 Welcome to our exam review for exam 3 in Math 1060 Trigonometry. As usual, I'll be your professor today, Dr. Andrew Missildine. If you're watching this video, it probably means this is not your first exam for this course. As such, I'm not going to talk about really the policies, time, place, and manner, and other semester-specific details about the exam in this review video. Please find that information in the syllabus, on Canvas, or contact me, your instructor. What I want to do in this video is very briefly talk about the topics that will be covered on exam 3, so you know which topics you should be studying for as you prepare for this exam. This exam will cover topics covered from lecture 21 through lecture 32, excuse me. So this will cover the topics of chapters 7. Chapter 7 was about solving trigonometric equations. It'll include chapter 8 about oblique triangles. It'll include chapter 9 about vectors. And technically speaking, section 10.1 about complex numbers is covered on this exam. This is sort of a timing anomaly. There will be one question about complex numbers. It will not be very intensive whatsoever. I will show it to you in the multiple choice section. So predominantly, as you're studying for this exam, focus on the topics of 7, 8, 9 solving trigonometric equations. That was chapter 7. Chapter 8 was solving oblique triangles like lost signs, lost cosines. And chapter 9 about vectors, both in geometric or algebraic form. This test, like usual, have 15 questions, and we broke up into two parts. The multiple choice section will actually consist of nine questions, which are worth five points each. Then we have six questions in the free response section. Most of them will be worth 10 points. Question number 10 is only worth five points. It's really a multiple choice-esque type question, but it's not really geared towards a multiple choice. It's on the easier side. You really don't need much to do it. But having six options doesn't really make sense for that question. We'll see what it is in a second, well, in a few minutes, I guess. And so let's go from there. In the multiple choice section, of course, you are not allowed any notes or a calculator. The only resource you'll have is yourself and, of course, the approved formula sheet that you hopefully have seen by now. In the free response section, you are allowed a scientific calculator. You are expected to prepare your note card. And these are things you're probably familiar with, so I won't say much more about that. So let's talk about the specific types of questions you might encounter when you're taking the exam. Let's start with the multiple choice section. Multiple choice section is actually going to be predominantly vector-type questions, chapter nine-type stuff. The vector questions, with the exception of maybe the story problem vector questions, which we saw, which actually we'll see one of those in the free response sections, they're very much simple calculations. And so you're going to see a lot of those in the multiple choice section. For example, question number one is going to ask you about vector magnitudes, vector direction, vector components. Can you compute the horizontal component? Can you convert the vertical component of a vector? This could be given algebraically. This could be given geometrically. And so a good place to consult here would be lectures 28, when we started our discussion about vectors, which focused on the geometric representation of vectors. Also lecture 30, which focused on the algebraic representation of a vector, which we can see as the example question. That's how it's given. Question number two is going to be very similar in that nature. Perform a calculation. Can we add vectors? Can we scale vectors? This is a lot easier in the algebraic form, but can you add and scale vectors if it's given geometrically as well? So those type of calculations are going to show up with vector quantities. Question number three you can also see as a vector question. This one, it's going to ask you to compute the dot product of two vectors. So you're going to prefer that be algebraic, but could you compute the dot product? If it was given geometric, you probably want to convert it to the algebraic representation. Or of course, you can use the fact that the dot product UV is equal to the magnitude of U times the magnitude of V times cosine of the angle between them. So could you compute the dot product using that formula? That's the first three questions you're going to see here. Let's take a break from vector quantities for a moment and talk about question number four. Given the timing of this exam given the end of the semester and the length of topics we did, I am going to throw one complex number question on exam three. Chapter 10 about complex numbers, and then the last chapter, chapter 11 of coordinates, will be covered on the final exam as exam three is our last midterm exam for this course. So we will see all of the questions about complex numbers, many of which you might have already seen as you're reviewing for this test. Because again, given the timing of the test and what we're learning in class, it's very likely you've seen many topics about complex numbers by now. On question number four, this will be the only question on this exam about complex numbers. And this is just going to be a basic arithmetic calculation of complex numbers where you're going to add, subtract, multiply, or divide, or some combination of them. Like this example right here, we have to divide two complex numbers. But to divide two complex numbers, you have to use the conjugate, you have to multiply the by the conjugate, so multiplication is necessary. You have to add and subtract the like term. So really, if you can divide two complex numbers, that's really the whole enchilada when it comes to complex arithmetic here. Now, you're going to do this in the Cartesian form, the trigonometric or polar form of a complex number, which you might be learning about presently, which we did introduce in lecture 32. That will not be covered on exam three. It will show up on the final exam, though. So question four, we'll just ask for the traditional Cartesian approach to complex numbers. The trigonometric or polar form will come later on. Don't worry about that for this exam. So like I said, that was lecture 32 that you'll see this question for. Question number five is going to be a question about area. You'll be given a triangle, and some data about that triangle will be provided to. So in this situation, we have a side, we have a side, we have an angle, so this is a side angle side situation. You should be able to compute the area of a triangle given whatever information's provided. So we're looking at all these formulas that we derived from the basic formula, one half base times height. All right. So what are some of the variations you could use? Well, there's the variation where area equals one half A times B times sine of C. So that was one version of the formula we've seen. We had another version where we had A squared times sine of B times sine of C sits above two sine of A, like so. So both of these versions are consequences of the law of sines that then you could use. Again, depending on the initial data, we'll determine which formula you want to use. And then lastly, we could use Huron's formula if appropriate, like we have a side, side, side situation where Huron's formula tells you s times s minus A times s minus B times s minus C. All of that product inside the square root gives us the area. So use one of those area formulas to compute the area of a triangle given some information. We learned about area, of course, in lecture. What was the lecture number again? Oh, I spaced it, I'm sorry. That was lecture 27. Sorry about the brain fart there. Moving on to the next page of the multiple choice, more stuff about vectors. So look at this one right here. You're given the magnitude, you're given the direction. Well, you're not given the direction per se, you're given sine theta equals three-fifths and you asked to find the horizontal components of these things. So some comments here, some important formulas for vector calculations here, like we talked about magnitude previously. So you should know that the magnitude here is equal to the square root of the horizontal component squared plus the vertical component squared. The direction theta is gonna equal, we'll say tangent theta is equal to the horizontal divided by the, excuse me, the vertical component divided by the horizontal component there. We also get that this horizontal component is the magnitude times cosine of theta. We're given that Vy equals the vertical component is equal to V times sine of theta. These formulas are the formulas we use to convert from the geometric form of a vector to a tau to break form or vice versa. These are also the formula we use for converting between the Cartesian form of complex number and the polar form of a complex number. Again, the polar form of complex number won't be covered on this exam, but these are the exact same formulas. These are derived from the typical Sokotoa right triangle trigonometry here, for which if we have a vector going in direction theta, the hypotenuse will be the length of the vector, it's magnitude, then you have its vertical component, it's horizontal component like so and this is associated to our vector V in standard position. So think of this right triangle here and be able to do a calculation like this on question number six. So again, I'm gonna cite sections 28 and 30 as ones to review. So 28, so when we talk about geometric vectors, 30 was when we talked about algebraic vectors, they're very related to each other, especially because of these formulas. 31 was about dot products. We saw a question about that one. There will be another one forthcoming that's actually here on the screen. Question number nine, we'll talk about that in a second. And then lecture 29 about vectors, that was exclusively about story problems. So you'll see that in the free response. So we'll talk some more about those as well, but like I already said, the multiple choice section, excuse me, multiple choice covers the vector chapter very extensively. The other topics that we're gonna see here, we did see a question mark area that came from chapter eight. That's gonna be the only question from chapter eight you should be prepared for on the multiple choice section. What about chapter seven about solving equations? You will see two questions from that unit. You see them right here, seven and eight. Can you solve a very basic linear trigonometric equation? All right, how does one do that? We learned about these, of course, at the start of chapter seven. So actually it was the first lecture for this unit three in our course. That goes all the way back to lecture 21. Can you solve a linear equation trigonometrically? Question eight, you'll do the same thing. It's just now you'll have a modified period. So that's what we did at the end of that unit in lecture 23. So they will both be linear equations. No identities will be necessary here. No quadratic formula, anything will be necessary. Just linear equations. And I mean, when you look at question number eight, you might be tempted to use the double angle identity, but you're better off just using a modified period. In fact, that's a two theta instead of a theta does make a slight change on how you approach question seven. Question eight will be similar except the period will be modified. Do pay attention to the instructions that are provided. So we might ask you to solve the equation in radians, solve it in degrees. As it's a multiple choice, you really don't get a lot of options here, but you'll see that if it's in radians, that all of the possible answers will be in radians. You should be prepared to do it in radians or in degrees. Question number seven, the way it's phrased, it's looking for the general solution. So like two pi thirds plus two pi K, four pi thirds plus two pi K. So you have those there. If this is if you're solving tangent C cant sign or cosecant, if it's tangent or cotangent, then the period is actually pi or 180 degrees. So do pay attention to that. Question number eight, you'll be modifying the period, right? So that would affect things. Like you have a two theta right here that affects the period. Instead of the period being two pi, it actually gets shrunken down to pi. That would change things if you're looking for the general solution. That could be a distract or watch out there. Another possible way that your answers could be expected is you'll be given an interval like this. Find all solutions in the interval zero to 360 degrees. So that one clearly wanna find degree measurements. We're not looking for the general solution, but just answers up to 360. Or this could be like zero theta to two pi. That would be the same thing in radians. So we should be prepared to find the general solution or specific solutions in a specific interval either in radians or in degrees. All right. And then I already mentioned this one briefly, but let's mention to get a last question in the multiple choice section. Question number nine, this is gonna be a question to ask you to, it is a story problem. This will be about force, right? This is using the idea that the force, excuse me, force, distance, right? Why did I start with a lowercase d? Who knows? Who knows anymore? Work was actually the work I was looking for there. So force and distance combined together and get work. And particularly the formula you should know was that work is the dot product of the force vector and the distance vector, sometimes called the displacement vector. You would use the dot product properly if these vectors were given in algebraic form. If it was given in geometric form, you're probably better off using the law of cosines variant that is the magnitude of f times the magnitude of d times cosine of the angle between them. That would also give you work. So be able to compute work. We did work in our section about dot products, which remember that was lecture 31. All right, that finishes up the multiple choice section. One other comment I do wanna mention on question number six, sort of a curious thing I wanna mention here is that in this question, you're not actually given the direction theta, you're given sine theta equals three fifths. Well, why is that relevant? If you're looking for the horizontal or vertical components, these are gonna be the magnitude of sine or cosine of theta. Now, if you wanna know sine theta, that's great. If you wanna know cosine theta, you of course can draw a right triangle here and go from there. Why are you told sine theta three fifths? Why not just told theta is whatever? Well, remember on the multiple choice section, you don't have a calculator. So you can't do things like arc, you can't do arc sine of three fifths to figure out the angle measurement. You couldn't do sine of 55 degrees. I wouldn't expect you to know that. So you might have to prepare yourself to use the right triangle diagram to help you convert from sine and cosine, like in this example, since you do need the horizontal component, be prepared for those type of calculations. Now let's talk about the free response section. The first question number 10, it is of course a multiple, excuse me, it's not a multiple choice question, it's a free response, but it's gonna be solving equations, solving trigonometric equations that's the word I was searching for, and it's gonna be on the easier side. So these are the types of equations we solved again in lecture 21. It's not gonna be too involved. Notice in this one, you can just, of course, you can solve it using some basic algebraic techniques. It will be a little bit more challenging than a linear equation, but some basic factoring, you could solve it. The quadratic formula, worst case scenario, you could solve this equation without too much effort, just some basic algebraic techniques, no trig identities will be necessary in question number 10, but unlike the multiple choice section, this question, when it asks for the solution, it could ask for the general solution, or it could ask for all solutions on an interval, and again, that could be given in degrees or radians. So unlike the multiple choice where it kind of formats it for you, this one does require you format it yourself, and that will be a big part of grading this question when I look at that one. Question number 11, because we have barely touched chapter eight whatsoever. We had one question about area and that's been it so far. Question number 11 will be one of the questions that covers the topics from chapter 11. So you will need to be prepared to answer questions involving the law of signs, like we saw in section 25. Question 12 will be similar. Can you answer a question involving the law of cosines? That is to say, can you solve an oblique triangle where some information is given like this is a side angle side situation? That's an appropriate use of the law of cosines. Like we saw in lecture 26, law of signs was in lecture 25, law of cosines was in lecture 26. You need to find all the missing parts. So like on this one, since you're given angle A, you're given side length B, you're given side length C, you do need to find side length A, angle C, angle B, like so. And so there will, you prefer to use the law of signs because that is the easier technique, but question number 12 will be set up in such a way that you do have to start off with the law of cosines, but you could finish using the law of signs if you prefer, which typically is the easier approach, I would say that. Don't get too paranoid if you see things like the square root of two as a side length. I will choose some irrational lengths actually to make the calculations easier, believe it or not. You'll, maybe in retrospect, you'll see this, but like the square root of two is actually a good friend of 45 degrees when you start thinking of trigonometry. So that actually does make things a little bit easier. You are allowed a calculator in this situation, so an approximate answer will be appropriate, but I promise you I've chosen things to the most part where you could probably get an exact answer, but I can't guarantee that. You do have a calculator, so don't worry too much about it. All right, question number 13, this will be another question about solving trigonometric equations, but this one will be sort of the capstone equation. Looking at this one right here, you'll notice there is this cosine of two theta, in which case a trigonometric identity would be appropriate for this question. Using the double angle identity for cosine, you can turn this into a quadratic equation. That quadratic equation could be solved using factoring, but you might have to use the quadratic formula as well. Again, this will be a more challenging one. So what are some identities that would be appropriate for these type of equations? Well, the double angle, like this example shows, would be appropriate. Some examples we saw in the homework and in our lectures, of course, the half angle would be a good one to use. You should use that. Of course, all the fundamental identities, reciprocal ratio, you definitely should know those ones. We have things like product to sum, and sum to product identities. Those might come into play as well. We had some examples of that. Now that's a lot of identities, right? You don't have a formula sheet for the free response, but you do have your note card, so put on your notes any identities you might think are relevant for question number 13. It could definitely help you out a lot there. So put on those identities that you need. So when you're preparing for question 13, look at the types of questions we saw in section 22, although 21 is a good reference. There are some quadratic equations there, but this one will have some identities of some kind, so be prepared for that. Moving to the last page of our test here on question number 14, this is gonna be a question about some story problem involving vectors. Static equilibrium we introduced at the end of lecture 28, but in that case, it was just a right triangle static equilibrium problem. That was the easiest case. In lecture 29, we introduced story problems when we used oblique triangles, that is the vectors formed oblique triangles, for which case you would use like the law of signs, to solve those ones. So static equilibrium is a good example to be prepared for. Some of these questions about headings and true course, those were some story problems we saw. What you should expect on question number 14 is a question similar to what we saw in lecture, what we saw in the homework. Be able to set up and work through these vector problems. Now when we did this in lecture 29 and in 28, we used the geometric approach, but by the time you're taking the test, you also have the algebraic approach to solve these problems, which actually could make it easier. So when it comes to solving question number 14, you can solve it geometrically, maybe using the law of signs or cosines, or you could solve it algebraically, which probably is easier, I would think, especially in the oblique case, but you have to convert. It's gonna be given geometrically, but you could solve it algebraically. You have that choice, choose for yourself what you think would be the best. And then question number 15, this is the last question on the exam. This is gonna be everyone's favorite to the ambiguous case, SSA. So you'll notice of course here we have our side, our side, our angle, this is the ambiguous case. So be prepared to solve the ambiguous case as we've learned about in lecture 26. Now in the ambiguous case, you can solve it using the law of cosines or you can use the law of signs, right? You can use the law of signs because you do have this angle opposite side pair, but also you have two sides and an angle. We don't know this, we don't know, excuse me, this side right here. You can use the law of signs, whichever you prefer. It doesn't really matter to me what's technique. Many of you will probably prefer the law of signs because that's how I demonstrated in class. That's my preferred way of doing it. Remember about the ambiguous case is that when you're solving the ambiguous case, there could be zero solutions, there could be one solution, there could be two solutions. Now if there's zero solutions, you can actually terminate the process much quicker, but if the answer turns out to be no solution, you need to provide the evidence that there was no solution. And that could have happened because you got something like sign equals three halves, three halves, right? Which is greater than one. That's not an acceptable sign ratio. That would be indicative that there's no solution. So that's fine. If it terminates with no solution, you can't just say there's no solution. You have to provide some evidence that there's no solution. And one of the best ways to say that is like, oh, this is an unacceptable sign ratio. Sign cannot equal three halves or something like that. That's if there's no solution. Now, what if there is a solution? Well, there's either one or two. Now, one of the solutions will come about by having angle B turn out to be acute. Again, this is if you do the sign approach. If you do the law of cosines, it's a little bit differently. If you do the law of cosines, it turns out to be like you're solving a quadratic equation without any real solutions. Now would be like that would be no solution right there. Then if you had one solution, that's because the quadratic equation has one solution or two solutions because the quadratic equation has two solutions. Again, I didn't demonstrate that in lecture. So many of you probably won't do it that way. But if you did, that's okay. Working with the law of cosines just means you have to solve quadratic equations which could have one, two or no solutions. If you're working with the law of signs, well, if there is a solution, you have to consider when B is acute, but you have to also consider when B is obtuse. Even if there's only one solution, if you do not consider the obtuse possibility, you wouldn't get full credit for this one. So when you solve this one, you have to show whether there's a solution or not. And if there are solutions, there needs to be an argument on why there is two solutions in which case you would provide both solutions. Or if there's only one solution, I do need to see the work that demonstrates why there's only one solution. Why wasn't there a second solution? That's what makes the ambiguous case the more challenging one because there are all these possibilities and you have to consider every path in order to truly say that you've solved the problem. And so that brings us to the end of exam three. We've talked about all the topics you're gonna see on this test. Of course, this practice test is just an example of what you could see. What you will see will be very similar to this practice test, I mind you. But there are, of course, our variations to questions, not just numbers and such that you should be prepared for and prepare those as well. Especially question number 14, which is a story problem. The example here only gives you one of the possibilities you could see. Another thing I should also warn you about is when you see these triangle diagrams on this exam, do not assume they're drawn to scale. They are intentionally not, mostly out of laziness on my part. But you can't infer any information from it. Like, you see this triangle on question number 15, then if I go back to question number 12, where are you? There you are. It's the exact same picture, but I just relabeled things. In one case, I said it was 30 degree angles and one I said it's 45 degree angles. These things are not drawn to scale. So if you do have any questions that you're preparing for this test, which you'll probably have many, please do reach out for help. Work with your classmates, form study groups. That's a great study tool. Post questions on Canvas on the discussion board for this test. That would be a great way to share your questions with me, but also so everyone can benefit from it. Also come by office hours to come talk to me, send me emails, go by the tutoring center. They're great resources to help you to prepare for this test. And best of luck. Like I said, let me know if you have any questions and I'll see you next time. Bye everyone.