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Determining the equation of a trig function Graphs of trig functions Trigonometry Khan Academy.mp3 | The midline is halfway between the maximum point and the minimum point. The maximum point right over here, it hits a value of y equals 1. At the minimum point, it hits a value of y is equal to negative 5. So halfway between those, the average of 1 and negative 5, 1 plus negative 5 is negative 4 divided by 2 is negative 2. So this right over here is the midline. So this thing is clearly, so this is y is equal to negative 2. This is y is equal to negative 2. |
Determining the equation of a trig function Graphs of trig functions Trigonometry Khan Academy.mp3 | So halfway between those, the average of 1 and negative 5, 1 plus negative 5 is negative 4 divided by 2 is negative 2. So this right over here is the midline. So this thing is clearly, so this is y is equal to negative 2. This is y is equal to negative 2. So it's clearly shifted down. So we're going to have, so actually I'll talk in a second about what type of an expression it might be. But now also let's think about its amplitude. |
Determining the equation of a trig function Graphs of trig functions Trigonometry Khan Academy.mp3 | This is y is equal to negative 2. So it's clearly shifted down. So we're going to have, so actually I'll talk in a second about what type of an expression it might be. But now also let's think about its amplitude. That's how far it might get away from the midline. We see here it went 3 above the midline, going from negative 2 to 1. It went 3 above the midline at the maximum point. |
Determining the equation of a trig function Graphs of trig functions Trigonometry Khan Academy.mp3 | But now also let's think about its amplitude. That's how far it might get away from the midline. We see here it went 3 above the midline, going from negative 2 to 1. It went 3 above the midline at the maximum point. And it can also go 3 below the midline at the minimum point. So this thing clearly has an amplitude of 3. So amplitude of 3. |
Determining the equation of a trig function Graphs of trig functions Trigonometry Khan Academy.mp3 | It went 3 above the midline at the maximum point. And it can also go 3 below the midline at the minimum point. So this thing clearly has an amplitude of 3. So amplitude of 3. So immediately we can say, well, look, this is going to have a form something like f of x is equal to the amplitude 3. We haven't figured out yet whether this is going to be a cosine function or a sine function. So I'll write cosine first. |
Determining the equation of a trig function Graphs of trig functions Trigonometry Khan Academy.mp3 | So amplitude of 3. So immediately we can say, well, look, this is going to have a form something like f of x is equal to the amplitude 3. We haven't figured out yet whether this is going to be a cosine function or a sine function. So I'll write cosine first. Cosine may be some coefficient times x plus the midline. The midline we already figured out was minus 2 or negative 2. So it could take that form, or it could take f of x is equal to 3 times, it could be sine of x, or sine of some coefficient times x. |
Determining the equation of a trig function Graphs of trig functions Trigonometry Khan Academy.mp3 | So I'll write cosine first. Cosine may be some coefficient times x plus the midline. The midline we already figured out was minus 2 or negative 2. So it could take that form, or it could take f of x is equal to 3 times, it could be sine of x, or sine of some coefficient times x. Sine of kx minus 2, plus the midline. So minus 2. So how do we figure out which of these are? |
Determining the equation of a trig function Graphs of trig functions Trigonometry Khan Academy.mp3 | So it could take that form, or it could take f of x is equal to 3 times, it could be sine of x, or sine of some coefficient times x. Sine of kx minus 2, plus the midline. So minus 2. So how do we figure out which of these are? Well, let's just think about the behavior of this function when x is equal to 0. When x is equal to 0, if this is kx, then the input into the cosine is going to be 0. Cosine of 0 is 1, whether you're talking about degrees or radians, cosine of 0 is 1. |
Determining the equation of a trig function Graphs of trig functions Trigonometry Khan Academy.mp3 | So how do we figure out which of these are? Well, let's just think about the behavior of this function when x is equal to 0. When x is equal to 0, if this is kx, then the input into the cosine is going to be 0. Cosine of 0 is 1, whether you're talking about degrees or radians, cosine of 0 is 1. While sine of 0, so if x is 0, k times 0 is going to be 0. Sine of 0 is 0. So what's this thing doing when x is equal to 0? |
Determining the equation of a trig function Graphs of trig functions Trigonometry Khan Academy.mp3 | Cosine of 0 is 1, whether you're talking about degrees or radians, cosine of 0 is 1. While sine of 0, so if x is 0, k times 0 is going to be 0. Sine of 0 is 0. So what's this thing doing when x is equal to 0? Well, when x is equal to 0, we are at the midline. If we're at the midline, that means that all of this stuff right over here evaluated to 0. So since when x equals 0, all of this stuff evaluated to 0, we can rule out the cosine function. |
Determining the equation of a trig function Graphs of trig functions Trigonometry Khan Academy.mp3 | So what's this thing doing when x is equal to 0? Well, when x is equal to 0, we are at the midline. If we're at the midline, that means that all of this stuff right over here evaluated to 0. So since when x equals 0, all of this stuff evaluated to 0, we can rule out the cosine function. When x equals 0 here, this stuff doesn't evaluate to 0. So we can rule out this one right over there. And so we are left with this. |
Determining the equation of a trig function Graphs of trig functions Trigonometry Khan Academy.mp3 | So since when x equals 0, all of this stuff evaluated to 0, we can rule out the cosine function. When x equals 0 here, this stuff doesn't evaluate to 0. So we can rule out this one right over there. And so we are left with this. And we just really need to figure out what could this constant actually be? And to think about that, let's look at the period of this function. So to go from, and we could, let's see, if we went from this point where we intersect the midline, we go this point to intersect the midline, and we have a positive slope, the next point that we do that is right over here. |
Determining the equation of a trig function Graphs of trig functions Trigonometry Khan Academy.mp3 | And so we are left with this. And we just really need to figure out what could this constant actually be? And to think about that, let's look at the period of this function. So to go from, and we could, let's see, if we went from this point where we intersect the midline, we go this point to intersect the midline, and we have a positive slope, the next point that we do that is right over here. So our period is 8. So what coefficient could we have here to make the period of this thing be equal to 8? Well, let's just remind ourselves what the period of sine of x is. |
Determining the equation of a trig function Graphs of trig functions Trigonometry Khan Academy.mp3 | So to go from, and we could, let's see, if we went from this point where we intersect the midline, we go this point to intersect the midline, and we have a positive slope, the next point that we do that is right over here. So our period is 8. So what coefficient could we have here to make the period of this thing be equal to 8? Well, let's just remind ourselves what the period of sine of x is. So the period of sine of x, so I'll write period right over here, is 2 pi. 2 pi, you increase your angle by 2 pi radians or decrease it. You're back at the same point on the unit circle. |
Determining the equation of a trig function Graphs of trig functions Trigonometry Khan Academy.mp3 | Well, let's just remind ourselves what the period of sine of x is. So the period of sine of x, so I'll write period right over here, is 2 pi. 2 pi, you increase your angle by 2 pi radians or decrease it. You're back at the same point on the unit circle. So what would be the period of sine of kx? Well, now your x, your input, is increasing k times faster. So you're going to get to the same point k times faster. |
Determining the equation of a trig function Graphs of trig functions Trigonometry Khan Academy.mp3 | You're back at the same point on the unit circle. So what would be the period of sine of kx? Well, now your x, your input, is increasing k times faster. So you're going to get to the same point k times faster. So your period is going to be 1 kth as long. So now your period is going to be 2 pi over k. Notice, you're increasing your argument as x increases. Your argument into the sine function is increasing k times as fast. |
Determining the equation of a trig function Graphs of trig functions Trigonometry Khan Academy.mp3 | So you're going to get to the same point k times faster. So your period is going to be 1 kth as long. So now your period is going to be 2 pi over k. Notice, you're increasing your argument as x increases. Your argument into the sine function is increasing k times as fast. You're multiplying it by k. So your period is going to be shorter. It's going to take you less distance for the whole argument to get to the same point on the unit circle. So let's think about it this way. |
Determining the equation of a trig function Graphs of trig functions Trigonometry Khan Academy.mp3 | Your argument into the sine function is increasing k times as fast. You're multiplying it by k. So your period is going to be shorter. It's going to take you less distance for the whole argument to get to the same point on the unit circle. So let's think about it this way. So if we wanted to say 2 pi over k is equal to 8, well, what is our k? Well, we could take the reciprocal of both sides. We get k over 2 pi is equal to 1 over 8. |
Determining the equation of a trig function Graphs of trig functions Trigonometry Khan Academy.mp3 | So let's think about it this way. So if we wanted to say 2 pi over k is equal to 8, well, what is our k? Well, we could take the reciprocal of both sides. We get k over 2 pi is equal to 1 over 8. Multiply both sides by 2 pi. And we get k is equal to pi over 4. And we are done. |
Non-linear systems of equations 3 Algebra II Khan Academy.mp3 | So we have y is equal to 2 times the quantity x minus 4 squared plus 3. We also have y is equal to negative x squared plus 2x minus 2. So the solution, or it might be 1, or it might be none, it might be two solutions, but the solutions to this system occur for the x values that generate the same y values. So the same x and y that satisfy both of these equations. So in order to find the x values, they need to equal the same y value. So this y has to be that y value. So the solution is going to occur when this guy right here, negative x squared plus 2x minus 2, is equal to that guy up there. |
Non-linear systems of equations 3 Algebra II Khan Academy.mp3 | So the same x and y that satisfy both of these equations. So in order to find the x values, they need to equal the same y value. So this y has to be that y value. So the solution is going to occur when this guy right here, negative x squared plus 2x minus 2, is equal to that guy up there. Is equal to 2 times x minus 4 squared plus 3. And now let's just try to solve for x. So the left-hand side, well, we're going to have to multiply this out. |
Non-linear systems of equations 3 Algebra II Khan Academy.mp3 | So the solution is going to occur when this guy right here, negative x squared plus 2x minus 2, is equal to that guy up there. Is equal to 2 times x minus 4 squared plus 3. And now let's just try to solve for x. So the left-hand side, well, we're going to have to multiply this out. So let's do that first. It's negative x squared plus 2x minus 2 is equal to, and on the right-hand side, 2 times x minus 4 squared is x squared minus 8x plus 16 plus 3. This is going to be equal to 2x squared, I'm just distributing the 2, minus 16x plus 32 plus 3, which is equal to 2x squared minus 16x plus 35. |
Non-linear systems of equations 3 Algebra II Khan Academy.mp3 | So the left-hand side, well, we're going to have to multiply this out. So let's do that first. It's negative x squared plus 2x minus 2 is equal to, and on the right-hand side, 2 times x minus 4 squared is x squared minus 8x plus 16 plus 3. This is going to be equal to 2x squared, I'm just distributing the 2, minus 16x plus 32 plus 3, which is equal to 2x squared minus 16x plus 35. And that's, of course, going to be equal to this thing on the left-hand side. Negative x squared plus 2x minus 2. And now let's just get rid of this whole thing from the left-hand side all at once. |
Non-linear systems of equations 3 Algebra II Khan Academy.mp3 | This is going to be equal to 2x squared, I'm just distributing the 2, minus 16x plus 32 plus 3, which is equal to 2x squared minus 16x plus 35. And that's, of course, going to be equal to this thing on the left-hand side. Negative x squared plus 2x minus 2. And now let's just get rid of this whole thing from the left-hand side all at once. By adding x squared to both sides, we could all do it in one step, we're going to add x squared to both sides. Let's subtract 2x from both sides. And let's add 2 to both sides. |
Non-linear systems of equations 3 Algebra II Khan Academy.mp3 | And now let's just get rid of this whole thing from the left-hand side all at once. By adding x squared to both sides, we could all do it in one step, we're going to add x squared to both sides. Let's subtract 2x from both sides. And let's add 2 to both sides. And we will get, on the left-hand side, those cancel out, those cancel out, those cancel out. You're left with 0 is equal to 2x squared plus x squared is 3x squared, negative 16x minus 2x is negative 18x, and then 35 plus 2 is 37. So we just have a plain, vanilla quadratic equation right here. |
Non-linear systems of equations 3 Algebra II Khan Academy.mp3 | And let's add 2 to both sides. And we will get, on the left-hand side, those cancel out, those cancel out, those cancel out. You're left with 0 is equal to 2x squared plus x squared is 3x squared, negative 16x minus 2x is negative 18x, and then 35 plus 2 is 37. So we just have a plain, vanilla quadratic equation right here. And we might as well apply the quadratic formula here to try to solve it. So our solutions are going to be x is equal to negative b. Well, b is negative 18, so negative b is positive 18. |
Non-linear systems of equations 3 Algebra II Khan Academy.mp3 | So we just have a plain, vanilla quadratic equation right here. And we might as well apply the quadratic formula here to try to solve it. So our solutions are going to be x is equal to negative b. Well, b is negative 18, so negative b is positive 18. So it's 18 plus or minus the square root of 18 squared minus 4 times 3 times c times 37. All of that over 2 times a, 2 times 3, which is 6. Now let's think about what this is going to be. |
Non-linear systems of equations 3 Algebra II Khan Academy.mp3 | Well, b is negative 18, so negative b is positive 18. So it's 18 plus or minus the square root of 18 squared minus 4 times 3 times c times 37. All of that over 2 times a, 2 times 3, which is 6. Now let's think about what this is going to be. Over here we have, so it's 18 plus or minus the square root of, well, let's just use our calculator. I could multiply it out, but I think, so we have 18 squared minus 4 times 3 times 37. Negative 120. |
Non-linear systems of equations 3 Algebra II Khan Academy.mp3 | Now let's think about what this is going to be. Over here we have, so it's 18 plus or minus the square root of, well, let's just use our calculator. I could multiply it out, but I think, so we have 18 squared minus 4 times 3 times 37. Negative 120. So it's 18 plus or minus the square root of negative 120. And you might even be able to figure out this is negative. 4 times 3 is 12. |
Non-linear systems of equations 3 Algebra II Khan Academy.mp3 | Negative 120. So it's 18 plus or minus the square root of negative 120. And you might even be able to figure out this is negative. 4 times 3 is 12. 12 times 37 is going to be a bigger number than 18. Although it's not 100% obvious, but you might be able to just get the intuition there. But we definitely end up with a negative number under the radical here. |
Non-linear systems of equations 3 Algebra II Khan Academy.mp3 | 4 times 3 is 12. 12 times 37 is going to be a bigger number than 18. Although it's not 100% obvious, but you might be able to just get the intuition there. But we definitely end up with a negative number under the radical here. Now, if we're dealing with real numbers, there is no square root of negative 120. So there is no solution to this quadratic equation. There is no solution. |
Non-linear systems of equations 3 Algebra II Khan Academy.mp3 | But we definitely end up with a negative number under the radical here. Now, if we're dealing with real numbers, there is no square root of negative 120. So there is no solution to this quadratic equation. There is no solution. And if we wanted to, we could have just looked at the discriminant. The discriminant is this part, b squared minus 4ac. We see the discriminant is negative. |
Non-linear systems of equations 3 Algebra II Khan Academy.mp3 | There is no solution. And if we wanted to, we could have just looked at the discriminant. The discriminant is this part, b squared minus 4ac. We see the discriminant is negative. There's no solution, which means that these two guys, these two equations, never intersect. There is no solution to the system. There are no x values that when you put into both of these equations give you the exact same y value. |
Non-linear systems of equations 3 Algebra II Khan Academy.mp3 | We see the discriminant is negative. There's no solution, which means that these two guys, these two equations, never intersect. There is no solution to the system. There are no x values that when you put into both of these equations give you the exact same y value. Now let's think a little bit about why that happened. This one is already in kind of our y-intercept form. And it's an upward opening parabola. |
Non-linear systems of equations 3 Algebra II Khan Academy.mp3 | There are no x values that when you put into both of these equations give you the exact same y value. Now let's think a little bit about why that happened. This one is already in kind of our y-intercept form. And it's an upward opening parabola. So it looks something like this. I'll do my best to draw it. Just a quick and dirty version of it. |
Non-linear systems of equations 3 Algebra II Khan Academy.mp3 | And it's an upward opening parabola. So it looks something like this. I'll do my best to draw it. Just a quick and dirty version of it. Let me draw my axes in a neutral color. So let's say that this right here is my y-axis. That right there is my x-axis, x and y. |
Non-linear systems of equations 3 Algebra II Khan Academy.mp3 | Just a quick and dirty version of it. Let me draw my axes in a neutral color. So let's say that this right here is my y-axis. That right there is my x-axis, x and y. This vertex, it's in the vertex form, occurs when x is equal to 4 and y is equal to 3. So x is equal to 4 and y is equal to 3. And it's an upward opening parabola. |
Non-linear systems of equations 3 Algebra II Khan Academy.mp3 | That right there is my x-axis, x and y. This vertex, it's in the vertex form, occurs when x is equal to 4 and y is equal to 3. So x is equal to 4 and y is equal to 3. And it's an upward opening parabola. We have a positive coefficient out here. So this will look something like this. It will look something like this. |
Non-linear systems of equations 3 Algebra II Khan Academy.mp3 | And it's an upward opening parabola. We have a positive coefficient out here. So this will look something like this. It will look something like this. I don't know the exact thing, but that's close enough. Now what will this thing look like? Well, it's a downward opening parabola. |
Non-linear systems of equations 3 Algebra II Khan Academy.mp3 | It will look something like this. I don't know the exact thing, but that's close enough. Now what will this thing look like? Well, it's a downward opening parabola. And we could actually put this in vertex form. Let me do that. Let me put the second equation in vertex form, just so we have it. |
Non-linear systems of equations 3 Algebra II Khan Academy.mp3 | Well, it's a downward opening parabola. And we could actually put this in vertex form. Let me do that. Let me put the second equation in vertex form, just so we have it. So we have a good sense. So y is equal to, we could factor out a negative 1, negative x squared minus 2x plus 2. x squared minus 2x plus 2. Actually, let me put the plus 2 further out. |
Non-linear systems of equations 3 Algebra II Khan Academy.mp3 | Let me put the second equation in vertex form, just so we have it. So we have a good sense. So y is equal to, we could factor out a negative 1, negative x squared minus 2x plus 2. x squared minus 2x plus 2. Actually, let me put the plus 2 further out. Plus 2, all the way up out there. And then we could say, OK, half of negative 2 is negative 1. You square it. |
Non-linear systems of equations 3 Algebra II Khan Academy.mp3 | Actually, let me put the plus 2 further out. Plus 2, all the way up out there. And then we could say, OK, half of negative 2 is negative 1. You square it. So you have a plus 1 and then a minus 1 there. And then this part right over here, we can rewrite as x minus 1 squared. So it becomes negative x minus 1 squared. |
Non-linear systems of equations 3 Algebra II Khan Academy.mp3 | You square it. So you have a plus 1 and then a minus 1 there. And then this part right over here, we can rewrite as x minus 1 squared. So it becomes negative x minus 1 squared. Let me just do it one step at a time. I don't want to skip steps. Negative x minus 1 squared minus 1 plus 2. |
Non-linear systems of equations 3 Algebra II Khan Academy.mp3 | So it becomes negative x minus 1 squared. Let me just do it one step at a time. I don't want to skip steps. Negative x minus 1 squared minus 1 plus 2. So that's plus 1 out here. Or if we want to distribute the negative, we get y is equal to negative x minus 1 squared minus 1. So here the vertex occurs at x is equal to 1, y is equal to negative 1. x is equal to 1, y is equal to negative 1. |
Non-linear systems of equations 3 Algebra II Khan Academy.mp3 | Negative x minus 1 squared minus 1 plus 2. So that's plus 1 out here. Or if we want to distribute the negative, we get y is equal to negative x minus 1 squared minus 1. So here the vertex occurs at x is equal to 1, y is equal to negative 1. x is equal to 1, y is equal to negative 1. The vertex is there. And this is a downward opening parabola. We have a negative coefficient out here on the second degree term, so it's going to look something like this. |
Non-linear systems of equations 3 Algebra II Khan Academy.mp3 | So here the vertex occurs at x is equal to 1, y is equal to negative 1. x is equal to 1, y is equal to negative 1. The vertex is there. And this is a downward opening parabola. We have a negative coefficient out here on the second degree term, so it's going to look something like this. It's going to look something like this. So as you see, they don't intersect. This vertex is above it and it opens upward. |
Non-linear systems of equations 3 Algebra II Khan Academy.mp3 | We have a negative coefficient out here on the second degree term, so it's going to look something like this. It's going to look something like this. So as you see, they don't intersect. This vertex is above it and it opens upward. This is its minimum point. And it's above this guy's maximum point. So they will never intersect. |
Matching ratios to trig functions Trigonometry Khan Academy.mp3 | And then over here on the left, we have the sine taken of angle MKJ, cosine of angle MKJ, and tangent of angle MKJ. And angle MKJ is this angle right over here, same thing as theta. So these two angles, these two angles have the same measure. We see that right over there. And what we wanna do is figure out which of these expressions are equivalent to which of these expressions right over here. And so I encourage you to pause the video and try to work this through on your own. So assuming you've had a go at it, so let's try to work this out. |
Matching ratios to trig functions Trigonometry Khan Academy.mp3 | We see that right over there. And what we wanna do is figure out which of these expressions are equivalent to which of these expressions right over here. And so I encourage you to pause the video and try to work this through on your own. So assuming you've had a go at it, so let's try to work this out. And when you look at this diagram, it looks like the intention here on the left is this evokes the unit circle definition of trig functions because this is really, this is a unit circle right over here. And this evokes kind of the Sokodowa definition because we're just kind of in a plain vanilla right triangle. And so just to remind ourselves, let's just remind ourselves of Sokodowa because we have a feeling it might be useful. |
Matching ratios to trig functions Trigonometry Khan Academy.mp3 | So assuming you've had a go at it, so let's try to work this out. And when you look at this diagram, it looks like the intention here on the left is this evokes the unit circle definition of trig functions because this is really, this is a unit circle right over here. And this evokes kind of the Sokodowa definition because we're just kind of in a plain vanilla right triangle. And so just to remind ourselves, let's just remind ourselves of Sokodowa because we have a feeling it might be useful. So sine is opposite over hypotenuse. Cosine is adjacent over hypotenuse. And tangent is opposite over adjacent. |
Matching ratios to trig functions Trigonometry Khan Academy.mp3 | And so just to remind ourselves, let's just remind ourselves of Sokodowa because we have a feeling it might be useful. So sine is opposite over hypotenuse. Cosine is adjacent over hypotenuse. And tangent is opposite over adjacent. So we can refer to this, and we can also refer to remind ourselves of the unit circle definition of trig functions, that the cosine of an angle is the x-coordinate, and that the sine of where this ray intersects the unit circle, and the sine of this angle is going to be the y-coordinate. And what we'll see through this video is that they're actually, the unit circle definition is just an extension of Sokodowa. So let's look first at x over one. |
Matching ratios to trig functions Trigonometry Khan Academy.mp3 | And tangent is opposite over adjacent. So we can refer to this, and we can also refer to remind ourselves of the unit circle definition of trig functions, that the cosine of an angle is the x-coordinate, and that the sine of where this ray intersects the unit circle, and the sine of this angle is going to be the y-coordinate. And what we'll see through this video is that they're actually, the unit circle definition is just an extension of Sokodowa. So let's look first at x over one. So we have x, x is the x-coordinate. That's also the length of this side right over here. Relative to this angle, theta, that is the adjacent side. |
Matching ratios to trig functions Trigonometry Khan Academy.mp3 | So let's look first at x over one. So we have x, x is the x-coordinate. That's also the length of this side right over here. Relative to this angle, theta, that is the adjacent side. So x is equal to the adjacent side. What is one? Well, this is a unit circle. |
Matching ratios to trig functions Trigonometry Khan Academy.mp3 | Relative to this angle, theta, that is the adjacent side. So x is equal to the adjacent side. What is one? Well, this is a unit circle. One is the length of the radius, which for this right triangle is also the hypotenuse. So if we apply the Sokodowa definition, x over one is adjacent over hypotenuse. Adjacent over hypotenuse, adjacent over hypotenuse, that's cosine. |
Matching ratios to trig functions Trigonometry Khan Academy.mp3 | Well, this is a unit circle. One is the length of the radius, which for this right triangle is also the hypotenuse. So if we apply the Sokodowa definition, x over one is adjacent over hypotenuse. Adjacent over hypotenuse, adjacent over hypotenuse, that's cosine. So that's going to be, this is equal to cosine of theta, but theta is the same thing as angle MKJ. They have the same measure. So cosine of angle MKJ is equal to cosine of theta, which is equal to x over one. |
Matching ratios to trig functions Trigonometry Khan Academy.mp3 | Adjacent over hypotenuse, adjacent over hypotenuse, that's cosine. So that's going to be, this is equal to cosine of theta, but theta is the same thing as angle MKJ. They have the same measure. So cosine of angle MKJ is equal to cosine of theta, which is equal to x over one. Now let's move over to y over one. Well, y is going to be the length of this side right over here. Y is going to be, let me do this in the blue. |
Matching ratios to trig functions Trigonometry Khan Academy.mp3 | So cosine of angle MKJ is equal to cosine of theta, which is equal to x over one. Now let's move over to y over one. Well, y is going to be the length of this side right over here. Y is going to be, let me do this in the blue. Y is going to be this length relative to angle theta. That is the opposite side. That is the opposite side. |
Matching ratios to trig functions Trigonometry Khan Academy.mp3 | Y is going to be, let me do this in the blue. Y is going to be this length relative to angle theta. That is the opposite side. That is the opposite side. Now which trig function is opposite over hypotenuse? Opposite over hypotenuse, that's sine of theta. Sine of theta. |
Matching ratios to trig functions Trigonometry Khan Academy.mp3 | That is the opposite side. Now which trig function is opposite over hypotenuse? Opposite over hypotenuse, that's sine of theta. Sine of theta. So sine of angle MKJ is the same thing as sine of theta. We see that they have the same measure. And now we see that's the same thing as y over one. |
Matching ratios to trig functions Trigonometry Khan Academy.mp3 | Sine of theta. So sine of angle MKJ is the same thing as sine of theta. We see that they have the same measure. And now we see that's the same thing as y over one. Now for both of these, I use the Sohcahtoa definition, but we could have also used the unit circle definition. X over one, that's the same thing. That's the same thing as x. |
Matching ratios to trig functions Trigonometry Khan Academy.mp3 | And now we see that's the same thing as y over one. Now for both of these, I use the Sohcahtoa definition, but we could have also used the unit circle definition. X over one, that's the same thing. That's the same thing as x. And the unit circle definition says, well, this x, this point, the x coordinate of where this, I guess you could say the terminal side of this angle, this ray right over here intersects the unit circle, that by definition, by the unit circle definition, is the cosine of this angle. X is equal to the cosine of this angle. And the unit circle definition, the y coordinate is equal to the sine of this angle. |
Matching ratios to trig functions Trigonometry Khan Academy.mp3 | That's the same thing as x. And the unit circle definition says, well, this x, this point, the x coordinate of where this, I guess you could say the terminal side of this angle, this ray right over here intersects the unit circle, that by definition, by the unit circle definition, is the cosine of this angle. X is equal to the cosine of this angle. And the unit circle definition, the y coordinate is equal to the sine of this angle. We could have written this as, instead of x comma y, we could have written this as cosine of theta, sine theta, just like that. But let's keep going. Now we have x over y. |
Matching ratios to trig functions Trigonometry Khan Academy.mp3 | And the unit circle definition, the y coordinate is equal to the sine of this angle. We could have written this as, instead of x comma y, we could have written this as cosine of theta, sine theta, just like that. But let's keep going. Now we have x over y. We have adjacent over opposite. So this is equal to adjacent over opposite. Tangent is opposite over adjacent, not adjacent over opposite. |
Matching ratios to trig functions Trigonometry Khan Academy.mp3 | Now we have x over y. We have adjacent over opposite. So this is equal to adjacent over opposite. Tangent is opposite over adjacent, not adjacent over opposite. So this is the reciprocal of tangent. So this right over here, if we had to, this is equal to one over tangent of theta. And we later learn about cotangent and all of that, which is essentially this. |
Matching ratios to trig functions Trigonometry Khan Academy.mp3 | Tangent is opposite over adjacent, not adjacent over opposite. So this is the reciprocal of tangent. So this right over here, if we had to, this is equal to one over tangent of theta. And we later learn about cotangent and all of that, which is essentially this. But it's not one of our choices. So we can rule this one out. But then we have y over x. |
Matching ratios to trig functions Trigonometry Khan Academy.mp3 | And we later learn about cotangent and all of that, which is essentially this. But it's not one of our choices. So we can rule this one out. But then we have y over x. Well, this is looking good. This is, y is opposite, opposite, x is adjacent relative to angle theta, adjacent. So this is the tangent of theta. |
Matching ratios to trig functions Trigonometry Khan Academy.mp3 | But then we have y over x. Well, this is looking good. This is, y is opposite, opposite, x is adjacent relative to angle theta, adjacent. So this is the tangent of theta. This is equal to tangent of theta. So tangent of angle mkj is the same thing as tangent of theta, which is equal to y over x. Now let's look at j over k. So j over k. Now we're moving over to this triangle. |
Matching ratios to trig functions Trigonometry Khan Academy.mp3 | So this is the tangent of theta. This is equal to tangent of theta. So tangent of angle mkj is the same thing as tangent of theta, which is equal to y over x. Now let's look at j over k. So j over k. Now we're moving over to this triangle. J over k. So relative to this angle, because this is the angle that we care about, j is the length of the adjacent side, and k is the length of the opposite side, of the opposite side. So this is adjacent over opposite. So this is equal to adjacent over opposite. |
Matching ratios to trig functions Trigonometry Khan Academy.mp3 | Now let's look at j over k. So j over k. Now we're moving over to this triangle. J over k. So relative to this angle, because this is the angle that we care about, j is the length of the adjacent side, and k is the length of the opposite side, of the opposite side. So this is adjacent over opposite. So this is equal to adjacent over opposite. Tangent is opposite over adjacent, not adjacent over opposite. So once again, this is one, this is a reciprocal of the tangent function, not one of the choices right over here. So we can rule that one out. |
Matching ratios to trig functions Trigonometry Khan Academy.mp3 | So this is equal to adjacent over opposite. Tangent is opposite over adjacent, not adjacent over opposite. So once again, this is one, this is a reciprocal of the tangent function, not one of the choices right over here. So we can rule that one out. Now k over j. Well now this is opposite over adjacent, opposite over adjacent. That is equal to tangent of theta. |
Matching ratios to trig functions Trigonometry Khan Academy.mp3 | So we can rule that one out. Now k over j. Well now this is opposite over adjacent, opposite over adjacent. That is equal to tangent of theta. This is equal to tangent of theta, or tangent of angle mkj. So this is equal to k over j. Now we have m over j. M over j. Hypotenuse over adjacent side. |
Matching ratios to trig functions Trigonometry Khan Academy.mp3 | That is equal to tangent of theta. This is equal to tangent of theta, or tangent of angle mkj. So this is equal to k over j. Now we have m over j. M over j. Hypotenuse over adjacent side. This of course, this of course is equal to the hypotenuse. Hypotenuse over adjacent. Well if it was adjacent over hypotenuse, we'd be dealing with cosine, but this is the reciprocal of that. |
Matching ratios to trig functions Trigonometry Khan Academy.mp3 | Now we have m over j. M over j. Hypotenuse over adjacent side. This of course, this of course is equal to the hypotenuse. Hypotenuse over adjacent. Well if it was adjacent over hypotenuse, we'd be dealing with cosine, but this is the reciprocal of that. So this is actually one over the cosine of theta, not one of our choices here, so I'll just rule that one out right over there. Then we have its reciprocal, j over m. That's adjacent over hypotenuse. Adjacent over hypotenuse is cosine. |
Matching ratios to trig functions Trigonometry Khan Academy.mp3 | Well if it was adjacent over hypotenuse, we'd be dealing with cosine, but this is the reciprocal of that. So this is actually one over the cosine of theta, not one of our choices here, so I'll just rule that one out right over there. Then we have its reciprocal, j over m. That's adjacent over hypotenuse. Adjacent over hypotenuse is cosine. So this is equal to cosine of theta, or cosine of angle mkj. So we could write it down. So this is equivalent to j over m. And then one last one, k over m. Well that's opposite over hypotenuse. |
Matching ratios to trig functions Trigonometry Khan Academy.mp3 | Adjacent over hypotenuse is cosine. So this is equal to cosine of theta, or cosine of angle mkj. So we could write it down. So this is equivalent to j over m. And then one last one, k over m. Well that's opposite over hypotenuse. Opposite over hypotenuse. That's going to be sine of theta. So this right over here is equal to sine of theta, which is the same thing as sine of angle mkj, which is the same thing as all of these expressions. |
Introduction to complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3 | Real numbers include things like zero and one and 0.3 repeating and pi and e, and I could keep listing real numbers. These are the numbers that you're kind of familiar with. And then we explored something interesting. We explored the notion of, well, what if there was a number that if I squared it, I would get the negative one? And we defined that thing, that if we squared it, we got negative one. We defined that thing as i. And so we defined a whole new class of numbers which you could really view as multiples of the imaginary unit. |
Introduction to complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3 | We explored the notion of, well, what if there was a number that if I squared it, I would get the negative one? And we defined that thing, that if we squared it, we got negative one. We defined that thing as i. And so we defined a whole new class of numbers which you could really view as multiples of the imaginary unit. So imaginary numbers would be i and negative i and pi times i and e times i. So this might raise another interesting question. What if I combined imaginary and real numbers? |
Introduction to complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3 | And so we defined a whole new class of numbers which you could really view as multiples of the imaginary unit. So imaginary numbers would be i and negative i and pi times i and e times i. So this might raise another interesting question. What if I combined imaginary and real numbers? What if I had numbers that were essentially sums or differences of real and imaginary numbers? For example, let's say that I had the number, let's say I call it z, and z tends to be what we, the most used variable when we're talking about what I'm about to talk about, complex numbers. Let's say that z is equal to the real number five plus the imaginary number three times i. |
Introduction to complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3 | What if I combined imaginary and real numbers? What if I had numbers that were essentially sums or differences of real and imaginary numbers? For example, let's say that I had the number, let's say I call it z, and z tends to be what we, the most used variable when we're talking about what I'm about to talk about, complex numbers. Let's say that z is equal to the real number five plus the imaginary number three times i. So this thing right over here, we have a real number plus an imaginary number. You might be tempted to add these two things, but you can't. It won't make any sense. |
Introduction to complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3 | Let's say that z is equal to the real number five plus the imaginary number three times i. So this thing right over here, we have a real number plus an imaginary number. You might be tempted to add these two things, but you can't. It won't make any sense. These are kind of going in different, well, we'll think about it visually in a second, but you can't simplify this anymore. You can't add this real number to this imaginary number. A number like this, let me make it clear, that's real and this is imaginary. |
Introduction to complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3 | It won't make any sense. These are kind of going in different, well, we'll think about it visually in a second, but you can't simplify this anymore. You can't add this real number to this imaginary number. A number like this, let me make it clear, that's real and this is imaginary. Imaginary. A number like this we call a complex number. A complex number. |
Introduction to complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3 | A number like this, let me make it clear, that's real and this is imaginary. Imaginary. A number like this we call a complex number. A complex number. It has a real part and an imaginary part. And sometimes you'll see notation like this where someone will say, well, what's the real part? What's the real part of our complex number z? |
Introduction to complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3 | A complex number. It has a real part and an imaginary part. And sometimes you'll see notation like this where someone will say, well, what's the real part? What's the real part of our complex number z? Well, that would be the five right over there. And then they might say, well, what's the imaginary part? What's the imaginary part of our complex number z? |
Introduction to complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3 | What's the real part of our complex number z? Well, that would be the five right over there. And then they might say, well, what's the imaginary part? What's the imaginary part of our complex number z? And then typically the way that this function is defined, they really wanna know, well, what multiple of i is this imaginary part right over here? And in this case, it is going to be three. And we can visualize this. |
Introduction to complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3 | What's the imaginary part of our complex number z? And then typically the way that this function is defined, they really wanna know, well, what multiple of i is this imaginary part right over here? And in this case, it is going to be three. And we can visualize this. We can visualize this in two dimensions. Instead of having the traditional two-dimensional Cartesian plane with real numbers on the horizontal and the vertical axis, what we do to plot complex numbers is we, on the vertical axis, we plot the imaginary part. So that's the imaginary part. |
Introduction to complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3 | And we can visualize this. We can visualize this in two dimensions. Instead of having the traditional two-dimensional Cartesian plane with real numbers on the horizontal and the vertical axis, what we do to plot complex numbers is we, on the vertical axis, we plot the imaginary part. So that's the imaginary part. And then on the horizontal axis, we plot the real part. We plot the real part just like that. We plot the real part. |
Introduction to complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3 | So that's the imaginary part. And then on the horizontal axis, we plot the real part. We plot the real part just like that. We plot the real part. So for example, z right over here, which is five plus three i, the real part is five. So we would go one, two, three, four, five. That's five right over there. |
Introduction to complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3 | We plot the real part. So for example, z right over here, which is five plus three i, the real part is five. So we would go one, two, three, four, five. That's five right over there. The imaginary part is three. One, two, three. And so on the complex plane, we would visualize that number right over here. |
Introduction to complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3 | That's five right over there. The imaginary part is three. One, two, three. And so on the complex plane, we would visualize that number right over here. This right over here is how we would visualize z on the complex plane. It's five, positive five in the real direction, positive three in the imaginary direction. We could plot other complex numbers. |
Introduction to complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3 | And so on the complex plane, we would visualize that number right over here. This right over here is how we would visualize z on the complex plane. It's five, positive five in the real direction, positive three in the imaginary direction. We could plot other complex numbers. Let's say we had the complex number a, which is equal to, let's say it's negative two plus i. Where would I plot that? Well, the real part is negative two. |
Introduction to complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3 | We could plot other complex numbers. Let's say we had the complex number a, which is equal to, let's say it's negative two plus i. Where would I plot that? Well, the real part is negative two. Negative two. And the imaginary part is going to be, you could imagine this is plus one i. So we go one up, it's going to be right over there. |
Introduction to complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3 | Well, the real part is negative two. Negative two. And the imaginary part is going to be, you could imagine this is plus one i. So we go one up, it's going to be right over there. So that right over there is our complex number, our complex number a, would be at that point of the complex, complex, let me write that, that point of the complex plane. And let me just do one more. Let's say you had the complex number b, which is going to be, let's say it is, let's say it's four minus three i. |
Introduction to complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3 | So we go one up, it's going to be right over there. So that right over there is our complex number, our complex number a, would be at that point of the complex, complex, let me write that, that point of the complex plane. And let me just do one more. Let's say you had the complex number b, which is going to be, let's say it is, let's say it's four minus three i. Where would we plot that? Well, one, two, three, four. And then, let's see, minus one, two, three, or negative three gets us right over there. |
Introduction to complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3 | Let's say you had the complex number b, which is going to be, let's say it is, let's say it's four minus three i. Where would we plot that? Well, one, two, three, four. And then, let's see, minus one, two, three, or negative three gets us right over there. So that right over there would be the complex number b. |
Introduction to complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3 | And then, let's see, minus one, two, three, or negative three gets us right over there. So that right over there would be the complex number b. |