 Hi, I'm Zor. Welcome to Unisor Education. Continuing talking about graphs of trigonometric functions, today is tangent. I do have notes obviously on Unisor.com, so I think you will be much better if you will review the notes, try to construct these graphs yourself, and then watch the video where I'm trying to do basically the same thing. All right, let's just start tangent, function tangent. First of all, this is a function which is odd, because as you remember, the definition of the tangent is sine over cosine, so let me just draw the graph. Now, wherever cosine is equal to zero, we have an asymptote, which is at pi over two and three pi over two, on the period from zero to two pi. And period is actually from zero to pi because the function repeats itself. So it goes from minus infinity through zero to plus infinity, and again from minus infinity through zero to plus infinity, etc. So, the length of the period is pi. That's the tangent, and I would like to start with this function and then draw all the different graphs which we're training. So important is that this function is odd because sine is odd and cosine is even. So sine changes the sine when the argument changes the sine. Cosine of the argument does not change the sine with argument changing the sine. That's why the whole ratio is changing the sine. So, its periodicity is pi less than one of those. Each one has a periodicity of two pi, but combined together, they have pi. And that's basically how graph looks. Acid totes at every pi over two plus pi plus two pi, etc. Now, let's go through examples. First is tangent of minus x. So, if argument changes the sine, odd function is changing the sine as well. What does it mean, basically? It means that if point AB belongs to the graph of the function tangent, which means tangent of A equals B, then point minus AB, which is symmetrical to this one relative to the y-axis, belongs to function tangent of minus x. Because if I will substitute instead of minus, well, I should say y is equal to tangent of x, minus x. And if I will substitute minus A instead of x, I will have, and B instead of y, I will have B is equal to tangent of minus negative A, which is A. And I know that this is B. So, again, with every point AB which belongs to this graph, there is a point minus AB which belongs to this graph. Now, this point is symmetrical to this relative to the y-axis, and that's why the whole graph would be symmetrical to this one relative to the y-axis, so it will go like this. So, the red one is y is equal to tangent of minus x, okay? Now, start from the beginning, tangent of 3x. Again, I'm basically repeating the same thing again and again for each trigonometric function. If point AB belongs to this, then one-third AB belongs to the graph of this function. Why? Because if I will substitute one-third instead of x, I will have tangent of 3 times one-third times A, which is tangent of A, which is B. So, I will have the correct equation. So, with each point of this graph, I have this point which is closer to the zero along the x-axis. So, it's like squeezing towards the y-axis along the x-axis. It's 3 times closer, so the whole graph will be squeezed, which means that instead of... Well, let me just draw it from minus pi over 2 to pi over 2, which is the period. So, my initial graph is this. Now, I will squeeze it three times, actually. So, it would be from minus pi over 6 to pi over 6, and the shape would be the same. And obviously, repeated, every blanks would be pi over 3, right? So, the next would be pi over 6 plus pi over 3 period is what? That's 2-6, that's 3-6, that's pi over... Okay, pi over 2. Okay, so the next would be pi over 2, etc. So, the whole graph would be squeezed three times. So, instead of big waves, or I don't know why, whether it's waves or not, big curves or wide-spread curves, I will have narrow-spread curves, and that would be the graph of this function. Next is a reverse operation. Instead of multiplying x by 3, I will divide x by 3. And for obvious reason, the graph instead of squeezing by the factor of 3 will be stretched by the factor of 3. So, again, if this is my initial graph from minus pi over 2 to pi over 2, and that's my initial graph, that's my tangent. Now, I will stretch it three times, so instead of pi over 2, it will be 3 pi over 2. Sometimes I'm saying p, sometimes pi. But that's the same letter. I should actually use the red one. So, that's the same tangent. And here as well, somewhere here. So, instead of this, I will have this. It's stretched three times. So, the period instead of pi would be 3 pi. In this case, it's from minus pi over 2 to pi over 2. In this case, the original one is minus pi over 2 to pi over 2. And this one is from minus 3 pi over 2 to plus 3 pi over 2. And then repeats every 3 pi over 2. Next is I'm multiplying not the argument, but the function itself by 3. For some reason, I consider this an easier exercise when I multiply the function itself instead of the argument. I don't know why, but it seems to be easier. Just every value of the function is increased three times, right? So, if original looks like this and this is tangent from minus 3 over 2 to plus pi over 2, then the new one would be stretched vertically up and down by three times, which means every value of the function will be increased by threefold. So, it would be something like this. It's steeper, actually. That's what it means. It's three times steeper. Well, in my case, it's not really three times. To get it more over three times would be something like this. Well, drawing is unimportant. What is important is that every segment from the original graph to the x-axis is supposed to be three times higher when it goes to the red line. All right, so that's easy. Now, how does this look like if you change something? Over 2 minus x. Well, let's just think about it. There are many ways, actually, to approach this. First of all, what we can do is we can transform it into tangent of minus x first and then tangent of minus x minus pi over 2. Two transformations. From original function, we transform original graph into this, which we have already done once. So let me draw the solution. So it's a reflection relative to the y. So instead of this, it will go like this. That's my tangent of minus x. Now, if I am adding or subtracting something to the argument, as we know, it shifts the graph in this particular case to the right by pi over 2. Why to the right? Well, because, again, it's obvious if point a, b belongs to tangent of minus x, and then point a plus pi over 2 b belongs to tangent of minus x minus pi over 2. If I will substitute x, substitute a plus pi over 2 instead of x in this formula, I will get minus a. So I have to transform this graph by shifting it to the right by pi over 2. Now, this is minus pi over 2, and this is pi over 2. So by shifting it by pi over 2 to the right, I will get this. That's my graph. And then it repeats left and right by periodicity, which is equal to pi. OK? So I transform twice my original graph of the tangent to get to this particular one. Now next is a combination of everything we have done already. Minus one-third tangent of minus 3x minus 3 pi over 2. OK. What's the steps of transformation which we can achieve this graph? Minus one-third. Now let me just start without minus-third. So it's tangent of minus x, then tangent of minus 3x, then tangent of minus 3x plus pi over 2. Then minus one-third of this tangent of minus 3x plus pi over 2. So that's the sequence of transformations of the graph which I have to do. All right, so let me start. Tangent of x, again, you know tangent of minus x is this. Minus pi over 2 pi over 2 xy. OK, done that. Then I multiply my argument by 3, which means the graph is squeezed in. So instead of pi over 2, I will have pi over 6. Right? That's this one. Now I have to shift by pi over 2 to the left. So this point 0 will be somewhere minus pi over 2. OK, so it would be something like this. So 0 would be at point minus pi over 2. And on the left and on the right would be pi over 6. So if I go from minus pi over 2, I subtract pi over 6. That would be what, pi over 3, right? Minus. And if I add pi over 6, I will get pi over 2 minus pi over 6 is equal to pi over 3, right? So this is 3, 6 minus 1, 6, it's 2, 6, right? And if I add, I would have 3 plus 4 to the 2 pi, right, 2 pi. So this is 2 pi minus 2 pi over 3. So the period would be from minus 2 pi over 3 to minus pi over 3 with 0 pointed minus pi over 2. And then repeat it every period times, which is pi over 3. Period is pi over 3, obviously. When we are shifting, the period is retained. Now, finally, I have to multiply by minus 1, 3. What does it mean? Well, it means that this particular graph should be number 1 reduced, every ordinate should be reduced by the factor of 3. So it would be something like this. But then there is a minus here, which means it will be inverted and it will be corresponding with this. So that's the final shape of the graph. And then repeat it every period, and the period is equal to pi over 3. That's it. OK, what's next? Next is a combination of two different tangents with two different arguments. Well, in this case, we just have to draw this graph and this graph and add them together. Well, let's try. So let's start with minus x. So it would be something like minus pi over 2 pi over 2. And since it's minus, it's this way. And then repeat it every pi. By the way, the periodicity is the same as this one, pi. Because this function has a periodicity smaller because this is a squeezed line, so it's pi over 2. But since this one is still pi, so the periodicity is maximum of these two periodicities. So now this one will be squeezed in tangent x. Now tangent x is from minus pi over 2 to pi over 2. But tangent 2x would be from minus pi over 4 to pi over 4. We already spoke about this. We are multiplying by 2. It means that the graph is squeezed in to the 0 point along the x-axis. So instead of this, I will have this. And then it repeats itself, which means it goes this way and here. So now let's choose one particular period, which is, let's say, from minus pi over 2 to pi over 2. So let's wipe out this. Wipe out this so it doesn't disturb too much. So this is the period from minus pi over 2 to pi over 2. Let's add these two graphs together. Well, let's talk about the points where we have asymptotes. One graph is 0, another is plus infinity. So we will get the function result is plus infinity. In this case, they are doubling each other. But now this one is asymptote as well. So the function will be somewhere from here down here and then go up again because now this is equal to infinity. Now in this case, on this side with asymptote, well, it obviously minus infinity. At some point, it might actually reach 0 now. But this is minus and this is plus. So we are subtracting. So it goes something like this to infinity again. And here, now this is asymptote as well. This is plus infinity. And this is a finite piece. So it goes this way. And finally, in this area, we will get something similar to this, but on a negative side. So it's something like this. So this is how the function behaves inside the period. And then it repeats itself. So basically, the next would be something like this, this, and this. That would be the next period, both ways. All right. And the final problem for tangent is tangent x plus tangent x plus pi over 2. OK, let's do it again. Function, tangent. Well, basically, these two functions have very similar graphs. But this one is shifted by pi over 2 to the left. So let's start with minus pi over 2 plus pi over 2. And this is tangent x. Now, well, let's just do it again because we are shifting. Now, we are shifting to the left by minus pi over 2, which is this distance. So the next graph would be this is minus pi. This is 0. So the next would be this. This is shifted to the left by pi over 2. And now we have to add them together. Basically, it doesn't really matter where exactly we will do it. But let's just do it whatever we have drawn, basically. Now, well, let's start from the right to left. Now, here, this function is having an asymptote. And this function, so this is tangent x. And this is tangent x plus pi over 2. All right, so let's consider on this interval, for instance. So here, we have minus infinity and 0. So the result will be minus infinity. And then it goes up, up, up. And here, we have a plus infinity on one function and 0 on another. So basically, the graph will behave like this. Now, here, we have a similar situation. We have one graph is minus infinity. Another is 0. So if we add them together, it will be minus infinity. And then it goes through 0. And then it goes up again. So that's basically the way how the graph would look like. And what's interesting is it looks like its periodicity is actually pi over 2. I mean, pi is obviously a period, because tangent has a pi as a period. But in this case, periodicity is pi over 2. Would be interesting to prove it, by the way, algebraically using the definition of the tangent sine over cosine. But most likely, that would be true. So that's the graph of the tangent plus tangent with a shift. OK, that's it for today. Thank you very much. So we covered lots of graphs, sine, cosine, tangent. I will continue these lectures with cotangent and secant and cosecant. And then there will be some calculation problems, never a second of rest. All right, good luck. And I would recommend you to read Xamarin to lecture again. And the graphs try to do it yourself and then compare with whatever I did. Thanks very much and good luck.