Theta squar over Theta which is Theta and if I plug in Infinity I get infinity so not you does the same rules apply here yes so this becomes Infinity so it's not you uh you become 2 root Theta over or 2 Theta to 4th over Theta the fourth you cancel out you become two can't really plug in Infinity because there's nothing to plug it into so I get two so not you you're my leading term you're my leading term so you become one over Theta to the 4th power if I plug in Infinity these I'm going to get like one over a really really really really really big number which is like 0 000000 Z oh it's zero you're my guy ah makes sense okay limits with polar functions behave very very very similarly as limits and regular old regular functions regular lots of regulars it's like it's like I own a restaurant lots of regulars there you are hello Mr calculator we've missed [Music] you all right graph the polar function in the calculator I love these These are going to be beautiful uh find R of 1.5 Pi so let's graph that in our calculator and see what we get so part one is to take this gross looking function and find out what 1.5 Pi is well before we can graph a polar graph we have to make sure our calculator is set to Polar so click mode mod go down the function and make sure it's polar like so then what we're going to do is we're going to hit Y equals and we're going to type out this thing right here five now this is where it gets crazy so bear with me here there's no sin squared button so this is how we're going to do it we're going to wrap s in parentheses so s Theta completely close it and then square that that is the same as s SAR Theta minus three do the same thing for cosine so wrap it in parentheses Co Theta make sure you close the whole thing so double up the parentheses square that as well and then plus one now let's hit graph and see what we get o if you needed to zoom in you could uh but I kind of like what we have let's leave it like that uh let's leave it just like that my job for part A is to find out 1.5 Pi so let's do second Cal and type out the value if you're like what are those what are those calculus you'll find out next year just hit value and we're doing one point 5 Pi second carat gives me pi hit enter ENT and we have six oh it's actually ne66 all right so negative six negative six find the average rate of change on the interval 75 Pi minus 1.25 Pi so what we're going to have to do is we're going to have to find the values of R of 1.25 Pi minus r of 75 PI right and then we're going to put that over uh 1.25 Pi minus 75 Pi so I don't know what the answer is yet whatever this is whatever grow gr number that ends up being is going to be over 0.5 Pi so let's see what happens all right so let's find the average rate of change of 7 pi and 1.25 Pi so we already did the part on the bottom let's do the part on the top uh let's do the second thing that we did here and calculate the value of 1.25 pi and we get uhga 1.414 214 so let's quit out of this and type that out [Music] 1.414 and then I'll hit minus and I'll do the other part when I get back to it let's go back here and calculate value now I'm going to do 1.25 Pi enter no way it's the same it's the same so since it's the same I'm gonna get zero up top zero over anything is zero oh it's zero the answer is zero and it's zero uh graph the function is it the same thing I just want to show it off okay so graph the function on the graphing calculator so um I already did so I'll probably take a screenshot of the uh graph and just show it here on the screen and I'm gonna have it right there and I'm gonna talk about it and say wow isn't that a cool picture what a really cool picture that is that picture right there what a neat picture it's like a weird deformed windmill type picture but isn't it such a pretty picture it is and speaking of pretty picture that's what you guys are to me pretty and pictures but uh this is the end of unit three why I sang that song at the very beginning uh so we're done we're done unit three uh good luck at this stage you should be if you've watched all of my videos you should be adequately prepared for the AP pre-calculus exam whenever that is for you probably sometime in May unless it's a different year and the world has changed drastically I mean we just had covid so you never know but uh thanks for watching my videos and good luck good luck love you bye hey hey hey hey hey [Music] heyyyyy hey