All right, hi everyone. Let's do some examples using some of the theory we saw in the last video. So I have the questions here on the screen, we're going to go through and work them out. We're going to go nice and slow and use our knowledge here. So keep the graphs, keep tables of values, keep those all handy as we go through and solve some of these problems. So if you want, pause the video and see if you can get the answer first, get as far as you can at least, and then we're going to watch me go through the solution. So take a second, pause the video and then try to work these out. Ready? First one says, sketch the graph of y equals negative 3 cosine x for x in the interval from negative 2pi to 2pi. All right, so here's the way I want you to start seeing these things. I want you to start seeing these functions as sort of building blocks as pieces. So what is the root function? What is the first graph that we're working with here? So in your mind, build this up. A lot of people, they just can't jump to the answer, but I want you to build it up. So obviously, this is some manipulation of the cosine graph. So in your mind, think of y equals cosine of x. Now I'll draw it again, but we saw last time, hopefully you have a handy. This is the graph and let's just do two periods of it. So minus 2pi to 2pi. I'll draw it small because this isn't the graph we want. This is just sort of a helper graph. Remember this graph starts at the point zero comma one, it goes down to negative one and then comes back up. There's a one-period, so looks like a little bit of a valley as I draw for the y-axis over on the positive x-axis and if I did, if I go left, now I get sort of the same thing and I get sort of two valleys. Looks like a bit of a W. Obviously, the maximum is one and negative one, fantastic. Okay, so this is my cosine function, this is what I'm supposed to be familiar with. Now let's look at what's the next thing that happens in this function. I have y equals negative 3x, so why do we multiply by three? What does multiplying by three do? We'll deal with that negative later, just do one thing at a time. What does three do? When you have a scalar in front, sometimes numbers in front they scale things. They're going to grow some values. So think of it like what would happen to some value. So if you plugged in zero, so what happens do zero? If you plug in zero, I get one normally for cosine of x, but now if I multiply by three and so now instead I get three. So your first point, your starting point is much, much higher at zero comma three instead of zero comma one and the same thing to, for the lowest point on the graph that was at pi, what happens if you plug in cosine of pi, you get negative one multiply it by three, and now you're going all the way down to like negative three on the graph and so what you're doing with this number in front is you're taking this wave and you're having it go much higher and much lower on the interval from zero to pi. I drew my axis a little too small there, but it's something like that. So I haven't changed the period of this thing, it still does one down and one up and goes back as well. But what is it called? What is the height of a wave called? Do you remember that? It's called the amplitude. So what I'm doing with the number in front is that I'm changing the amplitude of this function. So instead of a little w, now I have a nice big W with the max and min at three and negative three. So if you need to see some more values for this, certainly be my guest, do so. So now we're ready to finally do the last piece of this. So what happens, y equals negative 3 cosine x, so what happens now? What does a negative do to any function? I want to draw a nice big answer here, since this is our final answer. Think of the parabola, y equals x squared, what does that look like? What does y equals negative x squared looks like? What about y equals x compared to y equals negative x? What does negating a function do to the function? Well, it flips it over the x-axis. So once I know the general shape I'm working with, so this function starts off y equals 3 cosine x starts off high to low, so I'm going to flip it and go from low to high and you go all the way up to positive three and I'm going to come back down and because my intercept now, the y-intercept is going to be at zero negative three, and I have instead of a valley, almost like a hill and we're going to do one more since we're going all the way 2pi, something like that. So I have two little hills. I have the reflection of the graph about the x-axis. The max is at three, the minimum's at three, negative three and I have all my points, say the same, my maximum is at pi, my origin, my y-intercept zero, negative three, so this is the final graph. So yes, you can plug this into a calculator and just get this immediately, but I don't think that really gives you the big understanding of what the pieces are that go into a function. So try to see this is three steps and if you did another step if we added seven to the graph, what does that do, and again, we are sure to go back to the parabola, most people know what that does for the parabola. If we add seven it shifts the graph up, so I can move this thing up now. So you do whatever you need to do for these graphs as you go through and you graph them. Let's do another example. Now, let's do another example, find all values of t such that sine of 2t is negative one-half on zero pi over 2. So it's a little tricky part here where there's 2t. Where zero over pi over two, is saying 2t is allowed to go around the circle, one lap around the circle. So maybe we'll do the unit circle here, if you want. I don't know what the graph of sine of 2t looks like. So it's perhaps better to think about the unit circle one lap around. We've seen sine of something equal to negative one-half and that we've seen, we said in quadrant one. So think back sine of what is positive one-half, that's obviously going to play a role here, and hopefully you realize the answer is pi over 6. You can check your tables of values for that, but that's positive one-half. Where is sine? If the sine is the y-value, where is that negative? The y-value is negative in quadrants three, in quadrants four. So what this is asking is solve for t, but really going to focus on 2t for a minute. Where is negative one-half? It's down here. So what angle has reference angle pi over 6? That's the idea. So if I want negative one-half, hopefully, you see if you draw the unit circle and you label negative one-half, somewhere right in the middle, between zero and negative one, there's going to be two values. There's two options to draw a triangle. So y is your opposite side, they will have opposite side inscribed in the circle at negative one-half. So we're looking for two values for 2t. So let's see what happens there. So zero to 2 pi, so 2t, so if t is inside of zero to 2 pi, one way to think about 2t geometrically, is you're doing two laps around the circle. So you're doubling up t. So 2t is going to be an element of zero to 4 pi. So you imagine you're a rotation off the x-axis, you're going around twice. So let's get the first values here. So what values on my first lap around? So we'll start at the x-axis. We go all the way around to quadrant three. Why quadrant three? Again, because the sine has to be negative, so I can completely skip quadrants one and two. What angle has reference angle here? So we're looking for some value of 2t, is going to be pi plus pi over 6. Go pi plus a little more, you really move your hand around and see this. Do a little common denominator, put it over six over six, and you get 7 pi over 6. That's what 2t is equal to on the first lap around here. If you keep going to quadrant four, you're going to get your second answer. You get that 2t is going to be equal to, now, what's the best way to see this? Maybe we do 2t minus pi over 6, 2t minus pi over 6. We're going to need a common denominator over here. So let's do 6 times 6 upstairs and times 6 downstairs, 6 times 2 is 12, 12 minus one is 11. We have 11 pi over 6. That's my other answer. So this is my quadrant three answer and my quadrant four answer. But here's the kicker, 2t is like going around the circle twice. We just went around the circle one time. So let's go around again, because I already know one answer, going around again, is just adding 2 pi. So we're going to get another value, 2t is 7 pi over 6 six plus 2 pi. One more lap around we go. This will be the last lap, zero to 4 pi, is going around the circle twice. Can't go around anymore because the restriction of where t can live. So once again, adding fractions, hopefully we're getting pretty good at this, do a common denominator. Let's do times six, 6 times 2 is 12. So we get 12 plus 7, which is good old 19 over 6 pi. So that's pi one or two pi, around two more. Try that again, 2 pi plus 7 pi over 6. Lot's of pi's going on. Then for my fourth answer here, I have 11 pi over 6, which is my solution that I found in quadrant four. But then I go around the horn one more time. So I add 2 pi, a full rotation. Once again, common denominators is going to turn that into 12 pi over 6. Add that up and you get 23 pi over 6. Now, here's the important part. I get four answers. So I get two from one lap around and then I get another two from the second lap around for 2t, so as t travels. I don't want to 2t, the answers is self of t. So I have to take all these answers and divide by two. I have to take these answers and divide by two. I'll do that right below my image here. So t is, 2t equals seven pi over six. Divide both sides by two, I get 7 pi over 12. I get another solution from 2t equals 11 pi over six, and that gives me 11 pi over 12. My third solution 2t equals 19 pi over 6, that becomes 19 pi over 12. T equals 23 pi, my fourth and final answer, over 12. So I get four values of t. Again, I encourage you to check this. It's really nice to know how to do these things. If you notice I didn't rush to a calculator and try to find points of intersection. The challenge here is that it's a 2t. So we are increasing the frequency and we're going to get four solutions. If you plug this in, is going to give you some decimal. It's hard to get it in a closed form. Seven pi over 12. Sometimes, they'll ask you for a decimal and that's great, and sometimes they'll ask you for the closed form. So just be ready to do both. This is a tough one, make sure you understand that 2t. If you want to graph this as well and see the four solutions, write them as decimals, so you can see it. Let's do another one. Find the values of t in zero to two pi such that, ST is such that, two sine squared t plus three cosine t is zero. Now we're starting to get into scary functions involving both sine and cosine. Just as a big picture, before we get into this one, we're asked to find where something is equal to zero. It's a pretty common thing to do. If you remember we were doing that back. We have a quadratic formula, they gave you a function, said equal to zero and said solve. We're going to do all the same things, but now we're going to use sines and cosine here as well. One of the things, and this is a general rule, but of course there are exceptions to everything, is that when you have a function with sine and cosine, you're trying to do something that will put it in terms of just one. You ask yourself, is there a way to convert this into maybe just one of these functions? So I don't have to do that. Whenever you see sine squared, little warning signs should be going off in your head like yes, I can do that. Remember there's an identity. Sine squared t plus cosine squared t equals one. This is the fundamental Pythagorean identity. So I think we should use that identity here. There's probably more than one way to do this. So we rearrange, you get sine squared t is one minus that. So let's do that. Let's substitute one minus cosine squared of t plus three cosine of t is zero. I got to bring the two and of course to both pieces. So you get two minus two cosine squared of t plus three cosine of t is zero. Let's write this in a perhaps more familiar way. Normally when you write things, you write with the largest exponent first. So this is minus two cosine squared of t plus three cosine of t plus two equals zero. So I have the degree two term, degree one term, and then the constant term. Stare at this for a minute. This is basically a quadratic in disguise. Why is that? So the quadratic, you could do a couple of things. If you want to see this perhaps more clearly, we can make a substitution and just say u equals cosine of t. Just so we can get past the cosine function for a second. So if I do that, I get two u squared plus three u plus two is zero. If you want, you can call it x or something like that. This might look a little more, perhaps, familiar. What is this? Can you stare at this for a second? Do you see this? These factors. The beauty of this here is that it's got a factor, and this factors as two u minus u and then plus two and plus one. Stare at that for a second, foil that back. First, gives you negative two u squared. Outside, is four u minus one u is three u and then plus two. So this totally factors. Maybe you could have done that without doing the substitution. That's all fine, but the point is this factors and you get two cosine of t plus one times negative cosine of t plus two equals zero. That's nice. So now you have two things that multiply together to give you zero. So let's set each one of them equal to zero separately, just like we would do with any other terms. So this gives us negative cosine of t plus two is zero. So either the first one's zero or the second one's zero to give us that their product is in fact zero. Move some things around and you get cosine of t is negative one half or cosine of t is equal to two. Now, think of the graph of cosine t. Think of the boundedness of cosine of t. What values of t give cosine t as two. Hopefully, you're saying there's no such value. The maximum cosine takes is one. So there are no solutions on this side of the equation. Cosine t is never two. However, cosine t equals negative a half, and that's perfectly fine. This is a much easier problem to solve than the original one, although they are exactly the same. Cosine t is negative. So the negative is telling you what quadrant of the graph is going to live in. Cosine is negative. Cosine is the x value. So for it to be negative, we're going to live in quadrants two or three. So we're trying to find these things. Well, we want to use our reference angles. Cosine of what value is positive one half? Perhaps thinking quadrant one. So let's draw a picture of the unit circle. Okay. So what is cosine of what is a half? What is the value here? Hopefully, from your table, or you're getting to learn these things by now, it's pi over three. So cosine of pi over three is a half. So I want the value in quadrants two and quadrants three. So let's use the reference angle over here for that. So how do I find that one? Well, the one in quadrant two is like pi minus pi over three. Simply rotate back pi over three. So my first value of t in quadrant three is pi minus pi over three, that is of course two thirds pi. My other value where x is negative happens in the third quadrant, and that's going to be Pi plus Pi over 3. You can clean that up to get 4Pi over 3, and those are my two answers. There's two answers to this, you can leave them as a list or you can write them out, two Pi over three comma four Pi over three. Again, just because you found the answers, doesn't guarantee that you're right, go back and check, plug them in, and show that this in fact gives zero. The important piece here is the t was restricted to zero and a two Pi, so just one lap around the circle which has given me two answers there. Let's do another kind of problem. I have the function f of x equals Cosine of Sine of x. The question is, determine if that function, Cosine of Sine of x, is even, odd, or neither. A friendly reminder in terms of symmetry, even, means I have y axis symmetry, odd, means that I have rotational or origin symmetry, and neither means I have neither of those two, so nothing happens. Algebraically, now, I don't know the graph of Cosine of Sine of x. I don't know it. I can go look it up, but my eyes aren't good to figure that out. Maybe it's close, but not exact, who knows? Let's do an algebraic approach to this thing. Friendly reminder, even means that the value of a negative, is the same as the value of positive. In terms of algebraic relationship, I'm looking at f of negative x will be equal to f of x. If the function is even, this relationship will hold. Origin symmetry, says, that if I plug in a negative, then a negative will pop out. This will plug in the negative is negative f of x. These are your algebraic relations that you want to test. Let's see if it works. Let's test for, is it even? How do you test that? Well, you just throw in f of negative x, let's see what happens. This is Cosine of Sine of negative x. Remember, you're symmetry of Sine and Cosine, everything is built on these foundational pieces. Friendly reminder, that Cosine of negative x is just Cosine of x, and from the reminder, Sine of negative x is negative Sine of x. These should be on your small but growing list of things that we've seen, identities that we've seen, these are your even and odd identities. Here we go. Sine of negative x. The negative pops out front, and this gives me Cosine of negative Sine of x. The beautiful thing here is that Cosine of negative anything, x represents anything, the negative just goes away, so this does in fact turn into Cosine of Sine of x. I get back the original function that I've started with, and therefore, it is in fact even, so this is even. If we had just chosen, we'll go through it just to see it. If we had to ask ourselves, is this function odd? I would've looked at f of minus x, and I would have got back all the same thing. So it doesn't matter the question you asked, but the point is, it's not equal to negative of f of x. You don't get back just negative times the original function. You would have to get back negative Cosine of Sine of x, so the answer is, no it's not odd. Not odd, it is in fact an even function. Now, go grab a calculator, go to decimals or something like that, and graph this thing, and look for that y axis symmetry, you'll get that guaranteed to happen now. All right. Another one, write h of x, which is defined to be the Cosine squared of x squared minus 2x plus one, as a composition of the form h of x equals f of g of x. What I want do is break this apart. This is going to be a valuable skill going forward, we saw this a little bit when we graphed the first one, of what is happening to a function viewing it in terms of pieces? What is the recipe? What are the steps that go into this function? Again, pause the video and see if you can work it out. But I want to write h of x as, this one just asks for two steps, you can certainly break it up into more, but let's just do two. Find me an f and a g such that the function Cosine squared, parentheses x squared minus 2x plus one is f of g of x. So you're ready? What is f? What's the outside function? Hopefully, you can see the outside function is Cosine squared of x. G of x, my inside function, what's happening first, when we composition, you read it from right to left. What's happening first here? Well, this is x squared minus 2x plus one. What's happening, is I'm taking this quadratic, x squared minus 2x plus one, and I'm plugging it into Cosine squared and I'm getting my final result down. You got to remember, this is super important, this is a very, very common mistake. Cosine squared of, this is not multiplication, Cosine squared by itself is meaningless. It is a function you feed something into it, so we're composing these two functions, we are certainly not multiplying. Just a couple of examples, go do some more in the course. Work these out, check your answers graphically, check your answers algebraically, try to do as many as you can and go back and re-watch the video if you need to. We're going to do lots more examples, so if you get stuck on these, don't worry, keep at it, keep at it, keep at it. Great job on this video, and I'll see you next time.