 In this video, we're going to review the ideas of changing the amplitude in the period of a sine or cosine wave and actually combine them together. So recall that if we have numbers a and b, which for the sake of the moment, we're going to assume b as a positive number. If you consider the graphs of y equal a times sine of bx or y equals a times cosine of bx, then this coefficient in front determines the amplitude of the sine or cosine wave. In fact, the amplitude of sine or cosine will be the absolute value of this coefficient that sits in front. If that number is positive, then no reflections happened, but if that number was negative, the amplitude will be its absolute value, but the negative in front will cause the graph to be reflected across the x-axis. So you should have to pay attention to that. Now with respect to b, it turns out that b is this number inside of the function, it's inside the sine or cosine. This affects the period of the sine or cosine function. In fact, the period will be p is equal to 2 pi over b. 2 pi is the standard period and the factor of b has adapted, has modified the length of the period, elongating it or compressing it. But why does b have to be positive? Well, could b be negative like a is? Well, the answer is that if b is ever negative, we can fix it because of the symmetry identities we introduced previously. So if you had something like y equals a sine of, say, like negative bx where, again, b is a positive number. Well, because sine has the property of y, I should say sine of negative x, sine's an odd function, so this is equal to negative sine of x here. So if you have a negative sine inside of the sine, you can actually bring it out. And so you end up with negative a times sine of bx right here. So a negative inside of the sine can be brought out and it becomes a reflection across the x-axis. Well, with regard to sine, excuse me, just sine, if we with regard to cosine, if you have y equals a times cosine of negative bx. Well, the thing to remember here is that cosine of negative x is actually just equal to cosine of x itself. Cosine's an even function so that negative doesn't affect anything. So you can just discard that negative sine, a times cosine of bx. So use the symmetry identities to deal with having a negative inside of the sine or cosine function. So we don't have to worry about it affecting the period whatsoever. No reflections across the y-axis is necessary here. So let's look at an example of this. Let's consider graphing the function y equals 2 sine of negative pi x. Well, you have a negative inside. Just like we said a moment ago, you need to bring that negative out in front. And so we end up with y equals negative 2 sine of pi x. So let's see how this has affected the graph of things. So because we have this negative 2 in front, it turns out that the amplitude of the function is going to be 2. And because of the negative sine, we're going to have a reflection across the x-axis. Let's keep that in play there. Because of the pi in front, it turns, excuse me, the pi inside of the sine function. The period of our function is going to be 2 pi over pi for which the pi's cancel out and we get a period of just 2. So they ask us to graph the function on the interval negative 3 to 3. Now notice how long that interval is. You have 3 minus a negative 3. This is equal to 6. 6, of course, is 3 times 2. So it's asking us to graph three cycles of this function. So if we can graph one cycle, then we can just copy it for the other ones as well. So it makes sense that we should just graph it first from 0 to 2. Right? That's going to be the standard thing we want to do. Let's graph it from 0 to 2, because that's one full cycle. And then we'll just copy and paste for the other bits of the pieces as well. So when it comes to graphing a sine or cosine function, you want to think of the five quadrantal points. That is these angles that change the quadrants. So you want to think of 0, pi halves, where am I at? Pi, 3 pi halves, and 2 pi. Where happens to those points? Well, when you change the amplitude, when you change the period, that's going to affect some things. So for a standard sine wave, you should be 0. You're going to have 0 at 0. You have 1 at pi halves. You have 0 at where we at pi. You're going to have negative 1 at 3 pi halves, and then you get 0 again at 2 pi. But things aren't like that. We've changed the period. So notice that these are still the five points we want to use. They're equally spaced. So these points are going to be 0, 1 half, 1, 3 halves, and 2. So we basically just forgot the pi's. Oh, that's because we canceled the pi out earlier, and so our period is just a 2. So we have the five points we need. These are our five quadrantal points with our modified sine wave here. What are their amplitudes? Well, x intercepts don't get changed by this stretching. The amplitude is now 2, so things should be reflected. Not 1. Excuse me. They should now be 2. They got stretched. So we get something like this. So it's tempting to say this is our basic picture right here. But nope, we have to also remember the reflection. So what was positive now becomes negative. So because of the reflections, we end up with a point down here at negative 2. And then we have 1 up here at positive 2, like so. And if we connect those dots together in the typical sinusoidal fashion, this would be what our picture looks like from 0 to 2. That's one cycle. We want to go all the way to 3. So we need to continue our picture here. So just by copying and pasting, we should have a negative 2 here. We should have a 0 there at 3. So we need to go all the way up to 3. We have to also go down to negative 3. So again, just copy and paste this picture over and over and over again. Just repeat this pattern. So that gets us to negative 2. We have to continue on to negative 3. And so just connect these dots in the usual sinusoidal way, like so. And so this would be the graph of sine of 2, excuse me, y equals 2 sine of negative pi x from negative 3 to 3. But it turns out we can go the other way when it comes to graphing these things. We can actually be given the graph and asked to come up with the formula that determines the graph. So what we're looking for is a function of the form y equals a sine of bx. Or we take y equals a cosine of bx. It turns out there are some options, but without considering type of vertical or horizontal shifts, this one's going to look like a sine wave. Notice how it starts at the origin, then you come up to its maximum, then back to an x-intercept, down to its minimum, then back to an x-intercept again. And so that's one cycle of the graph you see right here. This is one cycle. So there's some information we can glean for that. First of all, we see that the period of this thing is going to be pi. So illustrating that observation right there, we get that the period is going to be pi. We'll come back to that in a second. What about the amplitude? How far above this middle line, this midline, can we get? Well, the distance we go above the midline, it looks like it's a 3. So we'd say the amplitude is equal to a 3. But is this thing a sine or a cosine? Well, sine starts on the midline, it goes to a max, then it goes to a min and repeats itself. That looks like a sine to me. A cosine on the other hand, cosine starts at the max, then it goes towards the midline, then it goes down and things like that. So are you starting in the middle? Are you starting at the extreme? Since you start in the middle, it turns out that we want to model this using a sine wave. So is there any reflections going on? Well, no, sine starts off by going up. So there's no reflection. So the reflection is now banned. No reflections are allowed on this graph. And so with the information we have, we can start graph or we can come up with a formula, right? We get y equals 3 sine of b. Well, we don't know what b is. We know what pi is. So we saw earlier that pi, excuse me, p is equal to 2 pi over b. So we know the period and we can compute the period if we have b. We have to go the way around. But guess what? We can times both sides of this equation by b so that these are canceled. And we can divide both sides by p. So those are canceled. And we end up with b equals 2 pi over p. So there's this inverse proportionality between b and p right here. And so we see that if the period is pi, that means b is going to equal 2 pi over pi. The pi is canceled. We get 2. And so we should then record this as y equals 3 sine of 2x. And this function right here would graph the function that we were given. Let's do one more example of this. You can see this right down here. We have one cycle of a graph, a trigonometric function is given. Can we come up with a function formula for this thing? Well, let's first decide. We want sine or cosine. This thing feels like a sine to me because the midline, the line that runs through the middle of this thing is right here, our midline. And since we start on the midline like this, we would call this a sine wave. So this looks like a sine. But notice how we're decreasing first, not increasing? That tells me there's some type of reflection that happened to my sine wave. So what I'm expecting here is y equals negative a times sine of bx. We then have to find a, we have to find b. A is going to be the amplitude. How far above the midline can we go? It looks like the amplitude here is going to be 5. The next thing to determine would then be our period. How long does it take to complete one cycle? So we can see that the period is going to turn out to be 8 pi, like so. Now, to go from the period p to the coefficient b, we have to take an inverse proportion b is equal to 2 pi over p, the period. But the period is an 8 pi. The pi's cancel out and this simplifies just to be 1 fourth, like so. And so we end up with our final form, y equals negative 5 times sine of 1 fourth x. And so this is then the correct graph. So understanding periods and amplitudes helps us to graph trigonometric functions given their formulas. But we can also take a trigonometric graph and look at its formula. That is look at the graph and get the formula as well.