diff --git "a/Trig04_24.json" "b/Trig04_24.json" new file mode 100644--- /dev/null +++ "b/Trig04_24.json" @@ -0,0 +1 @@ +[{"video_title": "Using right triangle ratios to approximate angle measure High school geometry Khan Academy.mp3", "Sentence": "We're told here are the approximate ratios for angle measures 25 degrees, 35 degrees, and 45 degrees. So what they're saying here is if you were to take the adjacent leg length over the hypotenuse leg length for 25 degree angle, it would be a ratio of approximately 0.91. For a 35 degree angle, it would be a ratio of 0.82, and then they do this for 45 degrees, and they do the different ratios right over here. So we're gonna use the table to approximate the measure of angle D in the triangle below. So pause this video and see if you can figure that out. All right, now let's work through this together. Now what information do they give us about angle D in this triangle?"}, {"video_title": "Using right triangle ratios to approximate angle measure High school geometry Khan Academy.mp3", "Sentence": "So we're gonna use the table to approximate the measure of angle D in the triangle below. So pause this video and see if you can figure that out. All right, now let's work through this together. Now what information do they give us about angle D in this triangle? Well, we are given the opposite length right over here. Let me label that. That is the opposite leg length, which is 3.4."}, {"video_title": "Using right triangle ratios to approximate angle measure High school geometry Khan Academy.mp3", "Sentence": "Now what information do they give us about angle D in this triangle? Well, we are given the opposite length right over here. Let me label that. That is the opposite leg length, which is 3.4. We're also given, what is this right over here? Is this adjacent or is this a hypotenuse? You might be tempted to say, well, this is right next to the angle, or this is one of the lines, or it's on the ray that helps form the angle, so maybe it's adjacent."}, {"video_title": "Using right triangle ratios to approximate angle measure High school geometry Khan Academy.mp3", "Sentence": "That is the opposite leg length, which is 3.4. We're also given, what is this right over here? Is this adjacent or is this a hypotenuse? You might be tempted to say, well, this is right next to the angle, or this is one of the lines, or it's on the ray that helps form the angle, so maybe it's adjacent. But remember, adjacent is the adjacent side that is not the hypotenuse. And this is clearly the hypotenuse. It is the longest side."}, {"video_title": "Using right triangle ratios to approximate angle measure High school geometry Khan Academy.mp3", "Sentence": "You might be tempted to say, well, this is right next to the angle, or this is one of the lines, or it's on the ray that helps form the angle, so maybe it's adjacent. But remember, adjacent is the adjacent side that is not the hypotenuse. And this is clearly the hypotenuse. It is the longest side. It is the side opposite the 90 degree angle. So this right over here is the hypotenuse. Hypotenuse."}, {"video_title": "Using right triangle ratios to approximate angle measure High school geometry Khan Academy.mp3", "Sentence": "It is the longest side. It is the side opposite the 90 degree angle. So this right over here is the hypotenuse. Hypotenuse. So we're given the opposite leg length and the hypotenuse length. And so let's see, which of these ratios deal with the opposite and the hypotenuse? And if we, let's see, this first one is adjacent and hypotenuse."}, {"video_title": "Using right triangle ratios to approximate angle measure High school geometry Khan Academy.mp3", "Sentence": "Hypotenuse. So we're given the opposite leg length and the hypotenuse length. And so let's see, which of these ratios deal with the opposite and the hypotenuse? And if we, let's see, this first one is adjacent and hypotenuse. The second one here is hypotenuse, sorry, opposite and hypotenuse. So that's exactly what we're talking about. We're talking about the opposite leg length over the hypotenuse, over the hypotenuse length."}, {"video_title": "Using right triangle ratios to approximate angle measure High school geometry Khan Academy.mp3", "Sentence": "And if we, let's see, this first one is adjacent and hypotenuse. The second one here is hypotenuse, sorry, opposite and hypotenuse. So that's exactly what we're talking about. We're talking about the opposite leg length over the hypotenuse, over the hypotenuse length. So in this case, what is going to be our opposite leg length over our hypotenuse leg length? It's going to be 3.4 over eight. 3.4 over eight, which is approximately going to be equal to, let me do this down here, this is eight goes into 3.4."}, {"video_title": "Using right triangle ratios to approximate angle measure High school geometry Khan Academy.mp3", "Sentence": "We're talking about the opposite leg length over the hypotenuse, over the hypotenuse length. So in this case, what is going to be our opposite leg length over our hypotenuse leg length? It's going to be 3.4 over eight. 3.4 over eight, which is approximately going to be equal to, let me do this down here, this is eight goes into 3.4. Eight goes in, doesn't go into three. Eight goes into 34 four times. Four times eight is 32."}, {"video_title": "Using right triangle ratios to approximate angle measure High school geometry Khan Academy.mp3", "Sentence": "3.4 over eight, which is approximately going to be equal to, let me do this down here, this is eight goes into 3.4. Eight goes in, doesn't go into three. Eight goes into 34 four times. Four times eight is 32. If I subtract, and I could scroll down a little bit, I get a two. I can bring down a zero. Eight goes into 20 two times."}, {"video_title": "Using right triangle ratios to approximate angle measure High school geometry Khan Academy.mp3", "Sentence": "Four times eight is 32. If I subtract, and I could scroll down a little bit, I get a two. I can bring down a zero. Eight goes into 20 two times. And that's about as much precision as any of these have. And so it looks like for this particular triangle and this angle of the triangle, if I were to take a ratio of the opposite length and the hypotenuse length, opposite over hypotenuse, I get 0.42. So that looks like this situation right over here."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "We use it in everyday language. We've done some examples on this playlist where if you had an angle like that, you might call that a 30-degree angle. If you have an angle like this, you could call that a 90-degree angle, and we'd often use this symbol just like that. If you were to go 180 degrees, you'd essentially form a straight line. Let me make these proper angles. If you go 360 degrees, you've essentially done one full rotation. If you watch figure skating on the Olympics and someone does a rotation, they'll say, oh, they did a 360, or especially in some skateboarding competitions and things like that."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "If you were to go 180 degrees, you'd essentially form a straight line. Let me make these proper angles. If you go 360 degrees, you've essentially done one full rotation. If you watch figure skating on the Olympics and someone does a rotation, they'll say, oh, they did a 360, or especially in some skateboarding competitions and things like that. But the one thing to realize, and it might not be obvious right from the get-go, is this whole notion of degrees, this is a human-constructed system. This is not the only way that you can measure angles. If you think about it, you say, well, why do we call a full rotation 360 degrees?"}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "If you watch figure skating on the Olympics and someone does a rotation, they'll say, oh, they did a 360, or especially in some skateboarding competitions and things like that. But the one thing to realize, and it might not be obvious right from the get-go, is this whole notion of degrees, this is a human-constructed system. This is not the only way that you can measure angles. If you think about it, you say, well, why do we call a full rotation 360 degrees? There are some possible theories, and I encourage you to think about them. Why does 360 degrees show up in our culture as a full rotation? Well, there's a couple of theories there."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "If you think about it, you say, well, why do we call a full rotation 360 degrees? There are some possible theories, and I encourage you to think about them. Why does 360 degrees show up in our culture as a full rotation? Well, there's a couple of theories there. One is ancient calendars, and even our calendar is close to this, but ancient calendars were based on 360 days in a year. Some ancient astronomers observed that things seemed to move 1 360th of the sky per day. Another theory is the ancient Babylonians liked equilateral triangles a lot, and they had a base 60 number system, so they had 60 symbols."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Well, there's a couple of theories there. One is ancient calendars, and even our calendar is close to this, but ancient calendars were based on 360 days in a year. Some ancient astronomers observed that things seemed to move 1 360th of the sky per day. Another theory is the ancient Babylonians liked equilateral triangles a lot, and they had a base 60 number system, so they had 60 symbols. We only have 10. We have a base 10. They had 60."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Another theory is the ancient Babylonians liked equilateral triangles a lot, and they had a base 60 number system, so they had 60 symbols. We only have 10. We have a base 10. They had 60. In our system, we like to divide things into 10. They probably like to divide things into 60. If you had a circle and you divided it into 6 equilateral triangles, and each of those equilateral triangles you divided into 60 sections because you have a base 60 number system, then you might end up with 360 degrees."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "They had 60. In our system, we like to divide things into 10. They probably like to divide things into 60. If you had a circle and you divided it into 6 equilateral triangles, and each of those equilateral triangles you divided into 60 sections because you have a base 60 number system, then you might end up with 360 degrees. What I want to think about in this video is an alternate way of measuring angles. That alternate way, even though it might not seem as intuitive to you from the get-go, in some ways is much more mathematically pure than degrees. It's not based on these cultural artifacts of base 60 number systems or astronomical patterns."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "If you had a circle and you divided it into 6 equilateral triangles, and each of those equilateral triangles you divided into 60 sections because you have a base 60 number system, then you might end up with 360 degrees. What I want to think about in this video is an alternate way of measuring angles. That alternate way, even though it might not seem as intuitive to you from the get-go, in some ways is much more mathematically pure than degrees. It's not based on these cultural artifacts of base 60 number systems or astronomical patterns. To some degree, an alien on another planet would not use degrees, especially if the degrees are motivated by these astronomical phenomena, but they might use what we're going to define as a radian. There's a certain degree of purity here. Radians."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "It's not based on these cultural artifacts of base 60 number systems or astronomical patterns. To some degree, an alien on another planet would not use degrees, especially if the degrees are motivated by these astronomical phenomena, but they might use what we're going to define as a radian. There's a certain degree of purity here. Radians. Let's just cut to the chase and define what a radian is. Let me draw a circle here. My best attempt at drawing a circle."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Radians. Let's just cut to the chase and define what a radian is. Let me draw a circle here. My best attempt at drawing a circle. Not bad. Let me draw the center of the circle. Now let me draw this radius."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "My best attempt at drawing a circle. Not bad. Let me draw the center of the circle. Now let me draw this radius. You might already notice the word radius is very close to the word radians, and that's not a coincidence. Let's say that this circle has a radius of length r. Let's construct an angle. I'll call that angle theta."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Now let me draw this radius. You might already notice the word radius is very close to the word radians, and that's not a coincidence. Let's say that this circle has a radius of length r. Let's construct an angle. I'll call that angle theta. Let's construct an angle theta. Let's call this angle right over here theta. Let's just say, for the sake of argument, that this angle is just the exact right measure."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "I'll call that angle theta. Let's construct an angle theta. Let's call this angle right over here theta. Let's just say, for the sake of argument, that this angle is just the exact right measure. If you look at the arc that subtends this angle, that seems like a very fancy word. Let me draw the angle. If you look at the arc that subtends the angle, that's a fancy word, but that's really just talking about the arc along the circle that intersects the two sides of the angles."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Let's just say, for the sake of argument, that this angle is just the exact right measure. If you look at the arc that subtends this angle, that seems like a very fancy word. Let me draw the angle. If you look at the arc that subtends the angle, that's a fancy word, but that's really just talking about the arc along the circle that intersects the two sides of the angles. This arc right over here subtends the angle theta. Let me write that down. Subtends this arc."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "If you look at the arc that subtends the angle, that's a fancy word, but that's really just talking about the arc along the circle that intersects the two sides of the angles. This arc right over here subtends the angle theta. Let me write that down. Subtends this arc. Subtends angle theta. Let's say theta is the exact right size. This arc is also the same length as the radius of the circle."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Subtends this arc. Subtends angle theta. Let's say theta is the exact right size. This arc is also the same length as the radius of the circle. This arc is also of length r. Given that, if you were defining a new type of angle measurement, and you wanted to call it a radian, which is very close to a radius, how many radians would you define this angle to be? The most obvious one, if you view a radian as another way of saying radiuses or radii, you say, look, this is subtended by an arc of one radius, so why don't we call this right over here one radian? Which is exactly how a radian is defined."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "This arc is also the same length as the radius of the circle. This arc is also of length r. Given that, if you were defining a new type of angle measurement, and you wanted to call it a radian, which is very close to a radius, how many radians would you define this angle to be? The most obvious one, if you view a radian as another way of saying radiuses or radii, you say, look, this is subtended by an arc of one radius, so why don't we call this right over here one radian? Which is exactly how a radian is defined. When you have a circle and you have an angle of one radian, the arc that subtends it is exactly one radius long, which you can imagine might be a little bit useful as we start to interpret more and more types of circles. When you give a degree, you really have to do a little bit of math and think about the circumference and all of that to think about how many radiuses are subtending that angle. Here, the angle in radians tells you exactly how many arc lengths that is subtending the angle."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Which is exactly how a radian is defined. When you have a circle and you have an angle of one radian, the arc that subtends it is exactly one radius long, which you can imagine might be a little bit useful as we start to interpret more and more types of circles. When you give a degree, you really have to do a little bit of math and think about the circumference and all of that to think about how many radiuses are subtending that angle. Here, the angle in radians tells you exactly how many arc lengths that is subtending the angle. Let's do a couple of thought experiments here. Given that, what would be the angle in radians if we were to go... Let me draw another circle here. Let me draw another circle here."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Here, the angle in radians tells you exactly how many arc lengths that is subtending the angle. Let's do a couple of thought experiments here. Given that, what would be the angle in radians if we were to go... Let me draw another circle here. Let me draw another circle here. That's the center. We'll start right over there. What would happen if I had an angle?"}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Let me draw another circle here. That's the center. We'll start right over there. What would happen if I had an angle? What angle, if I wanted to measure in radians, what angle would this be in radians? You can almost think of it as radiuses. What would that angle be?"}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "What would happen if I had an angle? What angle, if I wanted to measure in radians, what angle would this be in radians? You can almost think of it as radiuses. What would that angle be? Going one full revolution in degrees, that would be 360 degrees. Based on this definition, what would this be in radians? Let's think about the arc that subtends this angle."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "What would that angle be? Going one full revolution in degrees, that would be 360 degrees. Based on this definition, what would this be in radians? Let's think about the arc that subtends this angle. The arc that subtends this angle is the entire circumference of this circle. It's the entire circumference of this circle. What's the circumference of a circle in terms of radiuses?"}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Let's think about the arc that subtends this angle. The arc that subtends this angle is the entire circumference of this circle. It's the entire circumference of this circle. What's the circumference of a circle in terms of radiuses? If this has length r, if the radius is length r, what's the circumference of the circle in terms of r? We know that. That's going to be 2 pi r. Going back to this angle, the length of the arc that subtends this angle is how many radiuses this is?"}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "What's the circumference of a circle in terms of radiuses? If this has length r, if the radius is length r, what's the circumference of the circle in terms of r? We know that. That's going to be 2 pi r. Going back to this angle, the length of the arc that subtends this angle is how many radiuses this is? What's 2 pi radiuses this is? It's 2 pi times r. This angle right over here, I'll call this a different angle, x. x in this case is going to be 2 pi radians. It is subtended by an arc length of 2 pi radiuses."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "That's going to be 2 pi r. Going back to this angle, the length of the arc that subtends this angle is how many radiuses this is? What's 2 pi radiuses this is? It's 2 pi times r. This angle right over here, I'll call this a different angle, x. x in this case is going to be 2 pi radians. It is subtended by an arc length of 2 pi radiuses. If the radius was one unit, then this would be 2 pi times 1, 2 pi radiuses. Given that, let's start to think about how we can convert between radians and degrees and vice versa. If I were to have, and we can just follow up over here, if we do one full revolution, that is 2 pi radians, how many degrees is this going to be equal to?"}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "It is subtended by an arc length of 2 pi radiuses. If the radius was one unit, then this would be 2 pi times 1, 2 pi radiuses. Given that, let's start to think about how we can convert between radians and degrees and vice versa. If I were to have, and we can just follow up over here, if we do one full revolution, that is 2 pi radians, how many degrees is this going to be equal to? We already know this. A full revolution in degrees is 360 degrees. I could either write out the word degrees or I can use this little degree notation there."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "If I were to have, and we can just follow up over here, if we do one full revolution, that is 2 pi radians, how many degrees is this going to be equal to? We already know this. A full revolution in degrees is 360 degrees. I could either write out the word degrees or I can use this little degree notation there. Actually, let me write out the word degrees. It might make things a little bit clearer that we're using units in both cases. If we wanted to simplify this a little bit, we could divide both sides by 2, in which case we would get on the left-hand side, we would get pi radians would be equal to how many degrees?"}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "I could either write out the word degrees or I can use this little degree notation there. Actually, let me write out the word degrees. It might make things a little bit clearer that we're using units in both cases. If we wanted to simplify this a little bit, we could divide both sides by 2, in which case we would get on the left-hand side, we would get pi radians would be equal to how many degrees? It would be equal to 180 degrees. 180 degrees. I could write it that way or I could write it that way."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "If we wanted to simplify this a little bit, we could divide both sides by 2, in which case we would get on the left-hand side, we would get pi radians would be equal to how many degrees? It would be equal to 180 degrees. 180 degrees. I could write it that way or I could write it that way. You see over here, this is 180 degrees. You also see if you were to draw a circle around here, we've gone halfway around the circle. The arc length or the arc that subtends the angle is half the circumference."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "I could write it that way or I could write it that way. You see over here, this is 180 degrees. You also see if you were to draw a circle around here, we've gone halfway around the circle. The arc length or the arc that subtends the angle is half the circumference. Half the circumference are pi radiuses. We call this pi radians. Pi radians is 180 degrees."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "The arc length or the arc that subtends the angle is half the circumference. Half the circumference are pi radiuses. We call this pi radians. Pi radians is 180 degrees. From this, we can come up with conversions. One radian would be how many degrees? To do that, we would just have to divide both sides by pi."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Pi radians is 180 degrees. From this, we can come up with conversions. One radian would be how many degrees? To do that, we would just have to divide both sides by pi. On the left-hand side, you'd be left with 1. I'll just write it singular now. One radian is equal to, I'm just dividing both sides."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "To do that, we would just have to divide both sides by pi. On the left-hand side, you'd be left with 1. I'll just write it singular now. One radian is equal to, I'm just dividing both sides. Let me make it clear what I'm doing here just to show you. This isn't some voodoo. I'm just dividing both sides by pi here."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "One radian is equal to, I'm just dividing both sides. Let me make it clear what I'm doing here just to show you. This isn't some voodoo. I'm just dividing both sides by pi here. On the left-hand side, you're left with 1. On the right-hand side, you're left with 180 over pi degrees. One radian is equal to 180 over pi degrees, which is starting to make it an interesting way to convert them."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "I'm just dividing both sides by pi here. On the left-hand side, you're left with 1. On the right-hand side, you're left with 180 over pi degrees. One radian is equal to 180 over pi degrees, which is starting to make it an interesting way to convert them. Let's think about it the other way. If I were to have one degree, how many radians is that? Let's start off with, let me rewrite this thing over here."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "One radian is equal to 180 over pi degrees, which is starting to make it an interesting way to convert them. Let's think about it the other way. If I were to have one degree, how many radians is that? Let's start off with, let me rewrite this thing over here. We said pi radians is equal to 180 degrees. Now we want to think about one degree. Let's solve for one degree."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Let's start off with, let me rewrite this thing over here. We said pi radians is equal to 180 degrees. Now we want to think about one degree. Let's solve for one degree. One degree, we can divide both sides by 180. We are left with pi over 180 radians is equal to one degree. Pi over 180 radians is equal to one degree."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Let's solve for one degree. One degree, we can divide both sides by 180. We are left with pi over 180 radians is equal to one degree. Pi over 180 radians is equal to one degree. This might seem confusing and daunting, and it was for me the first time I was exposed to it, especially because we're not exposed to this in our everyday life. What we're going to see over the next few examples is that as long as we keep in mind this whole idea that 2 pi radians is equal to 360 degrees or that pi radians is equal to 180 degrees, which is the two things that I do keep in my mind, we can always re-derive these two things. You might say, hey, how do I remember if it's pi over 180 or 180 over pi to convert the two things?"}, {"video_title": "Inverse trig functions arccos Trigonometry Khan Academy.mp3", "Sentence": "I've already made videos on the arc sine and the arc tangent. So to kind of complete the trifecta, I might as well make a video on the arc cosine. And just like the other inverse trigonometric functions, the arc cosine is kind of the same thought process. If I were to tell you that the arc cosine of x is equal to theta, this is an equivalent statement to saying that the inverse cosine of x is equal to theta. These are just two different ways of writing the exact same thing. And as soon as I see either an arc anything or an inverse trig function in general, my brain immediately rearranges this, my brain immediately says, this is saying that if I take the cosine of some angle theta, that I'm going to get x. Or the same statement up here, either of these should boil down to this."}, {"video_title": "Inverse trig functions arccos Trigonometry Khan Academy.mp3", "Sentence": "If I were to tell you that the arc cosine of x is equal to theta, this is an equivalent statement to saying that the inverse cosine of x is equal to theta. These are just two different ways of writing the exact same thing. And as soon as I see either an arc anything or an inverse trig function in general, my brain immediately rearranges this, my brain immediately says, this is saying that if I take the cosine of some angle theta, that I'm going to get x. Or the same statement up here, either of these should boil down to this. If I say, what is the inverse cosine of x, my brain says, what angle can I take the cosine of to get x? So with that said, let's try it out on an example. Let's say that I have the arc."}, {"video_title": "Inverse trig functions arccos Trigonometry Khan Academy.mp3", "Sentence": "Or the same statement up here, either of these should boil down to this. If I say, what is the inverse cosine of x, my brain says, what angle can I take the cosine of to get x? So with that said, let's try it out on an example. Let's say that I have the arc. I'm told to evaluate the arc cosine of minus 1 half. My brain, let's say that this is going to be equal to some angle, and this is equivalent to saying that the cosine of my mystery angle is equal to minus 1 half. And as soon as you put it in this way, at least for my brain, it becomes a lot easier to process."}, {"video_title": "Inverse trig functions arccos Trigonometry Khan Academy.mp3", "Sentence": "Let's say that I have the arc. I'm told to evaluate the arc cosine of minus 1 half. My brain, let's say that this is going to be equal to some angle, and this is equivalent to saying that the cosine of my mystery angle is equal to minus 1 half. And as soon as you put it in this way, at least for my brain, it becomes a lot easier to process. So let's draw our unit circle and see if we can make some headway here. So that's my, let me see, I could draw a little straighter. Actually, maybe I could actually put rulers here."}, {"video_title": "Inverse trig functions arccos Trigonometry Khan Academy.mp3", "Sentence": "And as soon as you put it in this way, at least for my brain, it becomes a lot easier to process. So let's draw our unit circle and see if we can make some headway here. So that's my, let me see, I could draw a little straighter. Actually, maybe I could actually put rulers here. And if I put a ruler here, maybe I can draw a straight line, let me see. No, that's too hard. OK, so that is my y-axis."}, {"video_title": "Inverse trig functions arccos Trigonometry Khan Academy.mp3", "Sentence": "Actually, maybe I could actually put rulers here. And if I put a ruler here, maybe I can draw a straight line, let me see. No, that's too hard. OK, so that is my y-axis. That is my x-axis, not the most neatly drawn axes ever, but it'll do. And let me draw my unit circle. Looks more like a unit ellipse, but you get the idea."}, {"video_title": "Inverse trig functions arccos Trigonometry Khan Academy.mp3", "Sentence": "OK, so that is my y-axis. That is my x-axis, not the most neatly drawn axes ever, but it'll do. And let me draw my unit circle. Looks more like a unit ellipse, but you get the idea. And the cosine of an angle, it's defined on the unit circle definition, is the x value on the unit circle. So if we have some angle, the x value is going to be equal to minus 1 half. So we go to minus 1 half right here."}, {"video_title": "Inverse trig functions arccos Trigonometry Khan Academy.mp3", "Sentence": "Looks more like a unit ellipse, but you get the idea. And the cosine of an angle, it's defined on the unit circle definition, is the x value on the unit circle. So if we have some angle, the x value is going to be equal to minus 1 half. So we go to minus 1 half right here. And so the angle that we have to solve for, our theta, is the angle that when we intersect the unit circle, the x value is minus 1 half. So let me see, this is the angle that we're trying to figure out. This is theta that we need to determine."}, {"video_title": "Inverse trig functions arccos Trigonometry Khan Academy.mp3", "Sentence": "So we go to minus 1 half right here. And so the angle that we have to solve for, our theta, is the angle that when we intersect the unit circle, the x value is minus 1 half. So let me see, this is the angle that we're trying to figure out. This is theta that we need to determine. So how can we do that? So if this is minus 1 half right here, let's figure out these different angles. And the way I like to think about it, and it's not like to figure out this angle right here."}, {"video_title": "Inverse trig functions arccos Trigonometry Khan Academy.mp3", "Sentence": "This is theta that we need to determine. So how can we do that? So if this is minus 1 half right here, let's figure out these different angles. And the way I like to think about it, and it's not like to figure out this angle right here. And if I know that angle, I can just subtract that from 180 degrees to get this light blue angle that's kind of the solution to our problem. So let me make this triangle a little bit bigger. So that triangle looks something like this, where this distance right here is 1 half."}, {"video_title": "Inverse trig functions arccos Trigonometry Khan Academy.mp3", "Sentence": "And the way I like to think about it, and it's not like to figure out this angle right here. And if I know that angle, I can just subtract that from 180 degrees to get this light blue angle that's kind of the solution to our problem. So let me make this triangle a little bit bigger. So that triangle looks something like this, where this distance right here is 1 half. That distance right there is 1 half. This distance right here is 1. Hopefully you recognize that this is going to be a 30-60-90 triangle."}, {"video_title": "Inverse trig functions arccos Trigonometry Khan Academy.mp3", "Sentence": "So that triangle looks something like this, where this distance right here is 1 half. That distance right there is 1 half. This distance right here is 1. Hopefully you recognize that this is going to be a 30-60-90 triangle. You could actually solve for this other side. You'll get the square root of 3 over 2. And to solve for that other side, you just need to do the Pythagorean theorem."}, {"video_title": "Inverse trig functions arccos Trigonometry Khan Academy.mp3", "Sentence": "Hopefully you recognize that this is going to be a 30-60-90 triangle. You could actually solve for this other side. You'll get the square root of 3 over 2. And to solve for that other side, you just need to do the Pythagorean theorem. Actually, let me just do that. Let me just call this a. So you'd get a squared plus 1 half squared, which is 1 fourth, is equal to 1 squared, which is 1."}, {"video_title": "Inverse trig functions arccos Trigonometry Khan Academy.mp3", "Sentence": "And to solve for that other side, you just need to do the Pythagorean theorem. Actually, let me just do that. Let me just call this a. So you'd get a squared plus 1 half squared, which is 1 fourth, is equal to 1 squared, which is 1. You'd get a squared is equal to 3 fourths, or a is equal to the square root of 3 over 2. So you immediately notice it's a 30-60-90 triangle. And you know that because the sides of a 30-60-90 triangle, if the hypotenuse is 1, are 1 half and square root of 3 over 2."}, {"video_title": "Inverse trig functions arccos Trigonometry Khan Academy.mp3", "Sentence": "So you'd get a squared plus 1 half squared, which is 1 fourth, is equal to 1 squared, which is 1. You'd get a squared is equal to 3 fourths, or a is equal to the square root of 3 over 2. So you immediately notice it's a 30-60-90 triangle. And you know that because the sides of a 30-60-90 triangle, if the hypotenuse is 1, are 1 half and square root of 3 over 2. And you'll also know that the side opposite the square root of 3 over 2 side is the 60 degrees. That's 60. This is 90."}, {"video_title": "Inverse trig functions arccos Trigonometry Khan Academy.mp3", "Sentence": "And you know that because the sides of a 30-60-90 triangle, if the hypotenuse is 1, are 1 half and square root of 3 over 2. And you'll also know that the side opposite the square root of 3 over 2 side is the 60 degrees. That's 60. This is 90. This is the right angle. And this is 30 right up there. But this is the one we care about."}, {"video_title": "Inverse trig functions arccos Trigonometry Khan Academy.mp3", "Sentence": "This is 90. This is the right angle. And this is 30 right up there. But this is the one we care about. This angle right here. We just figured out is 60 degrees. So what's this?"}, {"video_title": "Inverse trig functions arccos Trigonometry Khan Academy.mp3", "Sentence": "But this is the one we care about. This angle right here. We just figured out is 60 degrees. So what's this? What's the bigger angle that we care about? What is 60 degrees supplementary to? It's supplementary to 180 degrees."}, {"video_title": "Inverse trig functions arccos Trigonometry Khan Academy.mp3", "Sentence": "So what's this? What's the bigger angle that we care about? What is 60 degrees supplementary to? It's supplementary to 180 degrees. So the arc cosine, or the inverse cosine, let me write that down, the arc cosine of minus 1 half is equal to 120 degrees. No, it's 180 minus the 60. This whole thing is 180."}, {"video_title": "Inverse trig functions arccos Trigonometry Khan Academy.mp3", "Sentence": "It's supplementary to 180 degrees. So the arc cosine, or the inverse cosine, let me write that down, the arc cosine of minus 1 half is equal to 120 degrees. No, it's 180 minus the 60. This whole thing is 180. So this is right here is 120 degrees. Right? 120 plus 60 is 180."}, {"video_title": "Inverse trig functions arccos Trigonometry Khan Academy.mp3", "Sentence": "This whole thing is 180. So this is right here is 120 degrees. Right? 120 plus 60 is 180. Or if we wanted to write that in radians, you just write 120 degrees times pi radian per 180 degrees. Degrees cancel out. 12 over 18 is 2 thirds."}, {"video_title": "Inverse trig functions arccos Trigonometry Khan Academy.mp3", "Sentence": "120 plus 60 is 180. Or if we wanted to write that in radians, you just write 120 degrees times pi radian per 180 degrees. Degrees cancel out. 12 over 18 is 2 thirds. So it equals 2 pi over 3 radians. So this right here is equal to 2 pi pi over 3 radians. Now, just like we saw in the arc sine and the arc tangent videos, you probably say, hey, OK, if I have 2 pi over 3 radians, that gives me a cosine of minus 1 half."}, {"video_title": "Inverse trig functions arccos Trigonometry Khan Academy.mp3", "Sentence": "12 over 18 is 2 thirds. So it equals 2 pi over 3 radians. So this right here is equal to 2 pi pi over 3 radians. Now, just like we saw in the arc sine and the arc tangent videos, you probably say, hey, OK, if I have 2 pi over 3 radians, that gives me a cosine of minus 1 half. And I could write that. Cosine of 2 pi over 3 is equal to minus 1 half. This gives you the same information as this statement up here."}, {"video_title": "Inverse trig functions arccos Trigonometry Khan Academy.mp3", "Sentence": "Now, just like we saw in the arc sine and the arc tangent videos, you probably say, hey, OK, if I have 2 pi over 3 radians, that gives me a cosine of minus 1 half. And I could write that. Cosine of 2 pi over 3 is equal to minus 1 half. This gives you the same information as this statement up here. But I could just keep going around the unit circle. For example, I could, well, this point over here, cosine of this angle, if I were to go this far, would also be minus 1 half. And then I could go 2 pi around and get back here."}, {"video_title": "Inverse trig functions arccos Trigonometry Khan Academy.mp3", "Sentence": "This gives you the same information as this statement up here. But I could just keep going around the unit circle. For example, I could, well, this point over here, cosine of this angle, if I were to go this far, would also be minus 1 half. And then I could go 2 pi around and get back here. So there's a lot of values that if I take the cosine of those angles, I'll get this minus 1 half. So we have to restrict ourselves. We have to restrict the values that the arc cosine function can take on."}, {"video_title": "Inverse trig functions arccos Trigonometry Khan Academy.mp3", "Sentence": "And then I could go 2 pi around and get back here. So there's a lot of values that if I take the cosine of those angles, I'll get this minus 1 half. So we have to restrict ourselves. We have to restrict the values that the arc cosine function can take on. So we're essentially restricting its range. What we do is we restrict the range to this upper hemisphere, the first and second quadrants. So if we make the statement that the arc cosine of x is equal to theta, we're going to restrict our range, theta, to that top."}, {"video_title": "Inverse trig functions arccos Trigonometry Khan Academy.mp3", "Sentence": "We have to restrict the values that the arc cosine function can take on. So we're essentially restricting its range. What we do is we restrict the range to this upper hemisphere, the first and second quadrants. So if we make the statement that the arc cosine of x is equal to theta, we're going to restrict our range, theta, to that top. So theta is going to be greater than or equal to 0 and less than or equal to pi, where this is also 0 degrees or 180 degrees. We're restricting ourselves to this part of the hemisphere right there. And so you can't do this."}, {"video_title": "Inverse trig functions arccos Trigonometry Khan Academy.mp3", "Sentence": "So if we make the statement that the arc cosine of x is equal to theta, we're going to restrict our range, theta, to that top. So theta is going to be greater than or equal to 0 and less than or equal to pi, where this is also 0 degrees or 180 degrees. We're restricting ourselves to this part of the hemisphere right there. And so you can't do this. This is the only point where the cosine of the angle is equal to minus 1 half. We can't take this angle because it's outside of our range. And what are the valid values for x?"}, {"video_title": "Inverse trig functions arccos Trigonometry Khan Academy.mp3", "Sentence": "And so you can't do this. This is the only point where the cosine of the angle is equal to minus 1 half. We can't take this angle because it's outside of our range. And what are the valid values for x? Well, any angle, if I take the cosine of it, it can be between minus 1 and plus 1. So x, the domain for the arc cosine function, is going to be x has to be less than or equal to 1 and greater than or equal to minus 1. And once again, let's just check our work."}, {"video_title": "Inverse trig functions arccos Trigonometry Khan Academy.mp3", "Sentence": "And what are the valid values for x? Well, any angle, if I take the cosine of it, it can be between minus 1 and plus 1. So x, the domain for the arc cosine function, is going to be x has to be less than or equal to 1 and greater than or equal to minus 1. And once again, let's just check our work. Let's see if the value I got here, that the arc cosine of minus 1 half, really is 2 pi over 3, as calculated by the TI-85. Let me turn it on. So I need to figure out the inverse cosine, which is the same thing as the arc cosine, of minus 1 half, of minus 0.5."}, {"video_title": "Inverse trig functions arccos Trigonometry Khan Academy.mp3", "Sentence": "And once again, let's just check our work. Let's see if the value I got here, that the arc cosine of minus 1 half, really is 2 pi over 3, as calculated by the TI-85. Let me turn it on. So I need to figure out the inverse cosine, which is the same thing as the arc cosine, of minus 1 half, of minus 0.5. It gives me that decimal, that strange number. Let's see if that's the same thing as 2 pi over 3. 2 times pi divided by 3 is equal to that exact same number, so the calculator gave me the same value I got."}, {"video_title": "Inverse trig functions arccos Trigonometry Khan Academy.mp3", "Sentence": "So I need to figure out the inverse cosine, which is the same thing as the arc cosine, of minus 1 half, of minus 0.5. It gives me that decimal, that strange number. Let's see if that's the same thing as 2 pi over 3. 2 times pi divided by 3 is equal to that exact same number, so the calculator gave me the same value I got. But this is kind of a useless, well, it's not a useless number, it's a valid, that is the answer. But it's not a nice, clean answer. I didn't know that this is 2 pi over 3 radians."}, {"video_title": "Inverse trig functions arccos Trigonometry Khan Academy.mp3", "Sentence": "2 times pi divided by 3 is equal to that exact same number, so the calculator gave me the same value I got. But this is kind of a useless, well, it's not a useless number, it's a valid, that is the answer. But it's not a nice, clean answer. I didn't know that this is 2 pi over 3 radians. And so when we did it using the unit circle, we were able to get that answer. So hopefully, and actually, let me ask you, let me just finish this up with an interesting question. And this applies to all of them."}, {"video_title": "Inverse trig functions arccos Trigonometry Khan Academy.mp3", "Sentence": "I didn't know that this is 2 pi over 3 radians. And so when we did it using the unit circle, we were able to get that answer. So hopefully, and actually, let me ask you, let me just finish this up with an interesting question. And this applies to all of them. If I were to ask you, say I were to take the arc cosine of x, and then I were to take the cosine of that, what is this going to be equal to? Well, this statement right here could be said, well, let's say that the arc cosine of x is equal to theta. That means that the cosine of theta is equal to x."}, {"video_title": "Inverse trig functions arccos Trigonometry Khan Academy.mp3", "Sentence": "And this applies to all of them. If I were to ask you, say I were to take the arc cosine of x, and then I were to take the cosine of that, what is this going to be equal to? Well, this statement right here could be said, well, let's say that the arc cosine of x is equal to theta. That means that the cosine of theta is equal to x. So if the arc cosine of x is equal to theta, we can replace this with theta. And then the cosine of theta, well, the cosine of theta is x. So this whole thing is going to be x. Hopefully, I didn't confuse you there."}, {"video_title": "Inverse trig functions arccos Trigonometry Khan Academy.mp3", "Sentence": "That means that the cosine of theta is equal to x. So if the arc cosine of x is equal to theta, we can replace this with theta. And then the cosine of theta, well, the cosine of theta is x. So this whole thing is going to be x. Hopefully, I didn't confuse you there. I'm just saying, look, arc cosine of x, let's just call that theta. Now, by definition, this means that the cosine of theta is equal to x. These are equivalent statements."}, {"video_title": "Inverse trig functions arccos Trigonometry Khan Academy.mp3", "Sentence": "So this whole thing is going to be x. Hopefully, I didn't confuse you there. I'm just saying, look, arc cosine of x, let's just call that theta. Now, by definition, this means that the cosine of theta is equal to x. These are equivalent statements. These are completely equivalent statements right here. So if we put a theta right there, we take the cosine of theta, it has to be equal to x. Now, let me ask you a bonus, slightly trickier question."}, {"video_title": "Inverse trig functions arccos Trigonometry Khan Academy.mp3", "Sentence": "These are equivalent statements. These are completely equivalent statements right here. So if we put a theta right there, we take the cosine of theta, it has to be equal to x. Now, let me ask you a bonus, slightly trickier question. What if I were to ask you, and this is true for any x that you put in here, this is true for any x, any value between negative 1 and 1, including those two endpoints. This is going to be true. Now, what if I were to ask you what the arc cosine of the cosine of theta is?"}, {"video_title": "Inverse trig functions arccos Trigonometry Khan Academy.mp3", "Sentence": "Now, let me ask you a bonus, slightly trickier question. What if I were to ask you, and this is true for any x that you put in here, this is true for any x, any value between negative 1 and 1, including those two endpoints. This is going to be true. Now, what if I were to ask you what the arc cosine of the cosine of theta is? What is this going to be equal to? My answer is, it depends on the theta. So if theta is in the range, if theta is between 0 and pi, so it's in our valid range for the product of the arc cosine, then this will be equal to theta, if this is true for theta."}, {"video_title": "Inverse trig functions arccos Trigonometry Khan Academy.mp3", "Sentence": "Now, what if I were to ask you what the arc cosine of the cosine of theta is? What is this going to be equal to? My answer is, it depends on the theta. So if theta is in the range, if theta is between 0 and pi, so it's in our valid range for the product of the arc cosine, then this will be equal to theta, if this is true for theta. But what if we take some theta out of that range? Let's try it out. So let me do it one with theta in that range."}, {"video_title": "Inverse trig functions arccos Trigonometry Khan Academy.mp3", "Sentence": "So if theta is in the range, if theta is between 0 and pi, so it's in our valid range for the product of the arc cosine, then this will be equal to theta, if this is true for theta. But what if we take some theta out of that range? Let's try it out. So let me do it one with theta in that range. Let's take the arc cosine of the cosine of, let's just do some one of them that we know, let's take the cosine of 2 pi over 3 radians. That's the same thing as the arc cosine of minus 1 half. Cosine of 2 pi over 3 is minus 1 half."}, {"video_title": "Inverse trig functions arccos Trigonometry Khan Academy.mp3", "Sentence": "So let me do it one with theta in that range. Let's take the arc cosine of the cosine of, let's just do some one of them that we know, let's take the cosine of 2 pi over 3 radians. That's the same thing as the arc cosine of minus 1 half. Cosine of 2 pi over 3 is minus 1 half. We just saw that in the earlier part of this video. And then we solved this. We said, oh, this is equal to 2 pi over 3."}, {"video_title": "Inverse trig functions arccos Trigonometry Khan Academy.mp3", "Sentence": "Cosine of 2 pi over 3 is minus 1 half. We just saw that in the earlier part of this video. And then we solved this. We said, oh, this is equal to 2 pi over 3. So if we're in the range, if theta is between 0 and pi, it worked. And that's because the arc cosine function can only produce values between 0 and pi. But what if I were to ask you, what is the arc cosine of 3 pi?"}, {"video_title": "Inverse trig functions arccos Trigonometry Khan Academy.mp3", "Sentence": "We said, oh, this is equal to 2 pi over 3. So if we're in the range, if theta is between 0 and pi, it worked. And that's because the arc cosine function can only produce values between 0 and pi. But what if I were to ask you, what is the arc cosine of 3 pi? So if I were to draw the unit circle here, let me draw the unit circle, a real quick one. And that's my axes. What's 3 pi?"}, {"video_title": "Inverse trig functions arccos Trigonometry Khan Academy.mp3", "Sentence": "But what if I were to ask you, what is the arc cosine of 3 pi? So if I were to draw the unit circle here, let me draw the unit circle, a real quick one. And that's my axes. What's 3 pi? 2 pi is if I go around once, and then I go around another pi. So I end up right here. So I've gone around 1 and 1 half times the unit circle."}, {"video_title": "Inverse trig functions arccos Trigonometry Khan Academy.mp3", "Sentence": "What's 3 pi? 2 pi is if I go around once, and then I go around another pi. So I end up right here. So I've gone around 1 and 1 half times the unit circle. So is it 3 pi? What's the x-coordinate here? It's minus 1."}, {"video_title": "Inverse trig functions arccos Trigonometry Khan Academy.mp3", "Sentence": "So I've gone around 1 and 1 half times the unit circle. So is it 3 pi? What's the x-coordinate here? It's minus 1. So cosine of 3 pi is minus 1. So what's arc cosine of minus 1? Arc cosine of minus 1."}, {"video_title": "Inverse trig functions arccos Trigonometry Khan Academy.mp3", "Sentence": "It's minus 1. So cosine of 3 pi is minus 1. So what's arc cosine of minus 1? Arc cosine of minus 1. Well, remember, the range or the set of values that arc cosine can evaluate to is in this upper hemisphere. This can only be between pi and 0. So arc cosine of negative 1 is just going to be pi."}, {"video_title": "Inverse trig functions arccos Trigonometry Khan Academy.mp3", "Sentence": "Arc cosine of minus 1. Well, remember, the range or the set of values that arc cosine can evaluate to is in this upper hemisphere. This can only be between pi and 0. So arc cosine of negative 1 is just going to be pi. So this is going to be pi. Arc cosine of negative 1 is pi. And that's a reasonable statement, because the difference between 3 pi and pi is just going around the unit circle a couple of times."}, {"video_title": "Inverse trig functions arccos Trigonometry Khan Academy.mp3", "Sentence": "So arc cosine of negative 1 is just going to be pi. So this is going to be pi. Arc cosine of negative 1 is pi. And that's a reasonable statement, because the difference between 3 pi and pi is just going around the unit circle a couple of times. And so you get an equivalent point on the unit circle. So I just thought I would throw those two at you. This one, I mean, this is a useful one."}, {"video_title": "Inverse trig functions arccos Trigonometry Khan Academy.mp3", "Sentence": "And that's a reasonable statement, because the difference between 3 pi and pi is just going around the unit circle a couple of times. And so you get an equivalent point on the unit circle. So I just thought I would throw those two at you. This one, I mean, this is a useful one. Well, actually, let me write it up here. This one is a useful one. The cosine of the arc cosine of x is always going to be x. I could also do that with sine."}, {"video_title": "Inverse trig functions arccos Trigonometry Khan Academy.mp3", "Sentence": "This one, I mean, this is a useful one. Well, actually, let me write it up here. This one is a useful one. The cosine of the arc cosine of x is always going to be x. I could also do that with sine. The sine of the arc sine of x is also going to be x. And these are just useful things to, you shouldn't just memorize them, because obviously you might memorize it the wrong way. But you should just think a little bit about it, and you'll never forget it."}, {"video_title": "Solving triangle in unit circle Trigonometry Khan Academy.mp3", "Sentence": "So the first thing we wanna figure out is what's the measure of angle, what's the radian measure, what's the radian measure, measure of angle ABD, of angle ABD. Actually, let's just do that first, and then I'll talk about the other things that we need to think about. So I assume you've paused the video and tried to do this on your own. So let's think about what ABD would be. We know two of the angles of this triangle, so if we know two of the angles of this triangle, you should be able to figure out the third. Now, it might be a little bit unfamiliar. We're used to saying that the sum of the interior angles of a triangle add up to 180 degrees, but now we're thinking in terms of radians, so we could say that the sum of the angles of a triangle add up to, instead of saying 180 degrees, 180 degrees is the same thing as pi radians."}, {"video_title": "Solving triangle in unit circle Trigonometry Khan Academy.mp3", "Sentence": "So let's think about what ABD would be. We know two of the angles of this triangle, so if we know two of the angles of this triangle, you should be able to figure out the third. Now, it might be a little bit unfamiliar. We're used to saying that the sum of the interior angles of a triangle add up to 180 degrees, but now we're thinking in terms of radians, so we could say that the sum of the angles of a triangle add up to, instead of saying 180 degrees, 180 degrees is the same thing as pi radians. So this angle plus that angle plus that angle are going to add up to pi. So let's just say that this right over here, let's just say measure of angle ABD in radians plus pi over four, plus pi over four, plus, this is a right angle, what would that be in radians? Well, a right angle in radians, a 90 degree angle in radians is pi over two radians."}, {"video_title": "Solving triangle in unit circle Trigonometry Khan Academy.mp3", "Sentence": "We're used to saying that the sum of the interior angles of a triangle add up to 180 degrees, but now we're thinking in terms of radians, so we could say that the sum of the angles of a triangle add up to, instead of saying 180 degrees, 180 degrees is the same thing as pi radians. So this angle plus that angle plus that angle are going to add up to pi. So let's just say that this right over here, let's just say measure of angle ABD in radians plus pi over four, plus pi over four, plus, this is a right angle, what would that be in radians? Well, a right angle in radians, a 90 degree angle in radians is pi over two radians. So plus pi over two. When you take the sum of them, the interior angles of this triangle, they're going to add up to pi radians, which is, of course, the same thing as 180 degrees. And now we can solve for the measure of angle ABD."}, {"video_title": "Solving triangle in unit circle Trigonometry Khan Academy.mp3", "Sentence": "Well, a right angle in radians, a 90 degree angle in radians is pi over two radians. So plus pi over two. When you take the sum of them, the interior angles of this triangle, they're going to add up to pi radians, which is, of course, the same thing as 180 degrees. And now we can solve for the measure of angle ABD. Measure of angle ABD is equal to pi minus pi over two minus pi over four. I just subtracted these two from both sides. So this is going to be equal to, as you could put a common denominator of four, then this is four pi over four."}, {"video_title": "Solving triangle in unit circle Trigonometry Khan Academy.mp3", "Sentence": "And now we can solve for the measure of angle ABD. Measure of angle ABD is equal to pi minus pi over two minus pi over four. I just subtracted these two from both sides. So this is going to be equal to, as you could put a common denominator of four, then this is four pi over four. This is minus two pi over four. And this, of course, is minus pi over four. So four minus two minus one is gonna get us to one."}, {"video_title": "Solving triangle in unit circle Trigonometry Khan Academy.mp3", "Sentence": "So this is going to be equal to, as you could put a common denominator of four, then this is four pi over four. This is minus two pi over four. And this, of course, is minus pi over four. So four minus two minus one is gonna get us to one. So this is going to give us two pi over four. So the measure of angle ABD is actually the same as the measure of angle BAD. It is pi over four."}, {"video_title": "Solving triangle in unit circle Trigonometry Khan Academy.mp3", "Sentence": "So four minus two minus one is gonna get us to one. So this is going to give us two pi over four. So the measure of angle ABD is actually the same as the measure of angle BAD. It is pi over four. So that angle right over there is pi over four. Now what does that help us with? So if we know that this is pi over four, and that is pi over four radians, and once again, we know this is a unit circle."}, {"video_title": "Solving triangle in unit circle Trigonometry Khan Academy.mp3", "Sentence": "It is pi over four. So that angle right over there is pi over four. Now what does that help us with? So if we know that this is pi over four, and that is pi over four radians, and once again, we know this is a unit circle. So we know that the length of segment AB, which is a radius of the circle, or is the radius of the circle, is length one. What else do we know about this triangle? Can we figure out the lengths of segment AD and the length of segment DB?"}, {"video_title": "Solving triangle in unit circle Trigonometry Khan Academy.mp3", "Sentence": "So if we know that this is pi over four, and that is pi over four radians, and once again, we know this is a unit circle. So we know that the length of segment AB, which is a radius of the circle, or is the radius of the circle, is length one. What else do we know about this triangle? Can we figure out the lengths of segment AD and the length of segment DB? Well, sure, because we have two base angles that have the same measure, that means that the corresponding sides are also going to have the same measure. That means that this side is going to be congruent to that side. I can reorient it in a way that might make it a little easier to realize."}, {"video_title": "Solving triangle in unit circle Trigonometry Khan Academy.mp3", "Sentence": "Can we figure out the lengths of segment AD and the length of segment DB? Well, sure, because we have two base angles that have the same measure, that means that the corresponding sides are also going to have the same measure. That means that this side is going to be congruent to that side. I can reorient it in a way that might make it a little easier to realize. If we were to flip it over, if we were to, not completely flip it over, but if we were to make it look like this, so the triangle, we could make it look like, let me make it look a little bit more like this. Actually, I want to make it look like a right angle, though. So my triangle, let me make it look like, there you go."}, {"video_title": "Solving triangle in unit circle Trigonometry Khan Academy.mp3", "Sentence": "I can reorient it in a way that might make it a little easier to realize. If we were to flip it over, if we were to, not completely flip it over, but if we were to make it look like this, so the triangle, we could make it look like, let me make it look a little bit more like this. Actually, I want to make it look like a right angle, though. So my triangle, let me make it look like, there you go. So if this is D, if this is D, this is B, this is A, this is our right angle. Now, this is pi over four radians, and this is also pi over four radians. When your two base angles are the same, you know you're dealing with an isosceles triangle, an isosceles, but because they're not all the same, you know it's not equilateral."}, {"video_title": "Solving triangle in unit circle Trigonometry Khan Academy.mp3", "Sentence": "So my triangle, let me make it look like, there you go. So if this is D, if this is D, this is B, this is A, this is our right angle. Now, this is pi over four radians, and this is also pi over four radians. When your two base angles are the same, you know you're dealing with an isosceles triangle, an isosceles, but because they're not all the same, you know it's not equilateral. If all of the angles were the same, this would be equilateral, but this is an isosceles non-equilateral triangle. So if your base angles are the same, then you also know that the corresponding sides are going to be the same. These two sides are the same."}, {"video_title": "Solving triangle in unit circle Trigonometry Khan Academy.mp3", "Sentence": "When your two base angles are the same, you know you're dealing with an isosceles triangle, an isosceles, but because they're not all the same, you know it's not equilateral. If all of the angles were the same, this would be equilateral, but this is an isosceles non-equilateral triangle. So if your base angles are the same, then you also know that the corresponding sides are going to be the same. These two sides are the same. This is an isosceles triangle. And so how does that help us figure out the lengths of the sides? Well, if you say that this side has length x, that means that this side has length x."}, {"video_title": "Solving triangle in unit circle Trigonometry Khan Academy.mp3", "Sentence": "These two sides are the same. This is an isosceles triangle. And so how does that help us figure out the lengths of the sides? Well, if you say that this side has length x, that means that this side has length x. If this side has length x, then this side has length x. And now we can use the Pythagorean theorem. We could say that x squared, this x squared plus this x squared, is equal to the hypotenuse squared, is equal to one squared."}, {"video_title": "Solving triangle in unit circle Trigonometry Khan Academy.mp3", "Sentence": "Well, if you say that this side has length x, that means that this side has length x. If this side has length x, then this side has length x. And now we can use the Pythagorean theorem. We could say that x squared, this x squared plus this x squared, is equal to the hypotenuse squared, is equal to one squared. Or we could write that two x squared is equal to one, or that x squared is equal to one over two, or just taking the principal root of both sides, we get x is equal to one over the square root of two. And a lot of folks don't like having a radical in the denominator. They don't like having a rational number in the denominator."}, {"video_title": "Solving triangle in unit circle Trigonometry Khan Academy.mp3", "Sentence": "We could say that x squared, this x squared plus this x squared, is equal to the hypotenuse squared, is equal to one squared. Or we could write that two x squared is equal to one, or that x squared is equal to one over two, or just taking the principal root of both sides, we get x is equal to one over the square root of two. And a lot of folks don't like having a radical in the denominator. They don't like having a rational number in the denominator. So we could rationalize the denominator by multiplying by the square root of two over the square root of two, which would be, you'll see the numerator will have the square root of two, and in the denominator, we're just going to have square root of two times square root of two is just two. So we've already been able to figure out several interesting things. We were able to figure out a measure of angle ABD in radians, we were able to figure out the lengths of segment AD and the length of segment BD."}, {"video_title": "Solving triangle in unit circle Trigonometry Khan Academy.mp3", "Sentence": "They don't like having a rational number in the denominator. So we could rationalize the denominator by multiplying by the square root of two over the square root of two, which would be, you'll see the numerator will have the square root of two, and in the denominator, we're just going to have square root of two times square root of two is just two. So we've already been able to figure out several interesting things. We were able to figure out a measure of angle ABD in radians, we were able to figure out the lengths of segment AD and the length of segment BD. Now what I want to do is figure out what are the sine, cosine, and tangent of pi over four radians, given all of the work that we have done. So let's first think about what is the sine. Let me do this in a color, do it in orange."}, {"video_title": "Solving triangle in unit circle Trigonometry Khan Academy.mp3", "Sentence": "We were able to figure out a measure of angle ABD in radians, we were able to figure out the lengths of segment AD and the length of segment BD. Now what I want to do is figure out what are the sine, cosine, and tangent of pi over four radians, given all of the work that we have done. So let's first think about what is the sine. Let me do this in a color, do it in orange. Given all the work we've done, what is the sine of pi over four radians? And I encourage you, once again, pause the video, think about the unit circle definition of trig functions, and think about what this is. Well, the unit circle definition of trig functions, and this angle, this pi over four radians, is forming an angle with a positive x-axis, and where its terminal ray intersects the unit circle, the x and y coordinates of this point are what specify the cosine and sine."}, {"video_title": "Solving triangle in unit circle Trigonometry Khan Academy.mp3", "Sentence": "Let me do this in a color, do it in orange. Given all the work we've done, what is the sine of pi over four radians? And I encourage you, once again, pause the video, think about the unit circle definition of trig functions, and think about what this is. Well, the unit circle definition of trig functions, and this angle, this pi over four radians, is forming an angle with a positive x-axis, and where its terminal ray intersects the unit circle, the x and y coordinates of this point are what specify the cosine and sine. So the coordinates of this point are going to be the cosine of pi over four radians is the x coordinate, and the sine of pi over four radians is the y coordinate, sine of pi over four. So what's the y coordinate going to be if we want the sine of pi over four? Well, it's going to be this length right over here, which is the same thing as this length, which is the length of x, which is square root of two over two."}, {"video_title": "Solving triangle in unit circle Trigonometry Khan Academy.mp3", "Sentence": "Well, the unit circle definition of trig functions, and this angle, this pi over four radians, is forming an angle with a positive x-axis, and where its terminal ray intersects the unit circle, the x and y coordinates of this point are what specify the cosine and sine. So the coordinates of this point are going to be the cosine of pi over four radians is the x coordinate, and the sine of pi over four radians is the y coordinate, sine of pi over four. So what's the y coordinate going to be if we want the sine of pi over four? Well, it's going to be this length right over here, which is the same thing as this length, which is the length of x, which is square root of two over two. Now what is the cosine of pi over four? And once again, I encourage you to pause the video, think about it. What's this x coordinate?"}, {"video_title": "Solving triangle in unit circle Trigonometry Khan Academy.mp3", "Sentence": "Well, it's going to be this length right over here, which is the same thing as this length, which is the length of x, which is square root of two over two. Now what is the cosine of pi over four? And once again, I encourage you to pause the video, think about it. What's this x coordinate? What's the x coordinate? The x coordinate is this distance right over here, which is once again going to be x, which is square root of two over two. Now, what is the tangent of pi over four going to be?"}, {"video_title": "Solving triangle in unit circle Trigonometry Khan Academy.mp3", "Sentence": "What's this x coordinate? What's the x coordinate? The x coordinate is this distance right over here, which is once again going to be x, which is square root of two over two. Now, what is the tangent of pi over four going to be? Well, the tangent of pi over four is just sine of pi over four over cosine of pi over four. Now, both of these are this exact same thing. They're both square root of two over two, so you're gonna have square root of two over two divided by square root of two over two."}, {"video_title": "Solving triangle in unit circle Trigonometry Khan Academy.mp3", "Sentence": "Now, what is the tangent of pi over four going to be? Well, the tangent of pi over four is just sine of pi over four over cosine of pi over four. Now, both of these are this exact same thing. They're both square root of two over two, so you're gonna have square root of two over two divided by square root of two over two. Well, that's just going to give you one. And that also makes sense, because remember, the tangent of this angle is the slope of this line. And we see the slope."}, {"video_title": "Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "What I want to attempt to do in this video is figure out what the sine of 7 pi over 12 is without using a calculator. And so let's just visualize 7 pi over 12 on the unit circle. So one side of the angle is going along the positive x-axis and let's see, if we go straight up, that's pi over 2, which is the same thing as 6 pi over 12, so then we essentially just have another pi over 12 to get right over there. This is the angle that we're talking about, that is 7 pi over 12 radians. And the sine of it, by the unit circle definition of sine, it's the y-coordinate of where this ray intersects the unit circle. So this is a unit circle, has radius 1. Where it intersects the y-coordinate is the sine."}, {"video_title": "Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "This is the angle that we're talking about, that is 7 pi over 12 radians. And the sine of it, by the unit circle definition of sine, it's the y-coordinate of where this ray intersects the unit circle. So this is a unit circle, has radius 1. Where it intersects the y-coordinate is the sine. So another way to think about it, it's the length of this line right over here. And I encourage you to pause the video right now and try to think about it on your own. See if you can use your powers of trigonometry to figure out what sine of 7 pi over 12 is, or essentially the length of this magenta line."}, {"video_title": "Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Where it intersects the y-coordinate is the sine. So another way to think about it, it's the length of this line right over here. And I encourage you to pause the video right now and try to think about it on your own. See if you can use your powers of trigonometry to figure out what sine of 7 pi over 12 is, or essentially the length of this magenta line. So I'm assuming you've given a go at it. And if you're like me, your first temptation might have been just to focus on this triangle right over here, that I kind of drew for you. So the triangle looks like this."}, {"video_title": "Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "See if you can use your powers of trigonometry to figure out what sine of 7 pi over 12 is, or essentially the length of this magenta line. So I'm assuming you've given a go at it. And if you're like me, your first temptation might have been just to focus on this triangle right over here, that I kind of drew for you. So the triangle looks like this. It looks like this, where that's what you're trying to figure out. This length right over here, sine of 7 pi over 12. We know the length of the hypotenuse is 1."}, {"video_title": "Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So the triangle looks like this. It looks like this, where that's what you're trying to figure out. This length right over here, sine of 7 pi over 12. We know the length of the hypotenuse is 1. It's a radius of the unit circle. It's a right triangle right over there. And we also know this angle right over here, which is this angle right over here."}, {"video_title": "Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "We know the length of the hypotenuse is 1. It's a radius of the unit circle. It's a right triangle right over there. And we also know this angle right over here, which is this angle right over here. This gets a 6 pi over 12, and then we have another pi over 12. So we know that that is pi over 12, not pi over 16. We know that this angle right over here is pi over 12."}, {"video_title": "Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And we also know this angle right over here, which is this angle right over here. This gets a 6 pi over 12, and then we have another pi over 12. So we know that that is pi over 12, not pi over 16. We know that this angle right over here is pi over 12. And so given this information, we can figure out this, or we can at least relate this side to these other sides using a trig function. Relative to this angle, this is the adjacent side. And so the cosine of pi over 12 is going to be this magenta side over 1."}, {"video_title": "Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "We know that this angle right over here is pi over 12. And so given this information, we can figure out this, or we can at least relate this side to these other sides using a trig function. Relative to this angle, this is the adjacent side. And so the cosine of pi over 12 is going to be this magenta side over 1. Or you could just say it's equal to this magenta side. So you could say this is cosine of pi over 12. So we just figured out that sine of 7 pi over 12 is the same thing as cosine of pi over 12."}, {"video_title": "Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And so the cosine of pi over 12 is going to be this magenta side over 1. Or you could just say it's equal to this magenta side. So you could say this is cosine of pi over 12. So we just figured out that sine of 7 pi over 12 is the same thing as cosine of pi over 12. But that still doesn't help me. I don't know offhand what the cosine of pi over 12 radians is without using a calculator. So instead of thinking about it this way, let's see if we can decompose it into some angles for which we do know the sine and cosine."}, {"video_title": "Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So we just figured out that sine of 7 pi over 12 is the same thing as cosine of pi over 12. But that still doesn't help me. I don't know offhand what the cosine of pi over 12 radians is without using a calculator. So instead of thinking about it this way, let's see if we can decompose it into some angles for which we do know the sine and cosine. And what angles are those? Well, those are the angles in special right triangles. So for example, we are very familiar with 30-60-90 triangles."}, {"video_title": "Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So instead of thinking about it this way, let's see if we can decompose it into some angles for which we do know the sine and cosine. And what angles are those? Well, those are the angles in special right triangles. So for example, we are very familiar with 30-60-90 triangles. 30-60-90 triangles look something like this. And this is my best attempt at hand drawing it. So instead of writing the 30-degree side, since we're thinking in radians, I'll write that as pi over 6 radians."}, {"video_title": "Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So for example, we are very familiar with 30-60-90 triangles. 30-60-90 triangles look something like this. And this is my best attempt at hand drawing it. So instead of writing the 30-degree side, since we're thinking in radians, I'll write that as pi over 6 radians. The 60-degree side, I'm going to write that as pi over 3 radians. And of course, this is the right angle. And if the hypotenuse here is 1, then the side opposite the 30-degree side, or the pi over 6 radian side, is going to be half the hypotenuse, which in this case is 1 half."}, {"video_title": "Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So instead of writing the 30-degree side, since we're thinking in radians, I'll write that as pi over 6 radians. The 60-degree side, I'm going to write that as pi over 3 radians. And of course, this is the right angle. And if the hypotenuse here is 1, then the side opposite the 30-degree side, or the pi over 6 radian side, is going to be half the hypotenuse, which in this case is 1 half. And then the other side that's opposite the 60-degree side, or the pi over 3 radian side, is going to be square root of 3 times the shorter side. So it's going to be square root of 3 over 2. And so we've used these types of triangles in the past to figure out the sine or cosine of 30 or 60, or in this case, pi over 6 or pi over 3."}, {"video_title": "Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And if the hypotenuse here is 1, then the side opposite the 30-degree side, or the pi over 6 radian side, is going to be half the hypotenuse, which in this case is 1 half. And then the other side that's opposite the 60-degree side, or the pi over 3 radian side, is going to be square root of 3 times the shorter side. So it's going to be square root of 3 over 2. And so we've used these types of triangles in the past to figure out the sine or cosine of 30 or 60, or in this case, pi over 6 or pi over 3. So we know about pi over 6 and pi over 3. We also know about 45-45-90 triangles. We know that they're isosceles right triangles."}, {"video_title": "Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And so we've used these types of triangles in the past to figure out the sine or cosine of 30 or 60, or in this case, pi over 6 or pi over 3. So we know about pi over 6 and pi over 3. We also know about 45-45-90 triangles. We know that they're isosceles right triangles. They look like this. My best attempt at drawing it. That one actually doesn't look that isosceles."}, {"video_title": "Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "We know that they're isosceles right triangles. They look like this. My best attempt at drawing it. That one actually doesn't look that isosceles. So let me make it a little bit more. So I don't know. That looks closer to being an isosceles right triangle."}, {"video_title": "Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "That one actually doesn't look that isosceles. So let me make it a little bit more. So I don't know. That looks closer to being an isosceles right triangle. And we know if the length of the hypotenuse is 1, and this comes straight out of Pythagorean theorem, then the length of each of the other two sides are going to be square root of 2 over 2 times the hypotenuse, which in this case is the square root of 2 over 2. Instead of describing these as 45-degree angles, we know that's the same thing as pi over 4 radians. And so if you give me pi over 6, pi over 3, pi over 4, I can use these triangles either using the classic definition, Sohcahtoa definitions, or I could stick them on the unit circle here to use the unit circle definition of trig functions to figure out what the sine, cosine, or tangent of these angles are."}, {"video_title": "Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "That looks closer to being an isosceles right triangle. And we know if the length of the hypotenuse is 1, and this comes straight out of Pythagorean theorem, then the length of each of the other two sides are going to be square root of 2 over 2 times the hypotenuse, which in this case is the square root of 2 over 2. Instead of describing these as 45-degree angles, we know that's the same thing as pi over 4 radians. And so if you give me pi over 6, pi over 3, pi over 4, I can use these triangles either using the classic definition, Sohcahtoa definitions, or I could stick them on the unit circle here to use the unit circle definition of trig functions to figure out what the sine, cosine, or tangent of these angles are. So can I decompose 7 pi over 12 into some combination of pi over 6's, pi over 3's, or pi over 4's? Well, to think about that, let me rewrite pi over 6, pi over 3, and pi over 4 with a denominator over 12. So let me write that."}, {"video_title": "Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And so if you give me pi over 6, pi over 3, pi over 4, I can use these triangles either using the classic definition, Sohcahtoa definitions, or I could stick them on the unit circle here to use the unit circle definition of trig functions to figure out what the sine, cosine, or tangent of these angles are. So can I decompose 7 pi over 12 into some combination of pi over 6's, pi over 3's, or pi over 4's? Well, to think about that, let me rewrite pi over 6, pi over 3, and pi over 4 with a denominator over 12. So let me write that. So pi over 6 is equal to 2 pi over 12. Pi over 3 is equal to 4 pi over 12. And pi over 4 is equal to 3 pi over 12."}, {"video_title": "Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So let me write that. So pi over 6 is equal to 2 pi over 12. Pi over 3 is equal to 4 pi over 12. And pi over 4 is equal to 3 pi over 12. So let's see. 2 plus 4 is not 7. 2 plus 3 is not 7."}, {"video_title": "Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And pi over 4 is equal to 3 pi over 12. So let's see. 2 plus 4 is not 7. 2 plus 3 is not 7. But 4 plus 3 is 7. So I could use this and this. 4 pi over 12 plus 3 pi over 12 is 7 pi over 12."}, {"video_title": "Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "2 plus 3 is not 7. But 4 plus 3 is 7. So I could use this and this. 4 pi over 12 plus 3 pi over 12 is 7 pi over 12. So I could rewrite this. This is the same thing as sine of 3 pi over 12 plus 4 pi over 12, which, of course, is the same thing. Sine of pi over 4."}, {"video_title": "Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "4 pi over 12 plus 3 pi over 12 is 7 pi over 12. So I could rewrite this. This is the same thing as sine of 3 pi over 12 plus 4 pi over 12, which, of course, is the same thing. Sine of pi over 4. I'll do this in another color. Sine of pi over 4 plus pi over 3. And now we can use our angle addition formula for sine in order to write this as the sum of products of cosines and sines of these angles."}, {"video_title": "Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Sine of pi over 4. I'll do this in another color. Sine of pi over 4 plus pi over 3. And now we can use our angle addition formula for sine in order to write this as the sum of products of cosines and sines of these angles. So let's actually do that. So this right over here is going to be equal to the sine of pi over 4 times the cosine of pi over 3 plus the other way around, cosine of pi over 4 times the sine of pi over 3. So now we just have to figure out these things."}, {"video_title": "Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And now we can use our angle addition formula for sine in order to write this as the sum of products of cosines and sines of these angles. So let's actually do that. So this right over here is going to be equal to the sine of pi over 4 times the cosine of pi over 3 plus the other way around, cosine of pi over 4 times the sine of pi over 3. So now we just have to figure out these things. And I've already set up the triangles to do it. What is sine of pi over 4? Well, let's think about this is pi over 4 right over here."}, {"video_title": "Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So now we just have to figure out these things. And I've already set up the triangles to do it. What is sine of pi over 4? Well, let's think about this is pi over 4 right over here. Sine is opposite over hypotenuse. Well, it's just going to be square root of 2 over 2. So this is square root of 2 over 2."}, {"video_title": "Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Well, let's think about this is pi over 4 right over here. Sine is opposite over hypotenuse. Well, it's just going to be square root of 2 over 2. So this is square root of 2 over 2. What is cosine of pi over 3? Well, this is a pi over 3 radian angle right over here. Cosine is adjacent over hypotenuse."}, {"video_title": "Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So this is square root of 2 over 2. What is cosine of pi over 3? Well, this is a pi over 3 radian angle right over here. Cosine is adjacent over hypotenuse. So this is going to be 1 half. What is cosine of pi over 4? Well, go back to pi over 4."}, {"video_title": "Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Cosine is adjacent over hypotenuse. So this is going to be 1 half. What is cosine of pi over 4? Well, go back to pi over 4. It's adjacent over hypotenuse. It's square root of 2 over 2. It is also square root of 2 over 2."}, {"video_title": "Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Well, go back to pi over 4. It's adjacent over hypotenuse. It's square root of 2 over 2. It is also square root of 2 over 2. And what's sine of pi over 3? Well, sine is opposite over hypotenuse. So it's square root of 3 over 2 over 1."}, {"video_title": "Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "It is also square root of 2 over 2. And what's sine of pi over 3? Well, sine is opposite over hypotenuse. So it's square root of 3 over 2 over 1. Square root of 3 over 2 divided by 1, which is square root of 3 over 2. And so now we just have to simplify all of this business. So this is going to be equal to the sum of this, or the product, I should say, is just square root of 2 over 4."}, {"video_title": "Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So it's square root of 3 over 2 over 1. Square root of 3 over 2 divided by 1, which is square root of 3 over 2. And so now we just have to simplify all of this business. So this is going to be equal to the sum of this, or the product, I should say, is just square root of 2 over 4. And then plus the product of these. Let's see. We could rewrite that as square root of 6 over 4."}, {"video_title": "Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So this is going to be equal to the sum of this, or the product, I should say, is just square root of 2 over 4. And then plus the product of these. Let's see. We could rewrite that as square root of 6 over 4. Or we could just rewrite this whole thing as we deserve a little bit of a drum roll at this point. This is equivalent to, let me scroll over to the right a little bit. This is equivalent to square root of 2 plus square root of 6."}, {"video_title": "Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy.mp3", "Sentence": "Determine the six trigonometric ratios for angle A in the right triangle below. So this right over here is angle A, it's at vertex A. And to help me remember the definitions of the trig ratios, and these are human constructed definitions that have ended up being very, very useful for analyzing a whole series of things in the world. But to help me remember them, I use the word SOH CAH TOA. So let me write that down. So SOH CAH TOA. Sometimes you can think of it as one word, but it's really the three parts that define at least three of the trig functions for you."}, {"video_title": "Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy.mp3", "Sentence": "But to help me remember them, I use the word SOH CAH TOA. So let me write that down. So SOH CAH TOA. Sometimes you can think of it as one word, but it's really the three parts that define at least three of the trig functions for you. And then we can get the other three by looking at the first three. So SOH tells us that sine of an angle, and in this case it's sine of A, so sine of A is equal to the opposite, opposite, that's the O, over the hypotenuse. Opposite over the hypotenuse."}, {"video_title": "Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy.mp3", "Sentence": "Sometimes you can think of it as one word, but it's really the three parts that define at least three of the trig functions for you. And then we can get the other three by looking at the first three. So SOH tells us that sine of an angle, and in this case it's sine of A, so sine of A is equal to the opposite, opposite, that's the O, over the hypotenuse. Opposite over the hypotenuse. Well in this context, what is the opposite side to angle A? Well we go across the triangle, it opens up onto side BC, it has length 12, so that is the opposite side. So this is going to be equal to 12."}, {"video_title": "Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy.mp3", "Sentence": "Opposite over the hypotenuse. Well in this context, what is the opposite side to angle A? Well we go across the triangle, it opens up onto side BC, it has length 12, so that is the opposite side. So this is going to be equal to 12. And what's the hypotenuse? Well the hypotenuse is the longest side of the triangle. It's opposite the 90 degree angle, and so we go opposite the 90 degree angle, longest side is side AB, it has length 13."}, {"video_title": "Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy.mp3", "Sentence": "So this is going to be equal to 12. And what's the hypotenuse? Well the hypotenuse is the longest side of the triangle. It's opposite the 90 degree angle, and so we go opposite the 90 degree angle, longest side is side AB, it has length 13. So this right over here is the hypotenuse. So the sine of A is 12 thirteens. Now let's go to CAH."}, {"video_title": "Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy.mp3", "Sentence": "It's opposite the 90 degree angle, and so we go opposite the 90 degree angle, longest side is side AB, it has length 13. So this right over here is the hypotenuse. So the sine of A is 12 thirteens. Now let's go to CAH. CAH defines cosine for us. It tells us that cosine of an angle, in this case cosine of A, is equal to the adjacent side, the adjacent side to the angle, over the hypotenuse. Over the hypotenuse."}, {"video_title": "Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy.mp3", "Sentence": "Now let's go to CAH. CAH defines cosine for us. It tells us that cosine of an angle, in this case cosine of A, is equal to the adjacent side, the adjacent side to the angle, over the hypotenuse. Over the hypotenuse. So what's the adjacent side to angle A? Well if we look at angle A, there's two sides that are next to it. One of them is the hypotenuse."}, {"video_title": "Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy.mp3", "Sentence": "Over the hypotenuse. So what's the adjacent side to angle A? Well if we look at angle A, there's two sides that are next to it. One of them is the hypotenuse. The other one has length five. The adjacent one is side CA. So it's five."}, {"video_title": "Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy.mp3", "Sentence": "One of them is the hypotenuse. The other one has length five. The adjacent one is side CA. So it's five. And what is the hypotenuse? Well we've already figured that out. The hypotenuse right over here, it's opposite the 90 degree angle, it's the longest side of the right triangle, it has length 13."}, {"video_title": "Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy.mp3", "Sentence": "So it's five. And what is the hypotenuse? Well we've already figured that out. The hypotenuse right over here, it's opposite the 90 degree angle, it's the longest side of the right triangle, it has length 13. So the cosine of A is 5 thirteens. And let me label this. This right over here is the adjacent side."}, {"video_title": "Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy.mp3", "Sentence": "The hypotenuse right over here, it's opposite the 90 degree angle, it's the longest side of the right triangle, it has length 13. So the cosine of A is 5 thirteens. And let me label this. This right over here is the adjacent side. And this is all specific to angle A. The hypotenuse would be the same regardless of what angle you pick. But the opposite and the adjacent is dependent on the angle that we choose in the right triangle."}, {"video_title": "Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy.mp3", "Sentence": "This right over here is the adjacent side. And this is all specific to angle A. The hypotenuse would be the same regardless of what angle you pick. But the opposite and the adjacent is dependent on the angle that we choose in the right triangle. Now let's go to TOA. TOA defines tangent for us. It tells us that the tangent, the tangent of an angle, is equal to the opposite, equal to the opposite side, over the adjacent side."}, {"video_title": "Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy.mp3", "Sentence": "But the opposite and the adjacent is dependent on the angle that we choose in the right triangle. Now let's go to TOA. TOA defines tangent for us. It tells us that the tangent, the tangent of an angle, is equal to the opposite, equal to the opposite side, over the adjacent side. So given this definition, what is the tangent of A? Well, the opposite side, we already figured out, has length 12, has length 12. And the adjacent side, we already figured out, has length five."}, {"video_title": "Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy.mp3", "Sentence": "It tells us that the tangent, the tangent of an angle, is equal to the opposite, equal to the opposite side, over the adjacent side. So given this definition, what is the tangent of A? Well, the opposite side, we already figured out, has length 12, has length 12. And the adjacent side, we already figured out, has length five. So the tangent of A, which is opposite over adjacent, is 12 fifths. Now we'll go to the other three trig ratios, which you could think of as the reciprocals of these right over here. But I'll define it."}, {"video_title": "Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy.mp3", "Sentence": "And the adjacent side, we already figured out, has length five. So the tangent of A, which is opposite over adjacent, is 12 fifths. Now we'll go to the other three trig ratios, which you could think of as the reciprocals of these right over here. But I'll define it. So first you have cosecant. And cosecant, it's always a little bit unintuitive why cosecant is the reciprocal of sine of A, even though it starts with a co, like cosine. But cosecant is the reciprocal of the sine of A."}, {"video_title": "Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy.mp3", "Sentence": "But I'll define it. So first you have cosecant. And cosecant, it's always a little bit unintuitive why cosecant is the reciprocal of sine of A, even though it starts with a co, like cosine. But cosecant is the reciprocal of the sine of A. So sine of A is opposite over hypotenuse. Cosecant of A is hypotenuse over opposite. And so what's the hypotenuse over the opposite?"}, {"video_title": "Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy.mp3", "Sentence": "But cosecant is the reciprocal of the sine of A. So sine of A is opposite over hypotenuse. Cosecant of A is hypotenuse over opposite. And so what's the hypotenuse over the opposite? Well, the hypotenuse is 13, and the opposite side is 12. And notice, 13 twelfths is the reciprocal of 12 thirteens. Now, secant of A is the reciprocal, so instead of being adjacent over hypotenuse, which we got from the co part of SOH CAH TOA, it's hypotenuse over adjacent."}, {"video_title": "Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy.mp3", "Sentence": "And so what's the hypotenuse over the opposite? Well, the hypotenuse is 13, and the opposite side is 12. And notice, 13 twelfths is the reciprocal of 12 thirteens. Now, secant of A is the reciprocal, so instead of being adjacent over hypotenuse, which we got from the co part of SOH CAH TOA, it's hypotenuse over adjacent. So what is the secant of A? Well, the hypotenuse, we've figured out multiple times already, is 13. And what is the adjacent side?"}, {"video_title": "Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy.mp3", "Sentence": "Now, secant of A is the reciprocal, so instead of being adjacent over hypotenuse, which we got from the co part of SOH CAH TOA, it's hypotenuse over adjacent. So what is the secant of A? Well, the hypotenuse, we've figured out multiple times already, is 13. And what is the adjacent side? It's five. So it's 13 fifths, which is once again the reciprocal of the cosine of A, five thirteens. Finally, let's get the cotangent."}, {"video_title": "Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy.mp3", "Sentence": "And what is the adjacent side? It's five. So it's 13 fifths, which is once again the reciprocal of the cosine of A, five thirteens. Finally, let's get the cotangent. And the cotangent is the reciprocal of tangent of A. Instead of being opposite over adjacent, it is adjacent over opposite. So what is the cotangent of A?"}, {"video_title": "Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy.mp3", "Sentence": "Finally, let's get the cotangent. And the cotangent is the reciprocal of tangent of A. Instead of being opposite over adjacent, it is adjacent over opposite. So what is the cotangent of A? Well, we figured out the adjacent side multiple times for angle A. It's length five. And the opposite side to angle A is 12."}, {"video_title": "How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "An expedition was sent to find how high the water had risen. The people measured the edge of the pyramid that's above the water and found it was 72 meters long. So this distance right over here is 72 meters. They knew that the entire length of the edge is 180 meters when it's not flooded, so this entire length is 180 meters. They also knew that the vertical height of the pyramid is 139 meters, so this is 139 meters. What is the level of the water above the ground? So the ground is right over here at the base of the pyramid, and so they want the level of the water above the ground."}, {"video_title": "How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "They knew that the entire length of the edge is 180 meters when it's not flooded, so this entire length is 180 meters. They also knew that the vertical height of the pyramid is 139 meters, so this is 139 meters. What is the level of the water above the ground? So the ground is right over here at the base of the pyramid, and so they want the level of the water above the ground. So that's this height right over here, so let's just call that h. We want to figure out what h is. Round your answer, if necessary, to two decimal places. So what do we know and what do we not know?"}, {"video_title": "How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So the ground is right over here at the base of the pyramid, and so they want the level of the water above the ground. So that's this height right over here, so let's just call that h. We want to figure out what h is. Round your answer, if necessary, to two decimal places. So what do we know and what do we not know? So they've labeled this little angle here theta, and this, of course, is a right angle. So this angle here at the base of the pyramid, this is going to be the complement of theta. It's going to be 90 degrees minus theta."}, {"video_title": "How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So what do we know and what do we not know? So they've labeled this little angle here theta, and this, of course, is a right angle. So this angle here at the base of the pyramid, this is going to be the complement of theta. It's going to be 90 degrees minus theta. And using that information, we can also figure out that this angle up here is also going to be theta. If that looks a little bit strange to you, let me just draw it out here and make it a little bit clearer. If we have a right triangle where this angle right over here is 90 minus theta, and we wanted to figure out what this is up here, let's say this is x, well, we could say x plus 90 minus theta plus 90 degrees is going to be equal to, well, the sum of the angles of a triangle are going to be 180 degrees."}, {"video_title": "How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "It's going to be 90 degrees minus theta. And using that information, we can also figure out that this angle up here is also going to be theta. If that looks a little bit strange to you, let me just draw it out here and make it a little bit clearer. If we have a right triangle where this angle right over here is 90 minus theta, and we wanted to figure out what this is up here, let's say this is x, well, we could say x plus 90 minus theta plus 90 degrees is going to be equal to, well, the sum of the angles of a triangle are going to be 180 degrees. Well, if we subtract 180 from both sides, so that's 180 from the left, 180 from the right, we get x minus theta is equal to 0, or if you add theta to both sides, x is equal to theta. So this thing up here is going to be theta as well. So this is also going to be theta."}, {"video_title": "How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "If we have a right triangle where this angle right over here is 90 minus theta, and we wanted to figure out what this is up here, let's say this is x, well, we could say x plus 90 minus theta plus 90 degrees is going to be equal to, well, the sum of the angles of a triangle are going to be 180 degrees. Well, if we subtract 180 from both sides, so that's 180 from the left, 180 from the right, we get x minus theta is equal to 0, or if you add theta to both sides, x is equal to theta. So this thing up here is going to be theta as well. So this is also going to be theta. And what else do we know? Well, we know that this is 72. We know that the whole thing is 180."}, {"video_title": "How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So this is also going to be theta. And what else do we know? Well, we know that this is 72. We know that the whole thing is 180. So this is 72 and the whole thing is 180. The part of this edge that's below the water, this distance right over here, let me draw it without cluttering the picture too much, doing the black color, this distance right over here is going to be 108. 108 plus 72 is 180."}, {"video_title": "How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "We know that the whole thing is 180. So this is 72 and the whole thing is 180. The part of this edge that's below the water, this distance right over here, let me draw it without cluttering the picture too much, doing the black color, this distance right over here is going to be 108. 108 plus 72 is 180. So what does this do for us? We need to figure out this height. We know that this right over here is a right triangle."}, {"video_title": "How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "108 plus 72 is 180. So what does this do for us? We need to figure out this height. We know that this right over here is a right triangle. This right over here is a right triangle. I could color this in just to make it a little bit clearer. This thing in yellow right over here is a right triangle."}, {"video_title": "How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "We know that this right over here is a right triangle. This right over here is a right triangle. I could color this in just to make it a little bit clearer. This thing in yellow right over here is a right triangle. If we look at that right triangle and if we wanted to solve for h and solve for h using a trig ratio based on this angle theta right over here, we know that relative to this angle theta, this side of length h is an adjacent side and this length of 108 right over here along the edge, that's the hypotenuse of this yellow triangle that I just highlighted in. So which trig ratio involves an adjacent side and a hypotenuse? We just write SOH CAH TOA."}, {"video_title": "How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "This thing in yellow right over here is a right triangle. If we look at that right triangle and if we wanted to solve for h and solve for h using a trig ratio based on this angle theta right over here, we know that relative to this angle theta, this side of length h is an adjacent side and this length of 108 right over here along the edge, that's the hypotenuse of this yellow triangle that I just highlighted in. So which trig ratio involves an adjacent side and a hypotenuse? We just write SOH CAH TOA. Sine is opposite over hypotenuse. That would be this distance over the hypotenuse. Cosine is adjacent over hypotenuse."}, {"video_title": "How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "We just write SOH CAH TOA. Sine is opposite over hypotenuse. That would be this distance over the hypotenuse. Cosine is adjacent over hypotenuse. We get the cosine of theta is going to be equal to the height that we care about, that's the adjacent side of this right triangle, over the length of the hypotenuse, over 108. That doesn't help us yet because we don't know what the cosine of theta is. But there's a clue here."}, {"video_title": "How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Cosine is adjacent over hypotenuse. We get the cosine of theta is going to be equal to the height that we care about, that's the adjacent side of this right triangle, over the length of the hypotenuse, over 108. That doesn't help us yet because we don't know what the cosine of theta is. But there's a clue here. Theta is also sitting up here, so maybe if we can figure out what cosine of theta is based up here, then we can solve for h. If we look at this theta, what is the cosine of theta? Now we're looking at a different right triangle. We're looking at this entire right triangle now."}, {"video_title": "How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "But there's a clue here. Theta is also sitting up here, so maybe if we can figure out what cosine of theta is based up here, then we can solve for h. If we look at this theta, what is the cosine of theta? Now we're looking at a different right triangle. We're looking at this entire right triangle now. Based on that entire right triangle, what is cosine of theta? Well, cosine of theta, once again, is equal to adjacent over hypotenuse. The adjacent length is this length right over here."}, {"video_title": "How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "We're looking at this entire right triangle now. Based on that entire right triangle, what is cosine of theta? Well, cosine of theta, once again, is equal to adjacent over hypotenuse. The adjacent length is this length right over here. We already know that's 139 meters, so it's going to be equal to 139 meters. And what's the length of the hypotenuse? Well, the hypotenuse is this length right over here."}, {"video_title": "How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "The adjacent length is this length right over here. We already know that's 139 meters, so it's going to be equal to 139 meters. And what's the length of the hypotenuse? Well, the hypotenuse is this length right over here. It's 72 plus 108, or we already have it labeled here. It's 180. We can assume that this pyramid is an isosceles triangle."}, {"video_title": "How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Well, the hypotenuse is this length right over here. It's 72 plus 108, or we already have it labeled here. It's 180. We can assume that this pyramid is an isosceles triangle. So 180 on that side, 180 on that side. So the cosine is adjacent, 139, over the hypotenuse, which is 180. These thetas are the same theta."}, {"video_title": "How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "We can assume that this pyramid is an isosceles triangle. So 180 on that side, 180 on that side. So the cosine is adjacent, 139, over the hypotenuse, which is 180. These thetas are the same theta. We just showed that. Now we have cosine of theta is h over 108. Cosine of theta is 139 over 180."}, {"video_title": "How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "These thetas are the same theta. We just showed that. Now we have cosine of theta is h over 108. Cosine of theta is 139 over 180. Or we could say that h over 108, which is cosine of theta, also is equal to 139 over 180. Both of these things are equal to cosine of theta. Now to solve for h, we just multiply both sides by 108."}, {"video_title": "How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Cosine of theta is 139 over 180. Or we could say that h over 108, which is cosine of theta, also is equal to 139 over 180. Both of these things are equal to cosine of theta. Now to solve for h, we just multiply both sides by 108. So h is equal to 139 times 108 over 180. So let's get our calculator out and calculate that. So that is going to be 139 times 108 divided by 180 gets us to 83.4 meters."}, {"video_title": "Sine and cosine of complements example Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "We are told that the cosine of 58 degrees is roughly equal to 0.53, and that's roughly equal to, because it just keeps going on and on, I just rounded it to the nearest hundredth. And then we're asked, what is the sine of 32 degrees? And I encourage you to pause this video and try it on your own. And a hint is to look at this right triangle, one of the angles is already labeled 32 degrees, figure out what all of the angles are, and then use the fundamental definitions, your SOH CAH TOA definitions to see if you could figure out what sine of 32 degrees is. So I'm assuming you've given a go at it. Let's work it through now. So we know that the sum of the angles of a triangle add up to 180."}, {"video_title": "Sine and cosine of complements example Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "And a hint is to look at this right triangle, one of the angles is already labeled 32 degrees, figure out what all of the angles are, and then use the fundamental definitions, your SOH CAH TOA definitions to see if you could figure out what sine of 32 degrees is. So I'm assuming you've given a go at it. Let's work it through now. So we know that the sum of the angles of a triangle add up to 180. Now in a right angle, one of the angles is 90 degrees. So that means that the other two must add up to 90. These two add up to 90, plus another 90 is going to be 180 degrees."}, {"video_title": "Sine and cosine of complements example Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So we know that the sum of the angles of a triangle add up to 180. Now in a right angle, one of the angles is 90 degrees. So that means that the other two must add up to 90. These two add up to 90, plus another 90 is going to be 180 degrees. Or another way to think about it is that the other two non-right angles are going to be complementary. So what plus 32 is equal to 90? Well, 90 minus 32 is 58."}, {"video_title": "Sine and cosine of complements example Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "These two add up to 90, plus another 90 is going to be 180 degrees. Or another way to think about it is that the other two non-right angles are going to be complementary. So what plus 32 is equal to 90? Well, 90 minus 32 is 58. So this right over here is going to be 58 degrees. Well, why is that interesting? Well, we already know what the cosine of 58 degrees is equal to."}, {"video_title": "Sine and cosine of complements example Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Well, 90 minus 32 is 58. So this right over here is going to be 58 degrees. Well, why is that interesting? Well, we already know what the cosine of 58 degrees is equal to. But let's think about it in terms of ratios of the lengths of sides of this right triangle. Let's just write down SOH CAH TOA. So sine is opposite over hypotenuse."}, {"video_title": "Sine and cosine of complements example Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Well, we already know what the cosine of 58 degrees is equal to. But let's think about it in terms of ratios of the lengths of sides of this right triangle. Let's just write down SOH CAH TOA. So sine is opposite over hypotenuse. CAH, cosine is adjacent over hypotenuse. TOA, tangent is opposite over adjacent. So we could write down the cosine of 58 degrees, which we already know."}, {"video_title": "Sine and cosine of complements example Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So sine is opposite over hypotenuse. CAH, cosine is adjacent over hypotenuse. TOA, tangent is opposite over adjacent. So we could write down the cosine of 58 degrees, which we already know. If we think about it in terms of these fundamental ratios, cosine is adjacent over hypotenuse. This is the 58 degree angle. The side that is adjacent to it is, let me do it in this color, is side BC, right over here."}, {"video_title": "Sine and cosine of complements example Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So we could write down the cosine of 58 degrees, which we already know. If we think about it in terms of these fundamental ratios, cosine is adjacent over hypotenuse. This is the 58 degree angle. The side that is adjacent to it is, let me do it in this color, is side BC, right over here. It's one of the sides of the angle, the side of the angle that is not the hypotenuse. The other side, this over here, is the hypotenuse. So this is going to be the adjacent, the length of the adjacent side, BC, over the length of the hypotenuse."}, {"video_title": "Sine and cosine of complements example Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "The side that is adjacent to it is, let me do it in this color, is side BC, right over here. It's one of the sides of the angle, the side of the angle that is not the hypotenuse. The other side, this over here, is the hypotenuse. So this is going to be the adjacent, the length of the adjacent side, BC, over the length of the hypotenuse. Over the length of the hypotenuse. The length of the hypotenuse, well that is AB. Now let's think about what the sine of 32 degrees would be."}, {"video_title": "Sine and cosine of complements example Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So this is going to be the adjacent, the length of the adjacent side, BC, over the length of the hypotenuse. Over the length of the hypotenuse. The length of the hypotenuse, well that is AB. Now let's think about what the sine of 32 degrees would be. So the sine of 32 degrees, well sine is opposite over hypotenuse. So now we're looking at this 32 degree angle. What side is opposite?"}, {"video_title": "Sine and cosine of complements example Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Now let's think about what the sine of 32 degrees would be. So the sine of 32 degrees, well sine is opposite over hypotenuse. So now we're looking at this 32 degree angle. What side is opposite? Well it opens up onto BC. It opens up onto BC, just like that. And what's the hypotenuse?"}, {"video_title": "Sine and cosine of complements example Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "What side is opposite? Well it opens up onto BC. It opens up onto BC, just like that. And what's the hypotenuse? Well we've already, or the length of the hypotenuse, it's AB. It's AB. Notice, the sine of 32 degrees is BC over AB."}, {"video_title": "Sine and cosine of complements example Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "And what's the hypotenuse? Well we've already, or the length of the hypotenuse, it's AB. It's AB. Notice, the sine of 32 degrees is BC over AB. The cosine of 58 degrees is BC over AB. Or another way of thinking about it, the sine of this angle is the same thing as the cosine of this angle. So we could literally write the sine, I'm gonna do that in that pink color, the sine, the sine of 32 degrees is equal to the cosine, cosine of 58 degrees, which is roughly equal to 0.53."}, {"video_title": "Sine and cosine of complements example Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Notice, the sine of 32 degrees is BC over AB. The cosine of 58 degrees is BC over AB. Or another way of thinking about it, the sine of this angle is the same thing as the cosine of this angle. So we could literally write the sine, I'm gonna do that in that pink color, the sine, the sine of 32 degrees is equal to the cosine, cosine of 58 degrees, which is roughly equal to 0.53. And this is a really, really useful property. The sine of an angle is equal to the cosine of its complement. So we could write this in general terms."}, {"video_title": "Sine and cosine of complements example Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So we could literally write the sine, I'm gonna do that in that pink color, the sine, the sine of 32 degrees is equal to the cosine, cosine of 58 degrees, which is roughly equal to 0.53. And this is a really, really useful property. The sine of an angle is equal to the cosine of its complement. So we could write this in general terms. We could write that the sine of some angle is equal to the cosine of its complement, is equal to the cosine of 90 minus theta. Think about it. I could have done, I could change this entire problem."}, {"video_title": "Sine and cosine of complements example Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So we could write this in general terms. We could write that the sine of some angle is equal to the cosine of its complement, is equal to the cosine of 90 minus theta. Think about it. I could have done, I could change this entire problem. Instead of making this the sine of 32 degrees, I could make this the sine of 25 degrees. And if someone gave you the cosine of what's 90 minus 25, if someone gave you the cosine of 65 degrees, then you could think about this as 25. The complement is going to be right over here."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Sort the expressions according to their values. You can put any number of cards in a category or leave a category empty. And so we have this diagram right over here. And then we have these cards that have these expressions. And we're supposed to sort these into different buckets. So we're trying to say, well, what is the length of segment AC over the length of segment BC equal to? Which of these expressions is it equal to?"}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "And then we have these cards that have these expressions. And we're supposed to sort these into different buckets. So we're trying to say, well, what is the length of segment AC over the length of segment BC equal to? Which of these expressions is it equal to? And then we should drag it into the appropriate buckets. So to figure these out, I've actually already redrawn this problem on my little, I guess you'd call it, scratch pad or blackboard, whatever you want to call it. This right over here is that same diagram blown up a little bit."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Which of these expressions is it equal to? And then we should drag it into the appropriate buckets. So to figure these out, I've actually already redrawn this problem on my little, I guess you'd call it, scratch pad or blackboard, whatever you want to call it. This right over here is that same diagram blown up a little bit. Here are the expressions that we need to drag into things. And here are the buckets that we need to see which of these expressions are equal to which of these expressions. So let's first look at this."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "This right over here is that same diagram blown up a little bit. Here are the expressions that we need to drag into things. And here are the buckets that we need to see which of these expressions are equal to which of these expressions. So let's first look at this. The length of segment AC over the length of segment BC. So let's think about what AC is. The length of segment AC."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So let's first look at this. The length of segment AC over the length of segment BC. So let's think about what AC is. The length of segment AC. AC is this right over here. So it's this length right over here in purple over the length of segment BC. Over this length right over here."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "The length of segment AC. AC is this right over here. So it's this length right over here in purple over the length of segment BC. Over this length right over here. So it's the ratio of the lengths of two sides of a right triangle. This is clearly a right triangle. Triangle ABC."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Over this length right over here. So it's the ratio of the lengths of two sides of a right triangle. This is clearly a right triangle. Triangle ABC. And I could color that in just so you know what triangle I'm talking about. Triangle ABC is this entire triangle that we could focus on. So you could imagine that it's reasonable that the ratio of two sides of a right triangle are going to be the sine of one of its angles."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Triangle ABC. And I could color that in just so you know what triangle I'm talking about. Triangle ABC is this entire triangle that we could focus on. So you could imagine that it's reasonable that the ratio of two sides of a right triangle are going to be the sine of one of its angles. And they give us one of the angles right over here. They give us this angle right over here. You say, well, no, all they did is mark that angle."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So you could imagine that it's reasonable that the ratio of two sides of a right triangle are going to be the sine of one of its angles. And they give us one of the angles right over here. They give us this angle right over here. You say, well, no, all they did is mark that angle. But notice, one arc is here. One arc is here. So anywhere we see only one arc, that's going to be 30 degrees."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "You say, well, no, all they did is mark that angle. But notice, one arc is here. One arc is here. So anywhere we see only one arc, that's going to be 30 degrees. So this is 30 degrees as well. You have two arcs here. That's 41 degrees."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So anywhere we see only one arc, that's going to be 30 degrees. So this is 30 degrees as well. You have two arcs here. That's 41 degrees. Two arcs here. This is going to be congruent to that. This over here is going to be 41 degrees."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "That's 41 degrees. Two arcs here. This is going to be congruent to that. This over here is going to be 41 degrees. This is three arcs. They don't tell us how many degrees that is. But this angle with the three arcs is congruent to this angle with the three arcs right over there."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "This over here is going to be 41 degrees. This is three arcs. They don't tell us how many degrees that is. But this angle with the three arcs is congruent to this angle with the three arcs right over there. So anyway, this yellow triangle, triangle ABC, we know the measure of this angle is 30 degrees. And then they give us these two sides. So how do these sides relate to this 30 degree angle?"}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "But this angle with the three arcs is congruent to this angle with the three arcs right over there. So anyway, this yellow triangle, triangle ABC, we know the measure of this angle is 30 degrees. And then they give us these two sides. So how do these sides relate to this 30 degree angle? Well, side AC is adjacent to it. It's literally one of the sides of the angle that is not the hypotenuse. So let me write that down."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So how do these sides relate to this 30 degree angle? Well, side AC is adjacent to it. It's literally one of the sides of the angle that is not the hypotenuse. So let me write that down. This is adjacent. And what is BC? Well, BC is the hypotenuse of this right triangle."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So let me write that down. This is adjacent. And what is BC? Well, BC is the hypotenuse of this right triangle. It's the side opposite the 90 degrees. So this is the hypotenuse. So what trig function, when applied to 30 degrees, is equal to the adjacent side over the hypotenuse?"}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Well, BC is the hypotenuse of this right triangle. It's the side opposite the 90 degrees. So this is the hypotenuse. So what trig function, when applied to 30 degrees, is equal to the adjacent side over the hypotenuse? Let's write down SOH CAH TOA just to remind ourselves. So SOH CAH TOA, sine of an angle is opposite over a hypotenuse. Cosine of an angle is adjacent over a hypotenuse."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So what trig function, when applied to 30 degrees, is equal to the adjacent side over the hypotenuse? Let's write down SOH CAH TOA just to remind ourselves. So SOH CAH TOA, sine of an angle is opposite over a hypotenuse. Cosine of an angle is adjacent over a hypotenuse. So cosine, let's write this down, cosine of 30 degrees is going to be equal to the length of the adjacent side, so that is AC, over the length of the hypotenuse, which is equal to BC. So this right over here is the same thing as the cosine of 30 degrees. So let's drag it in there."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Cosine of an angle is adjacent over a hypotenuse. So cosine, let's write this down, cosine of 30 degrees is going to be equal to the length of the adjacent side, so that is AC, over the length of the hypotenuse, which is equal to BC. So this right over here is the same thing as the cosine of 30 degrees. So let's drag it in there. This is equal to the cosine of 30 degrees. Now let's look at the next one. Cosine of angle DEC. Where is DEC?"}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So let's drag it in there. This is equal to the cosine of 30 degrees. Now let's look at the next one. Cosine of angle DEC. Where is DEC? So DEC. So that's this angle right over here. I'll put four arcs here so we don't get it confused."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Cosine of angle DEC. Where is DEC? So DEC. So that's this angle right over here. I'll put four arcs here so we don't get it confused. So this is angle DEC. So what is the cosine of DEC? Well, once again, cosine is adjacent over a hypotenuse."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "I'll put four arcs here so we don't get it confused. So this is angle DEC. So what is the cosine of DEC? Well, once again, cosine is adjacent over a hypotenuse. So cosine of angle DEC, the adjacent side to this, well, that's this right over here. You might say, well, isn't this side adjacent? Well, that side, side DE, that is the actual hypotenuse."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Well, once again, cosine is adjacent over a hypotenuse. So cosine of angle DEC, the adjacent side to this, well, that's this right over here. You might say, well, isn't this side adjacent? Well, that side, side DE, that is the actual hypotenuse. So that's not going to be the adjacent side. So the adjacent side is E. The adjacent side is, I could call it EC. It's the length of segment EC."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Well, that side, side DE, that is the actual hypotenuse. So that's not going to be the adjacent side. So the adjacent side is E. The adjacent side is, I could call it EC. It's the length of segment EC. And then the hypotenuse is this right over here. It's the length of the hypotenuse. The hypotenuse is side DE or ED, however you want to call it."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "It's the length of segment EC. And then the hypotenuse is this right over here. It's the length of the hypotenuse. The hypotenuse is side DE or ED, however you want to call it. And so the length of it is, we could just write it as DE. Now what is this also equal to? We don't see this choice over here."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "The hypotenuse is side DE or ED, however you want to call it. And so the length of it is, we could just write it as DE. Now what is this also equal to? We don't see this choice over here. We don't have the ratio EC over DE as one of these choices here. But what we do have is we do get one of the angles here. They give us this 41 degrees."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "We don't see this choice over here. We don't have the ratio EC over DE as one of these choices here. But what we do have is we do get one of the angles here. They give us this 41 degrees. And the ratio of this green side, the length of this green side over this orange side, what would that be in terms of if we wanted to apply a trig function to this angle? Well, relative to this angle, the green side is the opposite side. And the orange side is still the hypotenuse."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "They give us this 41 degrees. And the ratio of this green side, the length of this green side over this orange side, what would that be in terms of if we wanted to apply a trig function to this angle? Well, relative to this angle, the green side is the opposite side. And the orange side is still the hypotenuse. So relative to 41 degrees, so let's write this down. Relative to 41 degrees, this ratio is the opposite over the hypotenuse. It's the cosine of this angle."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "And the orange side is still the hypotenuse. So relative to 41 degrees, so let's write this down. Relative to 41 degrees, this ratio is the opposite over the hypotenuse. It's the cosine of this angle. But it's the sine of this angle right over here. So sine is opposite over hypotenuse. So this is equal to the sine of this angle right over here."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "It's the cosine of this angle. But it's the sine of this angle right over here. So sine is opposite over hypotenuse. So this is equal to the sine of this angle right over here. It's equal to the sine of 41 degrees. So that is this one right over here, the sine of 41 degrees. So let's drag that into the appropriate bucket."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So this is equal to the sine of this angle right over here. It's equal to the sine of 41 degrees. So that is this one right over here, the sine of 41 degrees. So let's drag that into the appropriate bucket. So let's sine of 41 degrees is the same thing as the cosine of angle DEC. Only have two left. So now we have to figure out what the sine of angle CDA is. So let's see, where is CDA?"}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So let's drag that into the appropriate bucket. So let's sine of 41 degrees is the same thing as the cosine of angle DEC. Only have two left. So now we have to figure out what the sine of angle CDA is. So let's see, where is CDA? CDA is this entire angle. It's this entire angle right over here. So I could put a bunch of arcs here if I want, just to show that it's different than all the other ones."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So let's see, where is CDA? CDA is this entire angle. It's this entire angle right over here. So I could put a bunch of arcs here if I want, just to show that it's different than all the other ones. So that's that angle right over there. So now we're really dealing with this larger right triangle. Let me highlight that in some."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So I could put a bunch of arcs here if I want, just to show that it's different than all the other ones. So that's that angle right over there. So now we're really dealing with this larger right triangle. Let me highlight that in some. Let me highlight it in this pink color. So we're now dealing with this larger right triangle right over here. We care about the sine of this whole thing."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Let me highlight that in some. Let me highlight it in this pink color. So we're now dealing with this larger right triangle right over here. We care about the sine of this whole thing. Remember, sine is opposite over hypotenuse. Sine is opposite over hypotenuse. So the opposite side is going to be side CA."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "We care about the sine of this whole thing. Remember, sine is opposite over hypotenuse. Sine is opposite over hypotenuse. So the opposite side is going to be side CA. So this is going to be equal to the length of CA over the hypotenuse, which is AD. So that is going to be over AD. Now once again, we don't see that as a choice here."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So the opposite side is going to be side CA. So this is going to be equal to the length of CA over the hypotenuse, which is AD. So that is going to be over AD. Now once again, we don't see that as a choice here. But maybe we can express this ratio. Maybe this ratio is a trig function applied to one of the other angles. And they give us one of the angles."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Now once again, we don't see that as a choice here. But maybe we can express this ratio. Maybe this ratio is a trig function applied to one of the other angles. And they give us one of the angles. They give us this angle right over here. I guess we could call this angle DAC. This is 30 degrees."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "And they give us one of the angles. They give us this angle right over here. I guess we could call this angle DAC. This is 30 degrees. So relative to this angle, what two sides are we taking the ratio of? We're taking now the ratio of, relative to this angle, the adjacent side over the hypotenuse. So this is the adjacent side over the hypotenuse."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "This is 30 degrees. So relative to this angle, what two sides are we taking the ratio of? We're taking now the ratio of, relative to this angle, the adjacent side over the hypotenuse. So this is the adjacent side over the hypotenuse. What deals with adjacent over hypotenuse? Well, cosine. So this is equal to the cosine of this angle."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So this is the adjacent side over the hypotenuse. What deals with adjacent over hypotenuse? Well, cosine. So this is equal to the cosine of this angle. So this is equal to cosine of 30 degrees. Sine of CDA is equal to the cosine of this angle right over here. So this one is equal to this right over here."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So this is equal to the cosine of this angle. So this is equal to cosine of 30 degrees. Sine of CDA is equal to the cosine of this angle right over here. So this one is equal to this right over here. So let me drag that in. So this one is equal to, so you can see that I just dragged it in, equal to that. And now we have one left."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So this one is equal to this right over here. So let me drag that in. So this one is equal to, so you can see that I just dragged it in, equal to that. And now we have one left. We have one left. Home stretch. We should be getting excited."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "And now we have one left. We have one left. Home stretch. We should be getting excited. AE over EB. AE, let me use this color. Length of segment AE."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "We should be getting excited. AE over EB. AE, let me use this color. Length of segment AE. That's this length right over here. Let me make that stand out more. Let me do it in this red."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Length of segment AE. That's this length right over here. Let me make that stand out more. Let me do it in this red. This color right over here. That's length of segment AE over length of segment EB. This is EB right over here."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Let me do it in this red. This color right over here. That's length of segment AE over length of segment EB. This is EB right over here. This is EB. So now we are focused on this triangle, this right triangle right over here. Well, we know the measure of this angle over here."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "This is EB right over here. This is EB. So now we are focused on this triangle, this right triangle right over here. Well, we know the measure of this angle over here. We have double arcs. We have double arcs right over here. And they say this is 41 degrees."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Well, we know the measure of this angle over here. We have double arcs. We have double arcs right over here. And they say this is 41 degrees. So we have double marks over here. And this is also going to be 41 degrees. So relative to this angle, what ratio is this?"}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "And they say this is 41 degrees. So we have double marks over here. And this is also going to be 41 degrees. So relative to this angle, what ratio is this? This is the opposite over the hypotenuse. Opposite over the hypotenuse. This right over here is going to be sine of that angle, sine of 41 degrees."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So relative to this angle, what ratio is this? This is the opposite over the hypotenuse. Opposite over the hypotenuse. This right over here is going to be sine of that angle, sine of 41 degrees. So it's equal to this first one right over there. So let's drag it. So this is going to be equal to sine of 41 degrees."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "This right over here is going to be sine of that angle, sine of 41 degrees. So it's equal to this first one right over there. So let's drag it. So this is going to be equal to sine of 41 degrees. So none of the ones actually ended up being equal to the tangent of 41 degrees. Now let's see if we actually got this right. I hope I did."}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So the first thing we have to ask ourselves is what does amplitude even refer to? Well, the amplitude of a periodic function is just half the difference between the minimum and maximum values it takes on. So if I were to draw a periodic function like this, and if we just go back and forth between two, let me draw it a little bit neater, it goes back and forth between two values like that. So between that value and that value, you take the difference between the two, and half of that is the amplitude. Another way of thinking about the amplitude is how much does it sway from its middle position? Right over here, we have y equals negative 1 half cosine of 3x. So what is going to be the amplitude of this?"}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So between that value and that value, you take the difference between the two, and half of that is the amplitude. Another way of thinking about the amplitude is how much does it sway from its middle position? Right over here, we have y equals negative 1 half cosine of 3x. So what is going to be the amplitude of this? Well, the easy way to think about it is just what is multiplying the cosine function? And you could do the same thing if it was a sine function. We have negative 1 half multiplying it."}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So what is going to be the amplitude of this? Well, the easy way to think about it is just what is multiplying the cosine function? And you could do the same thing if it was a sine function. We have negative 1 half multiplying it. So the amplitude in this situation is going to be the absolute value of negative 1 half, which is equal to 1 half. And you might say, well, why do I not care about the sine? Why do I take the absolute value of it?"}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "We have negative 1 half multiplying it. So the amplitude in this situation is going to be the absolute value of negative 1 half, which is equal to 1 half. And you might say, well, why do I not care about the sine? Why do I take the absolute value of it? Well, the negative just flips the function around. It's not going to change how much it sways between its minimum and maximum position. The other thing is, well, how is it just simply the absolute value of this thing?"}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Why do I take the absolute value of it? Well, the negative just flips the function around. It's not going to change how much it sways between its minimum and maximum position. The other thing is, well, how is it just simply the absolute value of this thing? And to realize the why, you just have to remember that a cosine function or a sine function varies between positive 1 and negative 1 if it's just a simple function. So this is just multiplying that positive 1 or negative 1. And so if normally the amplitude, if you didn't have any coefficient here, if the coefficient was positive or negative 1, the amplitude would just be 1."}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "The other thing is, well, how is it just simply the absolute value of this thing? And to realize the why, you just have to remember that a cosine function or a sine function varies between positive 1 and negative 1 if it's just a simple function. So this is just multiplying that positive 1 or negative 1. And so if normally the amplitude, if you didn't have any coefficient here, if the coefficient was positive or negative 1, the amplitude would just be 1. Now you're changing it or you're multiplying it by this amount. So the amplitude is 1 half. Now let's think about the period."}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And so if normally the amplitude, if you didn't have any coefficient here, if the coefficient was positive or negative 1, the amplitude would just be 1. Now you're changing it or you're multiplying it by this amount. So the amplitude is 1 half. Now let's think about the period. So the first thing I want to ask you is what does the period of a cyclical function, even or a periodic function, I should say, what does the period of a periodic function even refer to? Well, let me draw some axes on this function right over here. Let's say that this right over here is the y-axis."}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Now let's think about the period. So the first thing I want to ask you is what does the period of a cyclical function, even or a periodic function, I should say, what does the period of a periodic function even refer to? Well, let me draw some axes on this function right over here. Let's say that this right over here is the y-axis. That's the y-axis. And let's just say, for the sake of argument, this is the x-axis right over here. So the period of a periodic function is the length of the smallest interval that contains exactly one copy of the repeating pattern of that periodic function."}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Let's say that this right over here is the y-axis. That's the y-axis. And let's just say, for the sake of argument, this is the x-axis right over here. So the period of a periodic function is the length of the smallest interval that contains exactly one copy of the repeating pattern of that periodic function. So what do they mean here? Well, what's repeating? So we go down and then up, just like that."}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So the period of a periodic function is the length of the smallest interval that contains exactly one copy of the repeating pattern of that periodic function. So what do they mean here? Well, what's repeating? So we go down and then up, just like that. Then we go down and then we go up. So in this case, the length of the smallest interval that contains exactly one copy of the repeating pattern, this could be one of the smallest repeating patterns. And so this length between here and here would be one period."}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So we go down and then up, just like that. Then we go down and then we go up. So in this case, the length of the smallest interval that contains exactly one copy of the repeating pattern, this could be one of the smallest repeating patterns. And so this length between here and here would be one period. Then we could go between here and here is another period. And there's multiple. This isn't the only pattern that you could pick."}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And so this length between here and here would be one period. Then we could go between here and here is another period. And there's multiple. This isn't the only pattern that you could pick. You could say, well, I'm going to define my pattern starting here, going up, and then going down, like that. So you could say, that's my smallest length. And then you would see that, OK, well, if you go in the negative direction, the next repeating version of that pattern is right over there."}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "This isn't the only pattern that you could pick. You could say, well, I'm going to define my pattern starting here, going up, and then going down, like that. So you could say, that's my smallest length. And then you would see that, OK, well, if you go in the negative direction, the next repeating version of that pattern is right over there. But either way, you're going to get the same length that it takes to repeat that pattern. So given that, what is the period of this function right over here? Well, to figure out the period, we just take 2 pi and divide it by the absolute value of the coefficient right over here."}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And then you would see that, OK, well, if you go in the negative direction, the next repeating version of that pattern is right over there. But either way, you're going to get the same length that it takes to repeat that pattern. So given that, what is the period of this function right over here? Well, to figure out the period, we just take 2 pi and divide it by the absolute value of the coefficient right over here. So we divide it by the absolute value of 3, which is just 3. So we get 2 pi over 3. Now, we need to think about why does this work."}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Well, to figure out the period, we just take 2 pi and divide it by the absolute value of the coefficient right over here. So we divide it by the absolute value of 3, which is just 3. So we get 2 pi over 3. Now, we need to think about why does this work. Well, if you think about just a traditional cosine function or a traditional sine function, it has a period of 2 pi. If you think about the unit circle, 2 pi, if you start at 0, 2 pi radians later, you're back to where you started. 2 pi radians, another 2 pi, you're back to where you started."}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Now, we need to think about why does this work. Well, if you think about just a traditional cosine function or a traditional sine function, it has a period of 2 pi. If you think about the unit circle, 2 pi, if you start at 0, 2 pi radians later, you're back to where you started. 2 pi radians, another 2 pi, you're back to where you started. If you go in the negative direction, you go negative 2 pi, you're back to where you started. For any angle here, if you go 2 pi, you're back to where you were before. You go negative 2 pi, you're back to where you were before."}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "2 pi radians, another 2 pi, you're back to where you started. If you go in the negative direction, you go negative 2 pi, you're back to where you started. For any angle here, if you go 2 pi, you're back to where you were before. You go negative 2 pi, you're back to where you were before. So the periods for these are all 2 pi. And the reason why this makes sense is that this coefficient makes you get to 2 pi or negative, in this case, 2 pi, it's going to make you get to 2 pi all that much faster. And so your period is going to be a lower number."}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "You go negative 2 pi, you're back to where you were before. So the periods for these are all 2 pi. And the reason why this makes sense is that this coefficient makes you get to 2 pi or negative, in this case, 2 pi, it's going to make you get to 2 pi all that much faster. And so your period is going to be a lower number. It takes less length. You're going to get to 2 pi 3 times as fast. Now, you might say, well, why are you taking the absolute value here?"}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And so your period is going to be a lower number. It takes less length. You're going to get to 2 pi 3 times as fast. Now, you might say, well, why are you taking the absolute value here? Well, if this was a negative number, it would get you to negative 2 pi all that much faster. But either way, you're going to be completing one cycle. So with that out of the way, let's visualize these two things."}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Now, you might say, well, why are you taking the absolute value here? Well, if this was a negative number, it would get you to negative 2 pi all that much faster. But either way, you're going to be completing one cycle. So with that out of the way, let's visualize these two things. Let's actually draw negative 1 half cosine of 3x. So let me draw my axes here, my best attempt. So this is my y-axis."}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So with that out of the way, let's visualize these two things. Let's actually draw negative 1 half cosine of 3x. So let me draw my axes here, my best attempt. So this is my y-axis. This is my x-axis. And then let me draw some. So this is 0 right over here."}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So this is my y-axis. This is my x-axis. And then let me draw some. So this is 0 right over here. x is equal to 0. And let me draw x is equal to positive 1 half. I'll draw it right over here."}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So this is 0 right over here. x is equal to 0. And let me draw x is equal to positive 1 half. I'll draw it right over here. So x is equal to positive 1 half. And we haven't shifted this function up or down any. Then if we wanted to, we could add a constant out here, outside of the cosine function."}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "I'll draw it right over here. So x is equal to positive 1 half. And we haven't shifted this function up or down any. Then if we wanted to, we could add a constant out here, outside of the cosine function. But this is positive 1 half. Or we could just write that as 1 half. And then down here, let's say that this is negative 1 half."}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Then if we wanted to, we could add a constant out here, outside of the cosine function. But this is positive 1 half. Or we could just write that as 1 half. And then down here, let's say that this is negative 1 half. And so let me draw that bound. I'm just drawing these dotted lines, so it'll become easy for me to draw. And what happens when this is 0?"}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And then down here, let's say that this is negative 1 half. And so let me draw that bound. I'm just drawing these dotted lines, so it'll become easy for me to draw. And what happens when this is 0? Well, cosine of 0 is 1. But we're going to multiply it by negative 1 half. So it's going to be negative 1 half right over here."}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And what happens when this is 0? Well, cosine of 0 is 1. But we're going to multiply it by negative 1 half. So it's going to be negative 1 half right over here. And then it's going to start going up. It can only go in that direction. It's bounded."}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So it's going to be negative 1 half right over here. And then it's going to start going up. It can only go in that direction. It's bounded. It's going to start going up. Then it'll come back down. And then it will get back to that original point right over here."}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "It's bounded. It's going to start going up. Then it'll come back down. And then it will get back to that original point right over here. And the question is, what is this distance? What is this length? What is this length going to be?"}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And then it will get back to that original point right over here. And the question is, what is this distance? What is this length? What is this length going to be? Well, we know what its period is. It's 2 pi over 3. It's going to get to this point 3 times as fast as a traditional cosine function."}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "What is this length going to be? Well, we know what its period is. It's 2 pi over 3. It's going to get to this point 3 times as fast as a traditional cosine function. So this is going to be 2 pi over 3. And then if you give it another 2 pi over 3, it's going to get back to that same point again. So if you go another 2 pi over 3, so in this case, you've now gone 4 pi over 3."}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "It's going to get to this point 3 times as fast as a traditional cosine function. So this is going to be 2 pi over 3. And then if you give it another 2 pi over 3, it's going to get back to that same point again. So if you go another 2 pi over 3, so in this case, you've now gone 4 pi over 3. 4 pi over 3, you've completed another cycle. So that length right over there is the period. And you could also do the same thing in the negative direction."}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So if you go another 2 pi over 3, so in this case, you've now gone 4 pi over 3. 4 pi over 3, you've completed another cycle. So that length right over there is the period. And you could also do the same thing in the negative direction. So this right over here would be negative 2 pi over 3. And to visualize the amplitude, you see that it can go 1 half. Well, there's two ways to think about it."}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And you could also do the same thing in the negative direction. So this right over here would be negative 2 pi over 3. And to visualize the amplitude, you see that it can go 1 half. Well, there's two ways to think about it. The difference between the maximum and the minimum point is 1. Half of that is 1 half. Or you could say that it's going 1 half in magnitude."}, {"video_title": "Interpreting trigonometric graphs in context Trigonometry Algebra Khan Academy.mp3", "Sentence": "Her height above the ground in meters is modeled by h of t, where t is the time in seconds, and we can see that right over here. Now what I wanna focus on this video is some features of this graph, and the features we're going to focus on, actually the first of them, is going to be the midline. So pause this video and see if you can figure out the midline of this graph, or the midline of this function, and then we're gonna think about what it actually represents. Well, Alexa starts off at five meters above the ground, and then she goes higher and higher and higher, gets as high as 25 meters, and then goes back as low as five meters above the ground, then as high as 25 meters, and what we can view the midline as is the midpoint between these extremes, or the average of these extremes. Well, the extremes are, she goes as low as five and as high as 25. So what's the average of five and 25? Well, that would be 15."}, {"video_title": "Interpreting trigonometric graphs in context Trigonometry Algebra Khan Academy.mp3", "Sentence": "Well, Alexa starts off at five meters above the ground, and then she goes higher and higher and higher, gets as high as 25 meters, and then goes back as low as five meters above the ground, then as high as 25 meters, and what we can view the midline as is the midpoint between these extremes, or the average of these extremes. Well, the extremes are, she goes as low as five and as high as 25. So what's the average of five and 25? Well, that would be 15. So the midline would look something like this, and I'm actually gonna keep going off the graph, and the reason is is to help us think about what does that midline even represent? And one way to think about it is it represents the center of our rotation in this situation, or how high above the ground is the center of our Ferris wheel? And to help us visualize that, let me draw a Ferris wheel."}, {"video_title": "Interpreting trigonometric graphs in context Trigonometry Algebra Khan Academy.mp3", "Sentence": "Well, that would be 15. So the midline would look something like this, and I'm actually gonna keep going off the graph, and the reason is is to help us think about what does that midline even represent? And one way to think about it is it represents the center of our rotation in this situation, or how high above the ground is the center of our Ferris wheel? And to help us visualize that, let me draw a Ferris wheel. So I'm going to draw a circle with this as the center, and so the Ferris wheel would look something like, would look something like this, and it has some type of maybe support structure. So the Ferris wheel might look something like that, and this height above the ground, that is 15 meters, that is what the midline is representing. Now, the next feature I want to explore is the amplitude."}, {"video_title": "Interpreting trigonometric graphs in context Trigonometry Algebra Khan Academy.mp3", "Sentence": "And to help us visualize that, let me draw a Ferris wheel. So I'm going to draw a circle with this as the center, and so the Ferris wheel would look something like, would look something like this, and it has some type of maybe support structure. So the Ferris wheel might look something like that, and this height above the ground, that is 15 meters, that is what the midline is representing. Now, the next feature I want to explore is the amplitude. Pause this video and think about what is the amplitude of this oscillating function right over here, and then we'll think about what does that represent in the real world, or where does it come from in the real world? Well, the amplitude is the maximum difference or the maximum magnitude away from that midline, and you can see it right over here, actually right when Alexa starts, we have starting 10 meters below the midline, 10 meters below the center, and this is when Alexa is right over here. She is 10 meters below the midline, and then after a, looks like 10 seconds, she is right at the midline, so that means that she's right over here."}, {"video_title": "Interpreting trigonometric graphs in context Trigonometry Algebra Khan Academy.mp3", "Sentence": "Now, the next feature I want to explore is the amplitude. Pause this video and think about what is the amplitude of this oscillating function right over here, and then we'll think about what does that represent in the real world, or where does it come from in the real world? Well, the amplitude is the maximum difference or the maximum magnitude away from that midline, and you can see it right over here, actually right when Alexa starts, we have starting 10 meters below the midline, 10 meters below the center, and this is when Alexa is right over here. She is 10 meters below the midline, and then after a, looks like 10 seconds, she is right at the midline, so that means that she's right over here. Maybe the Ferris wheel is going this way, at least in my imagination, it's going clockwise, and then after another 10 seconds, she is at 25 meters, so she is right over there, and you can see that. She is right over there. I drew that circle intentionally of that size, and so we see the amplitude in full effect, 10 meters below to begin the midline and 10 meters above, and so it's the maximum displacement or the maximum change from that midline, and so over here, it really represents the radius of our Ferris wheel, 10 meters, and then from this part, she starts going back down again, and then over here, she's back to where she started."}, {"video_title": "Interpreting trigonometric graphs in context Trigonometry Algebra Khan Academy.mp3", "Sentence": "She is 10 meters below the midline, and then after a, looks like 10 seconds, she is right at the midline, so that means that she's right over here. Maybe the Ferris wheel is going this way, at least in my imagination, it's going clockwise, and then after another 10 seconds, she is at 25 meters, so she is right over there, and you can see that. She is right over there. I drew that circle intentionally of that size, and so we see the amplitude in full effect, 10 meters below to begin the midline and 10 meters above, and so it's the maximum displacement or the maximum change from that midline, and so over here, it really represents the radius of our Ferris wheel, 10 meters, and then from this part, she starts going back down again, and then over here, she's back to where she started. Now, the last feature I want to explore is the notion of a period. What is the period of this periodic function? Pause this video and think about that."}, {"video_title": "Interpreting trigonometric graphs in context Trigonometry Algebra Khan Academy.mp3", "Sentence": "I drew that circle intentionally of that size, and so we see the amplitude in full effect, 10 meters below to begin the midline and 10 meters above, and so it's the maximum displacement or the maximum change from that midline, and so over here, it really represents the radius of our Ferris wheel, 10 meters, and then from this part, she starts going back down again, and then over here, she's back to where she started. Now, the last feature I want to explore is the notion of a period. What is the period of this periodic function? Pause this video and think about that. Well, the period is how much time does it take to complete one cycle? So here, she's starting at the bottom, and let's see, after 10 seconds, not at the bottom yet, after 20 seconds, not at the bottom yet, after 30 seconds, not at the bottom yet, and then here she is, after 40 seconds, she's back at the bottom and about to head up again, and so this time right over here, that 40 seconds, that is the period, and if you think about what's going on over here, she starts over here, five meters above the ground, after 10 seconds, she is right over here, and that corresponds to this point right over here, after 10 more seconds, she's right over there, that corresponds to that point, after 10 more seconds, she's over here, that corresponds to that, and after 10 more seconds, or a total of 40 seconds, she is back to where she started. So the period in this example shows how long does it take to complete one full rotation?"}, {"video_title": "Interpreting trigonometric graphs in context Trigonometry Algebra Khan Academy.mp3", "Sentence": "Pause this video and think about that. Well, the period is how much time does it take to complete one cycle? So here, she's starting at the bottom, and let's see, after 10 seconds, not at the bottom yet, after 20 seconds, not at the bottom yet, after 30 seconds, not at the bottom yet, and then here she is, after 40 seconds, she's back at the bottom and about to head up again, and so this time right over here, that 40 seconds, that is the period, and if you think about what's going on over here, she starts over here, five meters above the ground, after 10 seconds, she is right over here, and that corresponds to this point right over here, after 10 more seconds, she's right over there, that corresponds to that point, after 10 more seconds, she's over here, that corresponds to that, and after 10 more seconds, or a total of 40 seconds, she is back to where she started. So the period in this example shows how long does it take to complete one full rotation? Now we have to be careful sometimes when we're trying to visually inspect the period, because sometimes it might be tempting to say, start right over here and say, okay, we're 15 meters above the ground, all right, let's see, we're going down, now we're going up again, and look, we're 15 meters above the ground, maybe this 20 seconds is a period, but when you look at it over here, it's clear that that is not the case. This point represents this point at being 15 meters above the ground, going down, that's getting us to this point, and then after another 10 seconds, we get back over here. Notice, all this is measuring is half of a cycle, going halfway around."}, {"video_title": "Another example using angle addition formula with cosine Trigonometry Khan Academy.mp3", "Sentence": "We're told theta is between pi and 2 pi, and cosine of theta is equal to negative square root of 3 over 2. And phi is an acute angle, and we can assume it's a positive acute angle. So we can say an acute positive angle, or as a positive acute angle. And cosine of phi is equal to 7 25ths. Find cosine of phi plus theta exactly. So essentially, can we figure it out without a calculator? And I encourage you to pause this video and think about it on your own."}, {"video_title": "Another example using angle addition formula with cosine Trigonometry Khan Academy.mp3", "Sentence": "And cosine of phi is equal to 7 25ths. Find cosine of phi plus theta exactly. So essentially, can we figure it out without a calculator? And I encourage you to pause this video and think about it on your own. So let's see if we can work through it. So when we see, find cosine of phi plus theta, we're finding the cosine of the addition of two angles. So to me at least, that kind of screams out that maybe the angle addition formula can help us evaluate this, especially because we know what cosine of theta is, cosine of phi is, and then maybe we can also use those to figure out what sine of theta and sine of phi are."}, {"video_title": "Another example using angle addition formula with cosine Trigonometry Khan Academy.mp3", "Sentence": "And I encourage you to pause this video and think about it on your own. So let's see if we can work through it. So when we see, find cosine of phi plus theta, we're finding the cosine of the addition of two angles. So to me at least, that kind of screams out that maybe the angle addition formula can help us evaluate this, especially because we know what cosine of theta is, cosine of phi is, and then maybe we can also use those to figure out what sine of theta and sine of phi are. So let's just write out the angle addition formula. It tells us that cosine of phi, cosine of phi plus theta, is equal to cosine of both of those angles, the product of the cosines of both of those angles. So cosine phi times cosine theta minus, so this is positive, this is going to be negative, this was a negative, this would be a positive, minus the product of the sines of both of these angles."}, {"video_title": "Another example using angle addition formula with cosine Trigonometry Khan Academy.mp3", "Sentence": "So to me at least, that kind of screams out that maybe the angle addition formula can help us evaluate this, especially because we know what cosine of theta is, cosine of phi is, and then maybe we can also use those to figure out what sine of theta and sine of phi are. So let's just write out the angle addition formula. It tells us that cosine of phi, cosine of phi plus theta, is equal to cosine of both of those angles, the product of the cosines of both of those angles. So cosine phi times cosine theta minus, so this is positive, this is going to be negative, this was a negative, this would be a positive, minus the product of the sines of both of these angles. So sine of phi times sine of theta. Sine of theta. And we already know some of this information."}, {"video_title": "Another example using angle addition formula with cosine Trigonometry Khan Academy.mp3", "Sentence": "So cosine phi times cosine theta minus, so this is positive, this is going to be negative, this was a negative, this would be a positive, minus the product of the sines of both of these angles. So sine of phi times sine of theta. Sine of theta. And we already know some of this information. We know what cosine of phi is. Cosine of phi is 7 25ths. So that is 7 over 25."}, {"video_title": "Another example using angle addition formula with cosine Trigonometry Khan Academy.mp3", "Sentence": "And we already know some of this information. We know what cosine of phi is. Cosine of phi is 7 25ths. So that is 7 over 25. We know what cosine of theta is. Cosine of theta is negative square root of 3 over 2. So negative square root of 3 over 2, so we're going to take a product here for this term."}, {"video_title": "Another example using angle addition formula with cosine Trigonometry Khan Academy.mp3", "Sentence": "So that is 7 over 25. We know what cosine of theta is. Cosine of theta is negative square root of 3 over 2. So negative square root of 3 over 2, so we're going to take a product here for this term. Now we need to figure out what sine of phi and sine of theta are. Well, lucky for us, we have the Pythagorean identity. The Pythagorean identity tells us that sine squared theta plus cosine squared theta is equal to 1."}, {"video_title": "Another example using angle addition formula with cosine Trigonometry Khan Academy.mp3", "Sentence": "So negative square root of 3 over 2, so we're going to take a product here for this term. Now we need to figure out what sine of phi and sine of theta are. Well, lucky for us, we have the Pythagorean identity. The Pythagorean identity tells us that sine squared theta plus cosine squared theta is equal to 1. Or we could say that sine squared theta is equal to 1 minus cosine squared theta. Or that sine of theta is equal to the plus or minus square root of 1 minus cosine squared theta. So for example, we could use this now to figure out what sine of theta is."}, {"video_title": "Another example using angle addition formula with cosine Trigonometry Khan Academy.mp3", "Sentence": "The Pythagorean identity tells us that sine squared theta plus cosine squared theta is equal to 1. Or we could say that sine squared theta is equal to 1 minus cosine squared theta. Or that sine of theta is equal to the plus or minus square root of 1 minus cosine squared theta. So for example, we could use this now to figure out what sine of theta is. So we could say sine of theta is going to be equal to the plus or minus square root of 1 minus cosine squared theta. Well, cosine squared theta is negative square root of 3 over 2. If you square it, it's going to be positive."}, {"video_title": "Another example using angle addition formula with cosine Trigonometry Khan Academy.mp3", "Sentence": "So for example, we could use this now to figure out what sine of theta is. So we could say sine of theta is going to be equal to the plus or minus square root of 1 minus cosine squared theta. Well, cosine squared theta is negative square root of 3 over 2. If you square it, it's going to be positive. If you square the square root of 3, you're going to get 3. And if you square 2, you're going to get 4. So it's the plus or minus square root of 1 minus 3 fourths, which is equal to the plus or minus square root of 1 fourth, which is equal to plus or minus 1 half."}, {"video_title": "Another example using angle addition formula with cosine Trigonometry Khan Academy.mp3", "Sentence": "If you square it, it's going to be positive. If you square the square root of 3, you're going to get 3. And if you square 2, you're going to get 4. So it's the plus or minus square root of 1 minus 3 fourths, which is equal to the plus or minus square root of 1 fourth, which is equal to plus or minus 1 half. Now which one is it going to be? Is sine of theta going to be positive 1 half or negative 1 half? Well, to think about that, we could draw ourselves a little unit circle here."}, {"video_title": "Another example using angle addition formula with cosine Trigonometry Khan Academy.mp3", "Sentence": "So it's the plus or minus square root of 1 minus 3 fourths, which is equal to the plus or minus square root of 1 fourth, which is equal to plus or minus 1 half. Now which one is it going to be? Is sine of theta going to be positive 1 half or negative 1 half? Well, to think about that, we could draw ourselves a little unit circle here. So that's my y-axis. That is my x-axis. Let me draw a little unit circle here as neatly as I can."}, {"video_title": "Another example using angle addition formula with cosine Trigonometry Khan Academy.mp3", "Sentence": "Well, to think about that, we could draw ourselves a little unit circle here. So that's my y-axis. That is my x-axis. Let me draw a little unit circle here as neatly as I can. So a little unit circle right over there. Now what do they tell us about theta? They tell us that theta is between pi and 2 pi."}, {"video_title": "Another example using angle addition formula with cosine Trigonometry Khan Academy.mp3", "Sentence": "Let me draw a little unit circle here as neatly as I can. So a little unit circle right over there. Now what do they tell us about theta? They tell us that theta is between pi and 2 pi. So it's between pi and 2 pi. So our angle, our terminal, I guess the terminal ray of the angle, is going to sit, is going to be in the third or fourth quadrants. So we're saying sine of theta is equal to either positive 1 half or negative 1 half."}, {"video_title": "Another example using angle addition formula with cosine Trigonometry Khan Academy.mp3", "Sentence": "They tell us that theta is between pi and 2 pi. So it's between pi and 2 pi. So our angle, our terminal, I guess the terminal ray of the angle, is going to sit, is going to be in the third or fourth quadrants. So we're saying sine of theta is equal to either positive 1 half or negative 1 half. So it's either positive 1 half, which could mean it's one of these angles right over here, or it's negative 1 half, which means it's one of these angles right over here. Well, this tells us that we're in the third or fourth quadrant. So sine of theta, we don't know if theta is this angle or if theta is this angle right over here, but we know if it's in the third or fourth quadrant, the sine of it is going to be non-positive."}, {"video_title": "Another example using angle addition formula with cosine Trigonometry Khan Academy.mp3", "Sentence": "So we're saying sine of theta is equal to either positive 1 half or negative 1 half. So it's either positive 1 half, which could mean it's one of these angles right over here, or it's negative 1 half, which means it's one of these angles right over here. Well, this tells us that we're in the third or fourth quadrant. So sine of theta, we don't know if theta is this angle or if theta is this angle right over here, but we know if it's in the third or fourth quadrant, the sine of it is going to be non-positive. So we know that for this theta, sine of theta is going to be negative 1 half. So this right over here is negative 1 half. And let's think about sine of phi."}, {"video_title": "Another example using angle addition formula with cosine Trigonometry Khan Academy.mp3", "Sentence": "So sine of theta, we don't know if theta is this angle or if theta is this angle right over here, but we know if it's in the third or fourth quadrant, the sine of it is going to be non-positive. So we know that for this theta, sine of theta is going to be negative 1 half. So this right over here is negative 1 half. And let's think about sine of phi. So sine of phi is going to be equal to plus or minus the square root of 1 minus cosine of phi squared. Cosine of phi is 7 25ths, so that's 49 over 625. Let's see, what is that going to be?"}, {"video_title": "Another example using angle addition formula with cosine Trigonometry Khan Academy.mp3", "Sentence": "And let's think about sine of phi. So sine of phi is going to be equal to plus or minus the square root of 1 minus cosine of phi squared. Cosine of phi is 7 25ths, so that's 49 over 625. Let's see, what is that going to be? Actually, let me do it over here. So 625 over 625 minus 49 over 625. We wrote 1 as 625 over 625."}, {"video_title": "Another example using angle addition formula with cosine Trigonometry Khan Academy.mp3", "Sentence": "Let's see, what is that going to be? Actually, let me do it over here. So 625 over 625 minus 49 over 625. We wrote 1 as 625 over 625. That's going to be, see, 625 minus 50 would be 575. It's going to be 1 more than that. It's going to be 576 over 625."}, {"video_title": "Another example using angle addition formula with cosine Trigonometry Khan Academy.mp3", "Sentence": "We wrote 1 as 625 over 625. That's going to be, see, 625 minus 50 would be 575. It's going to be 1 more than that. It's going to be 576 over 625. So it's equal to the plus or minus square root of 576 over 625. And let's see, I know what the square root of 625 is. It's 25."}, {"video_title": "Another example using angle addition formula with cosine Trigonometry Khan Academy.mp3", "Sentence": "It's going to be 576 over 625. So it's equal to the plus or minus square root of 576 over 625. And let's see, I know what the square root of 625 is. It's 25. 576, is it 24? 24 times 24, yep, it is 576. So this is equal to the plus or minus 24 25ths."}, {"video_title": "Another example using angle addition formula with cosine Trigonometry Khan Academy.mp3", "Sentence": "It's 25. 576, is it 24? 24 times 24, yep, it is 576. So this is equal to the plus or minus 24 25ths. So sine of phi is 24 25ths. And remember, the sine of an angle is the y-coordinate of where the terminal ray intersects the unit circle. So we're either looking at one of these angles."}, {"video_title": "Another example using angle addition formula with cosine Trigonometry Khan Academy.mp3", "Sentence": "So this is equal to the plus or minus 24 25ths. So sine of phi is 24 25ths. And remember, the sine of an angle is the y-coordinate of where the terminal ray intersects the unit circle. So we're either looking at one of these angles. We're either looking at one of those angles if the sine is positive. So we're either looking at this angle or that angle, or we're looking at a terminal ray down here again. Now they tell us that phi is a positive acute angle."}, {"video_title": "Another example using angle addition formula with cosine Trigonometry Khan Academy.mp3", "Sentence": "So we're either looking at one of these angles. We're either looking at one of those angles if the sine is positive. So we're either looking at this angle or that angle, or we're looking at a terminal ray down here again. Now they tell us that phi is a positive acute angle. So we know that we're dealing actually with this scenario right over here. So sine of phi is going to be the positive 24 25ths. So it's 24 over 25."}, {"video_title": "Another example using angle addition formula with cosine Trigonometry Khan Academy.mp3", "Sentence": "Now they tell us that phi is a positive acute angle. So we know that we're dealing actually with this scenario right over here. So sine of phi is going to be the positive 24 25ths. So it's 24 over 25. And now we just have to multiply the numbers and then do the subtraction. So this is going to be equal to 7 25ths. Let me just write it down."}, {"video_title": "Another example using angle addition formula with cosine Trigonometry Khan Academy.mp3", "Sentence": "So it's 24 over 25. And now we just have to multiply the numbers and then do the subtraction. So this is going to be equal to 7 25ths. Let me just write it down. So this is going to be equal to negative 7 square roots of 3 over 25 times 2 is 50, over 50 minus, but then we're going to have a negative out here, so we could say plus, negative times negative is positive, and then 24 over 25 times 1 half is 12 over 25. So plus 12 over 25. But actually let me just write it over 50 since we have a 50 right over here."}, {"video_title": "Another example using angle addition formula with cosine Trigonometry Khan Academy.mp3", "Sentence": "Let me just write it down. So this is going to be equal to negative 7 square roots of 3 over 25 times 2 is 50, over 50 minus, but then we're going to have a negative out here, so we could say plus, negative times negative is positive, and then 24 over 25 times 1 half is 12 over 25. So plus 12 over 25. But actually let me just write it over 50 since we have a 50 right over here. So this is going to be plus 24 over 50. And so this is going to be equal to 24 minus 7 times the square root of 3 over 50. And we are done."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So we're going to include zero and two pi in the possible values for theta. So to do this, I've set up a little chart for theta, cosine theta, and sine theta. And we can use this to, and the unit circle, to hopefully quickly graph what the graphs of y equals sine theta and y equals cosine theta are. And then we can think about how many times they intersect, and maybe where they actually intersect. So let's get started. So first of all, just to be clear, this is a unit circle. This is the x-axis."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And then we can think about how many times they intersect, and maybe where they actually intersect. So let's get started. So first of all, just to be clear, this is a unit circle. This is the x-axis. This is the y-axis. Over here, we're going to graph these two graphs. So this is going to be the y-axis, and it's going to be a function of theta, not x, on the horizontal axis."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "This is the x-axis. This is the y-axis. Over here, we're going to graph these two graphs. So this is going to be the y-axis, and it's going to be a function of theta, not x, on the horizontal axis. So first, let's think about what happens when theta is equal to zero. So when theta is equal to zero, you're at this point right over here. Let me do it in a different color."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So this is going to be the y-axis, and it's going to be a function of theta, not x, on the horizontal axis. So first, let's think about what happens when theta is equal to zero. So when theta is equal to zero, you're at this point right over here. Let me do it in a different color. You're at this point right over here on the unit circle. And what coordinate is that? Well, that's the point one comma zero."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Let me do it in a different color. You're at this point right over here on the unit circle. And what coordinate is that? Well, that's the point one comma zero. And so based on that, what is cosine of theta when theta is equal to zero? Well, cosine of theta is one, and sine of theta is going to be zero. This is the x-axis at the point of intersection with the unit circle."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Well, that's the point one comma zero. And so based on that, what is cosine of theta when theta is equal to zero? Well, cosine of theta is one, and sine of theta is going to be zero. This is the x-axis at the point of intersection with the unit circle. This is the x-coordinate at the point of intersection with the unit circle. This is the y-coordinate. Let's keep going."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "This is the x-axis at the point of intersection with the unit circle. This is the x-coordinate at the point of intersection with the unit circle. This is the y-coordinate. Let's keep going. What about pi over two? So pi over two, we are right over here. What is that coordinate?"}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Let's keep going. What about pi over two? So pi over two, we are right over here. What is that coordinate? Well, that's now x is zero, y is one. So based on that, cosine of theta is zero, and what is sine of theta? Well, that's going to be one."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "What is that coordinate? Well, that's now x is zero, y is one. So based on that, cosine of theta is zero, and what is sine of theta? Well, that's going to be one. It's the y-coordinate right over here. Now let's go all the way to pi. We're at this point in the unit circle."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Well, that's going to be one. It's the y-coordinate right over here. Now let's go all the way to pi. We're at this point in the unit circle. What is the coordinate? Well, this is negative one comma zero. So what is cosine of theta?"}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "We're at this point in the unit circle. What is the coordinate? Well, this is negative one comma zero. So what is cosine of theta? Well, it's the x-coordinate here, which is negative one, and sine of theta is going to be the y-coordinate, which is zero. Now let's keep going. Now we're down here at three pi over two."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So what is cosine of theta? Well, it's the x-coordinate here, which is negative one, and sine of theta is going to be the y-coordinate, which is zero. Now let's keep going. Now we're down here at three pi over two. If we go all the way around to three pi over two, what is this coordinate? Well, this is zero, negative one. Cosine of theta is the x-coordinate here, so cosine of theta is going to be zero, and what is sine of theta going to be?"}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Now we're down here at three pi over two. If we go all the way around to three pi over two, what is this coordinate? Well, this is zero, negative one. Cosine of theta is the x-coordinate here, so cosine of theta is going to be zero, and what is sine of theta going to be? Well, it's going to be negative one. And then finally we go back to two pi, which is making a full revolution around the circle. We went all the way around, and we're back to this point right over here."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Cosine of theta is the x-coordinate here, so cosine of theta is going to be zero, and what is sine of theta going to be? Well, it's going to be negative one. And then finally we go back to two pi, which is making a full revolution around the circle. We went all the way around, and we're back to this point right over here. So the coordinate is the exact same thing as when the angle equals zero radians. And so what is cosine of theta? Well, that's one, and sine of theta is zero."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "We went all the way around, and we're back to this point right over here. So the coordinate is the exact same thing as when the angle equals zero radians. And so what is cosine of theta? Well, that's one, and sine of theta is zero. And from this we can make a rough sketch of the graph and think about where they might intersect. So first let's do cosine of theta. When theta is zero, and let me mark this off."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Well, that's one, and sine of theta is zero. And from this we can make a rough sketch of the graph and think about where they might intersect. So first let's do cosine of theta. When theta is zero, and let me mark this off. So this is going to be when y is equal to one, and this is when y is equal to negative one. So y equals cosine of theta. I'm going to graph, let's see, theta equals zero, cosine of theta equals one."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "When theta is zero, and let me mark this off. So this is going to be when y is equal to one, and this is when y is equal to negative one. So y equals cosine of theta. I'm going to graph, let's see, theta equals zero, cosine of theta equals one. So cosine of theta is equal to one. When theta is equal to pi over two, cosine of theta is zero. When theta is equal to pi, cosine of theta is negative one."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "I'm going to graph, let's see, theta equals zero, cosine of theta equals one. So cosine of theta is equal to one. When theta is equal to pi over two, cosine of theta is zero. When theta is equal to pi, cosine of theta is negative one. When theta is equal to three pi over two, cosine of theta is equal to zero. That's this right over here. And then finally when theta is two pi, cosine of theta is one again."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "When theta is equal to pi, cosine of theta is negative one. When theta is equal to three pi over two, cosine of theta is equal to zero. That's this right over here. And then finally when theta is two pi, cosine of theta is one again. And the curve will look something like this. My best attempt to draw it. Make it a nice, smooth curve."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And then finally when theta is two pi, cosine of theta is one again. And the curve will look something like this. My best attempt to draw it. Make it a nice, smooth curve. So it's going to look something like this. The look of these curves should look somewhat familiar at this point. So it should look something like this."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Make it a nice, smooth curve. So it's going to look something like this. The look of these curves should look somewhat familiar at this point. So it should look something like this. So this is the graph of y is equal to cosine of theta. Now let's do the same thing for sine theta. When theta is equal to zero, sine theta is zero."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So it should look something like this. So this is the graph of y is equal to cosine of theta. Now let's do the same thing for sine theta. When theta is equal to zero, sine theta is zero. When theta is pi over two, sine of theta is one. When theta is equal to pi, sine of theta is zero. When theta is equal to three pi over two, sine of theta is negative one."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "When theta is equal to zero, sine theta is zero. When theta is pi over two, sine of theta is one. When theta is equal to pi, sine of theta is zero. When theta is equal to three pi over two, sine of theta is negative one. When theta is equal to two pi, sine of theta is equal to zero. And so the graph of sine of theta is going to look something like this. My best attempt at drawing it is going to look something like this."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "When theta is equal to three pi over two, sine of theta is negative one. When theta is equal to two pi, sine of theta is equal to zero. And so the graph of sine of theta is going to look something like this. My best attempt at drawing it is going to look something like this. So just visually we can think about the question. At how many points do the graphs of y equals sine of theta and y equals cosine of theta intersect for this range for theta? For theta being between zero and two pi, including those two points."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "My best attempt at drawing it is going to look something like this. So just visually we can think about the question. At how many points do the graphs of y equals sine of theta and y equals cosine of theta intersect for this range for theta? For theta being between zero and two pi, including those two points. Well, you just look at this graph. You see there's two points of intersection. This point right over here and this point right over here."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "For theta being between zero and two pi, including those two points. Well, you just look at this graph. You see there's two points of intersection. This point right over here and this point right over here. Just over the, between zero and two pi. These are cyclical graphs. If we kept going, they would keep intersecting with each other."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "This point right over here and this point right over here. Just over the, between zero and two pi. These are cyclical graphs. If we kept going, they would keep intersecting with each other. But just over this two pi range for theta, you get two points of intersection. Now let's think about what they are. Because they look to be pretty close between, right between zero and pi over two and right between pi and three pi over two."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "If we kept going, they would keep intersecting with each other. But just over this two pi range for theta, you get two points of intersection. Now let's think about what they are. Because they look to be pretty close between, right between zero and pi over two and right between pi and three pi over two. So let's look at our unit circle if we can figure out what those values are. It looks like this is at pi over four. So let's verify that."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Because they look to be pretty close between, right between zero and pi over two and right between pi and three pi over two. So let's look at our unit circle if we can figure out what those values are. It looks like this is at pi over four. So let's verify that. So let's think about what these values are at pi over four. So pi over four is that angle, or that's the terminal side of it. So this is pi over four."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So let's verify that. So let's think about what these values are at pi over four. So pi over four is that angle, or that's the terminal side of it. So this is pi over four. Pi over four is the exact same thing as a 45 degree angle. So let's do pi over four right over here. So we have to figure out what this point is, what the coordinates are."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So this is pi over four. Pi over four is the exact same thing as a 45 degree angle. So let's do pi over four right over here. So we have to figure out what this point is, what the coordinates are. So let's make this a right triangle. And so what do we know about this right triangle? And I'm going to draw it right over here to make it a little clearer."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So we have to figure out what this point is, what the coordinates are. So let's make this a right triangle. And so what do we know about this right triangle? And I'm going to draw it right over here to make it a little clearer. This is a very typical type of right triangle, so it's good to get some familiarity with it. Let me draw my best attempt. All right."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And I'm going to draw it right over here to make it a little clearer. This is a very typical type of right triangle, so it's good to get some familiarity with it. Let me draw my best attempt. All right. So we know it's a right triangle. We know that this is 45 degrees. What is the length of the hypotenuse?"}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "All right. So we know it's a right triangle. We know that this is 45 degrees. What is the length of the hypotenuse? Well, this is a unit circle. It has radius one, so the length of the hypotenuse here is one. And what do we know about this angle right over here?"}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "What is the length of the hypotenuse? Well, this is a unit circle. It has radius one, so the length of the hypotenuse here is one. And what do we know about this angle right over here? Well, we know that it too must be 45 degrees because all of these angles have to add up to 180. And since these two angles are the same, we know that these two sides are going to be the same. And then we could use the Pythagorean theorem to think about the length of those sides."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And what do we know about this angle right over here? Well, we know that it too must be 45 degrees because all of these angles have to add up to 180. And since these two angles are the same, we know that these two sides are going to be the same. And then we could use the Pythagorean theorem to think about the length of those sides. So using the Pythagorean theorem, knowing that these two sides are equal, what do we get for the length of those sides? Well, let's call these, if this has length a, well, then this also has length a. And we can use the Pythagorean theorem, and we could say a squared plus a squared is equal to the hypotenuse squared is equal to one."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And then we could use the Pythagorean theorem to think about the length of those sides. So using the Pythagorean theorem, knowing that these two sides are equal, what do we get for the length of those sides? Well, let's call these, if this has length a, well, then this also has length a. And we can use the Pythagorean theorem, and we could say a squared plus a squared is equal to the hypotenuse squared is equal to one. Or 2a squared is equal to 1, a squared is equal to 1 half, take the principal root of both sides. a is equal to the square root of 1 half, which is the square root of 1, which is 1 over the square root of 2. We can rationalize the denominator here by multiplying by square root of 2 over square root of 2, which gives us a is equal to, in the numerator, square root of 2, and in the denominator, square root of 2 times square root of 2 is 2."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And we can use the Pythagorean theorem, and we could say a squared plus a squared is equal to the hypotenuse squared is equal to one. Or 2a squared is equal to 1, a squared is equal to 1 half, take the principal root of both sides. a is equal to the square root of 1 half, which is the square root of 1, which is 1 over the square root of 2. We can rationalize the denominator here by multiplying by square root of 2 over square root of 2, which gives us a is equal to, in the numerator, square root of 2, and in the denominator, square root of 2 times square root of 2 is 2. So this length is square root of 2 over 2, and this length is the same thing. So this length right over here is square root of 2 over 2, and this height right over here is also square root of 2 over 2. So based on that, what is this coordinate point?"}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "We can rationalize the denominator here by multiplying by square root of 2 over square root of 2, which gives us a is equal to, in the numerator, square root of 2, and in the denominator, square root of 2 times square root of 2 is 2. So this length is square root of 2 over 2, and this length is the same thing. So this length right over here is square root of 2 over 2, and this height right over here is also square root of 2 over 2. So based on that, what is this coordinate point? Well, it's square root of 2 over 2 to the right, in the positive direction, so x is equal to square root of 2 over 2, and y is square root of 2 over 2 in the upwards direction, the vertical direction, the positive vertical direction, so it's also square root of 2 over 2. Cosine of theta is just the x coordinate, so it's square root of 2 over 2. Sine of theta is just the y coordinate."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So based on that, what is this coordinate point? Well, it's square root of 2 over 2 to the right, in the positive direction, so x is equal to square root of 2 over 2, and y is square root of 2 over 2 in the upwards direction, the vertical direction, the positive vertical direction, so it's also square root of 2 over 2. Cosine of theta is just the x coordinate, so it's square root of 2 over 2. Sine of theta is just the y coordinate. So you see immediately that they are indeed equal at that point. So at this point, they are both equal to square root of 2 over 2. Now what about this point right over here, which looks right in between pi and 3 pi over 2?"}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Sine of theta is just the y coordinate. So you see immediately that they are indeed equal at that point. So at this point, they are both equal to square root of 2 over 2. Now what about this point right over here, which looks right in between pi and 3 pi over 2? So that's going to be, so this is pi, this is 3 pi over 2, it is right over here, so it's another pi over 4 plus pi, so pi plus pi over 4 is the same thing as 4 pi over 4 plus pi over 4, so this is the angle 5 pi over 4. So this is 5 pi over 4, so this is equal to 5 pi over 4, so that's what we're trying to figure out. What are the value of these functions at theta equal 5 pi over 4?"}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Now what about this point right over here, which looks right in between pi and 3 pi over 2? So that's going to be, so this is pi, this is 3 pi over 2, it is right over here, so it's another pi over 4 plus pi, so pi plus pi over 4 is the same thing as 4 pi over 4 plus pi over 4, so this is the angle 5 pi over 4. So this is 5 pi over 4, so this is equal to 5 pi over 4, so that's what we're trying to figure out. What are the value of these functions at theta equal 5 pi over 4? Well, there's multiple ways to think about it. You can even use a little bit of geometry to say, well, if this is a 45 degree angle, then this right over here is also a 45 degree angle. You could say that the reference angle in terms of degrees is 45 degrees."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "What are the value of these functions at theta equal 5 pi over 4? Well, there's multiple ways to think about it. You can even use a little bit of geometry to say, well, if this is a 45 degree angle, then this right over here is also a 45 degree angle. You could say that the reference angle in terms of degrees is 45 degrees. And we could do a very similar thing. We can draw a right triangle. We know the hypotenuse is 1."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "You could say that the reference angle in terms of degrees is 45 degrees. And we could do a very similar thing. We can draw a right triangle. We know the hypotenuse is 1. We know that if this is a right angle, this is 45 degrees. If that's 45 degrees, then this is also 45 degrees. And we have a triangle that's very similar."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "We know the hypotenuse is 1. We know that if this is a right angle, this is 45 degrees. If that's 45 degrees, then this is also 45 degrees. And we have a triangle that's very similar. They're actually congruent triangles. So hypotenuse is 1, 45, 45, 90. We then know that the length of this side is square root of 2 over 2, and the length of this side is square root of 2 over 2."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And we have a triangle that's very similar. They're actually congruent triangles. So hypotenuse is 1, 45, 45, 90. We then know that the length of this side is square root of 2 over 2, and the length of this side is square root of 2 over 2. The exact same logic we used over here. So based on that, what is the coordinate of that point? Well, let's think about the x value."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "We then know that the length of this side is square root of 2 over 2, and the length of this side is square root of 2 over 2. The exact same logic we used over here. So based on that, what is the coordinate of that point? Well, let's think about the x value. It's square root of 2 over 2 in the negative direction. We have to go square root of 2 over 2 to the left of the origin. So it's negative square root of 2 over 2."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Well, let's think about the x value. It's square root of 2 over 2 in the negative direction. We have to go square root of 2 over 2 to the left of the origin. So it's negative square root of 2 over 2. This point on the x-axis is negative square root of 2 over 2. What about the y value? We have to go square root of 2 over 2 down, in the downward direction from the origin."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So it's negative square root of 2 over 2. This point on the x-axis is negative square root of 2 over 2. What about the y value? We have to go square root of 2 over 2 down, in the downward direction from the origin. So it's also negative square root of 2 over 2. So cosine of theta is negative square root of 2 over 2, and sine of theta is also negative square root of 2 over 2. So we see that we do indeed have the same value for cosine of theta and sine of theta right there."}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3", "Sentence": "Give the lengths to the nearest tenth. So when they say solve the right triangle, we can assume that they're saying, hey, figure out the lengths of all the sides, so whatever a is equal to, whatever b is equal to, and also what are all the angles of the right triangle? They've given two of them, we might have to figure out this third right over here. So there's multiple ways to tackle this, but we'll just go and we'll just try to tackle side xw first, try to figure out what a is, and I'll give you a hint. You can use a calculator, and using a calculator, you can use your trigonometric functions that we've looked at a good bit now. So I'll give you a few seconds to think about how to figure out what a is. Well, what do we know?"}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3", "Sentence": "So there's multiple ways to tackle this, but we'll just go and we'll just try to tackle side xw first, try to figure out what a is, and I'll give you a hint. You can use a calculator, and using a calculator, you can use your trigonometric functions that we've looked at a good bit now. So I'll give you a few seconds to think about how to figure out what a is. Well, what do we know? We know this angle y right over here, we know the side adjacent to angle y, and length a, this is the side, that's the length of the side that is opposite, that is opposite to angle y. So what trigonometric ratio, what trigonometric ratio deals with the opposite and the adjacent? So if we're looking at angle y, relative to angle y, this is the opposite, and this right over here is the adjacent."}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3", "Sentence": "Well, what do we know? We know this angle y right over here, we know the side adjacent to angle y, and length a, this is the side, that's the length of the side that is opposite, that is opposite to angle y. So what trigonometric ratio, what trigonometric ratio deals with the opposite and the adjacent? So if we're looking at angle y, relative to angle y, this is the opposite, and this right over here is the adjacent. Well, if we don't remember, we can go back to SOHCAHTOA. SOHCAHTOA. Sine deals with opposite and hypotenuse, cosine deals with adjacent and hypotenuse, tangent deals with opposite over adjacent, opposite over adjacent."}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3", "Sentence": "So if we're looking at angle y, relative to angle y, this is the opposite, and this right over here is the adjacent. Well, if we don't remember, we can go back to SOHCAHTOA. SOHCAHTOA. Sine deals with opposite and hypotenuse, cosine deals with adjacent and hypotenuse, tangent deals with opposite over adjacent, opposite over adjacent. So we can say that the tangent, the tangent of 65 degrees, of that angle of 65 degrees, is equal to the opposite, the length of the opposite side, which we know has length a, over the length of the adjacent side, which they gave us in the diagram, which has length, which has length five. And you might say, well, how do I figure out a? Well, we can use our calculator to evaluate what the tangent of 65 degrees are, and then we can solve for a."}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3", "Sentence": "Sine deals with opposite and hypotenuse, cosine deals with adjacent and hypotenuse, tangent deals with opposite over adjacent, opposite over adjacent. So we can say that the tangent, the tangent of 65 degrees, of that angle of 65 degrees, is equal to the opposite, the length of the opposite side, which we know has length a, over the length of the adjacent side, which they gave us in the diagram, which has length, which has length five. And you might say, well, how do I figure out a? Well, we can use our calculator to evaluate what the tangent of 65 degrees are, and then we can solve for a. And actually, if we just wanna get the expression, explicitly solving for a, we can just multiply both sides of this equation times five. So let's do that. Five times, times five, these cancel out, and we are left with, if we flip the equal around, we're left with a is equal to five times the tangent of 65, of 65 degrees."}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3", "Sentence": "Well, we can use our calculator to evaluate what the tangent of 65 degrees are, and then we can solve for a. And actually, if we just wanna get the expression, explicitly solving for a, we can just multiply both sides of this equation times five. So let's do that. Five times, times five, these cancel out, and we are left with, if we flip the equal around, we're left with a is equal to five times the tangent of 65, of 65 degrees. So now we can get our calculator out and figure out what this is to the nearest 10th. My handy TI-85 out, and I have five times the, not the, the tangent, no, I didn't need to press that second right over there, just the regular tangent, of 65 degrees. And I am, I get, if I round to the nearest 10th like they asked me to, I get 10.7."}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3", "Sentence": "Five times, times five, these cancel out, and we are left with, if we flip the equal around, we're left with a is equal to five times the tangent of 65, of 65 degrees. So now we can get our calculator out and figure out what this is to the nearest 10th. My handy TI-85 out, and I have five times the, not the, the tangent, no, I didn't need to press that second right over there, just the regular tangent, of 65 degrees. And I am, I get, if I round to the nearest 10th like they asked me to, I get 10.7. So this is, so a is approximately equal to 10.7. I say approximately, because I rounded it, I rounded it down. I didn't, this is not the exact number, but a is equal to 10.7."}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3", "Sentence": "And I am, I get, if I round to the nearest 10th like they asked me to, I get 10.7. So this is, so a is approximately equal to 10.7. I say approximately, because I rounded it, I rounded it down. I didn't, this is not the exact number, but a is equal to 10.7. So we now know that this has length 10.7 approximately. There's several ways that we can try to tackle b, and I'll let you pick the way you want to, but then I'll just do it the way I would like to. So my next question to you is, what is the length of the side yw, or what is the value of b?"}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3", "Sentence": "I didn't, this is not the exact number, but a is equal to 10.7. So we now know that this has length 10.7 approximately. There's several ways that we can try to tackle b, and I'll let you pick the way you want to, but then I'll just do it the way I would like to. So my next question to you is, what is the length of the side yw, or what is the value of b? Well, there's several ways to do it. This is a hypotenuse, so we could use trigonometric functions that deal with adjacent over hypotenuse or opposite over hypotenuse, or we could just use the Pythagorean theorem. We know two sides of a right triangle."}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3", "Sentence": "So my next question to you is, what is the length of the side yw, or what is the value of b? Well, there's several ways to do it. This is a hypotenuse, so we could use trigonometric functions that deal with adjacent over hypotenuse or opposite over hypotenuse, or we could just use the Pythagorean theorem. We know two sides of a right triangle. We can come up with the third side. I will go with using trigonometric ratios, since that's what we've been working on a good bit. So this length b, that's the length of the hypotenuse."}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3", "Sentence": "We know two sides of a right triangle. We can come up with the third side. I will go with using trigonometric ratios, since that's what we've been working on a good bit. So this length b, that's the length of the hypotenuse. So this side wy is the hypotenuse. And so what trigonometric ratios, and we can decide what we want to use. We could use opposite and hypotenuse."}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3", "Sentence": "So this length b, that's the length of the hypotenuse. So this side wy is the hypotenuse. And so what trigonometric ratios, and we can decide what we want to use. We could use opposite and hypotenuse. We could use adjacent and hypotenuse. Since we know that xy is exactly five, we don't have to deal with this approximation. Let's use that side."}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3", "Sentence": "We could use opposite and hypotenuse. We could use adjacent and hypotenuse. Since we know that xy is exactly five, we don't have to deal with this approximation. Let's use that side. So what trigonometric ratios deal with adjacent and hypotenuse? Well, we see from Sohcahtoa, cosine deals with adjacent over hypotenuse. So we could say that the cosine of 65 degrees, cosine of 65 degrees, is equal to the length of the adjacent side, which is five, over the length of the hypotenuse, which has length b."}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3", "Sentence": "Let's use that side. So what trigonometric ratios deal with adjacent and hypotenuse? Well, we see from Sohcahtoa, cosine deals with adjacent over hypotenuse. So we could say that the cosine of 65 degrees, cosine of 65 degrees, is equal to the length of the adjacent side, which is five, over the length of the hypotenuse, which has length b. And then we can try to solve for b. You multiply both sides times b. You're left with b times cosine of 65 degrees is equal to five."}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3", "Sentence": "So we could say that the cosine of 65 degrees, cosine of 65 degrees, is equal to the length of the adjacent side, which is five, over the length of the hypotenuse, which has length b. And then we can try to solve for b. You multiply both sides times b. You're left with b times cosine of 65 degrees is equal to five. And then to solve for b, you can divide both sides by cosine of 65 degrees. This is just a number here. So we're just dividing."}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3", "Sentence": "You're left with b times cosine of 65 degrees is equal to five. And then to solve for b, you can divide both sides by cosine of 65 degrees. This is just a number here. So we're just dividing. We have to figure it out with our calculator, but this is just going to evaluate to some number. So we can divide both sides by that, by cosine of 65 degrees, cosine of 65 degrees. And we're left with b is equal to five over the cosine of 65 degrees."}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3", "Sentence": "So we're just dividing. We have to figure it out with our calculator, but this is just going to evaluate to some number. So we can divide both sides by that, by cosine of 65 degrees, cosine of 65 degrees. And we're left with b is equal to five over the cosine of 65 degrees. So let us now use our calculator to figure out the length of b. Length of b is five divided by cosine of 65 degrees. And I get, if I round to the nearest tenth, 11.8."}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3", "Sentence": "And we're left with b is equal to five over the cosine of 65 degrees. So let us now use our calculator to figure out the length of b. Length of b is five divided by cosine of 65 degrees. And I get, if I round to the nearest tenth, 11.8. So b is approximately equal to, rounded to the nearest tenth, 11.8. So b is equal to 11.8. And then we're almost done solving this right triangle."}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3", "Sentence": "And I get, if I round to the nearest tenth, 11.8. So b is approximately equal to, rounded to the nearest tenth, 11.8. So b is equal to 11.8. And then we're almost done solving this right triangle. And you could have figured this out using the Pythagorean theorem as well, saying that five squared plus 10.7 squared should be equal to b squared. And hopefully you would get the exact same answer. And the last thing we have to figure out is the measure of angle w. Angle w right over here."}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3", "Sentence": "And then we're almost done solving this right triangle. And you could have figured this out using the Pythagorean theorem as well, saying that five squared plus 10.7 squared should be equal to b squared. And hopefully you would get the exact same answer. And the last thing we have to figure out is the measure of angle w. Angle w right over here. So I'll give you a few seconds to think about what the measure of angle w is. Well, here we just have to remember that the sum of the angles of a triangle add up to 180 degrees. So angle w plus 65 degrees, that's this angle right up here, plus the right angle, this is a right triangle, they're going to add up to 180 degrees."}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3", "Sentence": "And the last thing we have to figure out is the measure of angle w. Angle w right over here. So I'll give you a few seconds to think about what the measure of angle w is. Well, here we just have to remember that the sum of the angles of a triangle add up to 180 degrees. So angle w plus 65 degrees, that's this angle right up here, plus the right angle, this is a right triangle, they're going to add up to 180 degrees. So all we need to do is, well, we could simplify the left-hand side right over here. 65 plus 90 is 155. So angle w plus 155 degrees is equal to 180 degrees."}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "We've got two right triangles here. Let's say we also know that they both have an angle whose measure is equal to theta. So angle A is congruent to angle D. What do we now know about these two triangles? Well, for any right, or any triangle, if you know two of the angles, you're going to know the third angle because the sum of the angles of a triangle add up to 180 degrees. So if you have two angles in common, that means you're going to have three angles in common. And if you have three angles in common, you are dealing with similar triangles. Let me make that a little bit clearer."}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Well, for any right, or any triangle, if you know two of the angles, you're going to know the third angle because the sum of the angles of a triangle add up to 180 degrees. So if you have two angles in common, that means you're going to have three angles in common. And if you have three angles in common, you are dealing with similar triangles. Let me make that a little bit clearer. So if this angle is theta, this is 90, they all have to add up to 180 degrees. That means that this angle plus this angle up here have to add up to 90. We've already used up 90 right over here."}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Let me make that a little bit clearer. So if this angle is theta, this is 90, they all have to add up to 180 degrees. That means that this angle plus this angle up here have to add up to 90. We've already used up 90 right over here. So this angle, angle A and angle B need to be complements. So this angle right over here needs to be 90 minus theta. Well, we can use the same logic over here."}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "We've already used up 90 right over here. So this angle, angle A and angle B need to be complements. So this angle right over here needs to be 90 minus theta. Well, we can use the same logic over here. We already used up 90 degrees over here. So we have a remaining 90 degrees between theta and that angle. So this angle is going to be 90 degrees minus theta."}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Well, we can use the same logic over here. We already used up 90 degrees over here. So we have a remaining 90 degrees between theta and that angle. So this angle is going to be 90 degrees minus theta. 90 degrees minus theta. You have three congruent, three corresponding angles being congruent, you are dealing with similar triangles. Now, why is that interesting?"}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So this angle is going to be 90 degrees minus theta. 90 degrees minus theta. You have three congruent, three corresponding angles being congruent, you are dealing with similar triangles. Now, why is that interesting? Well, we know from geometry that the ratio of corresponding sides of a similar triangles are always going to be the same. So let's explore the corresponding sides here. Well, the side that jumps out when you're dealing with the right triangles the most is always the hypotenuse."}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Now, why is that interesting? Well, we know from geometry that the ratio of corresponding sides of a similar triangles are always going to be the same. So let's explore the corresponding sides here. Well, the side that jumps out when you're dealing with the right triangles the most is always the hypotenuse. So this right over here is the hypotenuse. This hypotenuse is going to correspond to this hypotenuse right over here. And we could write that down."}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Well, the side that jumps out when you're dealing with the right triangles the most is always the hypotenuse. So this right over here is the hypotenuse. This hypotenuse is going to correspond to this hypotenuse right over here. And we could write that down. This is the hypotenuse of this triangle. This is the hypotenuse of that triangle. Now, this side right over here, side BC, what side does that correspond to?"}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "And we could write that down. This is the hypotenuse of this triangle. This is the hypotenuse of that triangle. Now, this side right over here, side BC, what side does that correspond to? Well, if you look at this triangle, you can kind of view it as the side that is opposite this angle theta. So it's opposite. If you go across a triangle, you get there."}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Now, this side right over here, side BC, what side does that correspond to? Well, if you look at this triangle, you can kind of view it as the side that is opposite this angle theta. So it's opposite. If you go across a triangle, you get there. So let's go opposite angle D. If you go opposite angle A, you get to BC. Opposite angle D, you get to EF. You get EF, so it corresponds to this side right over here."}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "If you go across a triangle, you get there. So let's go opposite angle D. If you go opposite angle A, you get to BC. Opposite angle D, you get to EF. You get EF, so it corresponds to this side right over here. And then finally, side AC is the one remaining one. We could view it as, well, there's two sides that make up this angle A. One of them is a hypotenuse."}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "You get EF, so it corresponds to this side right over here. And then finally, side AC is the one remaining one. We could view it as, well, there's two sides that make up this angle A. One of them is a hypotenuse. We could call this maybe the adjacent side to it. And so D corresponds to A. And so this would be the side that corresponds."}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "One of them is a hypotenuse. We could call this maybe the adjacent side to it. And so D corresponds to A. And so this would be the side that corresponds. Now, the whole reason I did that is to leverage that corresponding sides, the ratio between corresponding sides of similar triangles is always going to be the same. So for example, the ratio between BC and the hypotenuse BA, so let me write that down. BC over BA is going to be equal to EF over ED."}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "And so this would be the side that corresponds. Now, the whole reason I did that is to leverage that corresponding sides, the ratio between corresponding sides of similar triangles is always going to be the same. So for example, the ratio between BC and the hypotenuse BA, so let me write that down. BC over BA is going to be equal to EF over ED. EF, the length of segment EF over the length of segment ED. Over the length of segment ED. Or we could also write that the length of segment AC, so AC over the hypotenuse, over this triangle's hypotenuse, over AB is equal to DF over DE."}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "BC over BA is going to be equal to EF over ED. EF, the length of segment EF over the length of segment ED. Over the length of segment ED. Or we could also write that the length of segment AC, so AC over the hypotenuse, over this triangle's hypotenuse, over AB is equal to DF over DE. Once again, this green side over the orange side, these are similar triangles. They're corresponding to each other. So this is equal to, this is equal to DF over DE."}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Or we could also write that the length of segment AC, so AC over the hypotenuse, over this triangle's hypotenuse, over AB is equal to DF over DE. Once again, this green side over the orange side, these are similar triangles. They're corresponding to each other. So this is equal to, this is equal to DF over DE. Over DE. Or we could also say, we could keep going, but I'll just do another one. Or we could say that the ratio of this side, right over this blue side to the green side, so of this triangle, BC, the length of BC over CA, over CA is going to be the same as the blue, the ratio between these two corresponding sides, the blue over the green."}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So this is equal to, this is equal to DF over DE. Over DE. Or we could also say, we could keep going, but I'll just do another one. Or we could say that the ratio of this side, right over this blue side to the green side, so of this triangle, BC, the length of BC over CA, over CA is going to be the same as the blue, the ratio between these two corresponding sides, the blue over the green. EF over DF. Over DF. And we got all of this from the fact that these are similar triangles."}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Or we could say that the ratio of this side, right over this blue side to the green side, so of this triangle, BC, the length of BC over CA, over CA is going to be the same as the blue, the ratio between these two corresponding sides, the blue over the green. EF over DF. Over DF. And we got all of this from the fact that these are similar triangles. So if this is true for all similar triangles, or this is true for any right triangle that has an angle theta, then those two triangles are going to be similar, and all of these ratios are going to be the same. Well, maybe we can give names to these ratios relative to the angle theta. So from angle theta's point of view, from theta's point of view, I'll write theta right over here, or we can just remember that."}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "And we got all of this from the fact that these are similar triangles. So if this is true for all similar triangles, or this is true for any right triangle that has an angle theta, then those two triangles are going to be similar, and all of these ratios are going to be the same. Well, maybe we can give names to these ratios relative to the angle theta. So from angle theta's point of view, from theta's point of view, I'll write theta right over here, or we can just remember that. What are these, what is the ratio of these two sides? Well, from theta's point of view, that blue side is the opposite side, it's opposite, so the opposite side of the right triangle. And then the orange side, we've already labeled the hypotenuse."}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So from angle theta's point of view, from theta's point of view, I'll write theta right over here, or we can just remember that. What are these, what is the ratio of these two sides? Well, from theta's point of view, that blue side is the opposite side, it's opposite, so the opposite side of the right triangle. And then the orange side, we've already labeled the hypotenuse. So from theta's point of view, this is the opposite side over the hypotenuse. And I keep stating from theta's point of view, because that wouldn't be the case for this other angle, for angle B. From angle B's point of view, this is the adjacent side over the hypotenuse."}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "And then the orange side, we've already labeled the hypotenuse. So from theta's point of view, this is the opposite side over the hypotenuse. And I keep stating from theta's point of view, because that wouldn't be the case for this other angle, for angle B. From angle B's point of view, this is the adjacent side over the hypotenuse. And we'll think about that relationship later on. But let's just all think of it from theta's point of view right over here. So from theta's point of view, what is this?"}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "From angle B's point of view, this is the adjacent side over the hypotenuse. And we'll think about that relationship later on. But let's just all think of it from theta's point of view right over here. So from theta's point of view, what is this? Well, theta's right over here. Clearly, A, B, and D, E, A, B, and D, E are still the hypotenuses, hypotenai, I don't know how to say that in plural again. And what is AC?"}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So from theta's point of view, what is this? Well, theta's right over here. Clearly, A, B, and D, E, A, B, and D, E are still the hypotenuses, hypotenai, I don't know how to say that in plural again. And what is AC? Or what is AC and what are DF? Well, these are adjacent to it. They're one of the sides, two sides that make up this angle that is not the hypotenuse."}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "And what is AC? Or what is AC and what are DF? Well, these are adjacent to it. They're one of the sides, two sides that make up this angle that is not the hypotenuse. So this we can view as the ratio in either of these triangles between the adjacent side. So this is relative, once again, this is opposite angle B, but we're only thinking about angle A right here, or the angle that measures theta, or angle D right over here. Relative to angle A, AC is adjacent."}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "They're one of the sides, two sides that make up this angle that is not the hypotenuse. So this we can view as the ratio in either of these triangles between the adjacent side. So this is relative, once again, this is opposite angle B, but we're only thinking about angle A right here, or the angle that measures theta, or angle D right over here. Relative to angle A, AC is adjacent. Relative to angle D, DF is adjacent. So this ratio right over here is the adjacent over the hypotenuse. And it's going to be the same for any right triangle that has an angle theta in it."}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Relative to angle A, AC is adjacent. Relative to angle D, DF is adjacent. So this ratio right over here is the adjacent over the hypotenuse. And it's going to be the same for any right triangle that has an angle theta in it. And then finally, this over here, this is going to be the opposite side. Once again, this was the opposite side over here. This ratio for either right triangle is going to be the opposite side over the adjacent side."}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "And it's going to be the same for any right triangle that has an angle theta in it. And then finally, this over here, this is going to be the opposite side. Once again, this was the opposite side over here. This ratio for either right triangle is going to be the opposite side over the adjacent side. Over the adjacent side. And I really wanna stress the importance, and we're gonna do many, many more examples of this to make this very concrete. But for any right triangle that has an angle theta, the ratio between its opposite side and its hypotenuse is going to be the same."}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "This ratio for either right triangle is going to be the opposite side over the adjacent side. Over the adjacent side. And I really wanna stress the importance, and we're gonna do many, many more examples of this to make this very concrete. But for any right triangle that has an angle theta, the ratio between its opposite side and its hypotenuse is going to be the same. That comes out of similar triangles, we just explored that. The ratio between the adjacent side to that angle that is theta and the hypotenuse is going to be the same for any of these triangles as long as it has that angle theta in it. And the ratio relative to the angle theta between the opposite side and the adjacent side, between the blue side and the green side, is always going to be the same."}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "But for any right triangle that has an angle theta, the ratio between its opposite side and its hypotenuse is going to be the same. That comes out of similar triangles, we just explored that. The ratio between the adjacent side to that angle that is theta and the hypotenuse is going to be the same for any of these triangles as long as it has that angle theta in it. And the ratio relative to the angle theta between the opposite side and the adjacent side, between the blue side and the green side, is always going to be the same. These are similar triangles. So given that, mathematicians decided to give these things names. Relative to the angle theta, this ratio is always going to be the same."}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "And the ratio relative to the angle theta between the opposite side and the adjacent side, between the blue side and the green side, is always going to be the same. These are similar triangles. So given that, mathematicians decided to give these things names. Relative to the angle theta, this ratio is always going to be the same. So they call this, the opposite of our hypotenuse, they call this the sine of the angle theta. So this is the sine, let me do this in a new color. This is by definition, and we're going to extend this definition in the future, this is sine of theta."}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Relative to the angle theta, this ratio is always going to be the same. So they call this, the opposite of our hypotenuse, they call this the sine of the angle theta. So this is the sine, let me do this in a new color. This is by definition, and we're going to extend this definition in the future, this is sine of theta. This right over here, by definition, is the cosine of theta. And this right over here, by definition, is the tangent. That, by definition, is the tangent of theta."}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "This is by definition, and we're going to extend this definition in the future, this is sine of theta. This right over here, by definition, is the cosine of theta. And this right over here, by definition, is the tangent. That, by definition, is the tangent of theta. And a mnemonic that will help you remember this, and these really are just definitions. People realize, wow, by similar triangles, for any angle theta, this ratio is always going to be the same. Because of similar triangles, for any angle theta, this ratio is always going to be same."}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "That, by definition, is the tangent of theta. And a mnemonic that will help you remember this, and these really are just definitions. People realize, wow, by similar triangles, for any angle theta, this ratio is always going to be the same. Because of similar triangles, for any angle theta, this ratio is always going to be same. This ratio is always going to be same, so let's make these definitions. And to help us remember it, there's the mnemonic SOH CAH TOA. So I'll write it like this."}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Because of similar triangles, for any angle theta, this ratio is always going to be same. This ratio is always going to be same, so let's make these definitions. And to help us remember it, there's the mnemonic SOH CAH TOA. So I'll write it like this. SOH, SOH, SOH is sine is opposite of our hypotenuse, CAH, cosine, cosine is adjacent over hypotenuse, cosine is adjacent over hypotenuse. And then finally, tangent, tangent is opposite over adjacent. Tangent is opposite, opposite over, opposite over adjacent."}, {"video_title": "Cosine equation solution set in an interval.mp3", "Sentence": "In a previous video, we established the entire solution set for the following equation. And we saw that all the Xs that can satisfy this equation are a combination of these Xs and these Xs here. The reason why I'm referring to each of them as numerous Xs is that for any integer value of N, you'll get another solution. For any integer value of N, you'll get another solution. What I wanna do in this video is to make things a little bit more concrete. And the way that we're going to do it is by exploring all of the X values that satisfy this equation that sit in the closed interval from negative pi over two to zero. So I encourage you, like always, pause this video and have a go at it by yourself before we work through it together."}, {"video_title": "Cosine equation solution set in an interval.mp3", "Sentence": "For any integer value of N, you'll get another solution. What I wanna do in this video is to make things a little bit more concrete. And the way that we're going to do it is by exploring all of the X values that satisfy this equation that sit in the closed interval from negative pi over two to zero. So I encourage you, like always, pause this video and have a go at it by yourself before we work through it together. All right, now let's work through this together. So the first helpful thing is we have these algebraic expressions. We have things written in terms of pi."}, {"video_title": "Cosine equation solution set in an interval.mp3", "Sentence": "So I encourage you, like always, pause this video and have a go at it by yourself before we work through it together. All right, now let's work through this together. So the first helpful thing is we have these algebraic expressions. We have things written in terms of pi. Let's approximate them all in terms of decimals. So even pi over two, we can approximate that. Let's see if pi is approximately 3.14."}, {"video_title": "Cosine equation solution set in an interval.mp3", "Sentence": "We have things written in terms of pi. Let's approximate them all in terms of decimals. So even pi over two, we can approximate that. Let's see if pi is approximately 3.14. Half of that is approximately 1.57. So we could say this is approximately the closed interval from negative 1.57 to zero. Negative 1.57 isn't exactly negative pi over two, but it'll hopefully be suitable for what we're trying to do here."}, {"video_title": "Cosine equation solution set in an interval.mp3", "Sentence": "Let's see if pi is approximately 3.14. Half of that is approximately 1.57. So we could say this is approximately the closed interval from negative 1.57 to zero. Negative 1.57 isn't exactly negative pi over two, but it'll hopefully be suitable for what we're trying to do here. And now let's see if we can write the different parts of these expressions, or at least approximate them as decimals. So this could be rewritten as X is approximately. If you were to take 1 8th times the inverse cosine of negative 1 6th, I encourage you to verify this on your own on a calculator, you would get that that's approximately 0.22."}, {"video_title": "Cosine equation solution set in an interval.mp3", "Sentence": "Negative 1.57 isn't exactly negative pi over two, but it'll hopefully be suitable for what we're trying to do here. And now let's see if we can write the different parts of these expressions, or at least approximate them as decimals. So this could be rewritten as X is approximately. If you were to take 1 8th times the inverse cosine of negative 1 6th, I encourage you to verify this on your own on a calculator, you would get that that's approximately 0.22. And then pi over four is approximately 0.785. So this expression would be approximately 0.22 minus 0.785 times n, where n could be any integer. And then this one over here on the right, let me do that in the yellow, X could be approximately equal to, well, if this evaluates to approximately 0.22, then this is just the negative of it."}, {"video_title": "Cosine equation solution set in an interval.mp3", "Sentence": "If you were to take 1 8th times the inverse cosine of negative 1 6th, I encourage you to verify this on your own on a calculator, you would get that that's approximately 0.22. And then pi over four is approximately 0.785. So this expression would be approximately 0.22 minus 0.785 times n, where n could be any integer. And then this one over here on the right, let me do that in the yellow, X could be approximately equal to, well, if this evaluates to approximately 0.22, then this is just the negative of it. So it's going to be negative 0.22. And then it's plus what approximately pi over four is. So 0.785 n. And now what we could do is just try different n's and see if we're starting above or below this interval, and then see which of the X values actually fall in this interval."}, {"video_title": "Cosine equation solution set in an interval.mp3", "Sentence": "And then this one over here on the right, let me do that in the yellow, X could be approximately equal to, well, if this evaluates to approximately 0.22, then this is just the negative of it. So it's going to be negative 0.22. And then it's plus what approximately pi over four is. So 0.785 n. And now what we could do is just try different n's and see if we're starting above or below this interval, and then see which of the X values actually fall in this interval. So let's just start here. If we just start at n equals zero, actually, why don't I set up a little table here. We have n here, and if we have the X value here, when n is zero, well, then you don't see this term and you just get approximately 0.22."}, {"video_title": "Cosine equation solution set in an interval.mp3", "Sentence": "So 0.785 n. And now what we could do is just try different n's and see if we're starting above or below this interval, and then see which of the X values actually fall in this interval. So let's just start here. If we just start at n equals zero, actually, why don't I set up a little table here. We have n here, and if we have the X value here, when n is zero, well, then you don't see this term and you just get approximately 0.22. Now let's compare that to the interval. The interval, the upper bound of that interval is zero. So this does not sit in the interval."}, {"video_title": "Cosine equation solution set in an interval.mp3", "Sentence": "We have n here, and if we have the X value here, when n is zero, well, then you don't see this term and you just get approximately 0.22. Now let's compare that to the interval. The interval, the upper bound of that interval is zero. So this does not sit in the interval. So this is too high. And we would want to define the Xs that sit in the interval. We want to find lower values."}, {"video_title": "Cosine equation solution set in an interval.mp3", "Sentence": "So this does not sit in the interval. So this is too high. And we would want to define the Xs that sit in the interval. We want to find lower values. So it's good that here we're subtracting 0.785. So I would use positive integer values of n to decrease this 0.22 here. So when n equals one, we would subtract 0.785 from that."}, {"video_title": "Cosine equation solution set in an interval.mp3", "Sentence": "We want to find lower values. So it's good that here we're subtracting 0.785. So I would use positive integer values of n to decrease this 0.22 here. So when n equals one, we would subtract 0.785 from that. And I'll round all of these to the hundredths place. And that would get us to negative 0.57. 0.57, and that does sit in the interval."}, {"video_title": "Cosine equation solution set in an interval.mp3", "Sentence": "So when n equals one, we would subtract 0.785 from that. And I'll round all of these to the hundredths place. And that would get us to negative 0.57. 0.57, and that does sit in the interval. So this looks good. So this would be a solution in that interval right over here. And let's try n equals two."}, {"video_title": "Cosine equation solution set in an interval.mp3", "Sentence": "0.57, and that does sit in the interval. So this looks good. So this would be a solution in that interval right over here. And let's try n equals two. So we would subtract 0.785 again, and that would get us to negative 1.35, not two five, three five, and that also sits in the interval. It's larger than negative 1.57, so that looks good. Let's subtract 0.785 again."}, {"video_title": "Cosine equation solution set in an interval.mp3", "Sentence": "And let's try n equals two. So we would subtract 0.785 again, and that would get us to negative 1.35, not two five, three five, and that also sits in the interval. It's larger than negative 1.57, so that looks good. Let's subtract 0.785 again. When n equals three, that would get us negative 2.14. Well, that's all of a sudden out of the interval because that's below the lower bound here. So this is too low."}, {"video_title": "Cosine equation solution set in an interval.mp3", "Sentence": "Let's subtract 0.785 again. When n equals three, that would get us negative 2.14. Well, that's all of a sudden out of the interval because that's below the lower bound here. So this is too low. So using this expression, we've been able to find two X values that sit in the interval that we cared about. Now let's use these X values right over here, and I'll set up another table. So let's see, we have our n and then we have our X values."}, {"video_title": "Cosine equation solution set in an interval.mp3", "Sentence": "So this is too low. So using this expression, we've been able to find two X values that sit in the interval that we cared about. Now let's use these X values right over here, and I'll set up another table. So let's see, we have our n and then we have our X values. So let's start with n equals zero because that's easy to compute. And then this term would go away and we'd have negative 0.22. And that's actually in this interval here."}, {"video_title": "Cosine equation solution set in an interval.mp3", "Sentence": "So let's see, we have our n and then we have our X values. So let's start with n equals zero because that's easy to compute. And then this term would go away and we'd have negative 0.22. And that's actually in this interval here. It's below zero, it's larger than negative 1.57. So that one checks out. But now to really explore, we have to go in both directions."}, {"video_title": "Cosine equation solution set in an interval.mp3", "Sentence": "And that's actually in this interval here. It's below zero, it's larger than negative 1.57. So that one checks out. But now to really explore, we have to go in both directions. We have to increase it or decrease it. So if we wanted to increase it, we could have a situation where n equals one. So if n equals one, we're gonna add 0.785 to this."}, {"video_title": "Cosine equation solution set in an interval.mp3", "Sentence": "But now to really explore, we have to go in both directions. We have to increase it or decrease it. So if we wanted to increase it, we could have a situation where n equals one. So if n equals one, we're gonna add 0.785 to this. Now you immediately know that that's going to be a positive value. If you computed it, it'd be 0.57, which is larger than zero. So this is too high."}, {"video_title": "Cosine equation solution set in an interval.mp3", "Sentence": "So if n equals one, we're gonna add 0.785 to this. Now you immediately know that that's going to be a positive value. If you computed it, it'd be 0.57, which is larger than zero. So this is too high. So now we could try going lower than negative 0.22 by having negative values of n. So if n is equal to negative one, that means we're subtracting 0.785 from this right over here which would get us to negative 1.01. Well, that one works out. So that's in our interval."}, {"video_title": "Cosine equation solution set in an interval.mp3", "Sentence": "So this is too high. So now we could try going lower than negative 0.22 by having negative values of n. So if n is equal to negative one, that means we're subtracting 0.785 from this right over here which would get us to negative 1.01. Well, that one works out. So that's in our interval. And now let's subtract 0.785 again. So I'll have n equals negative two. And so if I subtract 0.785 again, I could round that to negative 1.79, which is lower than negative 1.57."}, {"video_title": "Cosine equation solution set in an interval.mp3", "Sentence": "So that's in our interval. And now let's subtract 0.785 again. So I'll have n equals negative two. And so if I subtract 0.785 again, I could round that to negative 1.79, which is lower than negative 1.57. So it's out of our interval. So it's too low. So all of the x values that are in our interval that satisfy this equation are these two right over here and this one and this one."}, {"video_title": "Trigonometric system example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So this is a pretty daunting thing, and I just like to simplify it as much as possible. So what I want to do is, well, I want to do two things. All of the trig functions here are, they're all of, they're taking sine or the cosine of 3 theta, and they have theta between 0 and pi. I want to do a little substitution. This is just for my brain, because it simplifies my logic a little bit. Let's make a substitution that u is equal to 3 theta. And then u, when theta is 0, u is going to be, so u is still going to be greater than 0."}, {"video_title": "Trigonometric system example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "I want to do a little substitution. This is just for my brain, because it simplifies my logic a little bit. Let's make a substitution that u is equal to 3 theta. And then u, when theta is 0, u is going to be, so u is still going to be greater than 0. And when theta is pi, u is going to have to be less than 3 pi. And so this boils down to, let's find the number of u's between 0 and 3 pi, where y plus z cosine of u is equal to xyz sine of u, so on and so forth. So with that, let's do that substitution."}, {"video_title": "Trigonometric system example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And then u, when theta is 0, u is going to be, so u is still going to be greater than 0. And when theta is pi, u is going to have to be less than 3 pi. And so this boils down to, let's find the number of u's between 0 and 3 pi, where y plus z cosine of u is equal to xyz sine of u, so on and so forth. So with that, let's do that substitution. And also, I'm going to rearrange these so that they start to at least look a little familiar and see if we can somehow start manipulating with these equations and actually try to cancel out terms. So this first one up here, just to get into ways that I can recognize, let me distribute the cosine of 3 theta, or which we'll now call the cosine of u. So let me just write this."}, {"video_title": "Trigonometric system example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So with that, let's do that substitution. And also, I'm going to rearrange these so that they start to at least look a little familiar and see if we can somehow start manipulating with these equations and actually try to cancel out terms. So this first one up here, just to get into ways that I can recognize, let me distribute the cosine of 3 theta, or which we'll now call the cosine of u. So let me just write this. This is u. This is u. This is u now."}, {"video_title": "Trigonometric system example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So let me just write this. This is u. This is u. This is u now. This is u, u, u, u, and u. So this, if we distribute the cosine of u, this becomes cosine of u. Let me put the y out front."}, {"video_title": "Trigonometric system example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "This is u now. This is u, u, u, u, and u. So this, if we distribute the cosine of u, this becomes cosine of u. Let me put the y out front. So it's y cosine of u plus z cosine of u is equal to xyz sine of u. That's this first equation. This second equation, it looks like, well, if we multiply both sides of this equation by yz, we're going to have an xyz sine of u on the right-hand side, which is the exact same thing we have here."}, {"video_title": "Trigonometric system example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Let me put the y out front. So it's y cosine of u plus z cosine of u is equal to xyz sine of u. That's this first equation. This second equation, it looks like, well, if we multiply both sides of this equation by yz, we're going to have an xyz sine of u on the right-hand side, which is the exact same thing we have here. So let's multiply both sides of this equation by xyz, both sides. The right hand of the equation, not by xyz, just by yz. So we're going to multiply both sides of this equation by yz so that these leave the denominator."}, {"video_title": "Trigonometric system example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "This second equation, it looks like, well, if we multiply both sides of this equation by yz, we're going to have an xyz sine of u on the right-hand side, which is the exact same thing we have here. So let's multiply both sides of this equation by xyz, both sides. The right hand of the equation, not by xyz, just by yz. So we're going to multiply both sides of this equation by yz so that these leave the denominator. So we're also going to multiply the right side by yz. The left-hand side becomes xyz sine of u. And let me write it over here, just right under this."}, {"video_title": "Trigonometric system example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So we're going to multiply both sides of this equation by yz so that these leave the denominator. So we're also going to multiply the right side by yz. The left-hand side becomes xyz sine of u. And let me write it over here, just right under this. So this is xyz sine of u. And then yz times 2 cosine of u is going to be 2z cosine of u. So 2z cosine of u. I just swapped the sides."}, {"video_title": "Trigonometric system example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And let me write it over here, just right under this. So this is xyz sine of u. And then yz times 2 cosine of u is going to be 2z cosine of u. So 2z cosine of u. I just swapped the sides. And then yz times 2 sine of u is going to be 2y sine of u. So 2y sine of u. So this is the second equation."}, {"video_title": "Trigonometric system example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So 2z cosine of u. I just swapped the sides. And then yz times 2 sine of u is going to be 2y sine of u. So 2y sine of u. So this is the second equation. And now they don't look that different. When they're written like this, they look very different. Now let's think about this one."}, {"video_title": "Trigonometric system example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So this is the second equation. And now they don't look that different. When they're written like this, they look very different. Now let's think about this one. This has an xyz sine of u. I'll do it in magenta, this equation. xyz sine of u, I'll write it on this side. xyz sine of u is equal to, let's see, we have a 2z, let's distribute this cosine of u."}, {"video_title": "Trigonometric system example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Now let's think about this one. This has an xyz sine of u. I'll do it in magenta, this equation. xyz sine of u, I'll write it on this side. xyz sine of u is equal to, let's see, we have a 2z, let's distribute this cosine of u. We have a 2z cosine of u. So plus 2z cosine of u plus y cosine of u. So we have a plus y cosine of u."}, {"video_title": "Trigonometric system example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "xyz sine of u is equal to, let's see, we have a 2z, let's distribute this cosine of u. We have a 2z cosine of u. So plus 2z cosine of u plus y cosine of u. So we have a plus y cosine of u. That's that times that, plus y sine of u. So let me scroll to the left a little bit. y sine of u."}, {"video_title": "Trigonometric system example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So we have a plus y cosine of u. That's that times that, plus y sine of u. So let me scroll to the left a little bit. y sine of u. So I've rewritten these three equations. The problem looks a lot less daunting right now. And let's try to figure out the number of u's between 0 and 3 pi that will satisfy, that will give us a solution here."}, {"video_title": "Trigonometric system example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "y sine of u. So I've rewritten these three equations. The problem looks a lot less daunting right now. And let's try to figure out the number of u's between 0 and 3 pi that will satisfy, that will give us a solution here. So let's see. All of these three equations are equal to this expression right over here. So the left-hand sides of these equations all have to equal each other because they all equal the exact same value."}, {"video_title": "Trigonometric system example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And let's try to figure out the number of u's between 0 and 3 pi that will satisfy, that will give us a solution here. So let's see. All of these three equations are equal to this expression right over here. So the left-hand sides of these equations all have to equal each other because they all equal the exact same value. So let's do that. Let's see what we can do in the way of canceling things out. Let's see."}, {"video_title": "Trigonometric system example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So the left-hand sides of these equations all have to equal each other because they all equal the exact same value. So let's do that. Let's see what we can do in the way of canceling things out. Let's see. This is a plus right here. I don't know why I wrote an equal here. So let's see."}, {"video_title": "Trigonometric system example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Let's see. This is a plus right here. I don't know why I wrote an equal here. So let's see. If we write, this has got to be equal to that. So let me use these two first. So this thing, so we have 2y sine of u plus 2z cosine of u is equal to this, which is that, which has to be equal to this, is going to be equal to y sine of u plus y cosine of u plus 2z cosine of u."}, {"video_title": "Trigonometric system example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So let's see. If we write, this has got to be equal to that. So let me use these two first. So this thing, so we have 2y sine of u plus 2z cosine of u is equal to this, which is that, which has to be equal to this, is going to be equal to y sine of u plus y cosine of u plus 2z cosine of u. We have 2z cosine of u on both sides, so that gets rid of the z terms. And then we have 2y sine of u and we have a y sine of u. So if we subtract y sine of u from both sides, we end up with a y sine of u."}, {"video_title": "Trigonometric system example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So this thing, so we have 2y sine of u plus 2z cosine of u is equal to this, which is that, which has to be equal to this, is going to be equal to y sine of u plus y cosine of u plus 2z cosine of u. We have 2z cosine of u on both sides, so that gets rid of the z terms. And then we have 2y sine of u and we have a y sine of u. So if we subtract y sine of u from both sides, we end up with a y sine of u. Just subtracting this from both sides, 2y minus y sine of u which is going to be y sine of u is equal to y cosine of u. So in order to have a solution here, in order for this to be a sensical statement, and remember y cannot be equal to 0, in order for this to end up having a solution, the coefficients on y have to equal each other. Sine of u has to be equal to cosine of u."}, {"video_title": "Trigonometric system example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So if we subtract y sine of u from both sides, we end up with a y sine of u. Just subtracting this from both sides, 2y minus y sine of u which is going to be y sine of u is equal to y cosine of u. So in order to have a solution here, in order for this to be a sensical statement, and remember y cannot be equal to 0, in order for this to end up having a solution, the coefficients on y have to equal each other. Sine of u has to be equal to cosine of u. So that's one constraint. Sine of u has to be equal to cosine of u. And let's just think about the unit circle and think about how many times are the sine and the cosine equal to each other when you're going between 0 and 3 pi."}, {"video_title": "Trigonometric system example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Sine of u has to be equal to cosine of u. So that's one constraint. Sine of u has to be equal to cosine of u. And let's just think about the unit circle and think about how many times are the sine and the cosine equal to each other when you're going between 0 and 3 pi. So I got the unit circle right over here. Now clearly when we're at 45 degrees, sine and cosine are equal to each other, or 45 degrees is the same thing as pi over 4. We're at there, they're equal to each other."}, {"video_title": "Trigonometric system example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And let's just think about the unit circle and think about how many times are the sine and the cosine equal to each other when you're going between 0 and 3 pi. So I got the unit circle right over here. Now clearly when we're at 45 degrees, sine and cosine are equal to each other, or 45 degrees is the same thing as pi over 4. We're at there, they're equal to each other. You might be tempted to do this over here, but here the cosine is negative, sine is positive, so that won't work. They're both negative over here, but they're equal, so that's another value. And then this won't work."}, {"video_title": "Trigonometric system example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "We're at there, they're equal to each other. You might be tempted to do this over here, but here the cosine is negative, sine is positive, so that won't work. They're both negative over here, but they're equal, so that's another value. And then this won't work. So so far we've traveled 2 pi. We can go another half. We can go 3 pi."}, {"video_title": "Trigonometric system example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And then this won't work. So so far we've traveled 2 pi. We can go another half. We can go 3 pi. So we can go back to this one again. So we can go back to this value again. So there are 1, 2, and then when you go all the way around again, 3 values."}, {"video_title": "Trigonometric system example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "We can go 3 pi. So we can go back to this one again. So we can go back to this value again. So there are 1, 2, and then when you go all the way around again, 3 values. And then we can't go back to this one, because we can only go, let me just add another color, we can only go 3 pi for u's. So we can only go, that's 2 pi, and then go another time around. That is 3 pi."}, {"video_title": "Trigonometric system example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So there are 1, 2, and then when you go all the way around again, 3 values. And then we can't go back to this one, because we can only go, let me just add another color, we can only go 3 pi for u's. So we can only go, that's 2 pi, and then go another time around. That is 3 pi. So there's 1, 2, 3 values. So there's 3 possible u's. Just from this constraint, just when we used this blue equation and this magenta equation."}, {"video_title": "Trigonometric system example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "That is 3 pi. So there's 1, 2, 3 values. So there's 3 possible u's. Just from this constraint, just when we used this blue equation and this magenta equation. Now let's just make sure that there aren't any further constraints over here. So let's see if we can, let's use two of the other equations. And the ones that I would want to use, that seems like there might be some cancellation."}, {"video_title": "Trigonometric system example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Just from this constraint, just when we used this blue equation and this magenta equation. Now let's just make sure that there aren't any further constraints over here. So let's see if we can, let's use two of the other equations. And the ones that I would want to use, that seems like there might be some cancellation. Well, we could just use, we really could use, we could use either this guy and this guy. y cosine of u plus 2z. Yeah, why not?"}, {"video_title": "Trigonometric system example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And the ones that I would want to use, that seems like there might be some cancellation. Well, we could just use, we really could use, we could use either this guy and this guy. y cosine of u plus 2z. Yeah, why not? So let's use, so as long as we're using all three equations in our constraints, we will have kind of properly constrained all of the possible solutions. And so let's think about it. This equals that, which is equal to that, which is equal to that."}, {"video_title": "Trigonometric system example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Yeah, why not? So let's use, so as long as we're using all three equations in our constraints, we will have kind of properly constrained all of the possible solutions. And so let's think about it. This equals that, which is equal to that, which is equal to that. So we could write y cosine of u plus z cosine of u is equal to this whole thing over here, is equal to y sine of u plus y cosine of u plus 2z cosine of u. And then we have a y cosine of u on both sides. Those will cancel out."}, {"video_title": "Trigonometric system example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "This equals that, which is equal to that, which is equal to that. So we could write y cosine of u plus z cosine of u is equal to this whole thing over here, is equal to y sine of u plus y cosine of u plus 2z cosine of u. And then we have a y cosine of u on both sides. Those will cancel out. We can subtract a z cosine of u from both sides. A z cosine of u from both sides, and so we would get 0 is equal to y sine of u plus z cosine of u. And this seems like a pretty benign statement."}, {"video_title": "Trigonometric system example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Those will cancel out. We can subtract a z cosine of u from both sides. A z cosine of u from both sides, and so we would get 0 is equal to y sine of u plus z cosine of u. And this seems like a pretty benign statement. y sine of u plus z cosine of u is equal to 0. Let me make sure. Let's see, y sine of u plus z cosine of u is equal to 0."}, {"video_title": "Trigonometric system example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And this seems like a pretty benign statement. y sine of u plus z cosine of u is equal to 0. Let me make sure. Let's see, y sine of u plus z cosine of u is equal to 0. That means that 2 times this is also going to be 0. So this is equal to 2y sine of u plus 2z cosine of u is equal to 0. And the only reason why I multiplied it by 2 is because now this looks identical to this."}, {"video_title": "Trigonometric system example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Let's see, y sine of u plus z cosine of u is equal to 0. That means that 2 times this is also going to be 0. So this is equal to 2y sine of u plus 2z cosine of u is equal to 0. And the only reason why I multiplied it by 2 is because now this looks identical to this. So when I used this equation and this equation, I got a constraint that this expression right over here, this expression right over here, essentially needs to be equal to 0. That this expression over here, essentially, needs to be equal to 0, which is OK. Because u could make the sine equal to 0, or x can be equal to 0. Remember, they didn't put any constraints on x. x can be equal to 0."}, {"video_title": "Trigonometric system example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And the only reason why I multiplied it by 2 is because now this looks identical to this. So when I used this equation and this equation, I got a constraint that this expression right over here, this expression right over here, essentially needs to be equal to 0. That this expression over here, essentially, needs to be equal to 0, which is OK. Because u could make the sine equal to 0, or x can be equal to 0. Remember, they didn't put any constraints on x. x can be equal to 0. So this is really constraining us. x can clearly be equal to 0, which could make this thing equal to 0. So it's not limiting our constraints."}, {"video_title": "Trigonometric system example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Remember, they didn't put any constraints on x. x can be equal to 0. So this is really constraining us. x can clearly be equal to 0, which could make this thing equal to 0. So it's not limiting our constraints. So now we've used all of the information that's in the problem. We've used all three of these surfaces, essentially, to find out what are the constraints we have for an intersection of the three surfaces. And the only real constraint is that the sine of u has to be equal to the cosine of u."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "So let me do my best attempt at graphing that. And to start off, I'm going to graph with the simplest function, or the simplest version of this, or the root of this, which is just cosine of x. So let me just graph, and eventually you can kind of, let me just, let me graph cosine of x. So this is my y-axis. And I want to have some space here so I can eventually graph this entire thing. So let me, so let's say that this is negative 1, this is negative 2, this is positive 1, this is positive 2, and let's say that this right over here is, this right over here is 2 pi. 2 pi, and then of course that could be pi right over there."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "So this is my y-axis. And I want to have some space here so I can eventually graph this entire thing. So let me, so let's say that this is negative 1, this is negative 2, this is positive 1, this is positive 2, and let's say that this right over here is, this right over here is 2 pi. 2 pi, and then of course that could be pi right over there. Now, the first thing I'm going to do, let me copy this, because I could use it later. I could use it later to graph the whole thing. So, so let's start off."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "2 pi, and then of course that could be pi right over there. Now, the first thing I'm going to do, let me copy this, because I could use it later. I could use it later to graph the whole thing. So, so let's start off. So I'm just going to graph y is equal to cosine of x. So when x is equal to zero, and I'm just going to do it between the interval zero and 2 pi. Obviously it's a periodic function, it'll keep going in the negative and the positive directions."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "So, so let's start off. So I'm just going to graph y is equal to cosine of x. So when x is equal to zero, and I'm just going to do it between the interval zero and 2 pi. Obviously it's a periodic function, it'll keep going in the negative and the positive directions. So what happens when x is equal to zero? What is cosine of x? Well, cosine of zero is 1."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "Obviously it's a periodic function, it'll keep going in the negative and the positive directions. So what happens when x is equal to zero? What is cosine of x? Well, cosine of zero is 1. What about when, what about when x is equal to pi? What is cosine of pi? Well, cosine of pi is negative 1."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "Well, cosine of zero is 1. What about when, what about when x is equal to pi? What is cosine of pi? Well, cosine of pi is negative 1. Cosine of pi is negative 1. And then what's cosine of 2 pi? Well, that's 1 again."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "Well, cosine of pi is negative 1. Cosine of pi is negative 1. And then what's cosine of 2 pi? Well, that's 1 again. We get back, we've completed a period, or we've completed an entire cycle, and 2 pi is the period of cosine of x. So this is one cycle right over here. I could keep going if I wanted to, but the whole point, I just wanted to graph this one cycle between zero and 2 pi."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "Well, that's 1 again. We get back, we've completed a period, or we've completed an entire cycle, and 2 pi is the period of cosine of x. So this is one cycle right over here. I could keep going if I wanted to, but the whole point, I just wanted to graph this one cycle between zero and 2 pi. Now, what I want to think about is, what happens to this graph? Instead of graphing, instead of graphing y equals cosine of x, and let me draw some graph paper again. Instead of drawing y is equal to cosine of x, I'm going to draw y is equal to cosine of 1 3rd x."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "I could keep going if I wanted to, but the whole point, I just wanted to graph this one cycle between zero and 2 pi. Now, what I want to think about is, what happens to this graph? Instead of graphing, instead of graphing y equals cosine of x, and let me draw some graph paper again. Instead of drawing y is equal to cosine of x, I'm going to draw y is equal to cosine of 1 3rd x. So the only difference between that and that is now I'm multiplying the x by 1 3rd. What's going to happen to the graph over here? How is this going to change instead of being an x if it's a 1 3rd x?"}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "Instead of drawing y is equal to cosine of x, I'm going to draw y is equal to cosine of 1 3rd x. So the only difference between that and that is now I'm multiplying the x by 1 3rd. What's going to happen to the graph over here? How is this going to change instead of being an x if it's a 1 3rd x? What's going to happen over here? And now I'm going to do it over the entire interval between zero and 6 pi. So let me just make sure I have enough space."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "How is this going to change instead of being an x if it's a 1 3rd x? What's going to happen over here? And now I'm going to do it over the entire interval between zero and 6 pi. So let me just make sure I have enough space. 3 pi, 4 pi, 5 pi, and 6 pi. What's going to happen to this graph? Well, there's a couple of ways to think about it."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "So let me just make sure I have enough space. 3 pi, 4 pi, 5 pi, and 6 pi. What's going to happen to this graph? Well, there's a couple of ways to think about it. The easiest might just be to say, well, to complete an entire cycle, we're going to go 1 3rd as fast, or we're going to go 3 times slower. Or if you just want to think about the period here, what's the period of cosine of 1 3rd x? Well, the period is going to be 2 pi divided by the absolute value of this coefficient right over here."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "Well, there's a couple of ways to think about it. The easiest might just be to say, well, to complete an entire cycle, we're going to go 1 3rd as fast, or we're going to go 3 times slower. Or if you just want to think about the period here, what's the period of cosine of 1 3rd x? Well, the period is going to be 2 pi divided by the absolute value of this coefficient right over here. So it's the absolute value of 1 3rd, which is just 1 3rd. So the period is 2 pi over 1 3rd, which is the same thing as 2 pi times 3, which is 6 pi, which gels with the intuition. It's going to take 3 times as much time to get whatever we input into the cosine function to get back to 2 pi."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "Well, the period is going to be 2 pi divided by the absolute value of this coefficient right over here. So it's the absolute value of 1 3rd, which is just 1 3rd. So the period is 2 pi over 1 3rd, which is the same thing as 2 pi times 3, which is 6 pi, which gels with the intuition. It's going to take 3 times as much time to get whatever we input into the cosine function to get back to 2 pi. Because whatever we take x, we're taking 1 3rd of it. So to get to 2 pi, you can't just have x equals 2 pi. x now has to equal 6 pi to get 2 pi inputted into the cosine function."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "It's going to take 3 times as much time to get whatever we input into the cosine function to get back to 2 pi. Because whatever we take x, we're taking 1 3rd of it. So to get to 2 pi, you can't just have x equals 2 pi. x now has to equal 6 pi to get 2 pi inputted into the cosine function. So the period is now 6 pi. At x is equal to 0, 1 3rd times 0 is 0, and the cosine of 0 is 1. When x is equal to 6 pi, you have 6 pi divided by 3 is 2 pi."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "x now has to equal 6 pi to get 2 pi inputted into the cosine function. So the period is now 6 pi. At x is equal to 0, 1 3rd times 0 is 0, and the cosine of 0 is 1. When x is equal to 6 pi, you have 6 pi divided by 3 is 2 pi. Cosine of 2 pi is equal to 1. And if you want to go in between, over here, to go in between, we tried pi. But over here, we could try 3 pi."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "When x is equal to 6 pi, you have 6 pi divided by 3 is 2 pi. Cosine of 2 pi is equal to 1. And if you want to go in between, over here, to go in between, we tried pi. But over here, we could try 3 pi. When x is 3 pi, you have cosine of 1 3rd of 3 pi. That's cosine of pi. Cosine of pi is negative 1."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "But over here, we could try 3 pi. When x is 3 pi, you have cosine of 1 3rd of 3 pi. That's cosine of pi. Cosine of pi is negative 1. So when x is equal to 3 pi, we have cosine of 1 3rd times 3 pi is negative 1. So it's going to look something like this. It's going to look something like this."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "Cosine of pi is negative 1. So when x is equal to 3 pi, we have cosine of 1 3rd times 3 pi is negative 1. So it's going to look something like this. It's going to look something like this. Drawing my best attempt to draw it. So it's going to look something like this. So you see, to go from y equals cosine of x to y equals cosine of 1 3rd x, it essentially stretched out this function by a factor of 3."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "It's going to look something like this. Drawing my best attempt to draw it. So it's going to look something like this. So you see, to go from y equals cosine of x to y equals cosine of 1 3rd x, it essentially stretched out this function by a factor of 3. Or you could see that the period is 3 times longer. The period here was 2 pi. Period here was 2 pi."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "So you see, to go from y equals cosine of x to y equals cosine of 1 3rd x, it essentially stretched out this function by a factor of 3. Or you could see that the period is 3 times longer. The period here was 2 pi. Period here was 2 pi. All right. Well, there's only one more transformation we need in order to get to the function that they're asking us about. We just have to, instead of having a cosine of 1 3rd x, we just have to negative 2.5 cosine of 1 3rd x."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "Period here was 2 pi. All right. Well, there's only one more transformation we need in order to get to the function that they're asking us about. We just have to, instead of having a cosine of 1 3rd x, we just have to negative 2.5 cosine of 1 3rd x. So let's try to draw that. So let me put my axis here again. And let me label it."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "We just have to, instead of having a cosine of 1 3rd x, we just have to negative 2.5 cosine of 1 3rd x. So let's try to draw that. So let me put my axis here again. And let me label it. So that's cosine 2 pi, 3 pi, 4 pi, 5 pi, and 6 pi. And our goal now is to draw the graph of y is equal to. And we're just doing it over this between 0 and 6 pi here."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "And let me label it. So that's cosine 2 pi, 3 pi, 4 pi, 5 pi, and 6 pi. And our goal now is to draw the graph of y is equal to. And we're just doing it over this between 0 and 6 pi here. We only did it between 0 and 2 pi here. Obviously, they're all periodic. They all keep going on and on."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "And we're just doing it over this between 0 and 6 pi here. We only did it between 0 and 2 pi here. Obviously, they're all periodic. They all keep going on and on. But now we want to graph y is equal to negative 2.5 times cosine of 1 3rd x. So given this change, we're now multiplying by negative 2.5. What is going to be?"}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "They all keep going on and on. But now we want to graph y is equal to negative 2.5 times cosine of 1 3rd x. So given this change, we're now multiplying by negative 2.5. What is going to be? Well, actually, let's think about a few things. What was the amplitude in the first two graphs right over here? So what was the amplitude in these first two graphs?"}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "What is going to be? Well, actually, let's think about a few things. What was the amplitude in the first two graphs right over here? So what was the amplitude in these first two graphs? Well, there's two ways to think about it. You could say the amplitude is half the difference between the minimum and the maximum points. In either of these cases, the minimum is negative 1, maximum is 1."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "So what was the amplitude in these first two graphs? Well, there's two ways to think about it. You could say the amplitude is half the difference between the minimum and the maximum points. In either of these cases, the minimum is negative 1, maximum is 1. The difference is 2. Half of that is 1. Or you could just say it's the absolute value of the coefficient here, which is implicitly a 1."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "In either of these cases, the minimum is negative 1, maximum is 1. The difference is 2. Half of that is 1. Or you could just say it's the absolute value of the coefficient here, which is implicitly a 1. And the absolute value of 1 is once again 1. What's going to be the amplitude for this thing right over here? Well, the amplitude is going to be the absolute value of what's multiplying the cosine function."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "Or you could just say it's the absolute value of the coefficient here, which is implicitly a 1. And the absolute value of 1 is once again 1. What's going to be the amplitude for this thing right over here? Well, the amplitude is going to be the absolute value of what's multiplying the cosine function. So the amplitude in this case, I'll do it in green, is going to be equal to the absolute value of negative 2.5, which is equal to 2.5. So given that, how is multiplying by negative 2.5 going to transform this graph right over here? Well, let's think about it."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "Well, the amplitude is going to be the absolute value of what's multiplying the cosine function. So the amplitude in this case, I'll do it in green, is going to be equal to the absolute value of negative 2.5, which is equal to 2.5. So given that, how is multiplying by negative 2.5 going to transform this graph right over here? Well, let's think about it. If it was multiplying by just a positive 2.5, you would stretch it out. At each point, it would go up by a factor of 2 and 1 half. But it's a negative 2.5."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "Well, let's think about it. If it was multiplying by just a positive 2.5, you would stretch it out. At each point, it would go up by a factor of 2 and 1 half. But it's a negative 2.5. So at each point, you're going to stretch it out, and then you're going to flip it over the x-axis. So let's do that. So when x was 0, you got to 1 in this case."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "But it's a negative 2.5. So at each point, you're going to stretch it out, and then you're going to flip it over the x-axis. So let's do that. So when x was 0, you got to 1 in this case. But now we're going to multiply that by negative 2.5, which means you're going to get to negative 2.5. So let me draw negative 2.5 right over there. So that's negative 2.5."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "So when x was 0, you got to 1 in this case. But now we're going to multiply that by negative 2.5, which means you're going to get to negative 2.5. So let me draw negative 2.5 right over there. So that's negative 2.5. That would be negative. Let me make it clear. This would be negative 3 right over here."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "So that's negative 2.5. That would be negative. Let me make it clear. This would be negative 3 right over here. This would be positive 3. So that number right over there is negative 2.5. And let me draw a dotted line there."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "This would be negative 3 right over here. This would be positive 3. So that number right over there is negative 2.5. And let me draw a dotted line there. It could serve to be useful. Now, when cosine of 1 third x is 0, it doesn't matter what you multiply it by. You're still going to get 0 right over here."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "And let me draw a dotted line there. It could serve to be useful. Now, when cosine of 1 third x is 0, it doesn't matter what you multiply it by. You're still going to get 0 right over here. Now, when cosine of 1 third x was negative 1, which was the case when x is equal to 3 pi, what's going to happen over here? Well, cosine of 1 third x we see is negative 1. Negative 1 times negative 2.5 is positive 2.5."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "You're still going to get 0 right over here. Now, when cosine of 1 third x was negative 1, which was the case when x is equal to 3 pi, what's going to happen over here? Well, cosine of 1 third x we see is negative 1. Negative 1 times negative 2.5 is positive 2.5. So we're going to get to positive 2.5, which is right. Let me draw a dotted line over here. We're going to get to positive 2.5, which is right over there."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "Negative 1 times negative 2.5 is positive 2.5. So we're going to get to positive 2.5, which is right. Let me draw a dotted line over here. We're going to get to positive 2.5, which is right over there. And then when cosine of 1 third x is equal to 0, doesn't matter what we multiply it by. We get to 0. And then finally, when cosine of 1 third x, when x is at 6 pi, cosine of 1 third x is equal to 1, what's that going to be when you multiply it by negative 2.5?"}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "We're going to get to positive 2.5, which is right over there. And then when cosine of 1 third x is equal to 0, doesn't matter what we multiply it by. We get to 0. And then finally, when cosine of 1 third x, when x is at 6 pi, cosine of 1 third x is equal to 1, what's that going to be when you multiply it by negative 2.5? Well, it's going to be negative 2.5. So we're going to get back over here. So we're ready to draw our graph."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "And then finally, when cosine of 1 third x, when x is at 6 pi, cosine of 1 third x is equal to 1, what's that going to be when you multiply it by negative 2.5? Well, it's going to be negative 2.5. So we're going to get back over here. So we're ready to draw our graph. It looks something. Actually, let me do that magenta color, since that's the color I wrote this in. It will look like this."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "So we're ready to draw our graph. It looks something. Actually, let me do that magenta color, since that's the color I wrote this in. It will look like this. I can draw it as a solid line. So it will look like that. So you saw what happened."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "It will look like this. I can draw it as a solid line. So it will look like that. So you saw what happened. By putting this 1 third here, it stretched out the graph. It increased the period by a factor of 3. And then multiplying it by negative 2.5, if you just multiplied by 2.5, you would just multiply that out a little bit."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "So you saw what happened. By putting this 1 third here, it stretched out the graph. It increased the period by a factor of 3. And then multiplying it by negative 2.5, if you just multiplied by 2.5, you would just multiply that out a little bit. But now it's a negative. So not only do you increase the amplitude, but you flip it over. So it is indeed the case that the amplitude here is 2.5."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "And then multiplying it by negative 2.5, if you just multiplied by 2.5, you would just multiply that out a little bit. But now it's a negative. So not only do you increase the amplitude, but you flip it over. So it is indeed the case that the amplitude here is 2.5. We vary 2.5 from our middle position. Or you could say that the difference between the minimum and the maximum is 5. So half of that is 2.5."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "So it is indeed the case that the amplitude here is 2.5. We vary 2.5 from our middle position. Or you could say that the difference between the minimum and the maximum is 5. So half of that is 2.5. But it isn't just multiplying this graph by 2.5. If you multiply this graph by 2.5, you'd get something that looked something like that. But because we had a negative, we had to flip it over the x-axis."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "So half of that is 2.5. But it isn't just multiplying this graph by 2.5. If you multiply this graph by 2.5, you'd get something that looked something like that. But because we had a negative, we had to flip it over the x-axis. And we got this here. So this amplitude is 2.5. But it's a flipped over version of this graph."}, {"video_title": "Proof of angle addition formula for cosine Trigonometry Khan Academy.mp3", "Sentence": "So if we have a plus here, we're gonna have a minus here. Minus sine of x, sine of x, sine of y. And I'm gonna use a very similar technique to the way I proved it for sine. And so I encourage you to pause the video either now or at any time that you get the inspiration to see if you can do this proof on your own. So just like we thought about it for sine, what is the cosine of x plus y in this diagram right over here? Well, x plus y is this angle right over here. And if we look at the right triangle ADF, cosine of x plus y, well, cosine is adjacent over hypotenuse, segment AF over the hypotenuse."}, {"video_title": "Proof of angle addition formula for cosine Trigonometry Khan Academy.mp3", "Sentence": "And so I encourage you to pause the video either now or at any time that you get the inspiration to see if you can do this proof on your own. So just like we thought about it for sine, what is the cosine of x plus y in this diagram right over here? Well, x plus y is this angle right over here. And if we look at the right triangle ADF, cosine of x plus y, well, cosine is adjacent over hypotenuse, segment AF over the hypotenuse. Or, since the hypotenuse is just 1, AF divided by 1 is just going to be AF. So cosine of x plus y is just the length of segment. It's just the length of segment AF."}, {"video_title": "Proof of angle addition formula for cosine Trigonometry Khan Academy.mp3", "Sentence": "And if we look at the right triangle ADF, cosine of x plus y, well, cosine is adjacent over hypotenuse, segment AF over the hypotenuse. Or, since the hypotenuse is just 1, AF divided by 1 is just going to be AF. So cosine of x plus y is just the length of segment. It's just the length of segment AF. So that right over there is equivalent to this right over here. Let me actually write that down. So copy and paste."}, {"video_title": "Proof of angle addition formula for cosine Trigonometry Khan Academy.mp3", "Sentence": "It's just the length of segment AF. So that right over there is equivalent to this right over here. Let me actually write that down. So copy and paste. So length of segment AF is cosine of x plus y. So let's think about how we can get that. And the way I'm gonna think about it is, given the other right triangles we have in this diagram, if we could figure out that or AF, let me write it this way, this thing, which is the same thing as AF, is equal to, let me write it this way, it's equal to the length of segment AB."}, {"video_title": "Proof of angle addition formula for cosine Trigonometry Khan Academy.mp3", "Sentence": "So copy and paste. So length of segment AF is cosine of x plus y. So let's think about how we can get that. And the way I'm gonna think about it is, given the other right triangles we have in this diagram, if we could figure out that or AF, let me write it this way, this thing, which is the same thing as AF, is equal to, let me write it this way, it's equal to the length of segment AB. So it's equal to the length of segment AB, which is this entire segment right over here, minus the length of segment FB. So minus this segment right over here, minus the length of segment FB. And just from the way our angle addition formula looks for cosine, you might guess what's going to be AB and what's going to be FB."}, {"video_title": "Proof of angle addition formula for cosine Trigonometry Khan Academy.mp3", "Sentence": "And the way I'm gonna think about it is, given the other right triangles we have in this diagram, if we could figure out that or AF, let me write it this way, this thing, which is the same thing as AF, is equal to, let me write it this way, it's equal to the length of segment AB. So it's equal to the length of segment AB, which is this entire segment right over here, minus the length of segment FB. So minus this segment right over here, minus the length of segment FB. And just from the way our angle addition formula looks for cosine, you might guess what's going to be AB and what's going to be FB. If we can prove that AB is equal to this, and if we can prove that FB is equal to this, then we're done. Cuz we know that cosine of x plus y, which is AF, is equal to AB minus FB. Or if we can prove this, that it's equal to that minus that."}, {"video_title": "Proof of angle addition formula for cosine Trigonometry Khan Academy.mp3", "Sentence": "And just from the way our angle addition formula looks for cosine, you might guess what's going to be AB and what's going to be FB. If we can prove that AB is equal to this, and if we can prove that FB is equal to this, then we're done. Cuz we know that cosine of x plus y, which is AF, is equal to AB minus FB. Or if we can prove this, that it's equal to that minus that. So let's think about what these things actually are. What is AB? So let's look at right triangle ACB."}, {"video_title": "Proof of angle addition formula for cosine Trigonometry Khan Academy.mp3", "Sentence": "Or if we can prove this, that it's equal to that minus that. So let's think about what these things actually are. What is AB? So let's look at right triangle ACB. We know from the previous video that since AD has a hypotenuse, or since triangle ADC has a hypotenuse of 1, this length is 1, that AC is cosine of x. And so what is AB? Well, think about it."}, {"video_title": "Proof of angle addition formula for cosine Trigonometry Khan Academy.mp3", "Sentence": "So let's look at right triangle ACB. We know from the previous video that since AD has a hypotenuse, or since triangle ADC has a hypotenuse of 1, this length is 1, that AC is cosine of x. And so what is AB? Well, think about it. AB is adjacent to the angle that has measure y. Or we could say that the cosine, actually let me do it down here. We could say, since I've already looked at all of that, we could say that the cosine of y is equal to its adjacent side, the length of its adjacent side, so that is segment AB over the hypotenuse, over cosine of x, cosine of x."}, {"video_title": "Proof of angle addition formula for cosine Trigonometry Khan Academy.mp3", "Sentence": "Well, think about it. AB is adjacent to the angle that has measure y. Or we could say that the cosine, actually let me do it down here. We could say, since I've already looked at all of that, we could say that the cosine of y is equal to its adjacent side, the length of its adjacent side, so that is segment AB over the hypotenuse, over cosine of x, cosine of x. Or multiply both sides by cosine of x, we get that AB, segment AB is equal to cosine of x, cosine of x times cosine of y, cosine of y. Which is exactly what we set out to prove. We've just proven that AB is indeed, the length of segment AB is indeed equal to cosine x, cosine y."}, {"video_title": "Proof of angle addition formula for cosine Trigonometry Khan Academy.mp3", "Sentence": "We could say, since I've already looked at all of that, we could say that the cosine of y is equal to its adjacent side, the length of its adjacent side, so that is segment AB over the hypotenuse, over cosine of x, cosine of x. Or multiply both sides by cosine of x, we get that AB, segment AB is equal to cosine of x, cosine of x times cosine of y, cosine of y. Which is exactly what we set out to prove. We've just proven that AB is indeed, the length of segment AB is indeed equal to cosine x, cosine y. This whole thing is equal to cosine x, cosine y. So now we just have to prove that segment FB is equal to sine x, sine y. Well, this looks like a bit of a strange segment right over here."}, {"video_title": "Proof of angle addition formula for cosine Trigonometry Khan Academy.mp3", "Sentence": "We've just proven that AB is indeed, the length of segment AB is indeed equal to cosine x, cosine y. This whole thing is equal to cosine x, cosine y. So now we just have to prove that segment FB is equal to sine x, sine y. Well, this looks like a bit of a strange segment right over here. It's not part of any, at least right triangle, where we know that I've drawn where we know one of the angles. But we can see from this diagram, ECBF is a rectangle. We use that fact in the proof for the angle addition formula for sine."}, {"video_title": "Proof of angle addition formula for cosine Trigonometry Khan Academy.mp3", "Sentence": "Well, this looks like a bit of a strange segment right over here. It's not part of any, at least right triangle, where we know that I've drawn where we know one of the angles. But we can see from this diagram, ECBF is a rectangle. We use that fact in the proof for the angle addition formula for sine. We'll also use it now. Because that tells us that FB is the same as EC, is EC. And what is EC going to be equal to?"}, {"video_title": "Proof of angle addition formula for cosine Trigonometry Khan Academy.mp3", "Sentence": "We use that fact in the proof for the angle addition formula for sine. We'll also use it now. Because that tells us that FB is the same as EC, is EC. And what is EC going to be equal to? Well, we have this angle y right over here. And so what is, what is, what is, let's, let's see. This side is opposite the angle y, so we might want to involve sine."}, {"video_title": "Proof of angle addition formula for cosine Trigonometry Khan Academy.mp3", "Sentence": "And what is EC going to be equal to? Well, we have this angle y right over here. And so what is, what is, what is, let's, let's see. This side is opposite the angle y, so we might want to involve sine. So we know that sine of y, sine of y, I'm looking right over here, is equal to the length of the opposite side, which is the length of EC, which is the length of EC over the hypotenuse, which is sine of x. Sine of x. We figured that out from the last video."}, {"video_title": "Proof of angle addition formula for cosine Trigonometry Khan Academy.mp3", "Sentence": "This side is opposite the angle y, so we might want to involve sine. So we know that sine of y, sine of y, I'm looking right over here, is equal to the length of the opposite side, which is the length of EC, which is the length of EC over the hypotenuse, which is sine of x. Sine of x. We figured that out from the last video. If this is x, opposite over hypotenuse is sine of x. Well, the opposite is just 1, so the opposite is equal to sine of x. But over here, multiply both sides by sine of x, and we get what we were looking for."}, {"video_title": "Proof of angle addition formula for cosine Trigonometry Khan Academy.mp3", "Sentence": "We figured that out from the last video. If this is x, opposite over hypotenuse is sine of x. Well, the opposite is just 1, so the opposite is equal to sine of x. But over here, multiply both sides by sine of x, and we get what we were looking for. EC is equal to sine of x times sine of y. Once again, EC was the exact same thing, has the same length as segment FB. So we have just shown that segment FB is equal to sine of x times sine of y."}, {"video_title": "Proof of angle addition formula for cosine Trigonometry Khan Academy.mp3", "Sentence": "But over here, multiply both sides by sine of x, and we get what we were looking for. EC is equal to sine of x times sine of y. Once again, EC was the exact same thing, has the same length as segment FB. So we have just shown that segment FB is equal to sine of x times sine of y. This is equal to that right over there. So once again, cosine of x plus y, which is equal to segment AF, is equal to segment AB minus the length of segment FB, which is equal to, we've proven the length of segment AB is cosine x, cosine y, minus the length of segment FB, which is sine of x, sine of y. And we are done."}, {"video_title": "Rotation by radians and quadrants Trigonometry Khan Academy.mp3", "Sentence": "So we're going to start with this magenta ray. And we're going to rotate it around the origin counterclockwise by different angle measures and think about what quadrant do we fall into if we start with this and we were to rotate by counterclockwise by 3 pi over 5 radians. And then if we start with this and we were to rotate counterclockwise by 2 pi over 7 radians. Or if we were to start with this and then rotate counterclockwise by 3 radians. So I encourage you to pause the video and think about starting with this, if we were to rotate counterclockwise by each of these, what quadrant are we going to end up in? So assume you have had, you paused the video and you tried it out on your own. So let's try this first one, 3 pi over 5."}, {"video_title": "Rotation by radians and quadrants Trigonometry Khan Academy.mp3", "Sentence": "Or if we were to start with this and then rotate counterclockwise by 3 radians. So I encourage you to pause the video and think about starting with this, if we were to rotate counterclockwise by each of these, what quadrant are we going to end up in? So assume you have had, you paused the video and you tried it out on your own. So let's try this first one, 3 pi over 5. So 3 pi over 5, so we're going to start rotating. So if we go straight up, if we rotate it essentially, if we want to think in degrees, if you rotate it counterclockwise 90 degrees, that is going to get us 2 pi over 2. So that would have been a counterclockwise rotation of pi over 2 radians."}, {"video_title": "Rotation by radians and quadrants Trigonometry Khan Academy.mp3", "Sentence": "So let's try this first one, 3 pi over 5. So 3 pi over 5, so we're going to start rotating. So if we go straight up, if we rotate it essentially, if we want to think in degrees, if you rotate it counterclockwise 90 degrees, that is going to get us 2 pi over 2. So that would have been a counterclockwise rotation of pi over 2 radians. Now is 3 pi over 5 greater or less than that? Well 3 pi over 5, 3 pi over 5 is greater than, or I guess another way to say it is, 3 pi over 6 is less than 3 pi over 5. You make the denominator smaller, making the fraction larger."}, {"video_title": "Rotation by radians and quadrants Trigonometry Khan Academy.mp3", "Sentence": "So that would have been a counterclockwise rotation of pi over 2 radians. Now is 3 pi over 5 greater or less than that? Well 3 pi over 5, 3 pi over 5 is greater than, or I guess another way to say it is, 3 pi over 6 is less than 3 pi over 5. You make the denominator smaller, making the fraction larger. So 3 pi over 6 is the same thing as pi over 2. So let me write it this way. So 3 pi over 2 is less than 3 pi over 5."}, {"video_title": "Rotation by radians and quadrants Trigonometry Khan Academy.mp3", "Sentence": "You make the denominator smaller, making the fraction larger. So 3 pi over 6 is the same thing as pi over 2. So let me write it this way. So 3 pi over 2 is less than 3 pi over 5. So it's definitely past this, so we're going to go past this. Now does that get us all the way over here? If we were to go, essentially be pointed in the opposite direction, instead of being pointed to the right, making a full, I guess you could say, 180 degree counterclockwise rotation, that would be pi radians."}, {"video_title": "Rotation by radians and quadrants Trigonometry Khan Academy.mp3", "Sentence": "So 3 pi over 2 is less than 3 pi over 5. So it's definitely past this, so we're going to go past this. Now does that get us all the way over here? If we were to go, essentially be pointed in the opposite direction, instead of being pointed to the right, making a full, I guess you could say, 180 degree counterclockwise rotation, that would be pi radians. That would be pi radians. But this thing is less than pi. Pi would be 5 pi over 5."}, {"video_title": "Rotation by radians and quadrants Trigonometry Khan Academy.mp3", "Sentence": "If we were to go, essentially be pointed in the opposite direction, instead of being pointed to the right, making a full, I guess you could say, 180 degree counterclockwise rotation, that would be pi radians. That would be pi radians. But this thing is less than pi. Pi would be 5 pi over 5. So this is less than pi radians. So we are going to sit, we are going to sit someplace, someplace, and I'm just estimating it, we are going to sit someplace like that. And so we are going to sit in the second quadrant."}, {"video_title": "Rotation by radians and quadrants Trigonometry Khan Academy.mp3", "Sentence": "Pi would be 5 pi over 5. So this is less than pi radians. So we are going to sit, we are going to sit someplace, someplace, and I'm just estimating it, we are going to sit someplace like that. And so we are going to sit in the second quadrant. Now let's think about 2 pi over 7. So 2 pi over 7, do we even get past pi over 2? Well, pi over 2 here would be 3.5 pi over 7."}, {"video_title": "Rotation by radians and quadrants Trigonometry Khan Academy.mp3", "Sentence": "And so we are going to sit in the second quadrant. Now let's think about 2 pi over 7. So 2 pi over 7, do we even get past pi over 2? Well, pi over 2 here would be 3.5 pi over 7. So we don't even get to pi over 2. We're going to end up, we're going to end up someplace, someplace over here. This thing is greater than zero, so we're going to definitely start moving counterclockwise."}, {"video_title": "Rotation by radians and quadrants Trigonometry Khan Academy.mp3", "Sentence": "Well, pi over 2 here would be 3.5 pi over 7. So we don't even get to pi over 2. We're going to end up, we're going to end up someplace, someplace over here. This thing is greater than zero, so we're going to definitely start moving counterclockwise. But we're not even going to get to, this thing is less than pi over 2. So this is going to throw us in the first quadrant. Now what about 3 radians?"}, {"video_title": "Rotation by radians and quadrants Trigonometry Khan Academy.mp3", "Sentence": "This thing is greater than zero, so we're going to definitely start moving counterclockwise. But we're not even going to get to, this thing is less than pi over 2. So this is going to throw us in the first quadrant. Now what about 3 radians? So one way to think about it is 3 is a little bit less than pi. Right? 3 is less than pi, but it's greater than pi over 2."}, {"video_title": "Rotation by radians and quadrants Trigonometry Khan Academy.mp3", "Sentence": "Now what about 3 radians? So one way to think about it is 3 is a little bit less than pi. Right? 3 is less than pi, but it's greater than pi over 2. How do we know that? Well, pi is, pi is approximately 3.14159, and it just keeps going on and on and on forever. So 3 is definitely closer to that than it is to half of that."}, {"video_title": "Rotation by radians and quadrants Trigonometry Khan Academy.mp3", "Sentence": "3 is less than pi, but it's greater than pi over 2. How do we know that? Well, pi is, pi is approximately 3.14159, and it just keeps going on and on and on forever. So 3 is definitely closer to that than it is to half of that. So it's going to be between pi over 2 and pi. So it's going to be, if we start with this magenta, this magenta ray, and we rotate counterclockwise by 3 radians, we are going to get, actually it's probably going to be, it's going to look something, it's going to be something like this. But for the sake of this exercise, we have gotten ourselves, once again, into the second quadrant."}, {"video_title": "Pythagorean trig identity from soh cah toa Trigonometry Khan Academy.mp3", "Sentence": "And to do that, let's pick one of these non-right angles. So let's pick this angle right over here as theta. And let's just think about what the sine of theta is and what the cosine of theta is, and see if we can mess with them a little bit to somehow leverage the Pythagorean theorem. So before we do that, let's just write down SOH CAH TOA just so we remember the definitions of these trig functions. So sine is opposite of our hypotenuse. CAH, cosine is adjacent over hypotenuse. And TOA, tan is opposite over adjacent."}, {"video_title": "Pythagorean trig identity from soh cah toa Trigonometry Khan Academy.mp3", "Sentence": "So before we do that, let's just write down SOH CAH TOA just so we remember the definitions of these trig functions. So sine is opposite of our hypotenuse. CAH, cosine is adjacent over hypotenuse. And TOA, tan is opposite over adjacent. We won't be using tan, at least in this video. So let's think about sine of theta. So sine of theta, I will do it in this blue color."}, {"video_title": "Pythagorean trig identity from soh cah toa Trigonometry Khan Academy.mp3", "Sentence": "And TOA, tan is opposite over adjacent. We won't be using tan, at least in this video. So let's think about sine of theta. So sine of theta, I will do it in this blue color. So sine of theta is what? It is opposite over hypotenuse. So it is equal to the length of b, or it is equal to b. b is the length."}, {"video_title": "Pythagorean trig identity from soh cah toa Trigonometry Khan Academy.mp3", "Sentence": "So sine of theta, I will do it in this blue color. So sine of theta is what? It is opposite over hypotenuse. So it is equal to the length of b, or it is equal to b. b is the length. b over the length of the hypotenuse, which is c. Now what is cosine of theta? Cosine of theta, well, the adjacent side, the side of this angle that is not the hypotenuse, it has length a. So it's the length of the adjacent side over the length of the hypotenuse."}, {"video_title": "Pythagorean trig identity from soh cah toa Trigonometry Khan Academy.mp3", "Sentence": "So it is equal to the length of b, or it is equal to b. b is the length. b over the length of the hypotenuse, which is c. Now what is cosine of theta? Cosine of theta, well, the adjacent side, the side of this angle that is not the hypotenuse, it has length a. So it's the length of the adjacent side over the length of the hypotenuse. Now how could I relate these things? Well, it seems like if I square sine of theta, then I'm going to have sine squared theta is equal to b squared over c squared. And cosine squared theta is going to be a squared over c squared."}, {"video_title": "Pythagorean trig identity from soh cah toa Trigonometry Khan Academy.mp3", "Sentence": "So it's the length of the adjacent side over the length of the hypotenuse. Now how could I relate these things? Well, it seems like if I square sine of theta, then I'm going to have sine squared theta is equal to b squared over c squared. And cosine squared theta is going to be a squared over c squared. It seems like I might be able to add them to get something that's pretty close up with the Aggren theorem here. So let's try that out. So sine squared theta is equal to b squared over c squared."}, {"video_title": "Pythagorean trig identity from soh cah toa Trigonometry Khan Academy.mp3", "Sentence": "And cosine squared theta is going to be a squared over c squared. It seems like I might be able to add them to get something that's pretty close up with the Aggren theorem here. So let's try that out. So sine squared theta is equal to b squared over c squared. I just squared both sides. So over c squared. Cosine squared theta is equal to a squared over c squared."}, {"video_title": "Pythagorean trig identity from soh cah toa Trigonometry Khan Academy.mp3", "Sentence": "So sine squared theta is equal to b squared over c squared. I just squared both sides. So over c squared. Cosine squared theta is equal to a squared over c squared. So what's this sum? What's sine squared theta plus cosine squared theta? So sine squared theta plus cosine squared theta is going to be equal to what?"}, {"video_title": "Pythagorean trig identity from soh cah toa Trigonometry Khan Academy.mp3", "Sentence": "Cosine squared theta is equal to a squared over c squared. So what's this sum? What's sine squared theta plus cosine squared theta? So sine squared theta plus cosine squared theta is going to be equal to what? Sine squared theta is b squared over c squared plus a squared over c squared, which is going to be equal to, well, we have a common denominator of c squared. And the numerator, we have b squared plus a squared. Now what is b squared plus a squared?"}, {"video_title": "Pythagorean trig identity from soh cah toa Trigonometry Khan Academy.mp3", "Sentence": "So sine squared theta plus cosine squared theta is going to be equal to what? Sine squared theta is b squared over c squared plus a squared over c squared, which is going to be equal to, well, we have a common denominator of c squared. And the numerator, we have b squared plus a squared. Now what is b squared plus a squared? Well, we have it right over here. Pythagorean theorem tells us b squared plus a squared, or a squared plus b squared is going to be equal to c squared. So this numerator simplifies to c squared."}, {"video_title": "Pythagorean trig identity from soh cah toa Trigonometry Khan Academy.mp3", "Sentence": "Now what is b squared plus a squared? Well, we have it right over here. Pythagorean theorem tells us b squared plus a squared, or a squared plus b squared is going to be equal to c squared. So this numerator simplifies to c squared. And the whole expression is c squared over c squared, which is just equal to 1. So using the Soh Cah Toa definition, and in a future video, we'll use the unit circle definition. But you see, just even using the Soh Cah Toa definition of our trig functions, we see probably the most important of all the trig identities that the sine squared theta, sine squared of an angle, plus the cosine squared of that same angle."}, {"video_title": "Pythagorean trig identity from soh cah toa Trigonometry Khan Academy.mp3", "Sentence": "So this numerator simplifies to c squared. And the whole expression is c squared over c squared, which is just equal to 1. So using the Soh Cah Toa definition, and in a future video, we'll use the unit circle definition. But you see, just even using the Soh Cah Toa definition of our trig functions, we see probably the most important of all the trig identities that the sine squared theta, sine squared of an angle, plus the cosine squared of that same angle. I'm introducing orange unnecessarily. Plus the cosine squared of that same angle is going to be equal to 1. Now you might probably say, OK, Sal, that's kind of cool."}, {"video_title": "Pythagorean trig identity from soh cah toa Trigonometry Khan Academy.mp3", "Sentence": "But you see, just even using the Soh Cah Toa definition of our trig functions, we see probably the most important of all the trig identities that the sine squared theta, sine squared of an angle, plus the cosine squared of that same angle. I'm introducing orange unnecessarily. Plus the cosine squared of that same angle is going to be equal to 1. Now you might probably say, OK, Sal, that's kind of cool. But what's the big deal about this? Why should I care about this? Well, the big deal is now you give me the sine of an angle, and I can solve this equation for the cosine of that angle, or vice versa."}, {"video_title": "Law of cosines to determine grade Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Let's say you're studying some type of a little hill or rock formation right over here, and you're able to figure out the dimensions. You know that from this point to this point along the base, straight along level ground is 60 meters. You know the steeper side, the steeper, I guess, surface or edge of this cliff or whatever you wanna call it is 20 meters, and then the longer side here, I guess the less deep side, is 50 meters long. So you're able to measure that, but now what you wanna do is use your knowledge of trigonometry, given this information, to figure out how steep is this side, what is the actual inclination relative to level ground, or another way of thinking about it, what is this angle theta right over there? I'd encourage you to pause the video and think about it on your own. Well, it might be ringing a bell. Well, you know three sides of a triangle, and then we wanna figure out an angle, and so the thing that jumps out at my head is well, maybe the law of cosines could be useful."}, {"video_title": "Law of cosines to determine grade Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So you're able to measure that, but now what you wanna do is use your knowledge of trigonometry, given this information, to figure out how steep is this side, what is the actual inclination relative to level ground, or another way of thinking about it, what is this angle theta right over there? I'd encourage you to pause the video and think about it on your own. Well, it might be ringing a bell. Well, you know three sides of a triangle, and then we wanna figure out an angle, and so the thing that jumps out at my head is well, maybe the law of cosines could be useful. Let me just write out the law, let me just write out the law of cosines before we try to apply it to this triangle right over here. So the law of cosines tells us that c squared is equal to a squared plus b squared minus two a b times the cosine of theta. And just to remind ourselves what the a b's and c's are, c is the side that's opposite the angle theta."}, {"video_title": "Law of cosines to determine grade Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Well, you know three sides of a triangle, and then we wanna figure out an angle, and so the thing that jumps out at my head is well, maybe the law of cosines could be useful. Let me just write out the law, let me just write out the law of cosines before we try to apply it to this triangle right over here. So the law of cosines tells us that c squared is equal to a squared plus b squared minus two a b times the cosine of theta. And just to remind ourselves what the a b's and c's are, c is the side that's opposite the angle theta. So if I were to draw an arbitrary triangle right over here, and if this is our angle theta, then this determines that c is that side, and then a and b could be either of these two sides. So a could be that one, and b could be that one, or the other way around. As you can see, a and b essentially have the same role in this formula right over here."}, {"video_title": "Law of cosines to determine grade Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And just to remind ourselves what the a b's and c's are, c is the side that's opposite the angle theta. So if I were to draw an arbitrary triangle right over here, and if this is our angle theta, then this determines that c is that side, and then a and b could be either of these two sides. So a could be that one, and b could be that one, or the other way around. As you can see, a and b essentially have the same role in this formula right over here. This could be b, or this could be a. So what we wanna do is somehow relate this angle, we wanna figure out what theta is in our little hill example right over here. So if this is going to be theta, what is c going to be?"}, {"video_title": "Law of cosines to determine grade Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "As you can see, a and b essentially have the same role in this formula right over here. This could be b, or this could be a. So what we wanna do is somehow relate this angle, we wanna figure out what theta is in our little hill example right over here. So if this is going to be theta, what is c going to be? Well, c is going to be this 20 meter side, and then we could set either one of these to be a or b. We could say that this a is 50 meters, and b is 60 meters. And now we can just apply the law of cosines."}, {"video_title": "Law of cosines to determine grade Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So if this is going to be theta, what is c going to be? Well, c is going to be this 20 meter side, and then we could set either one of these to be a or b. We could say that this a is 50 meters, and b is 60 meters. And now we can just apply the law of cosines. So the law of cosines tells us that 20 squared is equal to a squared, so that's 50 squared, plus b squared, plus 60 squared, minus two, minus two, times a b, so minus two times 50, times 60, times 60, times the cosine of theta. Times the cosine of theta. This works out well for us, because they've given us everything, there's only one unknown."}, {"video_title": "Law of cosines to determine grade Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And now we can just apply the law of cosines. So the law of cosines tells us that 20 squared is equal to a squared, so that's 50 squared, plus b squared, plus 60 squared, minus two, minus two, times a b, so minus two times 50, times 60, times 60, times the cosine of theta. Times the cosine of theta. This works out well for us, because they've given us everything, there's only one unknown. There's theta here. So let's see if we can solve for theta. So 20 squared, that is 400."}, {"video_title": "Law of cosines to determine grade Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "This works out well for us, because they've given us everything, there's only one unknown. There's theta here. So let's see if we can solve for theta. So 20 squared, that is 400. 50 squared, 50 squared is, is 2,500. 60 squared is 3,600. And then 50 times, well let's see, two times 50 is 100, times 60, this is all equal to 6,000."}, {"video_title": "Law of cosines to determine grade Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So 20 squared, that is 400. 50 squared, 50 squared is, is 2,500. 60 squared is 3,600. And then 50 times, well let's see, two times 50 is 100, times 60, this is all equal to 6,000. So let's see, if we simplify this a little bit, we're going to get 400 is equal to 2,500 plus 3,600, let's see, that'd be 6,100, that's equal to 6,000, let me do this in a new color. So when I add, when I add these two, I get 6,100, did I do that right? Yeah, so it's 2,000 plus 3,000 is 5,000, 500 plus 6,000 is 1,100."}, {"video_title": "Law of cosines to determine grade Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And then 50 times, well let's see, two times 50 is 100, times 60, this is all equal to 6,000. So let's see, if we simplify this a little bit, we're going to get 400 is equal to 2,500 plus 3,600, let's see, that'd be 6,100, that's equal to 6,000, let me do this in a new color. So when I add, when I add these two, I get 6,100, did I do that right? Yeah, so it's 2,000 plus 3,000 is 5,000, 500 plus 6,000 is 1,100. So I get 6,100 minus 6,000, minus 6,000 times the cosine of theta, times the cosine of theta. And let's see, now we can subtract 6,100 from both sides. So I'm just gonna subtract 6,100 from both sides so that I get closer to isolating the theta."}, {"video_title": "Law of cosines to determine grade Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Yeah, so it's 2,000 plus 3,000 is 5,000, 500 plus 6,000 is 1,100. So I get 6,100 minus 6,000, minus 6,000 times the cosine of theta, times the cosine of theta. And let's see, now we can subtract 6,100 from both sides. So I'm just gonna subtract 6,100 from both sides so that I get closer to isolating the theta. So let's do that. So this is going to be, this is going to be negative 5,700, is that right? 5,700 plus, yes, that is right, right, because if this was the other way around, if this was 6,100 minus 400, it would be positive 5,700, all right?"}, {"video_title": "Law of cosines to determine grade Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So I'm just gonna subtract 6,100 from both sides so that I get closer to isolating the theta. So let's do that. So this is going to be, this is going to be negative 5,700, is that right? 5,700 plus, yes, that is right, right, because if this was the other way around, if this was 6,100 minus 400, it would be positive 5,700, all right? And then these two, of course, cancel out, and this is going to be equal to negative 6,000 times the cosine of theta, times the cosine of theta. Now we can divide both sides by negative 6,000, by negative 6,000, negative 6,000, and we get, and I'm just gonna swap the sides, we get cosine of theta is equal to, let's see, we could divide the numerator and the denominator by essentially negative 100, so these are both going to become positive. So cosine of theta is equal to 57, is equal to 57 over 60."}, {"video_title": "Law of cosines to determine grade Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "5,700 plus, yes, that is right, right, because if this was the other way around, if this was 6,100 minus 400, it would be positive 5,700, all right? And then these two, of course, cancel out, and this is going to be equal to negative 6,000 times the cosine of theta, times the cosine of theta. Now we can divide both sides by negative 6,000, by negative 6,000, negative 6,000, and we get, and I'm just gonna swap the sides, we get cosine of theta is equal to, let's see, we could divide the numerator and the denominator by essentially negative 100, so these are both going to become positive. So cosine of theta is equal to 57, is equal to 57 over 60. Actually, that can be simplified even more. Three goes into 57, is that 19 times? Yep, so this is actually, this could be simplified, this is equal to 19, 19 over 20."}, {"video_title": "Law of cosines to determine grade Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So cosine of theta is equal to 57, is equal to 57 over 60. Actually, that can be simplified even more. Three goes into 57, is that 19 times? Yep, so this is actually, this could be simplified, this is equal to 19, 19 over 20. And we actually didn't have to do that simplification step because we're about to use our calculators, but that makes the math a little bit, makes it a little bit more tractable, right? Three goes into 57, yeah, 19 times. And so now we can take the inverse cosine of both sides."}, {"video_title": "Law of cosines to determine grade Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Yep, so this is actually, this could be simplified, this is equal to 19, 19 over 20. And we actually didn't have to do that simplification step because we're about to use our calculators, but that makes the math a little bit, makes it a little bit more tractable, right? Three goes into 57, yeah, 19 times. And so now we can take the inverse cosine of both sides. So we could get theta is equal to the inverse cosine, or the arc cosine of 19 over 20, of 19 over, 19 over 20. So let's get our calculator out and see if we get something that makes sense. So we want to do the inverse cosine of 19 over 20, 19 over 20, and we deserve a drum roll."}, {"video_title": "Law of cosines to determine grade Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And so now we can take the inverse cosine of both sides. So we could get theta is equal to the inverse cosine, or the arc cosine of 19 over 20, of 19 over, 19 over 20. So let's get our calculator out and see if we get something that makes sense. So we want to do the inverse cosine of 19 over 20, 19 over 20, and we deserve a drum roll. We get 18.19 degrees. And I already verified that my calculator is in degree mode. So it gets 18.19 degrees."}, {"video_title": "Pythagorean trig identity from unit circle Trigonometry Khan Academy.mp3", "Sentence": "So let's review the unit circle definition of trig functions a little bit. Right over here, I've drawn a unit circle, and when we say unit circle, we're talking about a circle with radius one. So for example, this point right over here is the point one comma zero, x is equal to one, y is zero. This point is the point zero comma one. This is the point negative one comma zero, and this is the point zero comma negative one. That the radius over here, the distance from the center of the circle, which is at the origin, to any point in the circle, or any point on the circle, I should say, this radius is equal to one. So the unit circle definition of trig functions leverages this unit circle."}, {"video_title": "Pythagorean trig identity from unit circle Trigonometry Khan Academy.mp3", "Sentence": "This point is the point zero comma one. This is the point negative one comma zero, and this is the point zero comma negative one. That the radius over here, the distance from the center of the circle, which is at the origin, to any point in the circle, or any point on the circle, I should say, this radius is equal to one. So the unit circle definition of trig functions leverages this unit circle. That's why it's called the unit circle definition. And we saw that if we define an angle as the kind of bottom side of the angle being along the positive x-axis, and then the other side of that angle, thinking about where it intersects the unit circle, so let's say that this is the angle theta, we define sine of theta and cosine of theta, or cosine theta and sine of theta, as the x and y coordinates of this point at which this side of the angle, the side that is not the positive x-axis, where that intersects the unit circle. So for example, this point right over here, if we would call this the x-coordinate of this point, so this value right over here, we would call that cosine of theta, we would call that cosine of theta, and the y-coordinate of that point, which is this point right over here, we would call that sine of theta."}, {"video_title": "Pythagorean trig identity from unit circle Trigonometry Khan Academy.mp3", "Sentence": "So the unit circle definition of trig functions leverages this unit circle. That's why it's called the unit circle definition. And we saw that if we define an angle as the kind of bottom side of the angle being along the positive x-axis, and then the other side of that angle, thinking about where it intersects the unit circle, so let's say that this is the angle theta, we define sine of theta and cosine of theta, or cosine theta and sine of theta, as the x and y coordinates of this point at which this side of the angle, the side that is not the positive x-axis, where that intersects the unit circle. So for example, this point right over here, if we would call this the x-coordinate of this point, so this value right over here, we would call that cosine of theta, we would call that cosine of theta, and the y-coordinate of that point, which is this point right over here, we would call that sine of theta. And in previous videos on the unit circle, we talked about why this is really just a natural extension of the Sohcahtoa definition. What's useful is it starts to work for negative angles, it works for angles, it even works for 90-degree angles, it works for angles more than 90 degrees, it works for angles less than 90 degrees. So it's really, really, really useful."}, {"video_title": "Pythagorean trig identity from unit circle Trigonometry Khan Academy.mp3", "Sentence": "So for example, this point right over here, if we would call this the x-coordinate of this point, so this value right over here, we would call that cosine of theta, we would call that cosine of theta, and the y-coordinate of that point, which is this point right over here, we would call that sine of theta. And in previous videos on the unit circle, we talked about why this is really just a natural extension of the Sohcahtoa definition. What's useful is it starts to work for negative angles, it works for angles, it even works for 90-degree angles, it works for angles more than 90 degrees, it works for angles less than 90 degrees. So it's really, really, really useful. But what I wanna do is leverage what we already know about the unit circle definition of trig functions to help prove the Pythagorean identity. It almost falls out of the fact that this point right over here is on a circle, a circle of radius one. So what is the equation of a circle with radius one centered at the origin?"}, {"video_title": "Pythagorean trig identity from unit circle Trigonometry Khan Academy.mp3", "Sentence": "So it's really, really, really useful. But what I wanna do is leverage what we already know about the unit circle definition of trig functions to help prove the Pythagorean identity. It almost falls out of the fact that this point right over here is on a circle, a circle of radius one. So what is the equation of a circle with radius one centered at the origin? Well, the equation of that is x squared, x squared, we have other videos where we really prove this using the distance formula, which is really just an application of the Pythagorean theorem. The equation of a circle, of a unit circle centered at the origin is x squared plus y squared, plus y squared is equal to, is equal to one, is equal to the radius squared. This distance right over here is equal to one."}, {"video_title": "Pythagorean trig identity from unit circle Trigonometry Khan Academy.mp3", "Sentence": "So what is the equation of a circle with radius one centered at the origin? Well, the equation of that is x squared, x squared, we have other videos where we really prove this using the distance formula, which is really just an application of the Pythagorean theorem. The equation of a circle, of a unit circle centered at the origin is x squared plus y squared, plus y squared is equal to, is equal to one, is equal to the radius squared. This distance right over here is equal to one. Well, we've already said that we're defining cosine of theta as the x coordinate of this point, and we're defining sine of theta as the y coordinate of this point, and this point is sitting on the circle. It has to satisfy this relationship right over here. So that means, well, if we're defining cosine of theta to be the x, to be this x value, sine of theta to be the y value, and it has to satisfy this relationship, that means that cosine of theta squared plus sine squared of theta, plus sine squared of theta needs to be equal to, needs to be equal to one, or sine squared theta plus cosine squared of theta needs to be equal to one, and that's just from the point."}, {"video_title": "Pythagorean trig identity from unit circle Trigonometry Khan Academy.mp3", "Sentence": "This distance right over here is equal to one. Well, we've already said that we're defining cosine of theta as the x coordinate of this point, and we're defining sine of theta as the y coordinate of this point, and this point is sitting on the circle. It has to satisfy this relationship right over here. So that means, well, if we're defining cosine of theta to be the x, to be this x value, sine of theta to be the y value, and it has to satisfy this relationship, that means that cosine of theta squared plus sine squared of theta, plus sine squared of theta needs to be equal to, needs to be equal to one, or sine squared theta plus cosine squared of theta needs to be equal to one, and that's just from the point. This is the x, cosine theta is the x coordinate, sine theta is the y coordinate. They have to satisfy this relationship, which defines a circle, so cosine squared theta plus sine squared theta is one. Now, this is called, as we've seen in other videos, this is called the Pythagorean identity, and you say, why is that useful?"}, {"video_title": "Pythagorean trig identity from unit circle Trigonometry Khan Academy.mp3", "Sentence": "So that means, well, if we're defining cosine of theta to be the x, to be this x value, sine of theta to be the y value, and it has to satisfy this relationship, that means that cosine of theta squared plus sine squared of theta, plus sine squared of theta needs to be equal to, needs to be equal to one, or sine squared theta plus cosine squared of theta needs to be equal to one, and that's just from the point. This is the x, cosine theta is the x coordinate, sine theta is the y coordinate. They have to satisfy this relationship, which defines a circle, so cosine squared theta plus sine squared theta is one. Now, this is called, as we've seen in other videos, this is called the Pythagorean identity, and you say, why is that useful? Well, using this, if you know sine of theta, you can figure out what cosine of theta is going to be, or vice versa, and if you know one of cosine of theta, and then you can, say you know cosine theta, then you use this to figure out sine of theta, then you can figure out tangent of theta, because tangent of theta is just sine over cosine, but if you're a little bit confused as to why this is called the Pythagorean identity, well, it really just falls out of where even the equation of a circle even came from. If we look at this point right over here, if we look at this point right over here, which we're saying is the x coordinate is cosine theta and the y coordinate is sine of theta, what is the distance between that point and the origin? Well, to think about that, we can construct a right triangle, so this distance right over here, this distance right over here, and so that we can deal with any quadrant."}, {"video_title": "Pythagorean trig identity from unit circle Trigonometry Khan Academy.mp3", "Sentence": "Now, this is called, as we've seen in other videos, this is called the Pythagorean identity, and you say, why is that useful? Well, using this, if you know sine of theta, you can figure out what cosine of theta is going to be, or vice versa, and if you know one of cosine of theta, and then you can, say you know cosine theta, then you use this to figure out sine of theta, then you can figure out tangent of theta, because tangent of theta is just sine over cosine, but if you're a little bit confused as to why this is called the Pythagorean identity, well, it really just falls out of where even the equation of a circle even came from. If we look at this point right over here, if we look at this point right over here, which we're saying is the x coordinate is cosine theta and the y coordinate is sine of theta, what is the distance between that point and the origin? Well, to think about that, we can construct a right triangle, so this distance right over here, this distance right over here, and so that we can deal with any quadrant. I'll make it the absolute value of the cosine theta is this distance right over here, and this distance right over here is the absolute value of the sine of theta, the absolute value of the sine of theta, and I obviously don't have to take the absolute value for this first quadrant here, but if I went into the other quadrants and I were to set up a similar right triangle, then the absolute value is at play, and so what do we know from the Pythagorean theorem? This is a right triangle here. The hypotenuse has length one, so we know that this expression squared, the absolute value of cosine of theta squared plus this expression squared, which is this length, plus the absolute value of the sine of theta squared needs to be equal to the length of the hypotenuse squared, which is the same thing, which is going to be equal to one squared, or we could say, this is the same thing."}, {"video_title": "Pythagorean trig identity from unit circle Trigonometry Khan Academy.mp3", "Sentence": "Well, to think about that, we can construct a right triangle, so this distance right over here, this distance right over here, and so that we can deal with any quadrant. I'll make it the absolute value of the cosine theta is this distance right over here, and this distance right over here is the absolute value of the sine of theta, the absolute value of the sine of theta, and I obviously don't have to take the absolute value for this first quadrant here, but if I went into the other quadrants and I were to set up a similar right triangle, then the absolute value is at play, and so what do we know from the Pythagorean theorem? This is a right triangle here. The hypotenuse has length one, so we know that this expression squared, the absolute value of cosine of theta squared plus this expression squared, which is this length, plus the absolute value of the sine of theta squared needs to be equal to the length of the hypotenuse squared, which is the same thing, which is going to be equal to one squared, or we could say, this is the same thing. If you're gonna square something, the sine, if it's negative, it's gonna be negative times a negative, so it's just gonna be positive, so this is going to be the same thing as saying that the cosine squared theta plus sine squared theta, plus sine squared theta is equal to one. So once again, or I guess this is why it's called a Pythagorean identity, and actually that's even where the equation of a circle comes from. It comes straight out of the Pythagorean theorem where your hypotenuse has length one."}, {"video_title": "Cosine addition identity example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "We can verify that it is because it satisfies the Pythagorean theorem. 8 squared is 64 plus 15 squared is 225. 64 plus 225 is 289, which is 17 squared. So it satisfies the Pythagorean theorem. So we know that this right over here is a right triangle. But what they're asking us is what is the cosine of angle ABC? ABC, which is this angle right over here, plus 60 degrees."}, {"video_title": "Cosine addition identity example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So it satisfies the Pythagorean theorem. So we know that this right over here is a right triangle. But what they're asking us is what is the cosine of angle ABC? ABC, which is this angle right over here, plus 60 degrees. Now just like this, I don't have any clear way of evaluating it. But we do have some trig identities in our toolkit that we could use to express this in a way that we might be able to evaluate it. In particular, we know that the cosine of A plus B is equal to the cosine of A times the cosine of B minus the sine of A times the sine of B."}, {"video_title": "Cosine addition identity example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "ABC, which is this angle right over here, plus 60 degrees. Now just like this, I don't have any clear way of evaluating it. But we do have some trig identities in our toolkit that we could use to express this in a way that we might be able to evaluate it. In particular, we know that the cosine of A plus B is equal to the cosine of A times the cosine of B minus the sine of A times the sine of B. So we could do the exact same thing here with the cosine of angle ABC plus 60 degrees. This is going to be equal to, let me just write out the whole thing, the cosine of angle ABC times the cosine of 60 degrees minus the sine of angle ABC times the sine of 60 degrees. So this right over here is angle ABC."}, {"video_title": "Cosine addition identity example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "In particular, we know that the cosine of A plus B is equal to the cosine of A times the cosine of B minus the sine of A times the sine of B. So we could do the exact same thing here with the cosine of angle ABC plus 60 degrees. This is going to be equal to, let me just write out the whole thing, the cosine of angle ABC times the cosine of 60 degrees minus the sine of angle ABC times the sine of 60 degrees. So this right over here is angle ABC. This is angle ABC. And this is 60 degrees. And this is 60 degrees."}, {"video_title": "Cosine addition identity example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So this right over here is angle ABC. This is angle ABC. And this is 60 degrees. And this is 60 degrees. So let's evaluate each of these things. What is the cosine of angle ABC going to be equal to? I'll do that over here."}, {"video_title": "Cosine addition identity example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And this is 60 degrees. So let's evaluate each of these things. What is the cosine of angle ABC going to be equal to? I'll do that over here. The cosine of angle ABC. Well, we could go back to SOH CAH TOA. Let me write it down."}, {"video_title": "Cosine addition identity example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "I'll do that over here. The cosine of angle ABC. Well, we could go back to SOH CAH TOA. Let me write it down. SOH CAH TOA. Cosine is defined as the adjacent over the hypotenuse. Cosine is defined as the adjacent for this angle, the adjacent over the hypotenuse."}, {"video_title": "Cosine addition identity example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Let me write it down. SOH CAH TOA. Cosine is defined as the adjacent over the hypotenuse. Cosine is defined as the adjacent for this angle, the adjacent over the hypotenuse. So cosine of angle ABC is equal to 15 over 17. So this right over here is 15 over 17. What is the cosine of 60 degrees?"}, {"video_title": "Cosine addition identity example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Cosine is defined as the adjacent for this angle, the adjacent over the hypotenuse. So cosine of angle ABC is equal to 15 over 17. So this right over here is 15 over 17. What is the cosine of 60 degrees? Well, for there, we have to turn back to our knowledge of 30-60-90 triangles. So if I have a triangle like this, so let me do my best here, construct a 30-60-90 right triangle. So that's a 60-degree side."}, {"video_title": "Cosine addition identity example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "What is the cosine of 60 degrees? Well, for there, we have to turn back to our knowledge of 30-60-90 triangles. So if I have a triangle like this, so let me do my best here, construct a 30-60-90 right triangle. So that's a 60-degree side. This is a 30-degree side. We know, and we've seen this multiple times, if the hypotenuse has length 1, the side opposite the 30-degree side is 1 half. And the side opposite the 60-degree side is square root of 3 times this."}, {"video_title": "Cosine addition identity example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So that's a 60-degree side. This is a 30-degree side. We know, and we've seen this multiple times, if the hypotenuse has length 1, the side opposite the 30-degree side is 1 half. And the side opposite the 60-degree side is square root of 3 times this. So it's square root of 3 over 2. So the cosine of 60 degrees, if you're looking at this side right over here, let me write it in a color I haven't used yet. So I care about the cosine of 60 degrees."}, {"video_title": "Cosine addition identity example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And the side opposite the 60-degree side is square root of 3 times this. So it's square root of 3 over 2. So the cosine of 60 degrees, if you're looking at this side right over here, let me write it in a color I haven't used yet. So I care about the cosine of 60 degrees. The cosine of 60 degrees is going to be equal to, once again, adjacent over hypotenuse, 1 half over 1. It's going to be equal to 1 half over 1, which is equal to 1 half. Now let's think about the sine of angle ABC."}, {"video_title": "Cosine addition identity example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So I care about the cosine of 60 degrees. The cosine of 60 degrees is going to be equal to, once again, adjacent over hypotenuse, 1 half over 1. It's going to be equal to 1 half over 1, which is equal to 1 half. Now let's think about the sine of angle ABC. So that's this one right over here. So we have our triangle right here. Sine is opposite over hypotenuse."}, {"video_title": "Cosine addition identity example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Now let's think about the sine of angle ABC. So that's this one right over here. So we have our triangle right here. Sine is opposite over hypotenuse. So opposite has length 8 over hypotenuse is 17. So it's equal to 8 over 17. And then finally, we need to figure out, we need to think about what the sine of 60 degrees is."}, {"video_title": "Cosine addition identity example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Sine is opposite over hypotenuse. So opposite has length 8 over hypotenuse is 17. So it's equal to 8 over 17. And then finally, we need to figure out, we need to think about what the sine of 60 degrees is. Well, if we're at the 60-degree side of this right triangle, opposite over hypotenuse. So square root of 3 over 2 over 1, which is just square root of 3 over 2. So we have all the information we need to evaluate it."}, {"video_title": "Cosine addition identity example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And then finally, we need to figure out, we need to think about what the sine of 60 degrees is. Well, if we're at the 60-degree side of this right triangle, opposite over hypotenuse. So square root of 3 over 2 over 1, which is just square root of 3 over 2. So we have all the information we need to evaluate it. This whole thing, so this was the sine of 60 degrees. This whole thing is going to evaluate to cosine of angle ABC is 15 over 17 times cosine of 60 degrees is 1 half. So times 1 half."}, {"video_title": "Cosine addition identity example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So we have all the information we need to evaluate it. This whole thing, so this was the sine of 60 degrees. This whole thing is going to evaluate to cosine of angle ABC is 15 over 17 times cosine of 60 degrees is 1 half. So times 1 half. And then we're going to subtract sine of ABC, which is 8 over 17. And then times sine of 60, which is square root of 3 over 2. So times square root of 3 over 2."}, {"video_title": "Cosine addition identity example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So times 1 half. And then we're going to subtract sine of ABC, which is 8 over 17. And then times sine of 60, which is square root of 3 over 2. So times square root of 3 over 2. And now we just have to simplify a little bit. So this is going to be equal to, if I multiply 1 half times this, let's see, this is going to be 15 over 34 minus, and let's see, we're dividing by 2. So it's 4 17ths."}, {"video_title": "Cosine addition identity example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So times square root of 3 over 2. And now we just have to simplify a little bit. So this is going to be equal to, if I multiply 1 half times this, let's see, this is going to be 15 over 34 minus, and let's see, we're dividing by 2. So it's 4 17ths. I'll write this. 4 square roots of 3 over 17. And that's about as simple as I could do it."}, {"video_title": "Cosine addition identity example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So it's 4 17ths. I'll write this. 4 square roots of 3 over 17. And that's about as simple as I could do it. If I wanted, I could have a common denominator here of 34. And then I could add the 2. So it could be 8 square roots of 3 over 34."}, {"video_title": "Cosine addition identity example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And that's about as simple as I could do it. If I wanted, I could have a common denominator here of 34. And then I could add the 2. So it could be 8 square roots of 3 over 34. But that still won't simplify it that much. So this is a reasonable good answer for what they are asking for. 15 over 34 minus 4 square roots of 3 over 17."}, {"video_title": "IIT JEE trigonometric constraints Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So just from the get-go, it seems like we're going to have to solve for thetas and convert between sine and cosine, which is essentially what you're doing between tangent and cotangent. We'll do that in a little bit. And those of you who know me, if you're actually taking an exam under time pressure, I don't recommend you reproving your trig identities from first principles, but for education purposes, I always like to do it. So let's just, let me just draw a right triangle right over here. If you're actually going to take the IIT JEE exam, I recommend having most of the trig identities at your fingertips. But let's say, right to right triangle, and let's call that theta, and then this angle right over here is going to be pi over 2 minus theta. Pi over 2 minus theta."}, {"video_title": "IIT JEE trigonometric constraints Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So let's just, let me just draw a right triangle right over here. If you're actually going to take the IIT JEE exam, I recommend having most of the trig identities at your fingertips. But let's say, right to right triangle, and let's call that theta, and then this angle right over here is going to be pi over 2 minus theta. Pi over 2 minus theta. If we're in radians, right? 90 degrees right here is pi over 2. The whole triangle right over here is 180 degrees, which is pi."}, {"video_title": "IIT JEE trigonometric constraints Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Pi over 2 minus theta. If we're in radians, right? 90 degrees right here is pi over 2. The whole triangle right over here is 180 degrees, which is pi. So this is pi over 2, and so these two guys are going to have to add up to the other pi over 2, so this is pi over 2 minus theta. So let's think about what cosine of theta is. Cosine of theta, SOH CAH TOA."}, {"video_title": "IIT JEE trigonometric constraints Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "The whole triangle right over here is 180 degrees, which is pi. So this is pi over 2, and so these two guys are going to have to add up to the other pi over 2, so this is pi over 2 minus theta. So let's think about what cosine of theta is. Cosine of theta, SOH CAH TOA. Cosine of theta is this side, let's call this A, B, and C. It is going to be the adjacent side. So cosine is adjacent, oh let me write this down. SOH CAH TOA."}, {"video_title": "IIT JEE trigonometric constraints Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Cosine of theta, SOH CAH TOA. Cosine of theta is this side, let's call this A, B, and C. It is going to be the adjacent side. So cosine is adjacent, oh let me write this down. SOH CAH TOA. Cosine is adjacent over hypotenuse. So cosine of theta is B over C, adjacent over hypotenuse. It's equal to B over C. What's also equal to B over C?"}, {"video_title": "IIT JEE trigonometric constraints Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "SOH CAH TOA. Cosine is adjacent over hypotenuse. So cosine of theta is B over C, adjacent over hypotenuse. It's equal to B over C. What's also equal to B over C? If we look at this angle over here, from this angle's point of view, B is the opposite side. So it's opposite over hypotenuse for this angle's point of view. So it's also equal to the sine of pi over 2 minus theta."}, {"video_title": "IIT JEE trigonometric constraints Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "It's equal to B over C. What's also equal to B over C? If we look at this angle over here, from this angle's point of view, B is the opposite side. So it's opposite over hypotenuse for this angle's point of view. So it's also equal to the sine of pi over 2 minus theta. So we got our first main identity here, that the cosine of theta is equal to the sine of pi over 2 minus theta. And we could go the other way around, use the exact same logic to get that the sine of theta is equal to the cosine of pi over 2 minus theta. So that will be pretty useful when we try to solve this equation right over here."}, {"video_title": "IIT JEE trigonometric constraints Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So it's also equal to the sine of pi over 2 minus theta. So we got our first main identity here, that the cosine of theta is equal to the sine of pi over 2 minus theta. And we could go the other way around, use the exact same logic to get that the sine of theta is equal to the cosine of pi over 2 minus theta. So that will be pretty useful when we try to solve this equation right over here. And we can even use it when we solve this one. Because if I were to write down the cotangent of 5 theta, let me write this down. So the cotangent of 5 theta, this is the exact same thing, the cotangent of 5 theta is the exact same thing as the cosine of 5 theta over the sine of 5 theta."}, {"video_title": "IIT JEE trigonometric constraints Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So that will be pretty useful when we try to solve this equation right over here. And we can even use it when we solve this one. Because if I were to write down the cotangent of 5 theta, let me write this down. So the cotangent of 5 theta, this is the exact same thing, the cotangent of 5 theta is the exact same thing as the cosine of 5 theta over the sine of 5 theta. It's just 1 over the tangent. And this is the same thing, I can convert the cosine into sine using this identity up here. This is the sine of pi over 2 minus 5 theta over the cosine of pi over 2 minus 5 theta."}, {"video_title": "IIT JEE trigonometric constraints Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So the cotangent of 5 theta, this is the exact same thing, the cotangent of 5 theta is the exact same thing as the cosine of 5 theta over the sine of 5 theta. It's just 1 over the tangent. And this is the same thing, I can convert the cosine into sine using this identity up here. This is the sine of pi over 2 minus 5 theta over the cosine of pi over 2 minus 5 theta. For here, I'm just using this identity over here. And this is the exact same thing as the tangent of pi over 2 minus 5 theta. And since we're going to actually have to solve this equation eventually, and we want to make sure we get all of the solutions to this equation, one thing that you might want to remember is when you take a tangent of an angle, and I'll just draw the unit circle here."}, {"video_title": "IIT JEE trigonometric constraints Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "This is the sine of pi over 2 minus 5 theta over the cosine of pi over 2 minus 5 theta. For here, I'm just using this identity over here. And this is the exact same thing as the tangent of pi over 2 minus 5 theta. And since we're going to actually have to solve this equation eventually, and we want to make sure we get all of the solutions to this equation, one thing that you might want to remember is when you take a tangent of an angle, and I'll just draw the unit circle here. So if I have some angle, let me draw my axes. And let's say that this right here is theta. You probably remember that this theta is essentially the slope of the line formed."}, {"video_title": "IIT JEE trigonometric constraints Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And since we're going to actually have to solve this equation eventually, and we want to make sure we get all of the solutions to this equation, one thing that you might want to remember is when you take a tangent of an angle, and I'll just draw the unit circle here. So if I have some angle, let me draw my axes. And let's say that this right here is theta. You probably remember that this theta is essentially the slope of the line formed. It's opposite over adjacent. It's the slope of the line formed by the radius in the unit circle. So this theta is going to have the exact same tangent as if we add 180 degrees, or if we add pi to this."}, {"video_title": "IIT JEE trigonometric constraints Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "You probably remember that this theta is essentially the slope of the line formed. It's opposite over adjacent. It's the slope of the line formed by the radius in the unit circle. So this theta is going to have the exact same tangent as if we add 180 degrees, or if we add pi to this. So if you add pi, you get theta plus pi. So this whole angle right over here is theta plus pi. And its tangent is going to be the same thing."}, {"video_title": "IIT JEE trigonometric constraints Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So this theta is going to have the exact same tangent as if we add 180 degrees, or if we add pi to this. So if you add pi, you get theta plus pi. So this whole angle right over here is theta plus pi. And its tangent is going to be the same thing. The slope of the radius is going to be the same. So you can add multiples of pi to a tangent value, and you're going to get the exact same value. So let me put some multiples of pi over here so that we can make sure that we get all of the solutions."}, {"video_title": "IIT JEE trigonometric constraints Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And its tangent is going to be the same thing. The slope of the radius is going to be the same. So you can add multiples of pi to a tangent value, and you're going to get the exact same value. So let me put some multiples of pi over here so that we can make sure that we get all of the solutions. So with that said, with this written, we can now solve this first equation over here. And then we can worry about some of these other constraints later. So tangent of theta is equal to cotangent of 5 theta."}, {"video_title": "IIT JEE trigonometric constraints Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So let me put some multiples of pi over here so that we can make sure that we get all of the solutions. So with that said, with this written, we can now solve this first equation over here. And then we can worry about some of these other constraints later. So tangent of theta is equal to cotangent of 5 theta. Well, so we can write tangent of theta is equal to, instead of writing cotangent of 5 theta, we can write is equal to the tangent of pi over 2 minus 5 theta plus integer multiples of pi. And so the tangent of this is equal to the tangent of that. These things could be equal to each other."}, {"video_title": "IIT JEE trigonometric constraints Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So tangent of theta is equal to cotangent of 5 theta. Well, so we can write tangent of theta is equal to, instead of writing cotangent of 5 theta, we can write is equal to the tangent of pi over 2 minus 5 theta plus integer multiples of pi. And so the tangent of this is equal to the tangent of that. These things could be equal to each other. So you get theta is equal to pi over 2 minus 5 theta plus n pi. Or at minimum, this could be equal, when you take the tangent of this, it's equal to the tangent of either the same thing, or we could add multiples of pi to it. You could do it either way."}, {"video_title": "IIT JEE trigonometric constraints Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "These things could be equal to each other. So you get theta is equal to pi over 2 minus 5 theta plus n pi. Or at minimum, this could be equal, when you take the tangent of this, it's equal to the tangent of either the same thing, or we could add multiples of pi to it. You could do it either way. But now we just have to solve this. We can add 5 theta to both sides of this equation. We're adding this to both sides."}, {"video_title": "IIT JEE trigonometric constraints Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "You could do it either way. But now we just have to solve this. We can add 5 theta to both sides of this equation. We're adding this to both sides. So I get 6 theta is equal to pi over 2 minus n pi, minus multiples of pi. Sorry, plus multiples of pi. There's a plus right over there."}, {"video_title": "IIT JEE trigonometric constraints Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "We're adding this to both sides. So I get 6 theta is equal to pi over 2 minus n pi, minus multiples of pi. Sorry, plus multiples of pi. There's a plus right over there. Divide both sides by 6. I get theta is equal to pi over 12 plus n times pi over 6. Or just to make the fraction adding easier, this is the same thing as pi over 12 plus 2."}, {"video_title": "IIT JEE trigonometric constraints Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "There's a plus right over there. Divide both sides by 6. I get theta is equal to pi over 12 plus n times pi over 6. Or just to make the fraction adding easier, this is the same thing as pi over 12 plus 2. I'll write it this way, plus 2n pi over 12. This is the same thing as n pi over 6. I just multiply the numerator and denominator by 2."}, {"video_title": "IIT JEE trigonometric constraints Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Or just to make the fraction adding easier, this is the same thing as pi over 12 plus 2. I'll write it this way, plus 2n pi over 12. This is the same thing as n pi over 6. I just multiply the numerator and denominator by 2. And this is the exact same thing. You can imagine a 1 coefficient out here. This is the same thing as 2n plus 1 times pi over 12."}, {"video_title": "IIT JEE trigonometric constraints Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "I just multiply the numerator and denominator by 2. And this is the exact same thing. You can imagine a 1 coefficient out here. This is the same thing as 2n plus 1 times pi over 12. That's all of the solutions for this first equation right over there. Now let's do the same thing for this second equation. And then we'll see where they overlap in this range."}, {"video_title": "IIT JEE trigonometric constraints Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "This is the same thing as 2n plus 1 times pi over 12. That's all of the solutions for this first equation right over there. Now let's do the same thing for this second equation. And then we'll see where they overlap in this range. And we'll take out anything that satisfies this criteria right over here. The second equation, let me write it over here. We have sine of 2 theta."}, {"video_title": "IIT JEE trigonometric constraints Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And then we'll see where they overlap in this range. And we'll take out anything that satisfies this criteria right over here. The second equation, let me write it over here. We have sine of 2 theta. Let me write it over here first. Cosine of 4 theta. What does cosine of 4 theta equal to?"}, {"video_title": "IIT JEE trigonometric constraints Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "We have sine of 2 theta. Let me write it over here first. Cosine of 4 theta. What does cosine of 4 theta equal to? Cosine of 4 theta, using the same exact logic, is equal to sine of pi over 2 minus theta. And of course, when we take a sine or a cosine, or we take really any trig identity or any trig function of any angle, you're going to get the same value if you add multiples of 2 pi. So let me just add multiples of 2 pi here, just so that we can make sure that we get all the possible solutions."}, {"video_title": "IIT JEE trigonometric constraints Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "What does cosine of 4 theta equal to? Cosine of 4 theta, using the same exact logic, is equal to sine of pi over 2 minus theta. And of course, when we take a sine or a cosine, or we take really any trig identity or any trig function of any angle, you're going to get the same value if you add multiples of 2 pi. So let me just add multiples of 2 pi here, just so that we can make sure that we get all the possible solutions. So this is what cosine of 4 theta is equal to. So let us, let us, oh, let me be careful. Cosine of 4 theta is not equal to pi over 2 minus theta."}, {"video_title": "IIT JEE trigonometric constraints Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So let me just add multiples of 2 pi here, just so that we can make sure that we get all the possible solutions. So this is what cosine of 4 theta is equal to. So let us, let us, oh, let me be careful. Cosine of 4 theta is not equal to pi over 2 minus theta. That would be cosine of theta. If it's cosine of 4 theta, it's pi over 2 minus 4. Pi over 2 minus 4 theta, plus multiples of 2 pi."}, {"video_title": "IIT JEE trigonometric constraints Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Cosine of 4 theta is not equal to pi over 2 minus theta. That would be cosine of theta. If it's cosine of 4 theta, it's pi over 2 minus 4. Pi over 2 minus 4 theta, plus multiples of 2 pi. So plus 2 pi n. Because obviously you add multiples of 2 pi, you're going to go back to the same angle. So you go back to this equation up here. We get sine of 2 theta, sine of 2 theta, is equal to cosine of 4 theta, which is the same thing as this over here, which is equal to sine of pi over 2 minus 4 theta, plus, plus 2 pi n. And so, these evaluate to the same thing."}, {"video_title": "IIT JEE trigonometric constraints Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Pi over 2 minus 4 theta, plus multiples of 2 pi. So plus 2 pi n. Because obviously you add multiples of 2 pi, you're going to go back to the same angle. So you go back to this equation up here. We get sine of 2 theta, sine of 2 theta, is equal to cosine of 4 theta, which is the same thing as this over here, which is equal to sine of pi over 2 minus 4 theta, plus, plus 2 pi n. And so, these evaluate to the same thing. Let's set them equal to each other. 2 theta is equal to pi over 2 minus 4 pi plus 2 pi n. Let's add, let's add 4 pi to both sides of this equation. We get 6, we get 6 theta."}, {"video_title": "IIT JEE trigonometric constraints Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "We get sine of 2 theta, sine of 2 theta, is equal to cosine of 4 theta, which is the same thing as this over here, which is equal to sine of pi over 2 minus 4 theta, plus, plus 2 pi n. And so, these evaluate to the same thing. Let's set them equal to each other. 2 theta is equal to pi over 2 minus 4 pi plus 2 pi n. Let's add, let's add 4 pi to both sides of this equation. We get 6, we get 6 theta. 6 theta is equal to pi over 2, so this is cancelled out, plus 2 pi n. Plus 2 pi n. Let's divide both sides by 6. Let's divide both sides by 6. We get theta is equal to pi over 12, plus pi over 12, plus pi over 3n."}, {"video_title": "IIT JEE trigonometric constraints Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "We get 6, we get 6 theta. 6 theta is equal to pi over 2, so this is cancelled out, plus 2 pi n. Plus 2 pi n. Let's divide both sides by 6. Let's divide both sides by 6. We get theta is equal to pi over 12, plus pi over 12, plus pi over 3n. 2 divided by 6 is pi n over 3. Now we can put it over a common denominator just to simplify things. So we have over 12, we have this pi."}, {"video_title": "IIT JEE trigonometric constraints Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "We get theta is equal to pi over 12, plus pi over 12, plus pi over 3n. 2 divided by 6 is pi n over 3. Now we can put it over a common denominator just to simplify things. So we have over 12, we have this pi. Pi n over 3 is the same thing as plus 4n pi over 12. Or we could write this as being, this is equal to 4n plus, 4n plus 1, 4n plus 1 times pi over 12. So now we just need to see where there's overlap, where there's overlap between this solution and that solution."}, {"video_title": "IIT JEE trigonometric constraints Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So we have over 12, we have this pi. Pi n over 3 is the same thing as plus 4n pi over 12. Or we could write this as being, this is equal to 4n plus, 4n plus 1, 4n plus 1 times pi over 12. So now we just need to see where there's overlap, where there's overlap between this solution and that solution. Remember, we just have to count the number of solutions. We actually don't have to find the solutions. Actually, if you're pretty savvy and do things in your head, just at this point, you can think about how many of these you're going to have between negative pi over 2 and pi over 2."}, {"video_title": "IIT JEE trigonometric constraints Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So now we just need to see where there's overlap, where there's overlap between this solution and that solution. Remember, we just have to count the number of solutions. We actually don't have to find the solutions. Actually, if you're pretty savvy and do things in your head, just at this point, you can think about how many of these you're going to have between negative pi over 2 and pi over 2. If you remember, that was the range that we're working with, between negative pi over 2 and pi over 2, not including those two. Then you can figure out, and actually, every one of these is also going to be one of these. Because no matter what you set n equal to, if you set n equal to twice that, you're going to have the equivalent thing."}, {"video_title": "IIT JEE trigonometric constraints Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Actually, if you're pretty savvy and do things in your head, just at this point, you can think about how many of these you're going to have between negative pi over 2 and pi over 2. If you remember, that was the range that we're working with, between negative pi over 2 and pi over 2, not including those two. Then you can figure out, and actually, every one of these is also going to be one of these. Because no matter what you set n equal to, if you set n equal to twice that, you're going to have the equivalent thing. So any of these are going to be any of these. Anything that satisfies this equation will also satisfy this equation right here. So really, we can just count how many are in this one."}, {"video_title": "IIT JEE trigonometric constraints Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Because no matter what you set n equal to, if you set n equal to twice that, you're going to have the equivalent thing. So any of these are going to be any of these. Anything that satisfies this equation will also satisfy this equation right here. So really, we can just count how many are in this one. But just to make it clear, I'm going to find all of them that satisfy this equation, and then we're going to see how many of those satisfy this equation as well. But the fast way to do it, just find the ones that satisfy this one, or just count the ones that satisfy this one, and you've done the problem. So let's just start at n equals 0."}, {"video_title": "IIT JEE trigonometric constraints Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So really, we can just count how many are in this one. But just to make it clear, I'm going to find all of them that satisfy this equation, and then we're going to see how many of those satisfy this equation as well. But the fast way to do it, just find the ones that satisfy this one, or just count the ones that satisfy this one, and you've done the problem. So let's just start at n equals 0. If n is equal to 0, and I'm using this n up here, I won't even write that. If n is equal to 0, we just have pi over 12. If n is 1, we get 3 pi over 12."}, {"video_title": "IIT JEE trigonometric constraints Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So let's just start at n equals 0. If n is equal to 0, and I'm using this n up here, I won't even write that. If n is equal to 0, we just have pi over 12. If n is 1, we get 3 pi over 12. If n is 2, we get 5 pi over 12. And we can't make n equal 3, because if n was equal to 3, this would be 7 pi over 12, which is greater than pi over 2. That's greater than 6 pi over 12, so we can't have that."}, {"video_title": "IIT JEE trigonometric constraints Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "If n is 1, we get 3 pi over 12. If n is 2, we get 5 pi over 12. And we can't make n equal 3, because if n was equal to 3, this would be 7 pi over 12, which is greater than pi over 2. That's greater than 6 pi over 12, so we can't have that. We could also go to negative n. We could also have negative n. If n is negative 1, then this becomes negative pi over 12. If n is negative 2, this becomes negative 3 pi over 12. And if n is negative 3, then this becomes negative 5 pi over 12."}, {"video_title": "IIT JEE trigonometric constraints Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "That's greater than 6 pi over 12, so we can't have that. We could also go to negative n. We could also have negative n. If n is negative 1, then this becomes negative pi over 12. If n is negative 2, this becomes negative 3 pi over 12. And if n is negative 3, then this becomes negative 5 pi over 12. And once again, we can't go to n is equal to negative 4, because then we get negative 7 pi over 12, which is out of our range. So this satisfies this equation up here. Now, which of these overlap with this over here?"}, {"video_title": "IIT JEE trigonometric constraints Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And if n is negative 3, then this becomes negative 5 pi over 12. And once again, we can't go to n is equal to negative 4, because then we get negative 7 pi over 12, which is out of our range. So this satisfies this equation up here. Now, which of these overlap with this over here? So let's set n equal to 0. We get pi over 12. If n is equal to 1, we get 5 pi over 12."}, {"video_title": "IIT JEE trigonometric constraints Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Now, which of these overlap with this over here? So let's set n equal to 0. We get pi over 12. If n is equal to 1, we get 5 pi over 12. We can't set n equal to 2, because then we'll get out of our range. We'll go above pi over 2. If n is equal to negative 1, we get negative 3 pi over 12."}, {"video_title": "IIT JEE trigonometric constraints Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "If n is equal to 1, we get 5 pi over 12. We can't set n equal to 2, because then we'll get out of our range. We'll go above pi over 2. If n is equal to negative 1, we get negative 3 pi over 12. If n is equal to negative 2, this is negative 8, then we get negative 7 pi over 12, which is smaller than negative pi over 2. So that doesn't count. So there are 3 solutions."}, {"video_title": "IIT JEE trigonometric constraints Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "If n is equal to negative 1, we get negative 3 pi over 12. If n is equal to negative 2, this is negative 8, then we get negative 7 pi over 12, which is smaller than negative pi over 2. So that doesn't count. So there are 3 solutions. If we look over here, we have 3 solutions. We have 1, 2, 3 solutions. And none of them satisfy this."}, {"video_title": "IIT JEE trigonometric constraints Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So there are 3 solutions. If we look over here, we have 3 solutions. We have 1, 2, 3 solutions. And none of them satisfy this. None of them are multiples of n pi over 5. So the answer to our problem is 3. And you really could have just done the second equation right over here and realized that anything that's a solution to this top equation would be a solution to this bottom equation, and you would have just counted these solutions, and you would have been done."}, {"video_title": "Example Graph of cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "The horizontal axis is the theta axis, and the vertical axis is the y-axis. And so this is the graph of y is equal to cosine of theta. And it makes sense with our unit circle definition, and I'll just make sure that we're comfortable with that, because with our unit circle definition, let me draw ourselves a unit circle. And I'm just going to draw it very roughly, just so that we get the general idea of what's going on here. When theta is equal to zero, we're at this point right over here on the unit circle. Well, what's the x-coordinate of that point? Well, it's one."}, {"video_title": "Example Graph of cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And I'm just going to draw it very roughly, just so that we get the general idea of what's going on here. When theta is equal to zero, we're at this point right over here on the unit circle. Well, what's the x-coordinate of that point? Well, it's one. And you see, when theta is equal to zero on this graph, cosine of theta is equal to one. When theta is equal to pi over two, we're at this point on the unit circle, and the x-coordinate is what? Well, the x-coordinate there is zero."}, {"video_title": "Example Graph of cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Well, it's one. And you see, when theta is equal to zero on this graph, cosine of theta is equal to one. When theta is equal to pi over two, we're at this point on the unit circle, and the x-coordinate is what? Well, the x-coordinate there is zero. And you see, once again, when we're at pi over two, the x-coordinate is zero. So this is completely consistent with our unit circle definition. As we move in the rightward direction, we're moving counterclockwise around the unit circle."}, {"video_title": "Example Graph of cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Well, the x-coordinate there is zero. And you see, once again, when we're at pi over two, the x-coordinate is zero. So this is completely consistent with our unit circle definition. As we move in the rightward direction, we're moving counterclockwise around the unit circle. And as we move in the leftward direction, we're moving counterclockwise. If we move in the rightward direction, we're moving counterclockwise. And as we're moving in the leftward direction along the axis in the negative angles, we're moving in the clockwise direction around our unit circle."}, {"video_title": "Example Graph of cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "As we move in the rightward direction, we're moving counterclockwise around the unit circle. And as we move in the leftward direction, we're moving counterclockwise. If we move in the rightward direction, we're moving counterclockwise. And as we're moving in the leftward direction along the axis in the negative angles, we're moving in the clockwise direction around our unit circle. So let's answer their question. For what values of theta does cosine of theta equal one? Well, we can just read the graph right over here."}, {"video_title": "Example Graph of cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And as we're moving in the leftward direction along the axis in the negative angles, we're moving in the clockwise direction around our unit circle. So let's answer their question. For what values of theta does cosine of theta equal one? Well, we can just read the graph right over here. It equals one. So cosine of theta equals one at theta is equal to. Well, we see it right over here."}, {"video_title": "Example Graph of cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Well, we can just read the graph right over here. It equals one. So cosine of theta equals one at theta is equal to. Well, we see it right over here. Theta is equal to zero. Theta is equal to, well, we've got to go all the way again to 2 pi. But then it just keeps going on and on."}, {"video_title": "Example Graph of cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Well, we see it right over here. Theta is equal to zero. Theta is equal to, well, we've got to go all the way again to 2 pi. But then it just keeps going on and on. And it makes sense. Cosine of theta, the x-coordinate on this unit circle, equaled one right when we were at zero angle. And we had to go all the way around the circle to get back to that point, 2 pi radians."}, {"video_title": "Example Graph of cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "But then it just keeps going on and on. And it makes sense. Cosine of theta, the x-coordinate on this unit circle, equaled one right when we were at zero angle. And we had to go all the way around the circle to get back to that point, 2 pi radians. But then it'll be again when we get to 4 pi radians. And then 6 pi radians. So 2 pi, 4 pi, 6 pi."}, {"video_title": "Example Graph of cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And we had to go all the way around the circle to get back to that point, 2 pi radians. But then it'll be again when we get to 4 pi radians. And then 6 pi radians. So 2 pi, 4 pi, 6 pi. And I guess you could see the pattern here. We're going to keep hitting cosine of theta equals one every 2 pi. So you could really kind of view this as every multiple of 2 pi, 2 pi n, where n is an integer."}, {"video_title": "Example Graph of cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So 2 pi, 4 pi, 6 pi. And I guess you could see the pattern here. We're going to keep hitting cosine of theta equals one every 2 pi. So you could really kind of view this as every multiple of 2 pi, 2 pi n, where n is an integer. n is an integer. And that applies also for negative values. If you're going the other way around, we don't get back until we get to negative 2 pi."}, {"video_title": "Example Graph of cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So you could really kind of view this as every multiple of 2 pi, 2 pi n, where n is an integer. n is an integer. And that applies also for negative values. If you're going the other way around, we don't get back until we get to negative 2 pi. Notice we were at zero. And then the next time we're at one again is at negative 2 pi, and then negative 4 pi, and then over and over and over again. But this applies."}, {"video_title": "Example Graph of cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "If you're going the other way around, we don't get back until we get to negative 2 pi. Notice we were at zero. And then the next time we're at one again is at negative 2 pi, and then negative 4 pi, and then over and over and over again. But this applies. If n is an integer, n could be a negative number. And so we get to all of the negative values of theta where cosine of theta is equal to one. Now let's think about when cosine of theta is equal to negative one."}, {"video_title": "Example Graph of cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "But this applies. If n is an integer, n could be a negative number. And so we get to all of the negative values of theta where cosine of theta is equal to one. Now let's think about when cosine of theta is equal to negative one. So cosine of theta is equal to negative one at theta is equal to, well, we can just look at this graph right over here. Well, when theta is equal to pi. And let's see."}, {"video_title": "Example Graph of cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Now let's think about when cosine of theta is equal to negative one. So cosine of theta is equal to negative one at theta is equal to, well, we can just look at this graph right over here. Well, when theta is equal to pi. And let's see. Well, it kind of goes off this graph, but this graph would keep going like this. And you'd see it would also be at 3 pi. And you can visualize it over here."}, {"video_title": "Example Graph of cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And let's see. Well, it kind of goes off this graph, but this graph would keep going like this. And you'd see it would also be at 3 pi. And you can visualize it over here. Theta, cosine of theta is equal to negative one when we're at this point on the unit circle. So that happens when we get to pi radians. And then it won't happen again until we get to 2 pi, 3 pi radians."}, {"video_title": "Example Graph of cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And you can visualize it over here. Theta, cosine of theta is equal to negative one when we're at this point on the unit circle. So that happens when we get to pi radians. And then it won't happen again until we get to 2 pi, 3 pi radians. And it won't happen again until we add another 2 pi, until we make one entire revolution. So then that's going to be 5 pi radians. And you can keep going on and on and on."}, {"video_title": "Example Graph of cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And then it won't happen again until we get to 2 pi, 3 pi radians. And it won't happen again until we add another 2 pi, until we make one entire revolution. So then that's going to be 5 pi radians. And you can keep going on and on and on. And that's also true in the negative direction. So if we take 2 pi away from this, so if we were here, and if we go all the way around back to negative pi, it should also be the case. And you actually see it right over here on the graph."}, {"video_title": "Example Graph of cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And you can keep going on and on and on. And that's also true in the negative direction. So if we take 2 pi away from this, so if we were here, and if we go all the way around back to negative pi, it should also be the case. And you actually see it right over here on the graph. So you could think about this as 2 pi n plus pi. Or you could view it as 2n plus 1 or 2n plus 1 times pi, where n is an integer. At every n, let me write that a little bit neater, n is integer."}, {"video_title": "Example Graph of cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And you actually see it right over here on the graph. So you could think about this as 2 pi n plus pi. Or you could view it as 2n plus 1 or 2n plus 1 times pi, where n is an integer. At every n, let me write that a little bit neater, n is integer. At every one of those points, for every one of these thetas, cosine of theta is going to keep hitting negative 1 over and over again. And you see it. It goes from one bottom where you can kind of valley to the next valley."}, {"video_title": "Example Graph of cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "At every n, let me write that a little bit neater, n is integer. At every one of those points, for every one of these thetas, cosine of theta is going to keep hitting negative 1 over and over again. And you see it. It goes from one bottom where you can kind of valley to the next valley. It takes 2 pi to get to the next valley. 2 pi to get to the next valley. And that was also the same thing for the peaks."}, {"video_title": "Restricting domain of trig function to make invertible Trigonometry Khan Academy.mp3", "Sentence": "To which intervals could we restrict f of x is equal to cosine of x minus pi over four, so that f of x is invertible, and they show us what cosine of x minus pi over four, what it looks like, the graph of it. So let's just think about what it means for a function to be invertible. So a function is a mapping from a set of elements that we would call the domain. So let me, my pen is a little off today, so let's see if it works okay. So this right over here is our domain, and this over here is our range, our range. And a function maps from an element in our domain to an element in our range, that's what a function does. Now the inverse of the function maps from that element in the range to the element in the domain, so that over there would be f inverse."}, {"video_title": "Restricting domain of trig function to make invertible Trigonometry Khan Academy.mp3", "Sentence": "So let me, my pen is a little off today, so let's see if it works okay. So this right over here is our domain, and this over here is our range, our range. And a function maps from an element in our domain to an element in our range, that's what a function does. Now the inverse of the function maps from that element in the range to the element in the domain, so that over there would be f inverse. So that's the direction of the function, that's the direction of f inverse. Now a function is not invertible. One of the situations in which a function is not invertible, you could have a function where two elements of the domain map to the same element of the range."}, {"video_title": "Restricting domain of trig function to make invertible Trigonometry Khan Academy.mp3", "Sentence": "Now the inverse of the function maps from that element in the range to the element in the domain, so that over there would be f inverse. So that's the direction of the function, that's the direction of f inverse. Now a function is not invertible. One of the situations in which a function is not invertible, you could have a function where two elements of the domain map to the same element of the range. So these, both of these elements map to that element of the range, but then, so both of these are the function, but then if this is the case, it's going to be, you're not going to be able to create a function that maps the other way, because if you input this into the inverse function, where do you go? Do you go to that element of the domain, or do you go to that element right over there? And so one way to think about it is you need a one-to-one mapping."}, {"video_title": "Restricting domain of trig function to make invertible Trigonometry Khan Academy.mp3", "Sentence": "One of the situations in which a function is not invertible, you could have a function where two elements of the domain map to the same element of the range. So these, both of these elements map to that element of the range, but then, so both of these are the function, but then if this is the case, it's going to be, you're not going to be able to create a function that maps the other way, because if you input this into the inverse function, where do you go? Do you go to that element of the domain, or do you go to that element right over there? And so one way to think about it is you need a one-to-one mapping. For each element of the domain range, there's only one element of the domain that gets you there. Or another way to think about it, you could try to draw a horizontal line on the graph of the function, and see if it crosses through the function more than once. And you could see that this is indeed the case for this function right over here."}, {"video_title": "Restricting domain of trig function to make invertible Trigonometry Khan Academy.mp3", "Sentence": "And so one way to think about it is you need a one-to-one mapping. For each element of the domain range, there's only one element of the domain that gets you there. Or another way to think about it, you could try to draw a horizontal line on the graph of the function, and see if it crosses through the function more than once. And you could see that this is indeed the case for this function right over here. If I did a horizontal line right over here, now why is this the issue? Well, this is showing us, actually let me show, do a number that's a little bit easier to look at. So let's say I drew the horizontal line right over here."}, {"video_title": "Restricting domain of trig function to make invertible Trigonometry Khan Academy.mp3", "Sentence": "And you could see that this is indeed the case for this function right over here. If I did a horizontal line right over here, now why is this the issue? Well, this is showing us, actually let me show, do a number that's a little bit easier to look at. So let's say I drew the horizontal line right over here. Now why is this horizontal line an issue? Well, it's showing us that just even what the part, the part of the domain that's being graphed here, that there's several points that map to the same element of the range. They're mapping to 0.5."}, {"video_title": "Restricting domain of trig function to make invertible Trigonometry Khan Academy.mp3", "Sentence": "So let's say I drew the horizontal line right over here. Now why is this horizontal line an issue? Well, it's showing us that just even what the part, the part of the domain that's being graphed here, that there's several points that map to the same element of the range. They're mapping to 0.5. 0.5. If you, this value right over here, when you take, when you input f of that is equal to 0.5, f of this right over here is equal to 0.5, f of this right over here is 0.5. So if you have that, if you have multiple elements of your domain mapping to the same element of the range, then the function will not be invertible for that domain."}, {"video_title": "Restricting domain of trig function to make invertible Trigonometry Khan Academy.mp3", "Sentence": "They're mapping to 0.5. 0.5. If you, this value right over here, when you take, when you input f of that is equal to 0.5, f of this right over here is equal to 0.5, f of this right over here is 0.5. So if you have that, if you have multiple elements of your domain mapping to the same element of the range, then the function will not be invertible for that domain. So really what we're gonna do is we're gonna try to restrict the domain so that for that domain, if I were to essentially apply this, what I'd call the horizontal line test, I'd only intersect the function once. So let's look at the, or the graph of the function once. So let's look at these choices."}, {"video_title": "Restricting domain of trig function to make invertible Trigonometry Khan Academy.mp3", "Sentence": "So if you have that, if you have multiple elements of your domain mapping to the same element of the range, then the function will not be invertible for that domain. So really what we're gonna do is we're gonna try to restrict the domain so that for that domain, if I were to essentially apply this, what I'd call the horizontal line test, I'd only intersect the function once. So let's look at the, or the graph of the function once. So let's look at these choices. So the first one is an open set from negative five pi over four, negative five pi over four. So that's pi, that's negative pi and another fourth of pi. So that's, I think, starting right over here, going all the way to negative 1 4th pi."}, {"video_title": "Restricting domain of trig function to make invertible Trigonometry Khan Academy.mp3", "Sentence": "So let's look at these choices. So the first one is an open set from negative five pi over four, negative five pi over four. So that's pi, that's negative pi and another fourth of pi. So that's, I think, starting right over here, going all the way to negative 1 4th pi. So that's this domain right over here. Let me do this in a new color. So that's this."}, {"video_title": "Restricting domain of trig function to make invertible Trigonometry Khan Academy.mp3", "Sentence": "So that's, I think, starting right over here, going all the way to negative 1 4th pi. So that's this domain right over here. Let me do this in a new color. So that's this. And this does not include, this does not include the two endpoints. So here I can still apply the horizontal line and in that domain, I can show that there's two members of the domain that are mapping to the same element in the range. And so if I'm trying to construct the inverse of that, what would that, what would this element, which is I guess it's negative.6, what would that F inverse of negative.6 be?"}, {"video_title": "Restricting domain of trig function to make invertible Trigonometry Khan Academy.mp3", "Sentence": "So that's this. And this does not include, this does not include the two endpoints. So here I can still apply the horizontal line and in that domain, I can show that there's two members of the domain that are mapping to the same element in the range. And so if I'm trying to construct the inverse of that, what would that, what would this element, which is I guess it's negative.6, what would that F inverse of negative.6 be? Would it be this value here or would it be this value here? So I would rule this one out. So let's see, negative pi to pi."}, {"video_title": "Restricting domain of trig function to make invertible Trigonometry Khan Academy.mp3", "Sentence": "And so if I'm trying to construct the inverse of that, what would that, what would this element, which is I guess it's negative.6, what would that F inverse of negative.6 be? Would it be this value here or would it be this value here? So I would rule this one out. So let's see, negative pi to pi. So negative pi to pi, I'll do this in, I'll do this in this color right over here. Negative pi to pi. This is a fairly, so once again, over, this is a closed, so we're including the two boundaries, we're including negative pi and pi in the domain."}, {"video_title": "Restricting domain of trig function to make invertible Trigonometry Khan Academy.mp3", "Sentence": "So let's see, negative pi to pi. So negative pi to pi, I'll do this in, I'll do this in this color right over here. Negative pi to pi. This is a fairly, so once again, over, this is a closed, so we're including the two boundaries, we're including negative pi and pi in the domain. But once again, over that interval, I could apply my horizontal line here and notice, or actually I could even apply the original one that I did, that I did in blue, and notice there's multiple elements in the domain that map to, say, 0.5. So what would F inverse of 0.5 be? You can't construct a function where it maps only to one element of the domain."}, {"video_title": "Restricting domain of trig function to make invertible Trigonometry Khan Academy.mp3", "Sentence": "This is a fairly, so once again, over, this is a closed, so we're including the two boundaries, we're including negative pi and pi in the domain. But once again, over that interval, I could apply my horizontal line here and notice, or actually I could even apply the original one that I did, that I did in blue, and notice there's multiple elements in the domain that map to, say, 0.5. So what would F inverse of 0.5 be? You can't construct a function where it maps only to one element of the domain. So we could rule this one out right as well. Now, negative 1.5 pi to positive 1.5 pi. So negative 1.5 pi, so that is, so let me, so I'm running out of colors."}, {"video_title": "Restricting domain of trig function to make invertible Trigonometry Khan Academy.mp3", "Sentence": "You can't construct a function where it maps only to one element of the domain. So we could rule this one out right as well. Now, negative 1.5 pi to positive 1.5 pi. So negative 1.5 pi, so that is, so let me, so I'm running out of colors. So negative 1.5 pi to positive 1.5 pi. This one is interesting. If I apply a horizontal line there, there, there, so let's see."}, {"video_title": "Restricting domain of trig function to make invertible Trigonometry Khan Academy.mp3", "Sentence": "So negative 1.5 pi, so that is, so let me, so I'm running out of colors. So negative 1.5 pi to positive 1.5 pi. This one is interesting. If I apply a horizontal line there, there, there, so let's see. But if I apply a horizontal line right over here, I do intersect the function twice. So I have two members of this domain mapping to the same element of the range. So I could rule that one out as well."}, {"video_title": "Restricting domain of trig function to make invertible Trigonometry Khan Academy.mp3", "Sentence": "If I apply a horizontal line there, there, there, so let's see. But if I apply a horizontal line right over here, I do intersect the function twice. So I have two members of this domain mapping to the same element of the range. So I could rule that one out as well. And I'm left with one last choice, so I'm hoping this one will work out. So 1.5 pi, so that's an open set, so 1.5 pi right over there, to 5 pi over four. 5 pi over four, so that's pi and another 1.4, so that's right over there."}, {"video_title": "Restricting domain of trig function to make invertible Trigonometry Khan Academy.mp3", "Sentence": "So I could rule that one out as well. And I'm left with one last choice, so I'm hoping this one will work out. So 1.5 pi, so that's an open set, so 1.5 pi right over there, to 5 pi over four. 5 pi over four, so that's pi and another 1.4, so that's right over there. And let's see, this is, if I were to look at the graph here, it seems like it would pass the horizontal line test. At any point here, I could make a horizontal line over that domain. Actually, let me do it for the whole domain."}, {"video_title": "Restricting domain of trig function to make invertible Trigonometry Khan Academy.mp3", "Sentence": "5 pi over four, so that's pi and another 1.4, so that's right over there. And let's see, this is, if I were to look at the graph here, it seems like it would pass the horizontal line test. At any point here, I could make a horizontal line over that domain. Actually, let me do it for the whole domain. So you see that, for the whole domain, and I'm only intersecting the function once. So for every element of the range that we're mapping to, there's only one element in our domain that is mapping to it. It's passing our horizontal line test, so I would check this one right over there."}, {"video_title": "Trigonometry identity review fun Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And I have better recording software now, so I thought, two birds with one stone, let me rerecord a video and kind of refresh things in my own mind. So the trig identities that I'm going to assume that we know, because I've already made videos on them and they're a little bit involved to remember or to prove. Are that the sine of a plus b is equal to the sine of a times the cosine of b plus the sine of b times the cosine of a. That's the first one I assume going into this video we know. And then if we want to know the sine of, well I'll just write it a little differently. What if I wanted to figure out the sine of a plus, I'll write it this way, minus c. Which is the same thing as a minus c, right? Well, we could just use this formula up here to say, well that's equal to the sine of a times the cosine of minus c plus the sine of minus c times the cosine of a."}, {"video_title": "Trigonometry identity review fun Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "That's the first one I assume going into this video we know. And then if we want to know the sine of, well I'll just write it a little differently. What if I wanted to figure out the sine of a plus, I'll write it this way, minus c. Which is the same thing as a minus c, right? Well, we could just use this formula up here to say, well that's equal to the sine of a times the cosine of minus c plus the sine of minus c times the cosine of a. And we know, and I guess this is another assumption that we're going to have to have going into this video, that the cosine of minus c is equal to just the cosine of c. That the cosine is an even function. And you could look at that by looking at the graph of the cosine function or even at the unit circle itself. And that the sine is an odd function."}, {"video_title": "Trigonometry identity review fun Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Well, we could just use this formula up here to say, well that's equal to the sine of a times the cosine of minus c plus the sine of minus c times the cosine of a. And we know, and I guess this is another assumption that we're going to have to have going into this video, that the cosine of minus c is equal to just the cosine of c. That the cosine is an even function. And you could look at that by looking at the graph of the cosine function or even at the unit circle itself. And that the sine is an odd function. That the sine of minus c is actually equal to minus sine of c. So we could use both of that information to rewrite the second line up here to say that the sine of a minus c is equal to the sine of a times the cosine of c. Because the cosine of minus c is the same thing as the cosine of c. And then minus the sine of c. Instead of writing this, I could write this. Minus the sine of c times the cosine of a. So that we kind of pseudo-prove this by knowing this and this ahead of time."}, {"video_title": "Trigonometry identity review fun Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And that the sine is an odd function. That the sine of minus c is actually equal to minus sine of c. So we could use both of that information to rewrite the second line up here to say that the sine of a minus c is equal to the sine of a times the cosine of c. Because the cosine of minus c is the same thing as the cosine of c. And then minus the sine of c. Instead of writing this, I could write this. Minus the sine of c times the cosine of a. So that we kind of pseudo-prove this by knowing this and this ahead of time. Fair enough. And I'm going to use all of these to kind of prove a bunch of more trig identities that I'm going to need. So the other trig identity is that the cosine of a plus b is equal to the cosine of a."}, {"video_title": "Trigonometry identity review fun Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So that we kind of pseudo-prove this by knowing this and this ahead of time. Fair enough. And I'm going to use all of these to kind of prove a bunch of more trig identities that I'm going to need. So the other trig identity is that the cosine of a plus b is equal to the cosine of a. You don't mix up the cosines and the sines in this situation. Cosine of a times the sine of b. And this is minus."}, {"video_title": "Trigonometry identity review fun Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So the other trig identity is that the cosine of a plus b is equal to the cosine of a. You don't mix up the cosines and the sines in this situation. Cosine of a times the sine of b. And this is minus. Oh, sorry. I just said you don't mix it up. And then I mixed them up."}, {"video_title": "Trigonometry identity review fun Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And this is minus. Oh, sorry. I just said you don't mix it up. And then I mixed them up. Times the cosine of b minus sine of a times the sine of b. Now, if you want to know what the cosine of a minus b is, well, you use these same properties. Cosine of minus b, that's still going to be cosine of b."}, {"video_title": "Trigonometry identity review fun Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And then I mixed them up. Times the cosine of b minus sine of a times the sine of b. Now, if you want to know what the cosine of a minus b is, well, you use these same properties. Cosine of minus b, that's still going to be cosine of b. So that's going to be the cosine of a times the cosine. Cosine of minus b is the same thing as cosine of b. And then, but here you're going to have sine of minus b, which is the same thing as minus sine of b."}, {"video_title": "Trigonometry identity review fun Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Cosine of minus b, that's still going to be cosine of b. So that's going to be the cosine of a times the cosine. Cosine of minus b is the same thing as cosine of b. And then, but here you're going to have sine of minus b, which is the same thing as minus sine of b. And that minus will cancel that out. So it'll be plus sine of a times the sine of b. So it's a little tricky."}, {"video_title": "Trigonometry identity review fun Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And then, but here you're going to have sine of minus b, which is the same thing as minus sine of b. And that minus will cancel that out. So it'll be plus sine of a times the sine of b. So it's a little tricky. When you have a plus sign here, you get a minus there. When you have a minus sign there, you get a plus sign there. But fair enough."}, {"video_title": "Trigonometry identity review fun Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So it's a little tricky. When you have a plus sign here, you get a minus there. When you have a minus sign there, you get a plus sign there. But fair enough. I don't want to dwell on that too much, because we have many more identities to show. So what if I wanted an identity for, let's say, the cosine of 2a? Well, that's just the same thing as the cosine of a plus a."}, {"video_title": "Trigonometry identity review fun Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "But fair enough. I don't want to dwell on that too much, because we have many more identities to show. So what if I wanted an identity for, let's say, the cosine of 2a? Well, that's just the same thing as the cosine of a plus a. And then we could use this formula right up here. If my second a is just my b, then this is just equal to cosine of a times the cosine of a minus the sine of a times the sine of a. My b is also an a in this situation, which I could rewrite as, this is equal to the cosine squared of a. I just wrote cosine of a times itself twice, or times itself, minus sine squared of a."}, {"video_title": "Trigonometry identity review fun Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Well, that's just the same thing as the cosine of a plus a. And then we could use this formula right up here. If my second a is just my b, then this is just equal to cosine of a times the cosine of a minus the sine of a times the sine of a. My b is also an a in this situation, which I could rewrite as, this is equal to the cosine squared of a. I just wrote cosine of a times itself twice, or times itself, minus sine squared of a. So this is one, I guess, identity already. Cosine of 2a is equal to the cosine squared of a minus the sine squared of a. Let me box off my identities that we're showing in this video."}, {"video_title": "Trigonometry identity review fun Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "My b is also an a in this situation, which I could rewrite as, this is equal to the cosine squared of a. I just wrote cosine of a times itself twice, or times itself, minus sine squared of a. So this is one, I guess, identity already. Cosine of 2a is equal to the cosine squared of a minus the sine squared of a. Let me box off my identities that we're showing in this video. So I just showed you that one. What if I'm not satisfied? What if I just wanted in terms of cosines?"}, {"video_title": "Trigonometry identity review fun Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Let me box off my identities that we're showing in this video. So I just showed you that one. What if I'm not satisfied? What if I just wanted in terms of cosines? Well, we could break out the unit circle definition of our trig functions. This is kind of the most fundamental identity, that the sine squared of a plus the cosine squared of a is equal to 1. Or you could write that, let me think of the best way to do this."}, {"video_title": "Trigonometry identity review fun Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "What if I just wanted in terms of cosines? Well, we could break out the unit circle definition of our trig functions. This is kind of the most fundamental identity, that the sine squared of a plus the cosine squared of a is equal to 1. Or you could write that, let me think of the best way to do this. You could write that the sine squared of a is equal to 1 minus the cosine squared of a. And then we could take this and substitute it right here. So we could rewrite this identity as being equal to the cosine squared of a minus the sine squared of a."}, {"video_title": "Trigonometry identity review fun Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Or you could write that, let me think of the best way to do this. You could write that the sine squared of a is equal to 1 minus the cosine squared of a. And then we could take this and substitute it right here. So we could rewrite this identity as being equal to the cosine squared of a minus the sine squared of a. But the sine squared of a is this right there. So minus, I'll do it in a different color, minus 1 minus cosine squared of a. That's what I just substituted for the sine squared of a."}, {"video_title": "Trigonometry identity review fun Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So we could rewrite this identity as being equal to the cosine squared of a minus the sine squared of a. But the sine squared of a is this right there. So minus, I'll do it in a different color, minus 1 minus cosine squared of a. That's what I just substituted for the sine squared of a. So this is equal to the cosine squared of a minus 1 plus the cosine squared of a, which is equal to, let's see, we're just adding, I'll just continue on the right. We have 1 cosine squared of a plus another cosine squared of a, so it's 2 cosine squared of a minus 1. And all of that is equal to cosine of 2a."}, {"video_title": "Trigonometry identity review fun Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "That's what I just substituted for the sine squared of a. So this is equal to the cosine squared of a minus 1 plus the cosine squared of a, which is equal to, let's see, we're just adding, I'll just continue on the right. We have 1 cosine squared of a plus another cosine squared of a, so it's 2 cosine squared of a minus 1. And all of that is equal to cosine of 2a. Now what if I wanted to get an identity that gave me what cosine squared of a is in terms of this? So we could just solve for that. If we add 1 to both sides of this equation, actually let me write this, this is one of our other identities."}, {"video_title": "Trigonometry identity review fun Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And all of that is equal to cosine of 2a. Now what if I wanted to get an identity that gave me what cosine squared of a is in terms of this? So we could just solve for that. If we add 1 to both sides of this equation, actually let me write this, this is one of our other identities. But if we add 1 to both sides of that equation, we get 2 times the cosine squared of a is equal to cosine of 2a plus 1. And if we divide both sides of this by 2, we get the cosine squared of a is equal to 1 half, and we could rearrange these, just to do it, times 1 plus the cosine of 2a. And we're done."}, {"video_title": "Trigonometry identity review fun Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "If we add 1 to both sides of this equation, actually let me write this, this is one of our other identities. But if we add 1 to both sides of that equation, we get 2 times the cosine squared of a is equal to cosine of 2a plus 1. And if we divide both sides of this by 2, we get the cosine squared of a is equal to 1 half, and we could rearrange these, just to do it, times 1 plus the cosine of 2a. And we're done. And we have another identity, cosine squared of a. Sometimes it's called the power reduction identity right there. Now what if we wanted something in terms of the sine squared of a?"}, {"video_title": "Trigonometry identity review fun Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And we're done. And we have another identity, cosine squared of a. Sometimes it's called the power reduction identity right there. Now what if we wanted something in terms of the sine squared of a? Well then maybe we could go back up here, and we know from this identity that the sine squared of a is equal to 1 minus cosine squared of a. Or we could have gone the other way. We could have subtracted sine squared of a from both sides."}, {"video_title": "Trigonometry identity review fun Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Now what if we wanted something in terms of the sine squared of a? Well then maybe we could go back up here, and we know from this identity that the sine squared of a is equal to 1 minus cosine squared of a. Or we could have gone the other way. We could have subtracted sine squared of a from both sides. And we could have gotten, let me go down there, if I subtracted sine squared of a from both sides, you could get cosine squared of a is equal to 1 minus sine squared of a. And then we could go back into this formula right up here, and we could write down, and I'll do it in this blue color, we could write down that the cosine of 2a is equal to, instead of writing the cosine squared of a, I'll write this. Is equal to 1 minus sine squared of a minus sine squared of a."}, {"video_title": "Trigonometry identity review fun Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "We could have subtracted sine squared of a from both sides. And we could have gotten, let me go down there, if I subtracted sine squared of a from both sides, you could get cosine squared of a is equal to 1 minus sine squared of a. And then we could go back into this formula right up here, and we could write down, and I'll do it in this blue color, we could write down that the cosine of 2a is equal to, instead of writing the cosine squared of a, I'll write this. Is equal to 1 minus sine squared of a minus sine squared of a. So my cosine of 2a is equal to, see I have a minus sine squared of a minus another sine squared of a. So I have 1 minus 2 sine squared of a. So here's another identity."}, {"video_title": "Trigonometry identity review fun Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Is equal to 1 minus sine squared of a minus sine squared of a. So my cosine of 2a is equal to, see I have a minus sine squared of a minus another sine squared of a. So I have 1 minus 2 sine squared of a. So here's another identity. Another way to write my cosine of 2a. We're discovering a lot of ways to write our cosine of 2a. Now if we wanted to solve for sine squared of 2a, we could add it to both sides of the equation."}, {"video_title": "Trigonometry identity review fun Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So here's another identity. Another way to write my cosine of 2a. We're discovering a lot of ways to write our cosine of 2a. Now if we wanted to solve for sine squared of 2a, we could add it to both sides of the equation. So let me do that, and I'll just write it here for the sake of saving space. Let me scroll down a little bit. And we get, so I'm going to go here."}, {"video_title": "Trigonometry identity review fun Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Now if we wanted to solve for sine squared of 2a, we could add it to both sides of the equation. So let me do that, and I'll just write it here for the sake of saving space. Let me scroll down a little bit. And we get, so I'm going to go here. If I just add 2 sine squared of a to both sides of this, I get 2 sine squared of a plus cosine of 2a is equal to 1. Subtract cosine of 2a from both sides, you get 2 sine squared of a is equal to 1 minus cosine of 2a. Then you divide both sides of this by 2, and you get sine squared of a is equal to 1 half times 1 minus cosine of 2a."}, {"video_title": "Trigonometry identity review fun Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And we get, so I'm going to go here. If I just add 2 sine squared of a to both sides of this, I get 2 sine squared of a plus cosine of 2a is equal to 1. Subtract cosine of 2a from both sides, you get 2 sine squared of a is equal to 1 minus cosine of 2a. Then you divide both sides of this by 2, and you get sine squared of a is equal to 1 half times 1 minus cosine of 2a. And we have our other discovery, I guess we could call it, our finding. And this is interesting. It's always interesting to look at the symmetry."}, {"video_title": "Trigonometry identity review fun Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Then you divide both sides of this by 2, and you get sine squared of a is equal to 1 half times 1 minus cosine of 2a. And we have our other discovery, I guess we could call it, our finding. And this is interesting. It's always interesting to look at the symmetry. Cosine squared, they're identical, except for you have a plus 2a here for the cosine squared, and you have a minus cosine of 2a here for the sine squared. So we've already found a lot of interesting things. Let's see if we can do anything for the sine of 2a."}, {"video_title": "Trigonometry identity review fun Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "It's always interesting to look at the symmetry. Cosine squared, they're identical, except for you have a plus 2a here for the cosine squared, and you have a minus cosine of 2a here for the sine squared. So we've already found a lot of interesting things. Let's see if we can do anything for the sine of 2a. So let me pick a new color here that I haven't used. Well, I've pretty much used all my colors. So if I want to figure out the sine of 2a, this is equal to the sine of a plus a, which is equal to the sine of a times the cosine of a plus the sine of the second a times the cosine of the first a. I just wrote the same thing twice, so this is just equal to 2 sine of a cosine of a."}, {"video_title": "Trigonometry identity review fun Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Let's see if we can do anything for the sine of 2a. So let me pick a new color here that I haven't used. Well, I've pretty much used all my colors. So if I want to figure out the sine of 2a, this is equal to the sine of a plus a, which is equal to the sine of a times the cosine of a plus the sine of the second a times the cosine of the first a. I just wrote the same thing twice, so this is just equal to 2 sine of a cosine of a. That was a little bit easier. So sine of 2a is equal to that. So that's another result."}, {"video_title": "Trigonometry identity review fun Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So if I want to figure out the sine of 2a, this is equal to the sine of a plus a, which is equal to the sine of a times the cosine of a plus the sine of the second a times the cosine of the first a. I just wrote the same thing twice, so this is just equal to 2 sine of a cosine of a. That was a little bit easier. So sine of 2a is equal to that. So that's another result. I know I'm a little bit tired by playing with all of these sines and cosines, and I was able to get all the results that I needed for my calculus problem. So hopefully this was a good review for you, because it was a good review for me. And you can write these things down."}, {"video_title": "Trigonometry identity review fun Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So that's another result. I know I'm a little bit tired by playing with all of these sines and cosines, and I was able to get all the results that I needed for my calculus problem. So hopefully this was a good review for you, because it was a good review for me. And you can write these things down. You can memorize them if you want. But the really important takeaway is to realize that you really can derive all of these formulas really from these initial formulas that we just had. And even these, I also have proofs to show you how to get these from just the basic definitions of your trig functions."}, {"video_title": "Sine equation algebraic solution set Trigonometry Precalculus Khan Academy.mp3", "Sentence": "So I encourage you, like always, pause this video and see if you can work through this on your own before we work through it together. All right, now let's work through it together. Now, your intuition, which would be correct, might be, let's see if we can isolate the sine of x over four algebraically. And the first step I would do is subtract 11 from both sides. And if you do that, you would get eight sine of x over four is equal to three, just subtract 11 from both sides. Now, to isolate the sine, I would divide both sides by eight and I would get sine of x over four is equal to 3 8ths. Now, before I go further, let's think about whether this is the most general solution here or whether we're gonna find all of the solution set here."}, {"video_title": "Sine equation algebraic solution set Trigonometry Precalculus Khan Academy.mp3", "Sentence": "And the first step I would do is subtract 11 from both sides. And if you do that, you would get eight sine of x over four is equal to three, just subtract 11 from both sides. Now, to isolate the sine, I would divide both sides by eight and I would get sine of x over four is equal to 3 8ths. Now, before I go further, let's think about whether this is the most general solution here or whether we're gonna find all of the solution set here. Well, we have to remind ourselves, let me actually draw a little bit of a unit circle here. So that's my x-axis, that's my y-axis, then a circle. And if we have some angle theta right over here, so that's theta, we know that the sine of theta is equal to the y-coordinate of where this radius intersects the unit circle."}, {"video_title": "Sine equation algebraic solution set Trigonometry Precalculus Khan Academy.mp3", "Sentence": "Now, before I go further, let's think about whether this is the most general solution here or whether we're gonna find all of the solution set here. Well, we have to remind ourselves, let me actually draw a little bit of a unit circle here. So that's my x-axis, that's my y-axis, then a circle. And if we have some angle theta right over here, so that's theta, we know that the sine of theta is equal to the y-coordinate of where this radius intersects the unit circle. And we also know that if we add an arbitrary number of two pi's here, or if we subtract an arbitrary number of two pi's, we go all the way around the unit circle back to where we began, and so the sine of theta would be the same. So we know that sine of theta plus any integer multiple of two pi, that's going to be equal to the sine of theta. And so we can generalize this a little bit."}, {"video_title": "Sine equation algebraic solution set Trigonometry Precalculus Khan Academy.mp3", "Sentence": "And if we have some angle theta right over here, so that's theta, we know that the sine of theta is equal to the y-coordinate of where this radius intersects the unit circle. And we also know that if we add an arbitrary number of two pi's here, or if we subtract an arbitrary number of two pi's, we go all the way around the unit circle back to where we began, and so the sine of theta would be the same. So we know that sine of theta plus any integer multiple of two pi, that's going to be equal to the sine of theta. And so we can generalize this a little bit. We could, instead of just saying sine of x over four is equal to 3 8ths, we could write that sine of x over four plus any integer multiple of two pi is going to be equal to 3 8ths, where n is any integer. It could even be a negative one, a negative two, or of course it could be zero, one, two, three, so on and so forth. So is this, if we solve now for x, is this going to give us the most general solution set?"}, {"video_title": "Sine equation algebraic solution set Trigonometry Precalculus Khan Academy.mp3", "Sentence": "And so we can generalize this a little bit. We could, instead of just saying sine of x over four is equal to 3 8ths, we could write that sine of x over four plus any integer multiple of two pi is going to be equal to 3 8ths, where n is any integer. It could even be a negative one, a negative two, or of course it could be zero, one, two, three, so on and so forth. So is this, if we solve now for x, is this going to give us the most general solution set? Well, we can also remind ourselves that if I have theta here and sine of theta gets there, there's one other point on the unit circle where I get the same sine. It would be right over here. The y-coordinate would be the same."}, {"video_title": "Sine equation algebraic solution set Trigonometry Precalculus Khan Academy.mp3", "Sentence": "So is this, if we solve now for x, is this going to give us the most general solution set? Well, we can also remind ourselves that if I have theta here and sine of theta gets there, there's one other point on the unit circle where I get the same sine. It would be right over here. The y-coordinate would be the same. And one way to think about it is, if we start at pi radians, which would be right over there, and we were to subtract theta, we're going to get the same thing. So this angle right over here, you could view as pi minus theta. And you could keep trying it out for any theta, even the theta that put you in the second quadrant, third quadrant, or fourth quadrant."}, {"video_title": "Sine equation algebraic solution set Trigonometry Precalculus Khan Academy.mp3", "Sentence": "The y-coordinate would be the same. And one way to think about it is, if we start at pi radians, which would be right over there, and we were to subtract theta, we're going to get the same thing. So this angle right over here, you could view as pi minus theta. And you could keep trying it out for any theta, even the theta that put you in the second quadrant, third quadrant, or fourth quadrant. If you do pi minus theta, sine of pi minus theta, you're going to get the same sine value. So we also know that sine of pi minus theta is equal to sine of theta. And so let me write another expression over here."}, {"video_title": "Sine equation algebraic solution set Trigonometry Precalculus Khan Academy.mp3", "Sentence": "And you could keep trying it out for any theta, even the theta that put you in the second quadrant, third quadrant, or fourth quadrant. If you do pi minus theta, sine of pi minus theta, you're going to get the same sine value. So we also know that sine of pi minus theta is equal to sine of theta. And so let me write another expression over here. So it's not just sine of x over four is equal to 3 8ths. We could also write that sine of pi minus x over four, because x over four is the theta here, sine of pi minus x over four is equal to 3 8ths. And of course, we could also use the other principle that we could add two pi or subtract two pi from this an arbitrary number of times, and the sine of that will still be equal to 3 8ths."}, {"video_title": "Sine equation algebraic solution set Trigonometry Precalculus Khan Academy.mp3", "Sentence": "And so let me write another expression over here. So it's not just sine of x over four is equal to 3 8ths. We could also write that sine of pi minus x over four, because x over four is the theta here, sine of pi minus x over four is equal to 3 8ths. And of course, we could also use the other principle that we could add two pi or subtract two pi from this an arbitrary number of times, and the sine of that will still be equal to 3 8ths. So I could write it like this. Sine of pi minus x over four plus an integer multiple of two pi, that is going to be equal to 3 8ths. And if I solve both of these, the combination of them, the union of them would give me the broadest solution set."}, {"video_title": "Sine equation algebraic solution set Trigonometry Precalculus Khan Academy.mp3", "Sentence": "And of course, we could also use the other principle that we could add two pi or subtract two pi from this an arbitrary number of times, and the sine of that will still be equal to 3 8ths. So I could write it like this. Sine of pi minus x over four plus an integer multiple of two pi, that is going to be equal to 3 8ths. And if I solve both of these, the combination of them, the union of them would give me the broadest solution set. So let's do that. So over here, let me take the inverse sine of both sides. I get x over four plus two pi n is equal to the inverse sine of 3 8ths."}, {"video_title": "Sine equation algebraic solution set Trigonometry Precalculus Khan Academy.mp3", "Sentence": "And if I solve both of these, the combination of them, the union of them would give me the broadest solution set. So let's do that. So over here, let me take the inverse sine of both sides. I get x over four plus two pi n is equal to the inverse sine of 3 8ths. Now I could subtract two pi n from both sides. I get x over four is equal to the inverse sine of 3 8ths, minus two pi n. And if you think about it, because n can be any integer, this sine here in front of the two, this negative really doesn't matter. It could even be a positive."}, {"video_title": "Sine equation algebraic solution set Trigonometry Precalculus Khan Academy.mp3", "Sentence": "I get x over four plus two pi n is equal to the inverse sine of 3 8ths. Now I could subtract two pi n from both sides. I get x over four is equal to the inverse sine of 3 8ths, minus two pi n. And if you think about it, because n can be any integer, this sine here in front of the two, this negative really doesn't matter. It could even be a positive. And now let's multiply both sides by four. We get x is equal to four times the inverse sine of 3 8ths minus 8 pi n. And then if I work on this blue part right over here, same idea, take the inverse sine of both sides, we get pi minus x over four plus two pi n is equal to the inverse sine of 3 8ths. And then let's see, I could subtract pi from both sides and subtract two pi n from both sides."}, {"video_title": "Unit circle symmetries for tan Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "In the previous video, we explored how the cosine and sines of angles relate when we essentially take the terminal ray of the angle and we reflect it about the x or the y axis or both axes. What I want to do in this video is think a little bit about the tangent of these different angles. So just as a little bit of a reminder, we know that the tangent of theta is equal to the sine of an angle over the cosine of an angle. And by the unit circle definition, it's essentially saying what is the slope of the terminal ray right over here. If we remind ourselves slope, slope is rise over run. It is our change in the vertical axis over our change in the horizontal axis. If we're starting at the origin, what is our change in the vertical axis if we go from zero to sine theta?"}, {"video_title": "Unit circle symmetries for tan Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And by the unit circle definition, it's essentially saying what is the slope of the terminal ray right over here. If we remind ourselves slope, slope is rise over run. It is our change in the vertical axis over our change in the horizontal axis. If we're starting at the origin, what is our change in the vertical axis if we go from zero to sine theta? Well, our change in the vertical axis is sine theta. What is our change in the horizontal axis? It's cosine of theta."}, {"video_title": "Unit circle symmetries for tan Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "If we're starting at the origin, what is our change in the vertical axis if we go from zero to sine theta? Well, our change in the vertical axis is sine theta. What is our change in the horizontal axis? It's cosine of theta. So this is change in y over change in x for the terminal ray. So the tangent of theta is the sine of theta over cosine theta or you could view it as the slope of this ray right over here. So let's think about what other angles are going to have the exact same tangent of theta."}, {"video_title": "Unit circle symmetries for tan Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "It's cosine of theta. So this is change in y over change in x for the terminal ray. So the tangent of theta is the sine of theta over cosine theta or you could view it as the slope of this ray right over here. So let's think about what other angles are going to have the exact same tangent of theta. Well, this ray is collinear with this ray right over here. In fact, if you kind of put them together, you get a line. So the tangent of this angle right over here, this pink angle going all the way around, the tangent of pi plus theta or the tangent of theta plus pi."}, {"video_title": "Unit circle symmetries for tan Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So let's think about what other angles are going to have the exact same tangent of theta. Well, this ray is collinear with this ray right over here. In fact, if you kind of put them together, you get a line. So the tangent of this angle right over here, this pink angle going all the way around, the tangent of pi plus theta or the tangent of theta plus pi. Obviously, you could write theta plus pi instead of pi plus theta. This should be, just based on this kind of slope argument, this should be equal to the tangent of theta. Let's see if this actually is the case."}, {"video_title": "Unit circle symmetries for tan Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So the tangent of this angle right over here, this pink angle going all the way around, the tangent of pi plus theta or the tangent of theta plus pi. Obviously, you could write theta plus pi instead of pi plus theta. This should be, just based on this kind of slope argument, this should be equal to the tangent of theta. Let's see if this actually is the case. So these two things should be equal if we agree that the tangent of an angle is equal to the slope of the terminal ray. Of course, the other side of the angle is going to be the positive x-axis based on the conventions that we've set up. Well, let's think about what it is, what the tangent of theta plus pi is in terms of sine and cosine."}, {"video_title": "Unit circle symmetries for tan Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Let's see if this actually is the case. So these two things should be equal if we agree that the tangent of an angle is equal to the slope of the terminal ray. Of course, the other side of the angle is going to be the positive x-axis based on the conventions that we've set up. Well, let's think about what it is, what the tangent of theta plus pi is in terms of sine and cosine. So the tangent, let me write this down in pink color. The tangent, that's not pink, the tangent of pi plus theta, that's going to be equal to, put the parentheses to avoid ambiguity, that's equal to the sine of pi plus theta, or theta plus pi, over the cosine of theta plus pi. And in the previous video, we established that the sine of theta plus pi, that's the same thing as negative sine theta."}, {"video_title": "Unit circle symmetries for tan Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Well, let's think about what it is, what the tangent of theta plus pi is in terms of sine and cosine. So the tangent, let me write this down in pink color. The tangent, that's not pink, the tangent of pi plus theta, that's going to be equal to, put the parentheses to avoid ambiguity, that's equal to the sine of pi plus theta, or theta plus pi, over the cosine of theta plus pi. And in the previous video, we established that the sine of theta plus pi, that's the same thing as negative sine theta. So this is equal to negative sine theta. And the cosine of theta plus pi, we already established that's the same thing as negative cosine of theta. You have a negative divided by a negative, you can say the negatives cancel out, and we're left with sine theta over cosine theta, which is indeed tangent of theta."}, {"video_title": "Unit circle symmetries for tan Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And in the previous video, we established that the sine of theta plus pi, that's the same thing as negative sine theta. So this is equal to negative sine theta. And the cosine of theta plus pi, we already established that's the same thing as negative cosine of theta. You have a negative divided by a negative, you can say the negatives cancel out, and we're left with sine theta over cosine theta, which is indeed tangent of theta. So we could feel pretty good about that. Now what about the points, or the terminal rays, right over here? So let's think about this point."}, {"video_title": "Unit circle symmetries for tan Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "You have a negative divided by a negative, you can say the negatives cancel out, and we're left with sine theta over cosine theta, which is indeed tangent of theta. So we could feel pretty good about that. Now what about the points, or the terminal rays, right over here? So let's think about this point. What is the tangent of negative theta going to be? Well we know that the tangent of negative theta is the same thing as the sine of negative theta over the cosine of negative theta. And we already established the sine of negative theta, that's negative sine theta."}, {"video_title": "Unit circle symmetries for tan Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So let's think about this point. What is the tangent of negative theta going to be? Well we know that the tangent of negative theta is the same thing as the sine of negative theta over the cosine of negative theta. And we already established the sine of negative theta, that's negative sine theta. We see that right here, sine of negative theta, that's the negative, that's the opposite of the sine of theta. So we have that there. But the cosine of negative theta is the same thing as the cosine of theta."}, {"video_title": "Unit circle symmetries for tan Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And we already established the sine of negative theta, that's negative sine theta. We see that right here, sine of negative theta, that's the negative, that's the opposite of the sine of theta. So we have that there. But the cosine of negative theta is the same thing as the cosine of theta. So these things are the same. So we're left with negative sine theta over cosine of theta, which is the same thing, which is equal to negative tangent theta. And so we see here, if you take the negative of the angle, you're going to get the negative of the tangent, and that's because the sine, the numerator, in our definition of tangent, changes signs, but the denominator does not."}, {"video_title": "Unit circle symmetries for tan Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "But the cosine of negative theta is the same thing as the cosine of theta. So these things are the same. So we're left with negative sine theta over cosine of theta, which is the same thing, which is equal to negative tangent theta. And so we see here, if you take the negative of the angle, you're going to get the negative of the tangent, and that's because the sine, the numerator, in our definition of tangent, changes signs, but the denominator does not. So tangent of negative theta is the same thing as negative tangent of theta. Now what about this point right over here? Well over here, relative to theta, when we're looking at pi minus theta, so when we're looking at tangent of pi minus theta, so tangent of pi minus theta, that's sine of pi minus theta over cosine of pi minus theta."}, {"video_title": "Unit circle symmetries for tan Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And so we see here, if you take the negative of the angle, you're going to get the negative of the tangent, and that's because the sine, the numerator, in our definition of tangent, changes signs, but the denominator does not. So tangent of negative theta is the same thing as negative tangent of theta. Now what about this point right over here? Well over here, relative to theta, when we're looking at pi minus theta, so when we're looking at tangent of pi minus theta, so tangent of pi minus theta, that's sine of pi minus theta over cosine of pi minus theta. And we already established in the previous video that sine of pi minus theta is equal to sine of theta, and we see that right over here, they have the exact same signs, so this is equal to sine of theta, while cosine of pi minus theta, well, it's the opposite of cosine of theta, it's the negative of cosine of theta. So this is negative cosine theta, and so this once again is going to be equal to the negative sine over cosine, or the negative tangent of theta, which makes sense. This ray should have the same slope as this ray right over here, and we see that slope, we could view this as negative tangent of theta."}, {"video_title": "Unit circle symmetries for tan Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Well over here, relative to theta, when we're looking at pi minus theta, so when we're looking at tangent of pi minus theta, so tangent of pi minus theta, that's sine of pi minus theta over cosine of pi minus theta. And we already established in the previous video that sine of pi minus theta is equal to sine of theta, and we see that right over here, they have the exact same signs, so this is equal to sine of theta, while cosine of pi minus theta, well, it's the opposite of cosine of theta, it's the negative of cosine of theta. So this is negative cosine theta, and so this once again is going to be equal to the negative sine over cosine, or the negative tangent of theta, which makes sense. This ray should have the same slope as this ray right over here, and we see that slope, we could view this as negative tangent of theta. And we see just looking at these two, I guess if you combine the rays, that these two intersecting lines have the negative slope of each other. They're mirror images across the x-axis. And so we've just seen, if you take an angle, and you add pi to the angle, your tangent won't change, because you're going to essentially be sitting on the same line, pi, everything in degrees, you're going 180 degrees around, you're going in the opposite direction, but the slope of your ray has not changed."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy.mp3", "Sentence": "The hottest day of the year in Santiago, Chile, on average, is January 7th, when the average high temperature is 29 degrees Celsius. The coolest day of the year has an average high temperature of 14 degrees Celsius. Use a trigonometric function to model the temperature in Santiago, Chile, using 365 days as the length of a year. Remember that January 7th is the summer in Santiago. How many days after January 7th is the first spring day when the temperature reaches 20 degrees Celsius? So let's do this in two parts. So first, let's try to figure out a trigonometric function that models the temperature in Santiago, Chile."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Remember that January 7th is the summer in Santiago. How many days after January 7th is the first spring day when the temperature reaches 20 degrees Celsius? So let's do this in two parts. So first, let's try to figure out a trigonometric function that models the temperature in Santiago, Chile. So we'll have temperature as a function of days, where days are the number of days after January 7th. And once we have that trigonometric function to model that, then we can answer the second part, I guess the essential question, which is how many days after January 7th is the first spring day when the temperature reaches 20 degrees Celsius? So to think about this, let's graph it."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So first, let's try to figure out a trigonometric function that models the temperature in Santiago, Chile. So we'll have temperature as a function of days, where days are the number of days after January 7th. And once we have that trigonometric function to model that, then we can answer the second part, I guess the essential question, which is how many days after January 7th is the first spring day when the temperature reaches 20 degrees Celsius? So to think about this, let's graph it. Let's graph it, and it should become pretty apparent why they are suggesting that we use a trigonometric function to model this, because our seasonal variations, they're cyclical, they go up and down. And actually, if you look at the average temperature for any city over the course of the year, it really does look like a trigonometric function. So let's, so let's, this axis right over here, this is the passage of the days."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So to think about this, let's graph it. Let's graph it, and it should become pretty apparent why they are suggesting that we use a trigonometric function to model this, because our seasonal variations, they're cyclical, they go up and down. And actually, if you look at the average temperature for any city over the course of the year, it really does look like a trigonometric function. So let's, so let's, this axis right over here, this is the passage of the days. Let's do D for days, and that's going to be in days after January 7th. So this right over here would be January 7th. And the vertical axis, this is the horizontal, the vertical axis is going to be in terms of Celsius degrees."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So let's, so let's, this axis right over here, this is the passage of the days. Let's do D for days, and that's going to be in days after January 7th. So this right over here would be January 7th. And the vertical axis, this is the horizontal, the vertical axis is going to be in terms of Celsius degrees. So this, let's see, the high is 29, and I could write 29 degrees Celsius, and then, or the highest average day. And then if this is zero, then 14, which is the lowest average day, 14 degrees Celsius. And so our temperature will vary between these two extremes."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy.mp3", "Sentence": "And the vertical axis, this is the horizontal, the vertical axis is going to be in terms of Celsius degrees. So this, let's see, the high is 29, and I could write 29 degrees Celsius, and then, or the highest average day. And then if this is zero, then 14, which is the lowest average day, 14 degrees Celsius. And so our temperature will vary between these two extremes. Our temperature is going to vary, the highest average day, which they already told us is January 7th, we get to 29 degrees Celsius. And then the coldest day of the year, on average, you get to an average high of 14 degrees Celsius. So it looks like this."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy.mp3", "Sentence": "And so our temperature will vary between these two extremes. Our temperature is going to vary, the highest average day, which they already told us is January 7th, we get to 29 degrees Celsius. And then the coldest day of the year, on average, you get to an average high of 14 degrees Celsius. So it looks like this. We're talking about average highs on a given day. And the reason why a trigonometric function is a good idea is because it's cyclical. So if this is January 7th, if you go 365 days in the future, you're back at January 7th."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So it looks like this. We're talking about average highs on a given day. And the reason why a trigonometric function is a good idea is because it's cyclical. So if this is January 7th, if you go 365 days in the future, you're back at January 7th. So if the average high temperature is 29 degrees Celsius on that day, the average high temperature is going to be 29 degrees Celsius on that day. Now, we're using a trigonometric function, so we're gonna hit our low point exactly halfway in between. So we're gonna hit our low point exactly halfway in between, something right like that."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So if this is January 7th, if you go 365 days in the future, you're back at January 7th. So if the average high temperature is 29 degrees Celsius on that day, the average high temperature is going to be 29 degrees Celsius on that day. Now, we're using a trigonometric function, so we're gonna hit our low point exactly halfway in between. So we're gonna hit our low point exactly halfway in between, something right like that. And so our function is going to look like this. Our function, so let me see, let me just draw the low point right over there. And this is a high point."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So we're gonna hit our low point exactly halfway in between, something right like that. And so our function is going to look like this. Our function, so let me see, let me just draw the low point right over there. And this is a high point. It's a high point right over there. That looks pretty good. And then I have the high point right over here, and then I just need to connect them, and there you go."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy.mp3", "Sentence": "And this is a high point. It's a high point right over there. That looks pretty good. And then I have the high point right over here, and then I just need to connect them, and there you go. I've drawn one period of our trigonometric function. And our period is 365 days. If we go 365 days later, we're at the same point in the cycle."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy.mp3", "Sentence": "And then I have the high point right over here, and then I just need to connect them, and there you go. I've drawn one period of our trigonometric function. And our period is 365 days. If we go 365 days later, we're at the same point in the cycle. We are at the same point in the year. We're at the same point in the year. So what I want you to do right now is given what I've just drawn, try to model this right."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy.mp3", "Sentence": "If we go 365 days later, we're at the same point in the cycle. We are at the same point in the year. We're at the same point in the year. So what I want you to do right now is given what I've just drawn, try to model this right. So this right over here is, let's write this, this is capital T as a function of D. Try to figure out an expression for capital T as a function of D. And remember, it's going to be some trigonometric function. So I'm assuming you've given a go at it, and you might say, well, this looks like a cosine curve. Maybe it could be a sine curve."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So what I want you to do right now is given what I've just drawn, try to model this right. So this right over here is, let's write this, this is capital T as a function of D. Try to figure out an expression for capital T as a function of D. And remember, it's going to be some trigonometric function. So I'm assuming you've given a go at it, and you might say, well, this looks like a cosine curve. Maybe it could be a sine curve. Which one should I use? And you could actually use either one, but I always like to go with the simpler one. You just think about, well, if this were angles, either actually degrees or radians, which trigonometric function starts at your maximum point?"}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Maybe it could be a sine curve. Which one should I use? And you could actually use either one, but I always like to go with the simpler one. You just think about, well, if this were angles, either actually degrees or radians, which trigonometric function starts at your maximum point? Well, cosine of zero is one. The cosine starts at your maximum point. Sine of zero is zero."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy.mp3", "Sentence": "You just think about, well, if this were angles, either actually degrees or radians, which trigonometric function starts at your maximum point? Well, cosine of zero is one. The cosine starts at your maximum point. Sine of zero is zero. So I'm going to use cosine here. I'm going to use a cosine function. So temperature as a function of days is going to be some amplitude times our cosine function."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Sine of zero is zero. So I'm going to use cosine here. I'm going to use a cosine function. So temperature as a function of days is going to be some amplitude times our cosine function. And we're going to have some argument to our cosine function. And then I'm probably going to have to shift it. So let's think about how we would do that."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So temperature as a function of days is going to be some amplitude times our cosine function. And we're going to have some argument to our cosine function. And then I'm probably going to have to shift it. So let's think about how we would do that. Well, what's the midline here? Well, the midline's the halfway point between our high and our low. So our middle point, if we were to visualize it, looks just like that is our midline right over there."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So let's think about how we would do that. Well, what's the midline here? Well, the midline's the halfway point between our high and our low. So our middle point, if we were to visualize it, looks just like that is our midline right over there. And what value is this? Well, what's the average of 29 and 14? 29 plus 14 is 43 divided by 2 is 21.5 degrees Celsius."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So our middle point, if we were to visualize it, looks just like that is our midline right over there. And what value is this? Well, what's the average of 29 and 14? 29 plus 14 is 43 divided by 2 is 21.5 degrees Celsius. So that's our midline. So essentially, we've shifted up our function by that amount. If we just had a regular cosine function, our midline would be at 0."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy.mp3", "Sentence": "29 plus 14 is 43 divided by 2 is 21.5 degrees Celsius. So that's our midline. So essentially, we've shifted up our function by that amount. If we just had a regular cosine function, our midline would be at 0. But now we're at 21.5 degrees Celsius. So I'll just write plus 21.5. That's how much we've shifted it up."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy.mp3", "Sentence": "If we just had a regular cosine function, our midline would be at 0. But now we're at 21.5 degrees Celsius. So I'll just write plus 21.5. That's how much we've shifted it up. Now, what's the amplitude? Well, our amplitude is how much we diverge from the midline. So over here, we're 7.5 above the midline."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy.mp3", "Sentence": "That's how much we've shifted it up. Now, what's the amplitude? Well, our amplitude is how much we diverge from the midline. So over here, we're 7.5 above the midline. So that's plus 7.5. Here, we're 7.5 below the midline. So minus 7.5."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So over here, we're 7.5 above the midline. So that's plus 7.5. Here, we're 7.5 below the midline. So minus 7.5. So our amplitude is 7.5. So the maximum amount we go away from the midline is 7.5. So that's our amplitude."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So minus 7.5. So our amplitude is 7.5. So the maximum amount we go away from the midline is 7.5. So that's our amplitude. And now let's think about our argument to the cosine function right over here. So it's going to be a function of the days. And what do we want?"}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So that's our amplitude. And now let's think about our argument to the cosine function right over here. So it's going to be a function of the days. And what do we want? When 365 days have gone by, we want this entire argument to be 2 pi. So when d is 365, we want this whole thing to evaluate to 2 pi. So we could put 2 pi over 365 in here."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy.mp3", "Sentence": "And what do we want? When 365 days have gone by, we want this entire argument to be 2 pi. So when d is 365, we want this whole thing to evaluate to 2 pi. So we could put 2 pi over 365 in here. And you might remember your formulas. I always forget them. That's why I always try to reason through them again."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So we could put 2 pi over 365 in here. And you might remember your formulas. I always forget them. That's why I always try to reason through them again. The formula is, oh, you want 2 pi divided by your period and all the rest. But I just like to think, OK, look, after one period, which is 365 days, I want the whole argument over here to be 2 pi. I want to go around the unit circle once."}, {"video_title": "Double angle formula for cosine example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "We have triangle ABC here, which looks like a right triangle. We know it's a right triangle because 3 squared plus 4 squared is equal to 5 squared. And they want us to figure out what cosine of 2 times angle ABC is. So that's this angle, ABC. Well, we can't immediately evaluate that, but we do know what the cosine of angle ABC is. We know that the cosine of angle ABC, well, cosine is just adjacent over hypotenuse, is going to be equal to 3 fifths. And similarly, we know what the sine of angle ABC is."}, {"video_title": "Double angle formula for cosine example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So that's this angle, ABC. Well, we can't immediately evaluate that, but we do know what the cosine of angle ABC is. We know that the cosine of angle ABC, well, cosine is just adjacent over hypotenuse, is going to be equal to 3 fifths. And similarly, we know what the sine of angle ABC is. That's opposite over hypotenuse. That is 4 fifths. So if we could break this down into just cosines of ABC and sines of ABC, then we'll be able to evaluate it."}, {"video_title": "Double angle formula for cosine example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And similarly, we know what the sine of angle ABC is. That's opposite over hypotenuse. That is 4 fifths. So if we could break this down into just cosines of ABC and sines of ABC, then we'll be able to evaluate it. And lucky for us, we have a trig identity at our disposal that does exactly that. We know that the cosine of 2 times an angle is equal to cosine of that angle squared minus sine of that angle squared. We've proved this in other videos, but this becomes very helpful for us here."}, {"video_title": "Double angle formula for cosine example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So if we could break this down into just cosines of ABC and sines of ABC, then we'll be able to evaluate it. And lucky for us, we have a trig identity at our disposal that does exactly that. We know that the cosine of 2 times an angle is equal to cosine of that angle squared minus sine of that angle squared. We've proved this in other videos, but this becomes very helpful for us here. Because now we know that the cosine of angle ABC is going to be equal to the cosine of 2 times the angle ABC. That's what we care about. 2 times the angle ABC is going to be equal to the cosine of angle ABC squared minus sine of the angle ABC squared."}, {"video_title": "Double angle formula for cosine example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "We've proved this in other videos, but this becomes very helpful for us here. Because now we know that the cosine of angle ABC is going to be equal to the cosine of 2 times the angle ABC. That's what we care about. 2 times the angle ABC is going to be equal to the cosine of angle ABC squared minus sine of the angle ABC squared. And we know what these things are. This thing right over here is just going to be equal to 3 fifths squared. Cosine of angle ABC is 3 fifths, so we're going to square it."}, {"video_title": "Double angle formula for cosine example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "2 times the angle ABC is going to be equal to the cosine of angle ABC squared minus sine of the angle ABC squared. And we know what these things are. This thing right over here is just going to be equal to 3 fifths squared. Cosine of angle ABC is 3 fifths, so we're going to square it. And this right over here is just 4 fifths squared. So it's minus 4 fifths squared. And so this simplifies to 9 over 25 minus 16 over 25, which is equal to 7 over 25."}, {"video_title": "Double angle formula for cosine example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Cosine of angle ABC is 3 fifths, so we're going to square it. And this right over here is just 4 fifths squared. So it's minus 4 fifths squared. And so this simplifies to 9 over 25 minus 16 over 25, which is equal to 7 over 25. So this thing right over here is equal to 7. Sorry, it's negative. Got to be careful there."}, {"video_title": "Double angle formula for cosine example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And so this simplifies to 9 over 25 minus 16 over 25, which is equal to 7 over 25. So this thing right over here is equal to 7. Sorry, it's negative. Got to be careful there. 16 is larger than 9. Negative 7 over 25. Now one thing that might jump at you is, why did I get a negative value here when I doubled the angle here?"}, {"video_title": "Double angle formula for cosine example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Got to be careful there. 16 is larger than 9. Negative 7 over 25. Now one thing that might jump at you is, why did I get a negative value here when I doubled the angle here? Because the cosine was clearly a positive number. And there you just have to think of the unit circle, which we already know is an extension, the unit circle definition of trig functions, which is an extension of the Sohcahtoa definition. x-axis, y-axis."}, {"video_title": "Double angle formula for cosine example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Now one thing that might jump at you is, why did I get a negative value here when I doubled the angle here? Because the cosine was clearly a positive number. And there you just have to think of the unit circle, which we already know is an extension, the unit circle definition of trig functions, which is an extension of the Sohcahtoa definition. x-axis, y-axis. Let me draw a unit circle here. My best attempt. So that's our unit circle."}, {"video_title": "Double angle formula for cosine example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "x-axis, y-axis. Let me draw a unit circle here. My best attempt. So that's our unit circle. So this angle right over here looks like something like this. So it looks like something like this. And so you see its x-coordinate, which is the cosine of that angle, looks positive."}, {"video_title": "Double angle formula for cosine example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So that's our unit circle. So this angle right over here looks like something like this. So it looks like something like this. And so you see its x-coordinate, which is the cosine of that angle, looks positive. But then if you were to double this angle, it would take you out someplace like this. It would take you out someplace like this. And then you see by the unit circle definition, the x-coordinate is now, we are now sitting in the second quadrant, and the x-coordinate can be negative."}, {"video_title": "Examples using pythagorean identities to simplify trigonometric Trigonometry Khan Academy.mp3", "Sentence": "So let's say that I have 1 minus sine squared theta and this whole thing times cosine squared theta. So how could I simplify this? Well, the one thing that we do know, and this is the most fundamental trig identity, this comes straight out of the unit circle, is that cosine squared theta plus sine squared theta is equal to 1. And then if we subtract sine squared theta from both sides, we get cosine squared theta is equal to 1 minus sine squared theta. So we have two options. We could either replace this 1 minus sine squared theta with the cosine squared theta, or we could replace this cosine squared theta with the 1 minus sine squared theta. Well, I'd prefer to do the former because this is a more complicated expression."}, {"video_title": "Examples using pythagorean identities to simplify trigonometric Trigonometry Khan Academy.mp3", "Sentence": "And then if we subtract sine squared theta from both sides, we get cosine squared theta is equal to 1 minus sine squared theta. So we have two options. We could either replace this 1 minus sine squared theta with the cosine squared theta, or we could replace this cosine squared theta with the 1 minus sine squared theta. Well, I'd prefer to do the former because this is a more complicated expression. So if I can replace this with the cosine squared theta, then I think I'm simplifying this. So let's see, this will be cosine squared theta times another cosine squared theta. And so all of this is going to simplify to, this is cosine theta times cosine theta times cosine theta times cosine theta."}, {"video_title": "Examples using pythagorean identities to simplify trigonometric Trigonometry Khan Academy.mp3", "Sentence": "Well, I'd prefer to do the former because this is a more complicated expression. So if I can replace this with the cosine squared theta, then I think I'm simplifying this. So let's see, this will be cosine squared theta times another cosine squared theta. And so all of this is going to simplify to, this is cosine theta times cosine theta times cosine theta times cosine theta. Well, that's just going to be cosine to the fourth of theta. Let's do another example. Let's say that we have sine squared theta, all of that over 1 minus sine squared theta."}, {"video_title": "Examples using pythagorean identities to simplify trigonometric Trigonometry Khan Academy.mp3", "Sentence": "And so all of this is going to simplify to, this is cosine theta times cosine theta times cosine theta times cosine theta. Well, that's just going to be cosine to the fourth of theta. Let's do another example. Let's say that we have sine squared theta, all of that over 1 minus sine squared theta. What is this going to be equal to? Well, we already know that 1 minus sine squared theta is the same thing as cosine squared theta, so it's going to be sine squared theta over, this thing is the same thing as cosine squared theta, we just saw that, over cosine squared theta, which is going to be equal to, you could view this as sine theta over cosine theta, whole quantity squared. Well, what's sine over cosine?"}, {"video_title": "Examples using pythagorean identities to simplify trigonometric Trigonometry Khan Academy.mp3", "Sentence": "Let's say that we have sine squared theta, all of that over 1 minus sine squared theta. What is this going to be equal to? Well, we already know that 1 minus sine squared theta is the same thing as cosine squared theta, so it's going to be sine squared theta over, this thing is the same thing as cosine squared theta, we just saw that, over cosine squared theta, which is going to be equal to, you could view this as sine theta over cosine theta, whole quantity squared. Well, what's sine over cosine? That's tangent. So this is equal to tangent squared theta. Let's do one more example."}, {"video_title": "Examples using pythagorean identities to simplify trigonometric Trigonometry Khan Academy.mp3", "Sentence": "Well, what's sine over cosine? That's tangent. So this is equal to tangent squared theta. Let's do one more example. Let's say that we have cosine squared theta plus 1 minus sine squared theta. What is this going to be? Well, you might be tempted, especially with the way I wrote the colors, to think, hey, is there some identity for 1 plus sine squared theta?"}, {"video_title": "Examples using pythagorean identities to simplify trigonometric Trigonometry Khan Academy.mp3", "Sentence": "Let's do one more example. Let's say that we have cosine squared theta plus 1 minus sine squared theta. What is this going to be? Well, you might be tempted, especially with the way I wrote the colors, to think, hey, is there some identity for 1 plus sine squared theta? This is really all about rearranging it to realize that, gee, by the unit circle definition, I know what cosine squared theta plus sine squared theta is. Cosine squared theta plus sine squared theta for any given theta is going to be equal to 1. So this is going to be equal to 1 plus this one right over here, which is equal to 2."}, {"video_title": "Law of cosines for star distance Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "She knows the angle between these stars in the sky is three degrees. What is the width of Orion's belt? That is, what is the distance between Alnitak and Mintaka? And they want us to give us an answer in light years. So let's draw a little diagram to make sure we understand what's going on. Actually, even before we do that, I encourage you to pause this and try this on your own. Now let's make a diagram."}, {"video_title": "Law of cosines for star distance Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And they want us to give us an answer in light years. So let's draw a little diagram to make sure we understand what's going on. Actually, even before we do that, I encourage you to pause this and try this on your own. Now let's make a diagram. All right, so let's say that this is Artemis's house right over here. This is Artemis's house. I'll say that's A for Artemis's house."}, {"video_title": "Law of cosines for star distance Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Now let's make a diagram. All right, so let's say that this is Artemis's house right over here. This is Artemis's house. I'll say that's A for Artemis's house. And then, let me say H. Let me say this is home. This is home right over here. And we have these two stars."}, {"video_title": "Law of cosines for star distance Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "I'll say that's A for Artemis's house. And then, let me say H. Let me say this is home. This is home right over here. And we have these two stars. So she's looking out into the night sky and she sees these stars, Alnitak, which is 736 light years away. And obviously I'm not going to draw this to scale. So this is Alnitak, Alnitak, and Mintaka."}, {"video_title": "Law of cosines for star distance Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And we have these two stars. So she's looking out into the night sky and she sees these stars, Alnitak, which is 736 light years away. And obviously I'm not going to draw this to scale. So this is Alnitak, Alnitak, and Mintaka. So let's say this is Mintaka right over here. Mintaka. And we know a few things."}, {"video_title": "Law of cosines for star distance Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So this is Alnitak, Alnitak, and Mintaka. So let's say this is Mintaka right over here. Mintaka. And we know a few things. We know that this distance between her home and Alnitak is 736 light years. So this distance right over here. So that right over there."}, {"video_title": "Law of cosines for star distance Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And we know a few things. We know that this distance between her home and Alnitak is 736 light years. So this distance right over here. So that right over there. Everything we'll do is in light years. That's 736. And the distance between her house and Mintaka is 915 light years."}, {"video_title": "Law of cosines for star distance Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So that right over there. Everything we'll do is in light years. That's 736. And the distance between her house and Mintaka is 915 light years. So it would take light 915 years to get from her house to Mintaka or from her Mintaka to her house. So this is 915 light years. And what we want to do is figure out the width of Orion's Belt, which is the distance between Alnitak and Mintaka."}, {"video_title": "Law of cosines for star distance Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And the distance between her house and Mintaka is 915 light years. So it would take light 915 years to get from her house to Mintaka or from her Mintaka to her house. So this is 915 light years. And what we want to do is figure out the width of Orion's Belt, which is the distance between Alnitak and Mintaka. So we need to figure out this distance right over here. And the one thing that they did give us is this angle. They did give us that angle right over there."}, {"video_title": "Law of cosines for star distance Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And what we want to do is figure out the width of Orion's Belt, which is the distance between Alnitak and Mintaka. So we need to figure out this distance right over here. And the one thing that they did give us is this angle. They did give us that angle right over there. They said that the angle between these stars and the sky is three degrees. So this is three degrees right over there. So how can we figure out the distance between Alnitak and Mintaka?"}, {"video_title": "Law of cosines for star distance Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "They did give us that angle right over there. They said that the angle between these stars and the sky is three degrees. So this is three degrees right over there. So how can we figure out the distance between Alnitak and Mintaka? And let's just say that this is equal to x. This is equal to x. How do we do that?"}, {"video_title": "Law of cosines for star distance Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So how can we figure out the distance between Alnitak and Mintaka? And let's just say that this is equal to x. This is equal to x. How do we do that? Well, if we have two sides and an angle between them, we could use the law of cosines to figure out the third side. So the law of cosines, so let's just apply it. So the law of cosines tells us that x squared is going to be equal to the sum of the squares of the other two sides."}, {"video_title": "Law of cosines for star distance Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "How do we do that? Well, if we have two sides and an angle between them, we could use the law of cosines to figure out the third side. So the law of cosines, so let's just apply it. So the law of cosines tells us that x squared is going to be equal to the sum of the squares of the other two sides. So it's going to be equal to 736 squared plus 915 squared minus two times 736 times 915 times the cosine of this angle, times the cosine of three degrees. Three degrees. So once again, we're trying to find the length of the side opposite the three degrees."}, {"video_title": "Law of cosines for star distance Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So the law of cosines tells us that x squared is going to be equal to the sum of the squares of the other two sides. So it's going to be equal to 736 squared plus 915 squared minus two times 736 times 915 times the cosine of this angle, times the cosine of three degrees. Three degrees. So once again, we're trying to find the length of the side opposite the three degrees. We know the other two sides. So the law of cosines, it essentially, sorry, I just had to cough off camera because I had some peanuts and my throat was dry. Where was I?"}, {"video_title": "Law of cosines for star distance Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So once again, we're trying to find the length of the side opposite the three degrees. We know the other two sides. So the law of cosines, it essentially, sorry, I just had to cough off camera because I had some peanuts and my throat was dry. Where was I? Oh, I was saying, if we know the angle and we know the two sides on either side of the angle, we can figure out the length of side opposite by the law of cosines, where it essentially starts off not too different than the Pythagorean theorem, but then we give an adjustment because this is not an actual right triangle. And the adjustment, so we have the 736 squared plus 915 squared minus two times the product of these sides times the cosine of this angle. Or another way we could say, think about it is, x, let me write that, x is equal to the square root of all of this stuff."}, {"video_title": "Law of cosines for star distance Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Where was I? Oh, I was saying, if we know the angle and we know the two sides on either side of the angle, we can figure out the length of side opposite by the law of cosines, where it essentially starts off not too different than the Pythagorean theorem, but then we give an adjustment because this is not an actual right triangle. And the adjustment, so we have the 736 squared plus 915 squared minus two times the product of these sides times the cosine of this angle. Or another way we could say, think about it is, x, let me write that, x is equal to the square root of all of this stuff. So I can just copy and paste that. So copy and paste. X is going to be equal to the square root of that."}, {"video_title": "Law of cosines for star distance Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Or another way we could say, think about it is, x, let me write that, x is equal to the square root of all of this stuff. So I can just copy and paste that. So copy and paste. X is going to be equal to the square root of that. And so let's get our calculator to calculate it. And let me verify that I'm in degree mode. Yes, I am indeed in degree mode."}, {"video_title": "Law of cosines for star distance Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "X is going to be equal to the square root of that. And so let's get our calculator to calculate it. And let me verify that I'm in degree mode. Yes, I am indeed in degree mode. And so let's exit that. And so I want to calculate the square root of 736 squared plus 915 squared minus two times 736 times 915 times cosine of three degrees. And we deserve a drum roll now."}, {"video_title": "Law of cosines for star distance Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Yes, I am indeed in degree mode. And so let's exit that. And so I want to calculate the square root of 736 squared plus 915 squared minus two times 736 times 915 times cosine of three degrees. And we deserve a drum roll now. X is 100 if we round, let's see, what do they want us to do? Round your answer to the nearest light year. So to the nearest light year, it's gonna be 184 light years."}, {"video_title": "Law of cosines for star distance Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And we deserve a drum roll now. X is 100 if we round, let's see, what do they want us to do? Round your answer to the nearest light year. So to the nearest light year, it's gonna be 184 light years. So x is approximately equal to 184 light years. So it would take light 184 years to get from Mintaka to Alnitak. And so hopefully this actually shows you, it's pretty, if you are going to do any astronomy, the law of cosines, law of sines, in fact all of trigonometry becomes quite, quite handy."}, {"video_title": "Inverse tangent scenario Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So that's saying that the tangent, let's say that that particular angle is theta, is equal to 1. What should Javier do to find the angle? I encourage you to pause this video and look at these choices and think about which of these should he do to find the angle. Let's look through each of them. Instead of looking at the choices, let's think about what we would do to find the angle. They're saying that the tangent of some angle is equal to 1. One thing that you might want to do is say, if we take the inverse tangent of the tangent of theta, so if we take the inverse tangent of both sides of this, we of course would get the inverse tangent of the tangent of theta."}, {"video_title": "Inverse tangent scenario Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Let's look through each of them. Instead of looking at the choices, let's think about what we would do to find the angle. They're saying that the tangent of some angle is equal to 1. One thing that you might want to do is say, if we take the inverse tangent of the tangent of theta, so if we take the inverse tangent of both sides of this, we of course would get the inverse tangent of the tangent of theta. If the domain over here is restricted appropriately, it's just going to be equal to theta. So we can say the tangent of theta is going to be the inverse tangent of 1. It might be tempting to just pick this one right over here, type inverse tangent of 1 into his calculator."}, {"video_title": "Inverse tangent scenario Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "One thing that you might want to do is say, if we take the inverse tangent of the tangent of theta, so if we take the inverse tangent of both sides of this, we of course would get the inverse tangent of the tangent of theta. If the domain over here is restricted appropriately, it's just going to be equal to theta. So we can say the tangent of theta is going to be the inverse tangent of 1. It might be tempting to just pick this one right over here, type inverse tangent of 1 into his calculator. Maybe this looks like the best choice. But remember, I said if we restrict the domain right over here, if we restrict the possible values of tangent of theta here appropriately, then this is going to simplify to this. But there is a scenario where this does not happen."}, {"video_title": "Inverse tangent scenario Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "It might be tempting to just pick this one right over here, type inverse tangent of 1 into his calculator. Maybe this looks like the best choice. But remember, I said if we restrict the domain right over here, if we restrict the possible values of tangent of theta here appropriately, then this is going to simplify to this. But there is a scenario where this does not happen. That's if we pick thetas that are outside of the range of the inverse tangent function. What do I mean by that? It's really just based on the idea that there are multiple angles whose tangent is 1."}, {"video_title": "Inverse tangent scenario Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "But there is a scenario where this does not happen. That's if we pick thetas that are outside of the range of the inverse tangent function. What do I mean by that? It's really just based on the idea that there are multiple angles whose tangent is 1. Let me draw that here with a unit circle. So let me draw a unit circle. That's my x-axis, that's my y-axis."}, {"video_title": "Inverse tangent scenario Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "It's really just based on the idea that there are multiple angles whose tangent is 1. Let me draw that here with a unit circle. So let me draw a unit circle. That's my x-axis, that's my y-axis. Let me draw my unit circle here. Actually, you probably don't even have to draw the unit circle because the tangent is really much more about the slope of the ray created by the angle than where it intersects the unit circle, as would be the case with sine and cosine. You could have this angle right over here."}, {"video_title": "Inverse tangent scenario Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "That's my x-axis, that's my y-axis. Let me draw my unit circle here. Actually, you probably don't even have to draw the unit circle because the tangent is really much more about the slope of the ray created by the angle than where it intersects the unit circle, as would be the case with sine and cosine. You could have this angle right over here. Let's say this is a candidate theta. Where the tangent of this theta is the slope of this line. And this terminal ray, you could say, of the angle."}, {"video_title": "Inverse tangent scenario Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "You could have this angle right over here. Let's say this is a candidate theta. Where the tangent of this theta is the slope of this line. And this terminal ray, you could say, of the angle. The other side, the initial ray, is along the positive x-axis. You could say the tangent of this theta is 1 because the slope of this line is 1. Let me scroll over a little bit."}, {"video_title": "Inverse tangent scenario Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And this terminal ray, you could say, of the angle. The other side, the initial ray, is along the positive x-axis. You could say the tangent of this theta is 1 because the slope of this line is 1. Let me scroll over a little bit. Let me write it this way. Tangent theta is equal to 1. But I could construct another theta whose tangent is equal to 1 by going all the way over here and essentially going in the opposite direction."}, {"video_title": "Inverse tangent scenario Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Let me scroll over a little bit. Let me write it this way. Tangent theta is equal to 1. But I could construct another theta whose tangent is equal to 1 by going all the way over here and essentially going in the opposite direction. But the slope of this line, let's call this theta 2, tangent of theta 2 is also going to be equal to 1. Of course, you could go another pi radians and go back to the original angle, but that's functionally the same angle in terms of where it is relative to the positive x-axis or what direction it points into. But this one is fundamentally a different angle."}, {"video_title": "Inverse tangent scenario Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "But I could construct another theta whose tangent is equal to 1 by going all the way over here and essentially going in the opposite direction. But the slope of this line, let's call this theta 2, tangent of theta 2 is also going to be equal to 1. Of course, you could go another pi radians and go back to the original angle, but that's functionally the same angle in terms of where it is relative to the positive x-axis or what direction it points into. But this one is fundamentally a different angle. So we do not know, we do not have enough information, just given what we've been told, to know exactly which theta we're talking about, whether we're talking about this orange theta or this mauve theta. So I would say get more information. There are multiple angles which fit this description."}, {"video_title": "Law of cosines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And that we want to figure out the length of this side. And this side has length a, so we need to figure out what a is going to be equal to. Now, we won't be able to figure this out unless we also know the angle here, because you could bring the blue side and the green side close together, and then a would be small. But if this angle was larger, then a would be larger. So we need to know what this angle is as well. So let's say that we know that this angle, which we will call theta, is equal to 87 degrees. So how can we figure out a?"}, {"video_title": "Law of cosines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "But if this angle was larger, then a would be larger. So we need to know what this angle is as well. So let's say that we know that this angle, which we will call theta, is equal to 87 degrees. So how can we figure out a? I encourage you to pause this and try this on your own. Well, lucky for us, we have the law of cosines, which gives us a way for determining a third side if we know two of the sides and the angle between them. The law of cosines tells us that a squared is going to be equal to b squared plus c squared."}, {"video_title": "Law of cosines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So how can we figure out a? I encourage you to pause this and try this on your own. Well, lucky for us, we have the law of cosines, which gives us a way for determining a third side if we know two of the sides and the angle between them. The law of cosines tells us that a squared is going to be equal to b squared plus c squared. Now, if we were dealing with a pure right triangle, if this was 90 degrees, then a would be the hypotenuse, and we would be done. This would be the Pythagorean theorem. But the law of cosines gives us an adjustment to the Pythagorean theorem so that we can do this for any arbitrary angle."}, {"video_title": "Law of cosines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "The law of cosines tells us that a squared is going to be equal to b squared plus c squared. Now, if we were dealing with a pure right triangle, if this was 90 degrees, then a would be the hypotenuse, and we would be done. This would be the Pythagorean theorem. But the law of cosines gives us an adjustment to the Pythagorean theorem so that we can do this for any arbitrary angle. So the law of cosines tells us a squared is going to be b squared plus c squared minus 2 times bc times the cosine of theta. And this theta is the angle that opens up to the side that we care about. So we can use theta because we're looking for a."}, {"video_title": "Law of cosines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "But the law of cosines gives us an adjustment to the Pythagorean theorem so that we can do this for any arbitrary angle. So the law of cosines tells us a squared is going to be b squared plus c squared minus 2 times bc times the cosine of theta. And this theta is the angle that opens up to the side that we care about. So we can use theta because we're looking for a. If they gave us another angle right over here, that's not the angle that we would use. We care about the angle that opens up into the side that we are going to solve for. So now let's solve for a because we know what bc and theta actually are."}, {"video_title": "Law of cosines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So we can use theta because we're looking for a. If they gave us another angle right over here, that's not the angle that we would use. We care about the angle that opens up into the side that we are going to solve for. So now let's solve for a because we know what bc and theta actually are. So a squared is going to be equal to b squared. So it's going to be equal to 144 plus c squared, which is 81, minus 2 times b times c. So that's minus 2 times 12 times 9 times the cosine of 87 degrees. And this is going to be equal to, let's see, this is 225 minus, let's see, 12 times 9 is 108."}, {"video_title": "Law of cosines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So now let's solve for a because we know what bc and theta actually are. So a squared is going to be equal to b squared. So it's going to be equal to 144 plus c squared, which is 81, minus 2 times b times c. So that's minus 2 times 12 times 9 times the cosine of 87 degrees. And this is going to be equal to, let's see, this is 225 minus, let's see, 12 times 9 is 108. 108 times 2 is 216 minus 216 times the cosine of 87 degrees. Now let's get our calculator out in order to approximate this. And remember, this is a squared."}, {"video_title": "Law of cosines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And this is going to be equal to, let's see, this is 225 minus, let's see, 12 times 9 is 108. 108 times 2 is 216 minus 216 times the cosine of 87 degrees. Now let's get our calculator out in order to approximate this. And remember, this is a squared. Actually, before I get my calculator out, let's just solve for a. So a is just going to be the square root of this. So a is going to be equal to the square root of all of this business, which I can just copy and paste."}, {"video_title": "Law of cosines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And remember, this is a squared. Actually, before I get my calculator out, let's just solve for a. So a is just going to be the square root of this. So a is going to be equal to the square root of all of this business, which I can just copy and paste. It's going to be equal to the square root of that. So let me copy and paste it. So a is going to be equal to the square root of that, which we can now use the calculator to figure out."}, {"video_title": "Law of cosines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So a is going to be equal to the square root of all of this business, which I can just copy and paste. It's going to be equal to the square root of that. So let me copy and paste it. So a is going to be equal to the square root of that, which we can now use the calculator to figure out. So let me increase this radical a little bit so that we make sure we're taking the square root of this whole thing. So let me get my calculator out. So I want to find the square root of 220."}, {"video_title": "Law of cosines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So a is going to be equal to the square root of that, which we can now use the calculator to figure out. So let me increase this radical a little bit so that we make sure we're taking the square root of this whole thing. So let me get my calculator out. So I want to find the square root of 220. Actually, before I do that, let me just make sure I'm in degree mode. And I am in degree mode because we're obviously finding the, we're evaluating a trig function in degrees here. So that's fine."}, {"video_title": "Law of cosines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So I want to find the square root of 220. Actually, before I do that, let me just make sure I'm in degree mode. And I am in degree mode because we're obviously finding the, we're evaluating a trig function in degrees here. So that's fine. So let me exit. So it's going to be 225 minus 216 times cosine of 87 degrees. Not 88 degrees, 87 degrees."}, {"video_title": "Law of cosines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So that's fine. So let me exit. So it's going to be 225 minus 216 times cosine of 87 degrees. Not 88 degrees, 87 degrees. And we deserve a drum roll now. This is going to be 14.618. If, say, we wanted to round to the nearest tenth just to get an approximation, it would be approximately 14.6."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "It's 1,096.5 minutes long. Half a year later, when the days are at their shortest, the days are about 382.5 minutes long. If it's not a leap year, the year is 365 days long. And June 21st is the 172nd day of the year. Write a trigonometric function that models the length, L, of the t-th day of the year. So it's going to be L as a function of t, assuming it's not a leap year. So I encourage you to pause this video and try to do this on your own before I try to work through it."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And June 21st is the 172nd day of the year. Write a trigonometric function that models the length, L, of the t-th day of the year. So it's going to be L as a function of t, assuming it's not a leap year. So I encourage you to pause this video and try to do this on your own before I try to work through it. So let me give a go at it. So instead of first starting at L of t, I'm going to actually start with L as a function of u, where u is another variable. I'll just use this kind of an intermediary variable that'll help set it up in a little bit of a simpler way."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So I encourage you to pause this video and try to do this on your own before I try to work through it. So let me give a go at it. So instead of first starting at L of t, I'm going to actually start with L as a function of u, where u is another variable. I'll just use this kind of an intermediary variable that'll help set it up in a little bit of a simpler way. Where u is days after June 21st. So let's just think about this a little bit. June 21st, if we're thinking about in terms of u, u is going to be equal to 0, because it's 0 days after June 21st."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "I'll just use this kind of an intermediary variable that'll help set it up in a little bit of a simpler way. Where u is days after June 21st. So let's just think about this a little bit. June 21st, if we're thinking about in terms of u, u is going to be equal to 0, because it's 0 days after June 21st. But if we're thinking about in terms of t, June 21st is the 172nd day of the year. So 172. So what's the relationship between u and t?"}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "June 21st, if we're thinking about in terms of u, u is going to be equal to 0, because it's 0 days after June 21st. But if we're thinking about in terms of t, June 21st is the 172nd day of the year. So 172. So what's the relationship between u and t? Well, it's shifted by 172 days. u is going to be equal to t minus 172. Notice when t is 172, u is equal to 0."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So what's the relationship between u and t? Well, it's shifted by 172 days. u is going to be equal to t minus 172. Notice when t is 172, u is equal to 0. So let's figure out L of u first, and then later we can just substitute u with t minus 172. So first of all, what's happening when u is equal to 0? Let me write all this down."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Notice when t is 172, u is equal to 0. So let's figure out L of u first, and then later we can just substitute u with t minus 172. So first of all, what's happening when u is equal to 0? Let me write all this down. So what is happening when u is equal to 0? Well, u equals 0 is June 21st, and that's the maximum point. So what trig function hits its maximum point when the input into the trig function is 0?"}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Let me write all this down. So what is happening when u is equal to 0? Well, u equals 0 is June 21st, and that's the maximum point. So what trig function hits its maximum point when the input into the trig function is 0? Well, sine of 0 is 0, while cosine of 0 is 1. Cosine hits its maximum point. So it seems a little bit easier to model this with a cosine."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So what trig function hits its maximum point when the input into the trig function is 0? Well, sine of 0 is 0, while cosine of 0 is 1. Cosine hits its maximum point. So it seems a little bit easier to model this with a cosine. So it's going to be some amplitude times cosine of, and let's say I'll write some coefficient c right over here. Actually, let me just use a b, since I already used an a. Some coefficient over here times our u plus some constant."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So it seems a little bit easier to model this with a cosine. So it's going to be some amplitude times cosine of, and let's say I'll write some coefficient c right over here. Actually, let me just use a b, since I already used an a. Some coefficient over here times our u plus some constant. That'll shift the entire function up or down. So this is the form that our function of u is going to take. And now we just have to figure out what each of these parts are."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Some coefficient over here times our u plus some constant. That'll shift the entire function up or down. So this is the form that our function of u is going to take. And now we just have to figure out what each of these parts are. So first let's think about the amplitude and what the midline is going to be. The midline is essentially how much we're shifting the function up. So let's get our calculator out."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And now we just have to figure out what each of these parts are. So first let's think about the amplitude and what the midline is going to be. The midline is essentially how much we're shifting the function up. So let's get our calculator out. So the midline is going to be halfway between these two numbers, so we could say 1,096.5 plus 382.5 divided by 2 gets us to 739.5. So that's what c is equal to. c is equal to 739.5."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So let's get our calculator out. So the midline is going to be halfway between these two numbers, so we could say 1,096.5 plus 382.5 divided by 2 gets us to 739.5. So that's what c is equal to. c is equal to 739.5. Now the amplitude is how much do we vary from that midline? So we could take 1,096 minus this, or we could take this minus 382.5. So let's do that."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "c is equal to 739.5. Now the amplitude is how much do we vary from that midline? So we could take 1,096 minus this, or we could take this minus 382.5. So let's do that. So let's take 1,096.5 minus what we just got, 739.5. And we get 357. So this is how much we vary from that midline."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So let's do that. So let's take 1,096.5 minus what we just got, 739.5. And we get 357. So this is how much we vary from that midline. So a is equal to 357. So this right over here is equal to 357. So what's b going to be equal to?"}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So this is how much we vary from that midline. So a is equal to 357. So this right over here is equal to 357. So what's b going to be equal to? And for that, I always think about, well, what's the behavior of the function? What's the period of the function going to be? Let me make a little table here."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So what's b going to be equal to? And for that, I always think about, well, what's the behavior of the function? What's the period of the function going to be? Let me make a little table here. So when u, let's put some different inputs from u. When u is 0, we're 0 days after June 21. We're at our maximum point."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Let me make a little table here. So when u, let's put some different inputs from u. When u is 0, we're 0 days after June 21. We're at our maximum point. And we already said that what we want the cosine function to evaluate to at that point is essentially we want it to evaluate as 357 times cosine of 0 plus 739.5. Now what happens? What's a full period?"}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "We're at our maximum point. And we already said that what we want the cosine function to evaluate to at that point is essentially we want it to evaluate as 357 times cosine of 0 plus 739.5. Now what happens? What's a full period? Well, a full period is a year. At a year, we get to the same point in the year, which is, I guess, a little bit of common sense. So you go all the way to 365."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "What's a full period? Well, a full period is a year. At a year, we get to the same point in the year, which is, I guess, a little bit of common sense. So you go all the way to 365. When u is 365, we should have completed a period. We should be back to that maximum point. So this should essentially be 357 times cosine of 2 pi."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So you go all the way to 365. When u is 365, we should have completed a period. We should be back to that maximum point. So this should essentially be 357 times cosine of 2 pi. If we were just thinking in terms of a traditional trig function, if we just had a theta in here, you complete a period every 2 pi. So this should be equivalent to what I'm writing out right over here, plus 739.5. So one way to think about it is b times 365 should be equal to 2 pi."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So this should essentially be 357 times cosine of 2 pi. If we were just thinking in terms of a traditional trig function, if we just had a theta in here, you complete a period every 2 pi. So this should be equivalent to what I'm writing out right over here, plus 739.5. So one way to think about it is b times 365 should be equal to 2 pi. Notice, this is going to be b times 365. So let's write that down. b times 365, that's the input into the cosine function, needs to be equal to 2 pi."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So one way to think about it is b times 365 should be equal to 2 pi. Notice, this is going to be b times 365. So let's write that down. b times 365, that's the input into the cosine function, needs to be equal to 2 pi. Or b is equal to 2 pi over 365. b is equal to 2 pi over 365. And we are almost done. We figured out what a, b, and c are."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "b times 365, that's the input into the cosine function, needs to be equal to 2 pi. Or b is equal to 2 pi over 365. b is equal to 2 pi over 365. And we are almost done. We figured out what a, b, and c are. Now we just have to substitute u with t minus 172 to get our function of t. So let's just do that. So we get, we deserve a little bit of a drum roll now, L of t is equal to a, which is 357, times cosine of b, 2 pi over 365, times not u, but now we're going to write it in terms of t. We want to think about day of the year, not days after June 21st. So times t minus 172, and then finally plus our midline, plus 739.5."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "We figured out what a, b, and c are. Now we just have to substitute u with t minus 172 to get our function of t. So let's just do that. So we get, we deserve a little bit of a drum roll now, L of t is equal to a, which is 357, times cosine of b, 2 pi over 365, times not u, but now we're going to write it in terms of t. We want to think about day of the year, not days after June 21st. So times t minus 172, and then finally plus our midline, plus 739.5. And we are done. So it seems like a very complicated expression, but if you just break it down and kind of think about it, make the point that we're talking about, the extreme point, either the minimum or the maximum, make that something, make that when the input into our function is 0 or 2 pi. 0 is actually the easiest one."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "Let a equal 0, 0 and b equal lowercase b, 2, b points on the coordinate plane. Let a, b, c, d, e, f be a convex equilateral hexagon. So convex means that it's not concave. A concave hexagon would look like this. So that's two sides, 3, 4, 5, 6. This would be a concave hexagon. So it's going to be popped out and all the sides are going to be equal."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "A concave hexagon would look like this. So that's two sides, 3, 4, 5, 6. This would be a concave hexagon. So it's going to be popped out and all the sides are going to be equal. So it's an equilateral hexagon. They're not telling us that it's a regular hexagon. So we don't know that all of the angles are going to be the same, but all the sides will be the same."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "So it's going to be popped out and all the sides are going to be equal. So it's an equilateral hexagon. They're not telling us that it's a regular hexagon. So we don't know that all of the angles are going to be the same, but all the sides will be the same. Such that f, a, b is equal to 120 degrees. Then they show us a bunch of sides that are parallel to each other. And then the y-coordinates of its vertices are distinct elements of the set 0, 2, 4, 6, 8, 10."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "So we don't know that all of the angles are going to be the same, but all the sides will be the same. Such that f, a, b is equal to 120 degrees. Then they show us a bunch of sides that are parallel to each other. And then the y-coordinates of its vertices are distinct elements of the set 0, 2, 4, 6, 8, 10. The area of the hexagon can be written in the form m square roots of n, where m and n are positive integers and n is not divisible by the square of any prime. That's just a fancy way of saying that we've simplified this radical as much as possible. Find m plus n. So really the first part, let's just make sure we can visualize this hexagon."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "And then the y-coordinates of its vertices are distinct elements of the set 0, 2, 4, 6, 8, 10. The area of the hexagon can be written in the form m square roots of n, where m and n are positive integers and n is not divisible by the square of any prime. That's just a fancy way of saying that we've simplified this radical as much as possible. Find m plus n. So really the first part, let's just make sure we can visualize this hexagon. So let me draw. We know one vertex for short, 0, 0. So let me draw my x-axis."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "Find m plus n. So really the first part, let's just make sure we can visualize this hexagon. So let me draw. We know one vertex for short, 0, 0. So let me draw my x-axis. That is my x-axis right over there. And then my y-axis. My y-axis would look like that."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "So let me draw my x-axis. That is my x-axis right over there. And then my y-axis. My y-axis would look like that. We know that the vertex a sits at the point 0, 0. That is vertex a. Now we know that all of the vertices have y-coordinates that are either 0, 2, 4, 6, 8, 10."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "My y-axis would look like that. We know that the vertex a sits at the point 0, 0. That is vertex a. Now we know that all of the vertices have y-coordinates that are either 0, 2, 4, 6, 8, 10. And they are distinct members of the set, which means no two of the vertices share the same y-coordinate. So they're not going to be on the same horizontal line. So let me draw these horizontal lines."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "Now we know that all of the vertices have y-coordinates that are either 0, 2, 4, 6, 8, 10. And they are distinct members of the set, which means no two of the vertices share the same y-coordinate. So they're not going to be on the same horizontal line. So let me draw these horizontal lines. The x-axis is 0. Then you have y is equal to 2. Then you have 4."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "So let me draw these horizontal lines. The x-axis is 0. Then you have y is equal to 2. Then you have 4. Then you have 6. And then you have 8. And then you have 10 up here."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "Then you have 4. Then you have 6. And then you have 8. And then you have 10 up here. Now, b we already know. So first of all, we've already used up the 0 for a. a is already using up the 0. b uses up the 2. They tell us that the y-coordinate of b is 2."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "And then you have 10 up here. Now, b we already know. So first of all, we've already used up the 0 for a. a is already using up the 0. b uses up the 2. They tell us that the y-coordinate of b is 2. So we use that as well. Let me see if I can draw b over here. It sits on this horizontal someplace."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "They tell us that the y-coordinate of b is 2. So we use that as well. Let me see if I can draw b over here. It sits on this horizontal someplace. And the hexagon has side length s. We don't know what that length is, but they're all the same. So let's just call this s. This is going to help me think about it now that I know that since it's an equilateral hexagon, all of the sides are going to be the same length. And so we're going to go out here to the coordinate b, 2."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "It sits on this horizontal someplace. And the hexagon has side length s. We don't know what that length is, but they're all the same. So let's just call this s. This is going to help me think about it now that I know that since it's an equilateral hexagon, all of the sides are going to be the same length. And so we're going to go out here to the coordinate b, 2. We don't know what b is, but that is our vertex b. Now, f is the other vertex that is connected to a. f cannot sit on this horizontal, cannot sit on y is equal to 2. It can't sit on y is equal to 6, because then this distance would be super far, clearly much further than this distance over here."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "And so we're going to go out here to the coordinate b, 2. We don't know what b is, but that is our vertex b. Now, f is the other vertex that is connected to a. f cannot sit on this horizontal, cannot sit on y is equal to 2. It can't sit on y is equal to 6, because then this distance would be super far, clearly much further than this distance over here. Or actually, you could have that, but then you wouldn't be able to draw really a convex hexagon. So the next vertex is just going to have to sit on this horizontal. So it's going to be s away."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "It can't sit on y is equal to 6, because then this distance would be super far, clearly much further than this distance over here. Or actually, you could have that, but then you wouldn't be able to draw really a convex hexagon. So the next vertex is just going to have to sit on this horizontal. So it's going to be s away. Maybe it will be something like that. So let me draw it. So that is the next vertex."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "So it's going to be s away. Maybe it will be something like that. So let me draw it. So that is the next vertex. That is vertex f, because we're going a, b, c, d, e, f, and then back to a. Fair enough. Now what about vertex c?"}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "So that is the next vertex. That is vertex f, because we're going a, b, c, d, e, f, and then back to a. Fair enough. Now what about vertex c? Well, vertex c can't be on the 4 horizontal, so it's going to have to be on the 6 horizontal. So vertex c is going to have to be someplace like that. That's vertex c. And once again, that length is s, this length is s. Now what about vertex e?"}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "Now what about vertex c? Well, vertex c can't be on the 4 horizontal, so it's going to have to be on the 6 horizontal. So vertex c is going to have to be someplace like that. That's vertex c. And once again, that length is s, this length is s. Now what about vertex e? It can't be on the 6 horizontal, already taken up by vertex c. So the 4 and the 6 are already taken up. So it has to be at the 8 horizontal. And so this is length s. And we also know that we're going back to the origin now."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "That's vertex c. And once again, that length is s, this length is s. Now what about vertex e? It can't be on the 6 horizontal, already taken up by vertex c. So the 4 and the 6 are already taken up. So it has to be at the 8 horizontal. And so this is length s. And we also know that we're going back to the origin now. So this is vertex e right here. We know that we're going back to the same x value. This is going to be on the y-intercept."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "And so this is length s. And we also know that we're going back to the origin now. So this is vertex e right here. We know that we're going back to the same x value. This is going to be on the y-intercept. And the reason why we know that is this is length s, and this is length s, and they both have to, both of these diagonals travel the same vertical distance. This base is 4, this base is 4. So you can kind of view this as two right triangles."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "This is going to be on the y-intercept. And the reason why we know that is this is length s, and this is length s, and they both have to, both of these diagonals travel the same vertical distance. This base is 4, this base is 4. So you can kind of view this as two right triangles. Both of them have base 4 and hypotenuse s, and so they share this side right over here. So they both, this one goes out to the left that distance, and then this one's going to have to come back that distance. Now by the same logic over here, this guy is going to have to come back."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "So you can kind of view this as two right triangles. Both of them have base 4 and hypotenuse s, and so they share this side right over here. So they both, this one goes out to the left that distance, and then this one's going to have to come back that distance. Now by the same logic over here, this guy is going to have to come back. So we can now use the 10 coordinate, the 10 y horizontal, or the y coordinate of 10. That's the only one we haven't used yet for d. And since we came out when we had a diagonal of length s traveling 4 up this time, same logic. We had a diagonal of length s. It traveled up 4 over here, and it moved out this distance."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "Now by the same logic over here, this guy is going to have to come back. So we can now use the 10 coordinate, the 10 y horizontal, or the y coordinate of 10. That's the only one we haven't used yet for d. And since we came out when we had a diagonal of length s traveling 4 up this time, same logic. We had a diagonal of length s. It traveled up 4 over here, and it moved out this distance. When we go back in the other direction and traveling up 4, you're going to go back in the same direction. So this is going to be directly on top of b. So this, the coordinate for d is actually going to be b, 10."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "We had a diagonal of length s. It traveled up 4 over here, and it moved out this distance. When we go back in the other direction and traveling up 4, you're going to go back in the same direction. So this is going to be directly on top of b. So this, the coordinate for d is actually going to be b, 10. The y coordinate here is 10. And there we have our hexagon. We're done drawing our actual hexagon."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "So this, the coordinate for d is actually going to be b, 10. The y coordinate here is 10. And there we have our hexagon. We're done drawing our actual hexagon. And all this parallel line information they told us, ab is parallel to de. So ab is parallel to de. And this is kind of obvious here."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "We're done drawing our actual hexagon. And all this parallel line information they told us, ab is parallel to de. So ab is parallel to de. And this is kind of obvious here. bc is parallel to ef. And then they say cd is parallel to fa. So cd is parallel to fa."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "And this is kind of obvious here. bc is parallel to ef. And then they say cd is parallel to fa. So cd is parallel to fa. And the way that we drew it, it looks pretty clear that that is the case. Now we need to find the area of this hexagon. And it seems like a good starting point would be to figure out what s is."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "So cd is parallel to fa. And the way that we drew it, it looks pretty clear that that is the case. Now we need to find the area of this hexagon. And it seems like a good starting point would be to figure out what s is. And to figure out what s is, it's really going to be a function of how much we've inclined this thing. So let's draw. And you can see that this isn't an equiangular hexagon, that this is kind of skewed."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "And it seems like a good starting point would be to figure out what s is. And to figure out what s is, it's really going to be a function of how much we've inclined this thing. So let's draw. And you can see that this isn't an equiangular hexagon, that this is kind of skewed. It's kind of, we distorted it a little bit. But all the sides are the same length. So let's just call this theta."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "And you can see that this isn't an equiangular hexagon, that this is kind of skewed. It's kind of, we distorted it a little bit. But all the sides are the same length. So let's just call this theta. Let's call that angle right over there theta. And then they tell us that angle fab is 120 degrees. That is 120 degrees."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "So let's just call this theta. Let's call that angle right over there theta. And then they tell us that angle fab is 120 degrees. That is 120 degrees. So this angle over here on the left is going to be 180 minus 120 minus theta. So 180 minus 120 is 60. So this angle over here is 60 minus theta."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "That is 120 degrees. So this angle over here on the left is going to be 180 minus 120 minus theta. So 180 minus 120 is 60. So this angle over here is 60 minus theta. Now the reason why I did that is because we have some information. We know that we traveled up 4 over here. And we know that we traveled up 2 over here."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "So this angle over here is 60 minus theta. Now the reason why I did that is because we have some information. We know that we traveled up 4 over here. And we know that we traveled up 2 over here. And maybe we can use that information to solve for s. Because s is the hypotenuse of both of these right triangles that I just constructed. Let me draw them. So this right triangle right over here, I could draw it like this."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "And we know that we traveled up 2 over here. And maybe we can use that information to solve for s. Because s is the hypotenuse of both of these right triangles that I just constructed. Let me draw them. So this right triangle right over here, I could draw it like this. I could draw it like this. So I have s, I have theta, and I have 2. That's this right triangle right over here."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "So this right triangle right over here, I could draw it like this. I could draw it like this. So I have s, I have theta, and I have 2. That's this right triangle right over here. This right triangle looks like this. It looks like this. This angle is 60 minus theta."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "That's this right triangle right over here. This right triangle looks like this. It looks like this. This angle is 60 minus theta. And this height over here is 4. So let's see what we can do to solve for s. This triangle on the left, or it was on the right over here, this triangle says, if you take the sine of theta, the sine of theta is equal to the opposite over the hypotenuse. It's equal to 2 over s. This triangle tells us that the sine, and remember this hypotenuse over here is also s. The sine of 60 minus theta is equal to 4 over s. And if we want to set these equal to each other, we could multiply this guy by 2 on both sides."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "This angle is 60 minus theta. And this height over here is 4. So let's see what we can do to solve for s. This triangle on the left, or it was on the right over here, this triangle says, if you take the sine of theta, the sine of theta is equal to the opposite over the hypotenuse. It's equal to 2 over s. This triangle tells us that the sine, and remember this hypotenuse over here is also s. The sine of 60 minus theta is equal to 4 over s. And if we want to set these equal to each other, we could multiply this guy by 2 on both sides. You could say 2 sine of theta is equal to 4 over s. Sine of 60 minus theta is also equal to 4 over s. We can set them equal to each other. So we have 2 sine of theta is equal to sine of 60 minus theta. And then we could use some of our trig identities."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "It's equal to 2 over s. This triangle tells us that the sine, and remember this hypotenuse over here is also s. The sine of 60 minus theta is equal to 4 over s. And if we want to set these equal to each other, we could multiply this guy by 2 on both sides. You could say 2 sine of theta is equal to 4 over s. Sine of 60 minus theta is also equal to 4 over s. We can set them equal to each other. So we have 2 sine of theta is equal to sine of 60 minus theta. And then we could use some of our trig identities. We know the sine of a minus b is the same thing. The sine of a minus b, this is equal to the sine of a times the cosine of b. Or I should say theta in this case."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "And then we could use some of our trig identities. We know the sine of a minus b is the same thing. The sine of a minus b, this is equal to the sine of a times the cosine of b. Or I should say theta in this case. So sine of 60 minus theta minus, this is just a standard trig identity. That's the difference, sum and difference identity. Minus cosine of 60 times the sine of theta."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "Or I should say theta in this case. So sine of 60 minus theta minus, this is just a standard trig identity. That's the difference, sum and difference identity. Minus cosine of 60 times the sine of theta. And all of this is equal to 2 sine of theta. Well sine of 60 degrees, this is square root of 3 over 2. Cosine of 60 degrees is 1 half."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "Minus cosine of 60 times the sine of theta. And all of this is equal to 2 sine of theta. Well sine of 60 degrees, this is square root of 3 over 2. Cosine of 60 degrees is 1 half. So we could add 1 half sine theta to both sides of this and what are we going to get? So we're going to add 1 half sine theta. Then this guy is going to go away."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "Cosine of 60 degrees is 1 half. So we could add 1 half sine theta to both sides of this and what are we going to get? So we're going to add 1 half sine theta. Then this guy is going to go away. And then you add 1 half sine theta to 2 sine of theta, which is really 4 halves sine of theta. So that's going to be 5 halves sine of theta. So I'm just adding 1 half sine theta to this."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "Then this guy is going to go away. And then you add 1 half sine theta to 2 sine of theta, which is really 4 halves sine of theta. So that's going to be 5 halves sine of theta. So I'm just adding 1 half sine theta to this. So that's 5 halves sine of theta is equal to square root of 3 over 2 cosine of theta. I added 1 half sine theta to both sides of this to get this. I can multiply both sides by 2 just to simplify it."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "So I'm just adding 1 half sine theta to this. So that's 5 halves sine of theta is equal to square root of 3 over 2 cosine of theta. I added 1 half sine theta to both sides of this to get this. I can multiply both sides by 2 just to simplify it. So I get 5 sine of theta is equal to the square root of 3 cosine of theta. Now I want to use the identity sine square root of theta plus cosine square root of theta is equal to 1. So let me just square both sides and that will also help us with this radical."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "I can multiply both sides by 2 just to simplify it. So I get 5 sine of theta is equal to the square root of 3 cosine of theta. Now I want to use the identity sine square root of theta plus cosine square root of theta is equal to 1. So let me just square both sides and that will also help us with this radical. So we'll get 25 sine square root of theta is equal to square root of 3 squared is 3. Instead of writing cosine square of theta, let's just write 1 minus sine square of theta. Cosine square theta is 1 minus sine square of theta."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "So let me just square both sides and that will also help us with this radical. So we'll get 25 sine square root of theta is equal to square root of 3 squared is 3. Instead of writing cosine square of theta, let's just write 1 minus sine square of theta. Cosine square theta is 1 minus sine square of theta. I just squared both sides. Let me just write what I just did. I just squared both sides."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "Cosine square theta is 1 minus sine square of theta. I just squared both sides. Let me just write what I just did. I just squared both sides. And so we get 25 sine square theta is equal to 3 minus 3 sine square theta. We can add 3 sine square theta to both sides. 28 sine squared theta is equal to 3, or that the sine squared of theta, home stretch, is equal to 3 over 28."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "I just squared both sides. And so we get 25 sine square theta is equal to 3 minus 3 sine square theta. We can add 3 sine square theta to both sides. 28 sine squared theta is equal to 3, or that the sine squared of theta, home stretch, is equal to 3 over 28. Or we could even write that sine of theta is equal to the square root of 3 over 28. So it's equal to the square root of 3 over 28. Now we could simplify that."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "28 sine squared theta is equal to 3, or that the sine squared of theta, home stretch, is equal to 3 over 28. Or we could even write that sine of theta is equal to the square root of 3 over 28. So it's equal to the square root of 3 over 28. Now we could simplify that. 28 is 4 times 7. We could take it out. But that's good enough for now."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "Now we could simplify that. 28 is 4 times 7. We could take it out. But that's good enough for now. So that maybe we'll simplify it later if we have to. Sometimes these are easier to deal with. So let's see."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "But that's good enough for now. So that maybe we'll simplify it later if we have to. Sometimes these are easier to deal with. So let's see. So we have the sine of theta. Now we can relate that actually to s over here. We know that before I messed with this thing, we know that the sine of theta is equal to 2 over s, or that s over 2 is equal to 1 over sine of theta, or that s is equal to 2 over the sine of theta."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "So let's see. So we have the sine of theta. Now we can relate that actually to s over here. We know that before I messed with this thing, we know that the sine of theta is equal to 2 over s, or that s over 2 is equal to 1 over sine of theta, or that s is equal to 2 over the sine of theta. Well, we know what sine of theta is. It's square root of 3 over 28. So s is equal to 2 divided by sine of theta."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "We know that before I messed with this thing, we know that the sine of theta is equal to 2 over s, or that s over 2 is equal to 1 over sine of theta, or that s is equal to 2 over the sine of theta. Well, we know what sine of theta is. It's square root of 3 over 28. So s is equal to 2 divided by sine of theta. That's like multiplying by the inverse of sine of theta. So that's 2 times the square root of 28 over 3. So that is, we figured out our s, 2 times this thing over here."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "So s is equal to 2 divided by sine of theta. That's like multiplying by the inverse of sine of theta. So that's 2 times the square root of 28 over 3. So that is, we figured out our s, 2 times this thing over here. Now, given that we know an s, let's see how we can figure out the area. Well, what immediately pops out is that we have this triangle over here that has height, or I should say maybe it's base if you view it sideways. Its base is 8."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "So that is, we figured out our s, 2 times this thing over here. Now, given that we know an s, let's see how we can figure out the area. Well, what immediately pops out is that we have this triangle over here that has height, or I should say maybe it's base if you view it sideways. Its base is 8. And this distance right over here we should be able to figure out. We should be able to figure out using the Pythagorean theorem. Because we know that this distance right over here is 4."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "Its base is 8. And this distance right over here we should be able to figure out. We should be able to figure out using the Pythagorean theorem. Because we know that this distance right over here is 4. We know that distance is 4. We know that this distance, the hypotenuse, is s. So we could call this the height of it right over here. We could say that h squared plus 4 squared plus 16 is equal to the hypotenuse squared, is equal to s squared."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "Because we know that this distance right over here is 4. We know that distance is 4. We know that this distance, the hypotenuse, is s. So we could call this the height of it right over here. We could say that h squared plus 4 squared plus 16 is equal to the hypotenuse squared, is equal to s squared. s is this thing over here. So if we want to square s, it becomes 4 times 28 over 3. And we just subtract 16 from both sides."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "We could say that h squared plus 4 squared plus 16 is equal to the hypotenuse squared, is equal to s squared. s is this thing over here. So if we want to square s, it becomes 4 times 28 over 3. And we just subtract 16 from both sides. So h is equal to 4 times 28 over 3 minus, if I want to write 16 over 3, or if I want to write 16 as something over 3, that's going to be minus 48 over 3. And let's see, I don't want to have to multiply 4 times 28. So we can write 48 as 4 times 12."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "And we just subtract 16 from both sides. So h is equal to 4 times 28 over 3 minus, if I want to write 16 over 3, or if I want to write 16 as something over 3, that's going to be minus 48 over 3. And let's see, I don't want to have to multiply 4 times 28. So we can write 48 as 4 times 12. So this numerator is going to be 4 times 28 minus 12 over, remember, that's h squared, I should say. h squared is going to be 4 times 28 minus 12 over 3, which is equal to 4 times 16 over 3, which is equal to 64 over 3. That's h squared."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "So we can write 48 as 4 times 12. So this numerator is going to be 4 times 28 minus 12 over, remember, that's h squared, I should say. h squared is going to be 4 times 28 minus 12 over 3, which is equal to 4 times 16 over 3, which is equal to 64 over 3. That's h squared. So h is going to be the square root of that, which is 8 over the square root of 3. So h right over here is 8 over the square root of 3. So if I want to find the area of this whole thing over here, well, first let's find the area of this small thing right over here."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "That's h squared. So h is going to be the square root of that, which is 8 over the square root of 3. So h right over here is 8 over the square root of 3. So if I want to find the area of this whole thing over here, well, first let's find the area of this small thing right over here. That's just going to be h times 4. So it's going to be, well, I could do it either way. But let's just say this is h times 4 times 1 half."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "So if I want to find the area of this whole thing over here, well, first let's find the area of this small thing right over here. That's just going to be h times 4. So it's going to be, well, I could do it either way. But let's just say this is h times 4 times 1 half. So it's going to be 2 times, so the area of this triangle, let me do it in blue right here. This triangle's area is going to be h, which is 8 over the square root of 3 times 4 times 1 half. So this guy right over here is just going to be 2 times 8 over the square root of 3, or that's going to be 16 over the square root of 3."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "But let's just say this is h times 4 times 1 half. So it's going to be 2 times, so the area of this triangle, let me do it in blue right here. This triangle's area is going to be h, which is 8 over the square root of 3 times 4 times 1 half. So this guy right over here is just going to be 2 times 8 over the square root of 3, or that's going to be 16 over the square root of 3. So this guy over here is 16 over the square root of 3. So that is 16 over the square root of 3. Now, we have a bunch of them."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "So this guy right over here is just going to be 2 times 8 over the square root of 3, or that's going to be 16 over the square root of 3. So this guy over here is 16 over the square root of 3. So that is 16 over the square root of 3. Now, we have a bunch of them. We have this guy, and now this guy's the exact same area right over there. And then you have this guy, who's going to have the exact same area again. Same exact logic, same base, same height."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "Now, we have a bunch of them. We have this guy, and now this guy's the exact same area right over there. And then you have this guy, who's going to have the exact same area again. Same exact logic, same base, same height. They're actually congruent. So you have four of these triangles. You're going to multiply by 4 if you want the area of this area that I've already shaded in."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "Same exact logic, same base, same height. They're actually congruent. So you have four of these triangles. You're going to multiply by 4 if you want the area of this area that I've already shaded in. So 4 times this, we're at 64 over the square root of 3. Now, the only area we have left to figure out is the area of this parallelogram in the middle. Now, we know the base of the parallelogram."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "You're going to multiply by 4 if you want the area of this area that I've already shaded in. So 4 times this, we're at 64 over the square root of 3. Now, the only area we have left to figure out is the area of this parallelogram in the middle. Now, we know the base of the parallelogram. The base of this parallelogram is 8. The base of the parallelogram is 8. We just have to figure out its height."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "Now, we know the base of the parallelogram. The base of this parallelogram is 8. The base of the parallelogram is 8. We just have to figure out its height. We just have to figure out its height. And once again, we can use the Pythagorean theorem. So I'll call this, I don't know, we already used h. Well, I'll use h again, but you just have to remember that this is a different height over here."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "We just have to figure out its height. We just have to figure out its height. And once again, we can use the Pythagorean theorem. So I'll call this, I don't know, we already used h. Well, I'll use h again, but you just have to remember that this is a different height over here. This base over here is of length 2. I know it's hard to read now. So we can now write that h squared plus 4 plus 2 squared is equal to s squared."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "So I'll call this, I don't know, we already used h. Well, I'll use h again, but you just have to remember that this is a different height over here. This base over here is of length 2. I know it's hard to read now. So we can now write that h squared plus 4 plus 2 squared is equal to s squared. Now, we already figured out what s squared was in the past. It's 4 times 28 over 3. Let's subtract 4 from both sides."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "So we can now write that h squared plus 4 plus 2 squared is equal to s squared. Now, we already figured out what s squared was in the past. It's 4 times 28 over 3. Let's subtract 4 from both sides. You subtract 4 there. So minus 12 over 3. And now, let's see, 12 is the same thing as 4 times 3."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "Let's subtract 4 from both sides. You subtract 4 there. So minus 12 over 3. And now, let's see, 12 is the same thing as 4 times 3. So this is equal to 4 times 28 minus 3. So that's 4 times 25 over 3, which is equal to 100 over 3. That's h squared."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "And now, let's see, 12 is the same thing as 4 times 3. So this is equal to 4 times 28 minus 3. So that's 4 times 25 over 3, which is equal to 100 over 3. That's h squared. So this h is going to be equal to the square root of this, which is 10 over the square root of 3. So this distance right over here is 10 over the square root of 3. So we want the area of this parallelogram is going to be that height times the base of the parallelogram."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "That's h squared. So this h is going to be equal to the square root of this, which is 10 over the square root of 3. So this distance right over here is 10 over the square root of 3. So we want the area of this parallelogram is going to be that height times the base of the parallelogram. So the parallelogram is going to be 8 times 10 square roots of 3 or 80 square roots of 3. Oh, no, no, let me be very careful. This was 10 over."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "So we want the area of this parallelogram is going to be that height times the base of the parallelogram. So the parallelogram is going to be 8 times 10 square roots of 3 or 80 square roots of 3. Oh, no, no, let me be very careful. This was 10 over. This is 10 over the square root of 3 is this height. So the whole area of this parallelogram is 8 times 10 over the square root of 3. It's 80 over the square root of 3."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "This was 10 over. This is 10 over the square root of 3 is this height. So the whole area of this parallelogram is 8 times 10 over the square root of 3. It's 80 over the square root of 3. So our entire area now, if we add everything together, we have 64 square roots of 3 for these four triangles plus 80 over square root of 3. So let's add it together. So we have 80 over square root of 3 for the parallelogram plus 64 over square root of 3 for the triangle parts."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "It's 80 over the square root of 3. So our entire area now, if we add everything together, we have 64 square roots of 3 for these four triangles plus 80 over square root of 3. So let's add it together. So we have 80 over square root of 3 for the parallelogram plus 64 over square root of 3 for the triangle parts. And this is equal to 100 over 3. So this is equal to 144 over the square root of 3. We can rationalize the denominator."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "So we have 80 over square root of 3 for the parallelogram plus 64 over square root of 3 for the triangle parts. And this is equal to 100 over 3. So this is equal to 144 over the square root of 3. We can rationalize the denominator. So times the square root of 3 over the square root of 3. And in the denominator now, we're going to get a 3. 144 over 3 is going to be what?"}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "We can rationalize the denominator. So times the square root of 3 over the square root of 3. And in the denominator now, we're going to get a 3. 144 over 3 is going to be what? That's 48, right? 3 times 40 is 120. 3 times 8 is 24."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "144 over 3 is going to be what? That's 48, right? 3 times 40 is 120. 3 times 8 is 24. So it's going to be 48 square roots of 3 for the area of our entire hexagon. And so we have it in the form, 48 square roots of 3. So if you want to find m plus n, it's 48 plus 3, which is 51."}, {"video_title": "Trig challenge problem area of a hexagon Math for fun and glory Khan Academy.mp3", "Sentence": "3 times 8 is 24. So it's going to be 48 square roots of 3 for the area of our entire hexagon. And so we have it in the form, 48 square roots of 3. So if you want to find m plus n, it's 48 plus 3, which is 51. That was a tiring problem. My brain started to fry near the end of it. I had trouble keeping track of things."}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And to do that, I'll set up a little unit circle so that we can visualize what the tangent of various thetas are. So let's say that's a y-axis, this is my x-axis, that is my x-axis, and the unit circle would look something like this. And we already know, this is all a refresher of the unit circle definition of trig functions, that if I have an angle, an angle theta, where one side is the positive x-axis, and then the other side, so this is the other side, so the angle is formed like this, that where this ray intersects the unit circle, the coordinates of that, the x and y coordinates, are the sine of theta, sorry, the x coordinate is the cosine of theta, so it's the cosine of theta comma sine of theta. So the x coordinate here is cosine theta, the y coordinate there is the sine of theta. But we're concerned about tangent of theta. Well, we know that tangent of theta is the same thing as the sine of theta over the cosine of theta. Or if you're starting from the origin, and you're going, and you're taking the value of essentially the y coordinate, the y coordinate over the x coordinate, it's essentially the slope of this line."}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So the x coordinate here is cosine theta, the y coordinate there is the sine of theta. But we're concerned about tangent of theta. Well, we know that tangent of theta is the same thing as the sine of theta over the cosine of theta. Or if you're starting from the origin, and you're going, and you're taking the value of essentially the y coordinate, the y coordinate over the x coordinate, it's essentially the slope of this line. It's going to be, this is going to be your change in y, change in y over change in x. This right over here is going to be the slope, the slope, I guess you could say, of this ray right over here. And so that's going to help us visualize what the tangents of different thetas are."}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Or if you're starting from the origin, and you're going, and you're taking the value of essentially the y coordinate, the y coordinate over the x coordinate, it's essentially the slope of this line. It's going to be, this is going to be your change in y, change in y over change in x. This right over here is going to be the slope, the slope, I guess you could say, of this ray right over here. And so that's going to help us visualize what the tangents of different thetas are. So let me clean up my unit circle a little bit, just so that we can, all right, there we go. So now let's make a, let's make a table. So let's make a table."}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And so that's going to help us visualize what the tangents of different thetas are. So let me clean up my unit circle a little bit, just so that we can, all right, there we go. So now let's make a, let's make a table. So let's make a table. So if, so for various thetas, let's think about what tangent of theta is going to be. Tangent of theta. So maybe the easiest one, if theta is zero radians."}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So let's make a table. So if, so for various thetas, let's think about what tangent of theta is going to be. Tangent of theta. So maybe the easiest one, if theta is zero radians. So if it's zero radians, what is the slope of this ray? Well, that ray is, the slope is zero. As x changes, y doesn't change at all."}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So maybe the easiest one, if theta is zero radians. So if it's zero radians, what is the slope of this ray? Well, that ray is, the slope is zero. As x changes, y doesn't change at all. Now let's think about, and I'm just going to pick values that are very easy for us to think about what the tangent of those values are, and they'll help us form, they'll help us think about the shape of the graph of y is equal to tangent of theta. So let's take, let's take pi over, pi over four radians. So this one right over here, theta is equal to pi over four."}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "As x changes, y doesn't change at all. Now let's think about, and I'm just going to pick values that are very easy for us to think about what the tangent of those values are, and they'll help us form, they'll help us think about the shape of the graph of y is equal to tangent of theta. So let's take, let's take pi over, pi over four radians. So this one right over here, theta is equal to pi over four. Now why is that interesting? And sometimes it's easier to think in degrees. That's a 45 degree angle."}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So this one right over here, theta is equal to pi over four. Now why is that interesting? And sometimes it's easier to think in degrees. That's a 45 degree angle. This here, your x coordinate and your y coordinate is the same. You might remember it's square root of two over two, but the important thing is, whatever you move in the x direction, you move the same in the y direction. So the slope of this ray right over here is going to be equal to one, or another way of thinking about it, tangent of theta is going to be equal to one."}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "That's a 45 degree angle. This here, your x coordinate and your y coordinate is the same. You might remember it's square root of two over two, but the important thing is, whatever you move in the x direction, you move the same in the y direction. So the slope of this ray right over here is going to be equal to one, or another way of thinking about it, tangent of theta is going to be equal to one. Or sine of theta over cosine of theta, they're the same thing. So you're going to get one. So if you put, so let me just clean that up here, just because I'm going to keep reusing the same unit circle."}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So the slope of this ray right over here is going to be equal to one, or another way of thinking about it, tangent of theta is going to be equal to one. Or sine of theta over cosine of theta, they're the same thing. So you're going to get one. So if you put, so let me just clean that up here, just because I'm going to keep reusing the same unit circle. So if I have theta is pi over four, then the tangent of theta is going to be equal to one. Now what if theta is equal to negative pi over four? So that is this right over here."}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So if you put, so let me just clean that up here, just because I'm going to keep reusing the same unit circle. So if I have theta is pi over four, then the tangent of theta is going to be equal to one. Now what if theta is equal to negative pi over four? So that is this right over here. So with x, so let me just draw a little triangle here. So when x, this x coordinate over here is square root of two over two, we know that, we've seen that multiple times, square root of two, actually let me label it a little bit better. So here our theta is equal to negative pi over four radians."}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So that is this right over here. So with x, so let me just draw a little triangle here. So when x, this x coordinate over here is square root of two over two, we know that, we've seen that multiple times, square root of two, actually let me label it a little bit better. So here our theta is equal to negative pi over four radians. Now, or you could, if you like to think in degrees, this would be negative 45 degrees. And now your sine and cosine of this angle are going to be the opposites of each other. The cosine is square root of two over two, the x coordinate of where this intersects is square root of two over two."}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So here our theta is equal to negative pi over four radians. Now, or you could, if you like to think in degrees, this would be negative 45 degrees. And now your sine and cosine of this angle are going to be the opposites of each other. The cosine is square root of two over two, the x coordinate of where this intersects is square root of two over two. The y coordinate here is negative square root of two. Negative square root of two over two. So what's the tangent?"}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "The cosine is square root of two over two, the x coordinate of where this intersects is square root of two over two. The y coordinate here is negative square root of two. Negative square root of two over two. So what's the tangent? Well, it's going to be your sine over your cosine, which is going to be negative one. And you see that. For however much you move in the x direction, you move the opposite of that, or you move the negative of that in the y direction."}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So what's the tangent? Well, it's going to be your sine over your cosine, which is going to be negative one. And you see that. For however much you move in the x direction, you move the opposite of that, or you move the negative of that in the y direction. And so let me clean this up a little bit because I want to keep reusing my unit circle. So there you go. And so this is going to be negative one."}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "For however much you move in the x direction, you move the opposite of that, or you move the negative of that in the y direction. And so let me clean this up a little bit because I want to keep reusing my unit circle. So there you go. And so this is going to be negative one. This is going to be negative one. And so actually let's just start plotting a few of these points. So if we assume that this is the theta axis, if you can see that, that's the theta axis, and if this is the y axis, that's the y axis, we immediately see tangent of zero is zero, tangent of pi over four is one, I'm looking in radians, tangent of negative pi over four is negative one."}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And so this is going to be negative one. This is going to be negative one. And so actually let's just start plotting a few of these points. So if we assume that this is the theta axis, if you can see that, that's the theta axis, and if this is the y axis, that's the y axis, we immediately see tangent of zero is zero, tangent of pi over four is one, I'm looking in radians, tangent of negative pi over four is negative one. Now let's think, right now if you just saw that, you might say, oh, maybe this is some type of a line, but we'll see very clearly it's not a line because what happens as our angle gets closer and closer to pi over two? What happens to the slope of this line? So that is theta."}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So if we assume that this is the theta axis, if you can see that, that's the theta axis, and if this is the y axis, that's the y axis, we immediately see tangent of zero is zero, tangent of pi over four is one, I'm looking in radians, tangent of negative pi over four is negative one. Now let's think, right now if you just saw that, you might say, oh, maybe this is some type of a line, but we'll see very clearly it's not a line because what happens as our angle gets closer and closer to pi over two? What happens to the slope of this line? So that is theta. We're getting closer and closer to pi over two. Well, this ray, I guess I should say, is getting closer and closer to approaching the vertical. So its slope is getting more and more and more positive, and if you go all the way to pi over two, the slope at that point is really undefined, but it's approaching, one way to think about it is, it is approaching infinity."}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So that is theta. We're getting closer and closer to pi over two. Well, this ray, I guess I should say, is getting closer and closer to approaching the vertical. So its slope is getting more and more and more positive, and if you go all the way to pi over two, the slope at that point is really undefined, but it's approaching, one way to think about it is, it is approaching infinity. So as you get closer and closer to pi over two, so I'm going to make a, I'm going to draw essentially a vertical asymptote right over here at pi over two because it's not going to be, I guess one way to think about it is approaching infinity there. So this is going to be looking something like this. It's going to be looking something like this."}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So its slope is getting more and more and more positive, and if you go all the way to pi over two, the slope at that point is really undefined, but it's approaching, one way to think about it is, it is approaching infinity. So as you get closer and closer to pi over two, so I'm going to make a, I'm going to draw essentially a vertical asymptote right over here at pi over two because it's not going to be, I guess one way to think about it is approaching infinity there. So this is going to be looking something like this. It's going to be looking something like this. The slope of the ray as you get closer and closer to pi over two is getting closer and closer to infinity. And what happens when the angle is getting closer and closer to negative pi over two? Is getting closer and closer to negative pi over two?"}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "It's going to be looking something like this. The slope of the ray as you get closer and closer to pi over two is getting closer and closer to infinity. And what happens when the angle is getting closer and closer to negative pi over two? Is getting closer and closer to negative pi over two? Well, then the slope is getting more and more and more negative. It's really approaching negative. It's approaching negative infinity."}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Is getting closer and closer to negative pi over two? Well, then the slope is getting more and more and more negative. It's really approaching negative. It's approaching negative infinity. So let me draw that. So once again, not quite defined right over there. We have a vertical asymptote, and we are approaching negative infinity."}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "It's approaching negative infinity. So let me draw that. So once again, not quite defined right over there. We have a vertical asymptote, and we are approaching negative infinity. We are approaching negative infinity. So that's what the graph of tangent of theta looks, just over this section of, I guess we could say, the theta axis, but then we could keep going. There we could keep going."}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "We have a vertical asymptote, and we are approaching negative infinity. We are approaching negative infinity. So that's what the graph of tangent of theta looks, just over this section of, I guess we could say, the theta axis, but then we could keep going. There we could keep going. Because if our angle, right after we cross pi over two, so let's say we've just crossed pi over two. So we went right across it. Now what is the slope of this thing?"}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "There we could keep going. Because if our angle, right after we cross pi over two, so let's say we've just crossed pi over two. So we went right across it. Now what is the slope of this thing? Well, the slope of this thing is hugely negative. It looks almost like what I just drew down here. It's hugely negative."}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Now what is the slope of this thing? Well, the slope of this thing is hugely negative. It looks almost like what I just drew down here. It's hugely negative. So then the graph jumps back down here, and it's hugely negative again. It's hugely negative, and then as we increase our theta, as we increase our theta, it becomes less and less and less negative, all the way to when we go to, what is this, all the way until we go to, let me plot this, this angle right over here. Now what is this angle?"}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "It's hugely negative. So then the graph jumps back down here, and it's hugely negative again. It's hugely negative, and then as we increase our theta, as we increase our theta, it becomes less and less and less negative, all the way to when we go to, what is this, all the way until we go to, let me plot this, this angle right over here. Now what is this angle? This I haven't told you yet. Let's say that this angle right over here is three pi over four. Now why did I pick three pi over four?"}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Now what is this angle? This I haven't told you yet. Let's say that this angle right over here is three pi over four. Now why did I pick three pi over four? Because that is pi over two, that is pi over two plus pi over four. Or you could say two pi's over four plus another pi over four is three pi over four. And the reason why this is interesting is because it is another, it's forming another, I guess you could say, pi over four, pi over four, pi over two triangle, or 45, 45, 90 triangle, where the x and y coordinates, or the x and y distances have the same magnitude, but now the x is going to be negative and the y is positive."}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Now why did I pick three pi over four? Because that is pi over two, that is pi over two plus pi over four. Or you could say two pi's over four plus another pi over four is three pi over four. And the reason why this is interesting is because it is another, it's forming another, I guess you could say, pi over four, pi over four, pi over two triangle, or 45, 45, 90 triangle, where the x and y coordinates, or the x and y distances have the same magnitude, but now the x is going to be negative and the y is positive. So the slope here is going to be the slope at the same slope as we had for negative pi over four radians. We're going to have a slope of negative one. So at three pi over four, we have a slope of negative one."}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And the reason why this is interesting is because it is another, it's forming another, I guess you could say, pi over four, pi over four, pi over two triangle, or 45, 45, 90 triangle, where the x and y coordinates, or the x and y distances have the same magnitude, but now the x is going to be negative and the y is positive. So the slope here is going to be the slope at the same slope as we had for negative pi over four radians. We're going to have a slope of negative one. So at three pi over four, we have a slope of negative one. Then we increase our angle all the way to pi. Now our slope is back to zero. Our slope is back to zero."}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So at three pi over four, we have a slope of negative one. Then we increase our angle all the way to pi. Now our slope is back to zero. Our slope is back to zero. And then as we go beyond that, as we go to, as we increase by another, as we increase by another pi over four, our slope goes back to being positive one. Our slope goes back to being positive one. And then once again, as we approach three pi over two, our slope is becoming more and more and more positive, getting, approaching positive, approaching positive infinity."}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Our slope is back to zero. And then as we go beyond that, as we go to, as we increase by another, as we increase by another pi over four, our slope goes back to being positive one. Our slope goes back to being positive one. And then once again, as we approach three pi over two, our slope is becoming more and more and more positive, getting, approaching positive, approaching positive infinity. The slope, notice if you move a little bit in the x direction, you're moving a lot up in the y direction. So once again, so now the graph is going to look like this. Let me do it in a color that you can actually see."}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And then once again, as we approach three pi over two, our slope is becoming more and more and more positive, getting, approaching positive, approaching positive infinity. The slope, notice if you move a little bit in the x direction, you're moving a lot up in the y direction. So once again, so now the graph is going to look like this. Let me do it in a color that you can actually see. The graph is going to look something, something like, something like this. And it will just continue to do this. It will just continue to do this every, every pi radians."}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Let me do it in a color that you can actually see. The graph is going to look something, something like, something like this. And it will just continue to do this. It will just continue to do this every, every pi radians. Every, actually maybe we do that as a dotted line. Every pi radians, over and over and over again. So let me go back pi, and I can draw these asymptotes."}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "It will just continue to do this every, every pi radians. Every, actually maybe we do that as a dotted line. Every pi radians, over and over and over again. So let me go back pi, and I can draw these asymptotes. I can draw these asymptotes. And so we draw that and that. And so the graph of tangent, the graph of tangent of theta is going to look, is going to look something, something like this."}, {"video_title": "Pi Is (still) Wrong..mp3", "Sentence": "This is not your fault. It's not even your teacher's fault. It's pi's fault because pi is wrong. I don't mean that pi is incorrect. The ratio of a circle's diameter to its circumference is still 3.14 and so on. I mean that pi, as a concept, is a terrible mistake that has gone uncorrected for thousands of years. The problem with pi and pi day is the same as the problem with Columbus and Columbus day."}, {"video_title": "Pi Is (still) Wrong..mp3", "Sentence": "I don't mean that pi is incorrect. The ratio of a circle's diameter to its circumference is still 3.14 and so on. I mean that pi, as a concept, is a terrible mistake that has gone uncorrected for thousands of years. The problem with pi and pi day is the same as the problem with Columbus and Columbus day. I don't know if Christopher Columbus was a real person who did some stuff but everything you learn about him in school is warped and overemphasized. He didn't discover America, he didn't discover the world was round, and he was a bit of a jerk. So why do we celebrate Columbus day?"}, {"video_title": "Pi Is (still) Wrong..mp3", "Sentence": "The problem with pi and pi day is the same as the problem with Columbus and Columbus day. I don't know if Christopher Columbus was a real person who did some stuff but everything you learn about him in school is warped and overemphasized. He didn't discover America, he didn't discover the world was round, and he was a bit of a jerk. So why do we celebrate Columbus day? Same with pi. You learned in school that pi is the all-important circle constant and had to memorize a whole bunch of equations involving it because that's the way it's been taught for a very long time. If you found any of these equations confusing, it's not your fault."}, {"video_title": "Pi Is (still) Wrong..mp3", "Sentence": "So why do we celebrate Columbus day? Same with pi. You learned in school that pi is the all-important circle constant and had to memorize a whole bunch of equations involving it because that's the way it's been taught for a very long time. If you found any of these equations confusing, it's not your fault. It's just that pi is wrong. Let me show you what I mean. Gradients, good system for measuring angles when it comes to mathematics."}, {"video_title": "Pi Is (still) Wrong..mp3", "Sentence": "If you found any of these equations confusing, it's not your fault. It's just that pi is wrong. Let me show you what I mean. Gradients, good system for measuring angles when it comes to mathematics. It should make sense, but it doesn't because pi messes it up. For example, how much pi is this? You might think this should be 1 pi, but it's not."}, {"video_title": "Pi Is (still) Wrong..mp3", "Sentence": "Gradients, good system for measuring angles when it comes to mathematics. It should make sense, but it doesn't because pi messes it up. For example, how much pi is this? You might think this should be 1 pi, but it's not. The full 360 degrees of pi is actually 2 pi. What? Say I ask you how much pi you want and you say pi over 8."}, {"video_title": "Pi Is (still) Wrong..mp3", "Sentence": "You might think this should be 1 pi, but it's not. The full 360 degrees of pi is actually 2 pi. What? Say I ask you how much pi you want and you say pi over 8. You'd think this should be an eighth of a pi, but it's not, it's a sixteenth of a pi. That's confusing. You may be thinking, come on Vi, it's a simple conversion."}, {"video_title": "Pi Is (still) Wrong..mp3", "Sentence": "Say I ask you how much pi you want and you say pi over 8. You'd think this should be an eighth of a pi, but it's not, it's a sixteenth of a pi. That's confusing. You may be thinking, come on Vi, it's a simple conversion. All you have to do is divide by 2 or multiply by 2 if you're going the other way, so you just have to make sure you pay attention to which way you're starting. No, you're making excuses for pi. Mathematics should be as elegant and beautiful as possible."}, {"video_title": "Pi Is (still) Wrong..mp3", "Sentence": "You may be thinking, come on Vi, it's a simple conversion. All you have to do is divide by 2 or multiply by 2 if you're going the other way, so you just have to make sure you pay attention to which way you're starting. No, you're making excuses for pi. Mathematics should be as elegant and beautiful as possible. When you complicate something that should be as simple as 1 pi equals 1 pi by adding all these conversions, something gets lost in translation. But Vi, you ask, is there a better way? Well, for this particular example, there's an easy answer for what you'd have to do to make a pi be 1 pi instead of 2 pi."}, {"video_title": "Pi Is (still) Wrong..mp3", "Sentence": "Mathematics should be as elegant and beautiful as possible. When you complicate something that should be as simple as 1 pi equals 1 pi by adding all these conversions, something gets lost in translation. But Vi, you ask, is there a better way? Well, for this particular example, there's an easy answer for what you'd have to do to make a pi be 1 pi instead of 2 pi. You could redefine pi to be 2 pi or 6.28 and so on. But I don't want to redefine pi because that would be confusing, so let's use a different letter, tau, because tau looks kind of like pi. A full circle would be 1 tau, a half circle would be half tau, or tau over 2."}, {"video_title": "Pi Is (still) Wrong..mp3", "Sentence": "Well, for this particular example, there's an easy answer for what you'd have to do to make a pi be 1 pi instead of 2 pi. You could redefine pi to be 2 pi or 6.28 and so on. But I don't want to redefine pi because that would be confusing, so let's use a different letter, tau, because tau looks kind of like pi. A full circle would be 1 tau, a half circle would be half tau, or tau over 2. And if you want one sixteenth of this pi, you want tau over 16. That would be simple. But Vi, you say, that seems rather arbitrary."}, {"video_title": "Pi Is (still) Wrong..mp3", "Sentence": "A full circle would be 1 tau, a half circle would be half tau, or tau over 2. And if you want one sixteenth of this pi, you want tau over 16. That would be simple. But Vi, you say, that seems rather arbitrary. Your tau makes radians easier, but it would be annoying to have to convert between tau and pi every time you want to work in radians. True, but the way of mathematics is to make stuff up and see what happens, so let's see what happens if we use tau in other equations. Math classes make you memorize stuff like this, so that you can draw graphs like this."}, {"video_title": "Pi Is (still) Wrong..mp3", "Sentence": "But Vi, you say, that seems rather arbitrary. Your tau makes radians easier, but it would be annoying to have to convert between tau and pi every time you want to work in radians. True, but the way of mathematics is to make stuff up and see what happens, so let's see what happens if we use tau in other equations. Math classes make you memorize stuff like this, so that you can draw graphs like this. I mean, sure, you could derive these values every time, but you don't because it's easier to just memorize it or use your calculator because pi and radians are confusing. This appalling notation makes us forget what the sine wave actually represents, which is how high this point is when you've gone however far around this unit circle. When your radians are notated horrifically, all of trigonometry becomes ugly."}, {"video_title": "Pi Is (still) Wrong..mp3", "Sentence": "Math classes make you memorize stuff like this, so that you can draw graphs like this. I mean, sure, you could derive these values every time, but you don't because it's easier to just memorize it or use your calculator because pi and radians are confusing. This appalling notation makes us forget what the sine wave actually represents, which is how high this point is when you've gone however far around this unit circle. When your radians are notated horrifically, all of trigonometry becomes ugly. But it doesn't have to be this way. What if we used tau? Let's make a sine wave starting with tau at 0."}, {"video_title": "Pi Is (still) Wrong..mp3", "Sentence": "When your radians are notated horrifically, all of trigonometry becomes ugly. But it doesn't have to be this way. What if we used tau? Let's make a sine wave starting with tau at 0. The height of sine tau is also 0. At tau over 4, we've gone a quarter of the way around the circle. The height, or y value of this point, is so obviously 1 when you don't have to do the extra step of the in-your-head conversion of pi over 2 is actually a quarter of a circle."}, {"video_title": "Pi Is (still) Wrong..mp3", "Sentence": "Let's make a sine wave starting with tau at 0. The height of sine tau is also 0. At tau over 4, we've gone a quarter of the way around the circle. The height, or y value of this point, is so obviously 1 when you don't have to do the extra step of the in-your-head conversion of pi over 2 is actually a quarter of a circle. Tau over 2, half a circle around, back at 0. 3 quarters tau, 3 quarters of the way around, negative 1. A full turn brings us all the way back to 0, and bam, that just makes sense."}, {"video_title": "Pi Is (still) Wrong..mp3", "Sentence": "The height, or y value of this point, is so obviously 1 when you don't have to do the extra step of the in-your-head conversion of pi over 2 is actually a quarter of a circle. Tau over 2, half a circle around, back at 0. 3 quarters tau, 3 quarters of the way around, negative 1. A full turn brings us all the way back to 0, and bam, that just makes sense. Why? Because we don't make circles using a diameter. We make circles using a radius."}, {"video_title": "Pi Is (still) Wrong..mp3", "Sentence": "A full turn brings us all the way back to 0, and bam, that just makes sense. Why? Because we don't make circles using a diameter. We make circles using a radius. The length of the radius is the fundamental thing that determines the circumference of a circle. So why would we define the circle constant as the ratio of the diameter to the circumference? Defining it by the ratio of the radius to the circumference makes much more sense, and that's how you get our lovely tau."}, {"video_title": "Pi Is (still) Wrong..mp3", "Sentence": "We make circles using a radius. The length of the radius is the fundamental thing that determines the circumference of a circle. So why would we define the circle constant as the ratio of the diameter to the circumference? Defining it by the ratio of the radius to the circumference makes much more sense, and that's how you get our lovely tau. There's a boatload of important equations and identities where 2 pi shows up, which could and should be simplified to tau. But why, you say? What about e to the i pi?"}, {"video_title": "Pi Is (still) Wrong..mp3", "Sentence": "Defining it by the ratio of the radius to the circumference makes much more sense, and that's how you get our lovely tau. There's a boatload of important equations and identities where 2 pi shows up, which could and should be simplified to tau. But why, you say? What about e to the i pi? Are you really suggesting we ruin it by making it e to the i tau over 2 equals negative 1? To which I respond, who do you think I am? I would never suggest doing something so ghastly as killing Euler's identity, which by the way comes from Euler's formula, which is e to the i theta equals cosine theta plus i sine theta."}, {"video_title": "Pi Is (still) Wrong..mp3", "Sentence": "What about e to the i pi? Are you really suggesting we ruin it by making it e to the i tau over 2 equals negative 1? To which I respond, who do you think I am? I would never suggest doing something so ghastly as killing Euler's identity, which by the way comes from Euler's formula, which is e to the i theta equals cosine theta plus i sine theta. Let's replace theta with tau. It's easy to remember that the sine, or y value, of a full tau turn of a unit circle is 0, so this is all 0. Cosine of a full turn is the x value, which is 1."}, {"video_title": "Pi Is (still) Wrong..mp3", "Sentence": "I would never suggest doing something so ghastly as killing Euler's identity, which by the way comes from Euler's formula, which is e to the i theta equals cosine theta plus i sine theta. Let's replace theta with tau. It's easy to remember that the sine, or y value, of a full tau turn of a unit circle is 0, so this is all 0. Cosine of a full turn is the x value, which is 1. So check this out. e to the i tau equals 1. What now?"}, {"video_title": "Pi Is (still) Wrong..mp3", "Sentence": "Cosine of a full turn is the x value, which is 1. So check this out. e to the i tau equals 1. What now? If you're still not convinced, I'd recommend reading the tau manifesto by Michael Hartle, who does a pretty thorough job addressing every possible complaint at tauday.com. If you still want to celebrate Pi Day, that's fine. You can have your pi and eat it."}, {"video_title": "Pi Is (still) Wrong..mp3", "Sentence": "What now? If you're still not convinced, I'd recommend reading the tau manifesto by Michael Hartle, who does a pretty thorough job addressing every possible complaint at tauday.com. If you still want to celebrate Pi Day, that's fine. You can have your pi and eat it. But I hope you'll all join me on June 28th, because I'll be making tau and eating 2. I've got pi here and I've got pi there. I'm pi winning."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Let's construct ourselves some right triangles. And I want to be very clear, the way I've defined it so far, this will only work in right triangles. So if you're trying to find the trig functions of angles that aren't part of right triangles, we're going to see that we're going to have to construct right triangles. But let's just focus on the right triangles for now. So let's say that I have a triangle where let's say this length down here is 7. And let's say the length of this side up here, let's say that that is 4. And let's figure out what the hypotenuse over here is going to be."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "But let's just focus on the right triangles for now. So let's say that I have a triangle where let's say this length down here is 7. And let's say the length of this side up here, let's say that that is 4. And let's figure out what the hypotenuse over here is going to be. So we know, let's call the hypotenuse h. We know that h squared is going to be equal to 7 squared plus 4 squared. We know that from the Pythagorean theorem. That the hypotenuse squared is equal to the sum of the squares of the other two sides."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "And let's figure out what the hypotenuse over here is going to be. So we know, let's call the hypotenuse h. We know that h squared is going to be equal to 7 squared plus 4 squared. We know that from the Pythagorean theorem. That the hypotenuse squared is equal to the sum of the squares of the other two sides. h squared is equal to 7 squared plus 4 squared. So this is equal to 49 plus 16. 49 plus 10 is 59."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "That the hypotenuse squared is equal to the sum of the squares of the other two sides. h squared is equal to 7 squared plus 4 squared. So this is equal to 49 plus 16. 49 plus 10 is 59. Plus 6 is 65. So this h squared, let me write h squared, it's a different shade of yellow. So we have h squared is equal to 65."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "49 plus 10 is 59. Plus 6 is 65. So this h squared, let me write h squared, it's a different shade of yellow. So we have h squared is equal to 65. Did I do that right? 49 plus 10 is 59. Plus another 6 is 65."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So we have h squared is equal to 65. Did I do that right? 49 plus 10 is 59. Plus another 6 is 65. Or we could say that h is equal to, if we take the square root of both sides, square root of 65. And we really can't simplify this at all. This is 13, this is the same thing as 13 times 5."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Plus another 6 is 65. Or we could say that h is equal to, if we take the square root of both sides, square root of 65. And we really can't simplify this at all. This is 13, this is the same thing as 13 times 5. Both of those are not perfect squares and they're both primes. You can't simplify this anymore. So this is equal to the square root of 65."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "This is 13, this is the same thing as 13 times 5. Both of those are not perfect squares and they're both primes. You can't simplify this anymore. So this is equal to the square root of 65. Now let's find the trig functions for this angle up here. Let's call that angle up there theta. So whenever you do it, you always want to write down, at least for me it works out to write down, SOH CAH TOA."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So this is equal to the square root of 65. Now let's find the trig functions for this angle up here. Let's call that angle up there theta. So whenever you do it, you always want to write down, at least for me it works out to write down, SOH CAH TOA. I have these vague memories of my trigonometry teacher. Maybe I read it in some book. I don't know about some type of Indian princess named SOH CAH TOA or whatever."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So whenever you do it, you always want to write down, at least for me it works out to write down, SOH CAH TOA. I have these vague memories of my trigonometry teacher. Maybe I read it in some book. I don't know about some type of Indian princess named SOH CAH TOA or whatever. But it's a very useful mnemonic. So we can apply SOH CAH TOA. Let's say we wanted to find the cosine."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "I don't know about some type of Indian princess named SOH CAH TOA or whatever. But it's a very useful mnemonic. So we can apply SOH CAH TOA. Let's say we wanted to find the cosine. We want to find the cosine of our angle. You say SOH CAH TOA. So the CAH tells us what to do with cosine."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Let's say we wanted to find the cosine. We want to find the cosine of our angle. You say SOH CAH TOA. So the CAH tells us what to do with cosine. The CAH part tells us that cosine is adjacent over hypotenuse. Cosine is equal to adjacent over hypotenuse. So let's look over here."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So the CAH tells us what to do with cosine. The CAH part tells us that cosine is adjacent over hypotenuse. Cosine is equal to adjacent over hypotenuse. So let's look over here. To theta, what side is adjacent? Well, we know that the hypotenuse is this side over here. So it can't be that side."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So let's look over here. To theta, what side is adjacent? Well, we know that the hypotenuse is this side over here. So it can't be that side. The only other side that's kind of adjacent to it, that isn't the hypotenuse, is this 4. So the adjacent side over here, that side is literally right next to the angle. It's one of the sides that kind of forms the angle."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So it can't be that side. The only other side that's kind of adjacent to it, that isn't the hypotenuse, is this 4. So the adjacent side over here, that side is literally right next to the angle. It's one of the sides that kind of forms the angle. It's 4 over the hypotenuse. The hypotenuse, we already know, is square root of 65. So it's 4 over the square root of 65."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "It's one of the sides that kind of forms the angle. It's 4 over the hypotenuse. The hypotenuse, we already know, is square root of 65. So it's 4 over the square root of 65. And sometimes people want you to rationalize the denominator, which means they don't like to have an irrational number in the denominator, like the square root of 65. And if you want to rewrite this without an irrational number in the denominator, you can multiply the numerator and the denominator by square root of 65. This clearly will not change the number because we're multiplying it by something over itself."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So it's 4 over the square root of 65. And sometimes people want you to rationalize the denominator, which means they don't like to have an irrational number in the denominator, like the square root of 65. And if you want to rewrite this without an irrational number in the denominator, you can multiply the numerator and the denominator by square root of 65. This clearly will not change the number because we're multiplying it by something over itself. So we're multiplying the number by 1. That won't change the number, but at least it gets rid of the irrational number in the denominator. So the numerator becomes 4 times the square root of 65."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "This clearly will not change the number because we're multiplying it by something over itself. So we're multiplying the number by 1. That won't change the number, but at least it gets rid of the irrational number in the denominator. So the numerator becomes 4 times the square root of 65. And the denominator, square root of 65 times square root of 65, is just going to be 65. We didn't get rid of the irrational number. It's still there, but it's now in the numerator."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So the numerator becomes 4 times the square root of 65. And the denominator, square root of 65 times square root of 65, is just going to be 65. We didn't get rid of the irrational number. It's still there, but it's now in the numerator. Now let's do the other trig functions, or at least the other core trig functions. We'll learn in the future that there's actually a ton of them, but they're all derived from these. So let's think about what the sine of theta is."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "It's still there, but it's now in the numerator. Now let's do the other trig functions, or at least the other core trig functions. We'll learn in the future that there's actually a ton of them, but they're all derived from these. So let's think about what the sine of theta is. Once again, go to SOH CAH TOA. The SOH tells us what to do with sine. Sine is opposite over hypotenuse."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So let's think about what the sine of theta is. Once again, go to SOH CAH TOA. The SOH tells us what to do with sine. Sine is opposite over hypotenuse. So for this angle, what side is opposite? Well, you just go opposite it. What it opens into, it's opposite the 7."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Sine is opposite over hypotenuse. So for this angle, what side is opposite? Well, you just go opposite it. What it opens into, it's opposite the 7. So the opposite side is the 7. This is right here. That is the opposite side."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "What it opens into, it's opposite the 7. So the opposite side is the 7. This is right here. That is the opposite side. And then the hypotenuse, it's opposite over hypotenuse. The hypotenuse is the square root of 65. And once again, if we wanted to rationalize this, we could multiply it times the square root of 65 over the square root of 65."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "That is the opposite side. And then the hypotenuse, it's opposite over hypotenuse. The hypotenuse is the square root of 65. And once again, if we wanted to rationalize this, we could multiply it times the square root of 65 over the square root of 65. In the numerator, we'll get 7 square roots of 65. And in the denominator, we will get just 65 again. Now let's do tangent."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "And once again, if we wanted to rationalize this, we could multiply it times the square root of 65 over the square root of 65. In the numerator, we'll get 7 square roots of 65. And in the denominator, we will get just 65 again. Now let's do tangent. So if I asked you the tangent of theta, once again, go back to SOH CAH TOA. The TOA part tells us what to do with tangent. It tells us that tangent is equal to opposite over adjacent."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Now let's do tangent. So if I asked you the tangent of theta, once again, go back to SOH CAH TOA. The TOA part tells us what to do with tangent. It tells us that tangent is equal to opposite over adjacent. So for this angle, what is opposite? We've already figured it out. It's 7."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "It tells us that tangent is equal to opposite over adjacent. So for this angle, what is opposite? We've already figured it out. It's 7. It opens into the 7. It's opposite the 7. So it's 7 over what side is adjacent?"}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "It's 7. It opens into the 7. It's opposite the 7. So it's 7 over what side is adjacent? Well, this 4 is adjacent. So the adjacent side is 4. So it's 7 over 4."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So it's 7 over what side is adjacent? Well, this 4 is adjacent. So the adjacent side is 4. So it's 7 over 4. And we're done. We figured out all of the trig ratios for theta. Let's do another one."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So it's 7 over 4. And we're done. We figured out all of the trig ratios for theta. Let's do another one. And I'll make it a little bit concrete. Because right now we've been saying, oh, what's tangent of x? What's tangent of theta?"}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Let's do another one. And I'll make it a little bit concrete. Because right now we've been saying, oh, what's tangent of x? What's tangent of theta? Let's make it a little bit more concrete. Let's say, let me draw another right triangle. Everything we're dealing with, these are going to be right triangles."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "What's tangent of theta? Let's make it a little bit more concrete. Let's say, let me draw another right triangle. Everything we're dealing with, these are going to be right triangles. Let's say the hypotenuse has length 4. Let's say that this side over here has length 2. And let's say that this length over here is going to be 2 times the square root of 3."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Everything we're dealing with, these are going to be right triangles. Let's say the hypotenuse has length 4. Let's say that this side over here has length 2. And let's say that this length over here is going to be 2 times the square root of 3. We can verify that this works. If you have this side squared, so you have, let me write it down, 2 times the square root of 3 squared plus 2 squared is equal to what? This is 2."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "And let's say that this length over here is going to be 2 times the square root of 3. We can verify that this works. If you have this side squared, so you have, let me write it down, 2 times the square root of 3 squared plus 2 squared is equal to what? This is 2. This is going to be 4 times 3. 4 times 3 plus 4. And this is going to be equal to 12 plus 4 is equal to 16."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "This is 2. This is going to be 4 times 3. 4 times 3 plus 4. And this is going to be equal to 12 plus 4 is equal to 16. And 16 is indeed 4 squared. So this does equal 4 squared. It does equal 4 squared."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "And this is going to be equal to 12 plus 4 is equal to 16. And 16 is indeed 4 squared. So this does equal 4 squared. It does equal 4 squared. It satisfies the Pythagorean theorem. And if you remember some of your work from 30-60-90 triangles that you might have learned in geometry, you might recognize that this is a 30-60-90 triangle. This right here is our right angle."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "It does equal 4 squared. It satisfies the Pythagorean theorem. And if you remember some of your work from 30-60-90 triangles that you might have learned in geometry, you might recognize that this is a 30-60-90 triangle. This right here is our right angle. I should have drawn it from the get-go to show that this is a right triangle. This angle right over here is our 30-degree angle. And then this angle up here is a 60-degree angle."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "This right here is our right angle. I should have drawn it from the get-go to show that this is a right triangle. This angle right over here is our 30-degree angle. And then this angle up here is a 60-degree angle. And it's a 30-60-90 because the side opposite the 30 degrees is half the hypotenuse. And then the side opposite the 60 degrees is the square root of 3 times the other side that's not the hypotenuse. So with that said, this isn't supposed to be a review of 30-60-90 triangles, although I just did it."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "And then this angle up here is a 60-degree angle. And it's a 30-60-90 because the side opposite the 30 degrees is half the hypotenuse. And then the side opposite the 60 degrees is the square root of 3 times the other side that's not the hypotenuse. So with that said, this isn't supposed to be a review of 30-60-90 triangles, although I just did it. Let's actually find the trig ratios for the different angles. So if I were to ask you, or if anyone were to ask you, what is the sine of 30 degrees? And remember, 30 degrees is one of the angles in this triangle, but it would apply whenever you have a 30-degree angle and you're dealing with a right triangle."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So with that said, this isn't supposed to be a review of 30-60-90 triangles, although I just did it. Let's actually find the trig ratios for the different angles. So if I were to ask you, or if anyone were to ask you, what is the sine of 30 degrees? And remember, 30 degrees is one of the angles in this triangle, but it would apply whenever you have a 30-degree angle and you're dealing with a right triangle. We'll have broader definitions in the future. But if you say sine of 30 degrees, hey, this angle right over here is 30 degrees, so I can use this right triangle. And we just have to remember SOH CAH TOA."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "And remember, 30 degrees is one of the angles in this triangle, but it would apply whenever you have a 30-degree angle and you're dealing with a right triangle. We'll have broader definitions in the future. But if you say sine of 30 degrees, hey, this angle right over here is 30 degrees, so I can use this right triangle. And we just have to remember SOH CAH TOA. Let me rewrite it. SOH CAH TOA. SOH tells us what to do with sine."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "And we just have to remember SOH CAH TOA. Let me rewrite it. SOH CAH TOA. SOH tells us what to do with sine. Sine is opposite over hypotenuse. Sine of 30 degrees is the opposite side. That is the opposite side, which is 2, over the hypotenuse."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "SOH tells us what to do with sine. Sine is opposite over hypotenuse. Sine of 30 degrees is the opposite side. That is the opposite side, which is 2, over the hypotenuse. The hypotenuse here is 4. It is 2 fourths, which is the same thing as 1 half. Sine of 30 degrees, you'll see, is always going to be equal to 1 half."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "That is the opposite side, which is 2, over the hypotenuse. The hypotenuse here is 4. It is 2 fourths, which is the same thing as 1 half. Sine of 30 degrees, you'll see, is always going to be equal to 1 half. Now, what is the cosine? What is the cosine of 30 degrees? Once again, go back to SOH CAH TOA."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Sine of 30 degrees, you'll see, is always going to be equal to 1 half. Now, what is the cosine? What is the cosine of 30 degrees? Once again, go back to SOH CAH TOA. The CAH tells us what to do with cosine. Cosine is adjacent over hypotenuse. So if we're looking at the 30-degree angle, it's the adjacent."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Once again, go back to SOH CAH TOA. The CAH tells us what to do with cosine. Cosine is adjacent over hypotenuse. So if we're looking at the 30-degree angle, it's the adjacent. This right over here is adjacent. It's right next to it. It's not the hypotenuse."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So if we're looking at the 30-degree angle, it's the adjacent. This right over here is adjacent. It's right next to it. It's not the hypotenuse. It's the adjacent over the hypotenuse. So it's 2 square roots of 3 adjacent over the hypotenuse, over 4. Or if we simplify that, we divide the numerator and denominator by 2, it's the square root of 3 over 2."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "It's not the hypotenuse. It's the adjacent over the hypotenuse. So it's 2 square roots of 3 adjacent over the hypotenuse, over 4. Or if we simplify that, we divide the numerator and denominator by 2, it's the square root of 3 over 2. Finally, let's do the tangent. The tangent of 30 degrees. We go back to SOH CAH TOA."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Or if we simplify that, we divide the numerator and denominator by 2, it's the square root of 3 over 2. Finally, let's do the tangent. The tangent of 30 degrees. We go back to SOH CAH TOA. TOA tells us what to do with tangent. It's opposite over adjacent. We go to the 30-degree angle, because that's what we care about."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "We go back to SOH CAH TOA. TOA tells us what to do with tangent. It's opposite over adjacent. We go to the 30-degree angle, because that's what we care about. Tangent of 30. Opposite is 2, and the adjacent is 2 square roots of 3. It's right next to it."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "We go to the 30-degree angle, because that's what we care about. Tangent of 30. Opposite is 2, and the adjacent is 2 square roots of 3. It's right next to it. It's adjacent to it. Adjacent means next to. So 2 square roots of 3."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "It's right next to it. It's adjacent to it. Adjacent means next to. So 2 square roots of 3. So this is equal to, the 2's cancel out, 1 over the square root of 3. Or we can multiply the numerator and the denominator by the square root of 3. So we have square root of 3 over square root of 3."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So 2 square roots of 3. So this is equal to, the 2's cancel out, 1 over the square root of 3. Or we can multiply the numerator and the denominator by the square root of 3. So we have square root of 3 over square root of 3. This is going to be equal to, the numerator is square root of 3, and then the denominator right over here is just going to be 3. So we've rationalized it. Square root of 3 over 3."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So we have square root of 3 over square root of 3. This is going to be equal to, the numerator is square root of 3, and then the denominator right over here is just going to be 3. So we've rationalized it. Square root of 3 over 3. Fair enough. Now let's use the same triangle to figure out the trig ratios for the 60 degrees, since we've already drawn it. So what is the sine of 60 degrees?"}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Square root of 3 over 3. Fair enough. Now let's use the same triangle to figure out the trig ratios for the 60 degrees, since we've already drawn it. So what is the sine of 60 degrees? I think you're hopefully getting the hang of it now. Sine is opposite over adjacent. So, from the Sohcahtoa, for the 60 degree angle, what side is opposite?"}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So what is the sine of 60 degrees? I think you're hopefully getting the hang of it now. Sine is opposite over adjacent. So, from the Sohcahtoa, for the 60 degree angle, what side is opposite? Well, it opens out into the 2 square roots of 3. So the opposite side is 2 square roots of 3. And for the 60 degree angle, the adjacent, or sorry, it's opposite over hypotenuse."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So, from the Sohcahtoa, for the 60 degree angle, what side is opposite? Well, it opens out into the 2 square roots of 3. So the opposite side is 2 square roots of 3. And for the 60 degree angle, the adjacent, or sorry, it's opposite over hypotenuse. Don't want to confuse you. So it's opposite over hypotenuse. So it's 2 square roots of 3 over 4."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "And for the 60 degree angle, the adjacent, or sorry, it's opposite over hypotenuse. Don't want to confuse you. So it's opposite over hypotenuse. So it's 2 square roots of 3 over 4. 4 is the hypotenuse. So it is equal to, this simplifies to square root of 3 over 2. What is the cosine of 60 degrees?"}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So it's 2 square roots of 3 over 4. 4 is the hypotenuse. So it is equal to, this simplifies to square root of 3 over 2. What is the cosine of 60 degrees? So remember, Sohcahtoa, cosine is adjacent over hypotenuse. Adjacent is the 2 side. It's right next to the 60 degree angle."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "What is the cosine of 60 degrees? So remember, Sohcahtoa, cosine is adjacent over hypotenuse. Adjacent is the 2 side. It's right next to the 60 degree angle. So it's 2 over the hypotenuse, which is 4. So this is equal to 1 half. And then finally, what is the tangent?"}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "It's right next to the 60 degree angle. So it's 2 over the hypotenuse, which is 4. So this is equal to 1 half. And then finally, what is the tangent? What is the tangent of 60 degrees? Well, tangent, Sohcahtoa, tangent is opposite over adjacent. Opposite the 60 degrees is 2 square roots of 3."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "And then finally, what is the tangent? What is the tangent of 60 degrees? Well, tangent, Sohcahtoa, tangent is opposite over adjacent. Opposite the 60 degrees is 2 square roots of 3. And adjacent to that is 2. Adjacent to 60 degrees is 2. So it's opposite over adjacent."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Opposite the 60 degrees is 2 square roots of 3. And adjacent to that is 2. Adjacent to 60 degrees is 2. So it's opposite over adjacent. 2 square roots of 3 over 2, which is just equal to the square root of 3. And I just want to, you know, look how these are related. The sine of 30 degrees is the same thing as the cosine of 60 degrees."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So it's opposite over adjacent. 2 square roots of 3 over 2, which is just equal to the square root of 3. And I just want to, you know, look how these are related. The sine of 30 degrees is the same thing as the cosine of 60 degrees. The cosine of 30 degrees is the same thing as the sine of 60 degrees. And then these guys are the inverse of each other. I think if you think a little bit about this triangle, it will start to make sense why."}, {"video_title": "Determining the equation of a trig function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So we have this clearly periodic function. So immediately you might say, well, this is either going to be a sine function or cosine function. But its midline and its amplitude are not just the plain vanilla sine or cosine function. And we can see that right over here. 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."}, {"video_title": "Determining the equation of a trig function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And we can see that right over here. 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."}, {"video_title": "Determining the equation of a trig function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "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. 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."}, {"video_title": "Determining the equation of a trig function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "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. 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."}, {"video_title": "Determining the equation of a trig function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "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. And it can also go 3 below the midline at the minimum point. So this thing clearly has an amplitude of 3."}, {"video_title": "Determining the equation of a trig function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "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. 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."}, {"video_title": "Determining the equation of a trig function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "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. Cosine may be some coefficient times x plus the midline. The midline we already figured out was minus 2 or negative 2."}, {"video_title": "Determining the equation of a trig function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "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. Sine of kx minus 2, plus the midline. So minus 2."}, {"video_title": "Determining the equation of a trig function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "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? 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."}, {"video_title": "Determining the equation of a trig function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "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. While sine of 0, so if x is 0, k times 0 is going to be 0. Sine of 0 is 0."}, {"video_title": "Determining the equation of a trig function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "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? 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."}, {"video_title": "Determining the equation of a trig function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "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. When x equals 0 here, this stuff doesn't evaluate to 0. So we can rule out this one right over there."}, {"video_title": "Determining the equation of a trig function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "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. 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."}, {"video_title": "Determining the equation of a trig function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "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. So our period is 8. So what coefficient could we have here to make the period of this thing be equal to 8?"}, {"video_title": "Determining the equation of a trig function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "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. 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."}, {"video_title": "Determining the equation of a trig function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "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. So what would be the period of sine of kx? Well, now your x, your input, is increasing k times faster."}, {"video_title": "Determining the equation of a trig function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "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. 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."}, {"video_title": "Determining the equation of a trig function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "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. 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."}, {"video_title": "Determining the equation of a trig function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "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. 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."}, {"video_title": "Determining the equation of a trig function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "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. Multiply both sides by 2 pi. And we get k is equal to pi over 4."}, {"video_title": "Proof of the tangent angle sum and difference identities.mp3", "Sentence": "In this video, I'm going to assume that you already know a few things, and we've covered this, we've proved this in other videos, that sine of x plus y is equal to sine of x cosine y, plus, and then you swap the cosines and the sines, cosine of x sine y, and that cosine of x plus y is equal to cosine x cosine y minus sine x sine y. Once again, we've proven these in other videos. And then there's some other properties we know of cosine and sine that we have looked at in other videos. Cosine of negative x is equal to cosine of x, and that sine of negative x is equal to negative sine of x. And that, of course, the tangent of something is defined as sine over cosine of that something. Now, with that out of the way, I wanna come up with a formula for tangent of x plus y expressed just in terms of tangent of x and tangent of y. You can view it as the analog for what we did up here for sine and cosine."}, {"video_title": "Proof of the tangent angle sum and difference identities.mp3", "Sentence": "Cosine of negative x is equal to cosine of x, and that sine of negative x is equal to negative sine of x. And that, of course, the tangent of something is defined as sine over cosine of that something. Now, with that out of the way, I wanna come up with a formula for tangent of x plus y expressed just in terms of tangent of x and tangent of y. You can view it as the analog for what we did up here for sine and cosine. Well, the immediate thing that you might recognize is that tangent of x plus y, based on the definition of tangent, is the same thing as sine of x plus y over cosine of x plus y. And what's that going to be equal to? Well, we know that sine of x plus y can be expressed this way, so let me write that down."}, {"video_title": "Proof of the tangent angle sum and difference identities.mp3", "Sentence": "You can view it as the analog for what we did up here for sine and cosine. Well, the immediate thing that you might recognize is that tangent of x plus y, based on the definition of tangent, is the same thing as sine of x plus y over cosine of x plus y. And what's that going to be equal to? Well, we know that sine of x plus y can be expressed this way, so let me write that down. So that's going to be sine of x cosine y plus cosine of x sine of y. And then, and actually, so that we can save a little bit of writing, I'm gonna awkwardly write, make the line down here because we're going to put something here in a second, but I think you'll get the idea. So that's going to be that over cosine of x plus y, which is this expression."}, {"video_title": "Proof of the tangent angle sum and difference identities.mp3", "Sentence": "Well, we know that sine of x plus y can be expressed this way, so let me write that down. So that's going to be sine of x cosine y plus cosine of x sine of y. And then, and actually, so that we can save a little bit of writing, I'm gonna awkwardly write, make the line down here because we're going to put something here in a second, but I think you'll get the idea. So that's going to be that over cosine of x plus y, which is this expression. When you just express it in terms of cosines of x and cosines of y and sines of x and sines of y. So let me write it here. So you're going to have cosine of x cosine y minus sine of x and then sine y."}, {"video_title": "Proof of the tangent angle sum and difference identities.mp3", "Sentence": "So that's going to be that over cosine of x plus y, which is this expression. When you just express it in terms of cosines of x and cosines of y and sines of x and sines of y. So let me write it here. So you're going to have cosine of x cosine y minus sine of x and then sine y. Now, we wanna express everything in terms of tangents of x's and y's. And so it might make sense here to say, all right, well, we know tangent is sine over cosine. So what if we were to divide both the numerator and the denominator by some expression that can start to make the numerator and denominator express in terms of tangents?"}, {"video_title": "Proof of the tangent angle sum and difference identities.mp3", "Sentence": "So you're going to have cosine of x cosine y minus sine of x and then sine y. Now, we wanna express everything in terms of tangents of x's and y's. And so it might make sense here to say, all right, well, we know tangent is sine over cosine. So what if we were to divide both the numerator and the denominator by some expression that can start to make the numerator and denominator express in terms of tangents? And I will cut a little bit to the chase here. So in the numerator, what I can do is, and I'm gonna do this just in the numerator and then I'm gonna do it in the denominator as well. I'm gonna divide the numerator by cosine of x cosine y. Cosine y."}, {"video_title": "Proof of the tangent angle sum and difference identities.mp3", "Sentence": "So what if we were to divide both the numerator and the denominator by some expression that can start to make the numerator and denominator express in terms of tangents? And I will cut a little bit to the chase here. So in the numerator, what I can do is, and I'm gonna do this just in the numerator and then I'm gonna do it in the denominator as well. I'm gonna divide the numerator by cosine of x cosine y. Cosine y. And of course, I can't just divide the numerator by cosine of x cosine y. That would change the value of this rational expression. I have to do that to the denominator as well."}, {"video_title": "Proof of the tangent angle sum and difference identities.mp3", "Sentence": "I'm gonna divide the numerator by cosine of x cosine y. Cosine y. And of course, I can't just divide the numerator by cosine of x cosine y. That would change the value of this rational expression. I have to do that to the denominator as well. So I know this is a very complex looking fraction here, but it's going to simplify in a second. So I'm also gonna divide the denominator by cosine of x cosine of y. And now let's see if we can simplify this in certain ways."}, {"video_title": "Proof of the tangent angle sum and difference identities.mp3", "Sentence": "I have to do that to the denominator as well. So I know this is a very complex looking fraction here, but it's going to simplify in a second. So I'm also gonna divide the denominator by cosine of x cosine of y. And now let's see if we can simplify this in certain ways. In the numerator, we can see that this cosine y cancels with this cosine y. And so that first term becomes, let's write it in another color here. So this sine of x over cosine of x."}, {"video_title": "Proof of the tangent angle sum and difference identities.mp3", "Sentence": "And now let's see if we can simplify this in certain ways. In the numerator, we can see that this cosine y cancels with this cosine y. And so that first term becomes, let's write it in another color here. So this sine of x over cosine of x. And so the numerator, I can say this is going to be equal to sine of x over cosine of x is tangent of x. And then this second term here, we can see that this cosine of x cancels with this cosine of x. And so we're left with sine of y over cosine of y, which is of course, tangent of y."}, {"video_title": "Proof of the tangent angle sum and difference identities.mp3", "Sentence": "So this sine of x over cosine of x. And so the numerator, I can say this is going to be equal to sine of x over cosine of x is tangent of x. And then this second term here, we can see that this cosine of x cancels with this cosine of x. And so we're left with sine of y over cosine of y, which is of course, tangent of y. So plus tangent of y. And then all of that is going to be over. Now we can look at the denominator."}, {"video_title": "Proof of the tangent angle sum and difference identities.mp3", "Sentence": "And so we're left with sine of y over cosine of y, which is of course, tangent of y. So plus tangent of y. And then all of that is going to be over. Now we can look at the denominator. So this first term here, we can see the cosine of x cancels with the cosine of x. And the cosine of y cancels out with the cosine of y. So you could view this first term here when you divide by this cosine of x cosine y, it just becomes one."}, {"video_title": "Proof of the tangent angle sum and difference identities.mp3", "Sentence": "Now we can look at the denominator. So this first term here, we can see the cosine of x cancels with the cosine of x. And the cosine of y cancels out with the cosine of y. So you could view this first term here when you divide by this cosine of x cosine y, it just becomes one. And then we're going to have the minus. And now this second term is interesting. We have sine of x over cosine of x, sine of y over cosine of y."}, {"video_title": "Proof of the tangent angle sum and difference identities.mp3", "Sentence": "So you could view this first term here when you divide by this cosine of x cosine y, it just becomes one. And then we're going to have the minus. And now this second term is interesting. We have sine of x over cosine of x, sine of y over cosine of y. So sine of x over cosine of x, that over there is tangent of x. And then sine of y over cosine of y, that's tangent of y. So this is going to be tangent of x times tangent of y."}, {"video_title": "Proof of the tangent angle sum and difference identities.mp3", "Sentence": "We have sine of x over cosine of x, sine of y over cosine of y. So sine of x over cosine of x, that over there is tangent of x. And then sine of y over cosine of y, that's tangent of y. So this is going to be tangent of x times tangent of y. And just like that, we have come up with an expression for tangent of x plus y that just deals with tangent of x's and tangent of y's. Now the next question you might say, well, all right, that's great for tangent of x plus y, but what about tangent of x minus y? Well, here we just have to recognize a little bit of what we've seen before."}, {"video_title": "Proof of the tangent angle sum and difference identities.mp3", "Sentence": "So this is going to be tangent of x times tangent of y. And just like that, we have come up with an expression for tangent of x plus y that just deals with tangent of x's and tangent of y's. Now the next question you might say, well, all right, that's great for tangent of x plus y, but what about tangent of x minus y? Well, here we just have to recognize a little bit of what we've seen before. Let me write it over here. Tangent of negative x is equal to sine of negative x of negative x over cosine of negative x. And what's that going to be equal to?"}, {"video_title": "Proof of the tangent angle sum and difference identities.mp3", "Sentence": "Well, here we just have to recognize a little bit of what we've seen before. Let me write it over here. Tangent of negative x is equal to sine of negative x of negative x over cosine of negative x. And what's that going to be equal to? And I know I'm running out of space. This is going to be equal to sine of negative x is the same thing as negative sine of x, negative sine of x. And then cosine of negative x is just cosine of x."}, {"video_title": "Proof of the tangent angle sum and difference identities.mp3", "Sentence": "And what's that going to be equal to? And I know I'm running out of space. This is going to be equal to sine of negative x is the same thing as negative sine of x, negative sine of x. And then cosine of negative x is just cosine of x. Well, this is just the negative of the tangent of x. So this is negative tangent of x. And the reason why that is useful is I can rewrite this as being, let me write it here."}, {"video_title": "Proof of the tangent angle sum and difference identities.mp3", "Sentence": "And then cosine of negative x is just cosine of x. Well, this is just the negative of the tangent of x. So this is negative tangent of x. And the reason why that is useful is I can rewrite this as being, let me write it here. This is the same thing as tangent of x plus negative y. So everywhere we saw a y here, we can replace it with a negative y. So this is going to be equal to tangent of x plus the tangent of negative y."}, {"video_title": "Proof of the tangent angle sum and difference identities.mp3", "Sentence": "And the reason why that is useful is I can rewrite this as being, let me write it here. This is the same thing as tangent of x plus negative y. So everywhere we saw a y here, we can replace it with a negative y. So this is going to be equal to tangent of x plus the tangent of negative y. All of that over one minus the tangent of x times the tangent of negative y. Well, we know the tangent of negative y is the same thing as the negative tangent of y. And we know that over here as well."}, {"video_title": "Proof of the tangent angle sum and difference identities.mp3", "Sentence": "So this is going to be equal to tangent of x plus the tangent of negative y. All of that over one minus the tangent of x times the tangent of negative y. Well, we know the tangent of negative y is the same thing as the negative tangent of y. And we know that over here as well. So this could just be, we could write the tangent of y here, and then the negative would turn this into a plus. And so just to write everything neatly, we know that also the tangent of x minus y can be rewritten as tangent of x minus tangent of y. All of that over one plus tangent of x and tangent of y."}, {"video_title": "Trig challenge problem area of a triangle Khan Academy.mp3", "Sentence": "Maybe we'll have to use some coordinates. So let's draw the right angle at the origin. So let's say that this is C right over here. This is C. Let's see, AC is equal to 7. So A we'll put over here. This distance over here is going to be 7. And then we have the hypotenuse of our triangle."}, {"video_title": "Trig challenge problem area of a triangle Khan Academy.mp3", "Sentence": "This is C. Let's see, AC is equal to 7. So A we'll put over here. This distance over here is going to be 7. And then we have the hypotenuse of our triangle. And then this can be B. And this distance over here, they tell us that BC is equal to 24. Alright, now point M is the midpoint of AB."}, {"video_title": "Trig challenge problem area of a triangle Khan Academy.mp3", "Sentence": "And then we have the hypotenuse of our triangle. And then this can be B. And this distance over here, they tell us that BC is equal to 24. Alright, now point M is the midpoint of AB. So point M is right over here. Let me do that in a different color. Point M is the midpoint."}, {"video_title": "Trig challenge problem area of a triangle Khan Academy.mp3", "Sentence": "Alright, now point M is the midpoint of AB. So point M is right over here. Let me do that in a different color. Point M is the midpoint. It's the midpoint of AB. So this distance is equal to that distance. And D is on the same side of AB."}, {"video_title": "Trig challenge problem area of a triangle Khan Academy.mp3", "Sentence": "Point M is the midpoint. It's the midpoint of AB. So this distance is equal to that distance. And D is on the same side of AB. So D is on the same side of AB as C. So C is on this side, I guess you'd call it the bottom left-hand side of AB. So that AD is equal to BD is equal to 15. So D is going to be some place over here."}, {"video_title": "Trig challenge problem area of a triangle Khan Academy.mp3", "Sentence": "And D is on the same side of AB. So D is on the same side of AB as C. So C is on this side, I guess you'd call it the bottom left-hand side of AB. So that AD is equal to BD is equal to 15. So D is going to be some place over here. It's going to be equal distance between A and B. Obviously all of the points equal distance between A and B are going to sit on a line that looks something like that. This is the midpoint of AB."}, {"video_title": "Trig challenge problem area of a triangle Khan Academy.mp3", "Sentence": "So D is going to be some place over here. It's going to be equal distance between A and B. Obviously all of the points equal distance between A and B are going to sit on a line that looks something like that. This is the midpoint of AB. So D is going to sit right over here and it's 15 away from both A and B. So it would look something like that. That distance over there is 15."}, {"video_title": "Trig challenge problem area of a triangle Khan Academy.mp3", "Sentence": "This is the midpoint of AB. So D is going to sit right over here and it's 15 away from both A and B. So it would look something like that. That distance over there is 15. And then this distance over here is also going to be 15. Given that the area of triangle CDM may be expressed as M times the square root of N over P, where M, N, and P are positive integers, and M and P are relatively prime, so that just means you can't simplify it, and N is not divisible by the square of any prime, so you've simplified the radical as much as possible. Find M plus N plus P. So we essentially need to find the area of this green triangle of CDM over here."}, {"video_title": "Trig challenge problem area of a triangle Khan Academy.mp3", "Sentence": "That distance over there is 15. And then this distance over here is also going to be 15. Given that the area of triangle CDM may be expressed as M times the square root of N over P, where M, N, and P are positive integers, and M and P are relatively prime, so that just means you can't simplify it, and N is not divisible by the square of any prime, so you've simplified the radical as much as possible. Find M plus N plus P. So we essentially need to find the area of this green triangle of CDM over here. So let's see what we can do to figure it out. We could try some of the coordinates for some of these points. So this point right over here, A, its x value is going to be 7."}, {"video_title": "Trig challenge problem area of a triangle Khan Academy.mp3", "Sentence": "Find M plus N plus P. So we essentially need to find the area of this green triangle of CDM over here. So let's see what we can do to figure it out. We could try some of the coordinates for some of these points. So this point right over here, A, its x value is going to be 7. I could draw the coordinate just so you see what I'm doing. That could be the x-axis, and then this side is on the y-axis. So the coordinate for A would be 7, 0."}, {"video_title": "Trig challenge problem area of a triangle Khan Academy.mp3", "Sentence": "So this point right over here, A, its x value is going to be 7. I could draw the coordinate just so you see what I'm doing. That could be the x-axis, and then this side is on the y-axis. So the coordinate for A would be 7, 0. The coordinate for C would be 0, 0. And the coordinate for B would be 0, 24. So the coordinate for M would just be the average of B and A."}, {"video_title": "Trig challenge problem area of a triangle Khan Academy.mp3", "Sentence": "So the coordinate for A would be 7, 0. The coordinate for C would be 0, 0. And the coordinate for B would be 0, 24. So the coordinate for M would just be the average of B and A. So the coordinate for M, the average of 0 and 7 is 7 halves. And the coordinate for the y-coordinate, the average of 24 and 0 is 12. That's fair enough."}, {"video_title": "Trig challenge problem area of a triangle Khan Academy.mp3", "Sentence": "So the coordinate for M would just be the average of B and A. So the coordinate for M, the average of 0 and 7 is 7 halves. And the coordinate for the y-coordinate, the average of 24 and 0 is 12. That's fair enough. Now let's see what we can figure out about the sides. So we know this is a right triangle, so our gut reaction is always to use the Pythagorean theorem. We know this side and that side."}, {"video_title": "Trig challenge problem area of a triangle Khan Academy.mp3", "Sentence": "That's fair enough. Now let's see what we can figure out about the sides. So we know this is a right triangle, so our gut reaction is always to use the Pythagorean theorem. We know this side and that side. So if we wanted to figure out AB, we could just say that 24 squared plus 7 squared is equal to AB squared. And 24 squared is 576 plus 49 is equal to AB squared. And let's see, 576 plus 49, if it was plus 50, it would get us to 626, but it's one less than that, so it's 625."}, {"video_title": "Trig challenge problem area of a triangle Khan Academy.mp3", "Sentence": "We know this side and that side. So if we wanted to figure out AB, we could just say that 24 squared plus 7 squared is equal to AB squared. And 24 squared is 576 plus 49 is equal to AB squared. And let's see, 576 plus 49, if it was plus 50, it would get us to 626, but it's one less than that, so it's 625. So 625 is equal to AB squared. So AB is going to equal 25. So the distance of this big hypotenuse here is 25."}, {"video_title": "Trig challenge problem area of a triangle Khan Academy.mp3", "Sentence": "And let's see, 576 plus 49, if it was plus 50, it would get us to 626, but it's one less than that, so it's 625. So 625 is equal to AB squared. So AB is going to equal 25. So the distance of this big hypotenuse here is 25. Or half of the distance, this is going to be 25 over 2 from V to M. From M to A is also going to be 25 over 2. Now the other thing we know is that M right over here, the triangle CMA is an isosceles triangle. How do we know that?"}, {"video_title": "Trig challenge problem area of a triangle Khan Academy.mp3", "Sentence": "So the distance of this big hypotenuse here is 25. Or half of the distance, this is going to be 25 over 2 from V to M. From M to A is also going to be 25 over 2. Now the other thing we know is that M right over here, the triangle CMA is an isosceles triangle. How do we know that? Well M, if you look at its X coordinate, its X coordinate is directly in between the X coordinates for C and A. 7 halves, it's the average. This is 7, this is 0, its M coordinate is right over there."}, {"video_title": "Trig challenge problem area of a triangle Khan Academy.mp3", "Sentence": "How do we know that? Well M, if you look at its X coordinate, its X coordinate is directly in between the X coordinates for C and A. 7 halves, it's the average. This is 7, this is 0, its M coordinate is right over there. It's directly above the midpoint of this base over here. So this is going to be an isosceles triangle, it's symmetric, you could flip the triangle over. So this length, and this seems useful because this is kind of the base of the area, the base of the triangle we care about."}, {"video_title": "Trig challenge problem area of a triangle Khan Academy.mp3", "Sentence": "This is 7, this is 0, its M coordinate is right over there. It's directly above the midpoint of this base over here. So this is going to be an isosceles triangle, it's symmetric, you could flip the triangle over. So this length, and this seems useful because this is kind of the base of the area, the base of the triangle we care about. We can kind of view this as the base of CDM. This is also going to be 25 over 2. It's an isosceles triangle."}, {"video_title": "Trig challenge problem area of a triangle Khan Academy.mp3", "Sentence": "So this length, and this seems useful because this is kind of the base of the area, the base of the triangle we care about. We can kind of view this as the base of CDM. This is also going to be 25 over 2. It's an isosceles triangle. This is going to be the same as that because we're symmetric around this right over here. So let's see, we know one side of this triangle. Let's see if we can figure out this side over here."}, {"video_title": "Trig challenge problem area of a triangle Khan Academy.mp3", "Sentence": "It's an isosceles triangle. This is going to be the same as that because we're symmetric around this right over here. So let's see, we know one side of this triangle. Let's see if we can figure out this side over here. It seems pretty straightforward. This is going to be a right triangle right over here because this line from D to M is going to be perpendicular to AB. All of the points that are equidistant between A and B are going to be on a line that is perpendicular to AB."}, {"video_title": "Trig challenge problem area of a triangle Khan Academy.mp3", "Sentence": "Let's see if we can figure out this side over here. It seems pretty straightforward. This is going to be a right triangle right over here because this line from D to M is going to be perpendicular to AB. All of the points that are equidistant between A and B are going to be on a line that is perpendicular to AB. So this is going to be a right triangle. So we can figure out DM using the Pythagorean theorem again. We get 25 over 2 squared plus DM squared is going to be equal to 15 squared."}, {"video_title": "Trig challenge problem area of a triangle Khan Academy.mp3", "Sentence": "All of the points that are equidistant between A and B are going to be on a line that is perpendicular to AB. So this is going to be a right triangle. So we can figure out DM using the Pythagorean theorem again. We get 25 over 2 squared plus DM squared is going to be equal to 15 squared. It's going to equal to the hypotenuse of this triangle. So it's going to equal 225. And so what do we get?"}, {"video_title": "Trig challenge problem area of a triangle Khan Academy.mp3", "Sentence": "We get 25 over 2 squared plus DM squared is going to be equal to 15 squared. It's going to equal to the hypotenuse of this triangle. So it's going to equal 225. And so what do we get? We get DM squared is equal to 225 minus 625 over 4. So let's put this over 4. So 225 with 4 as the denominator is the same thing as 900 over 4."}, {"video_title": "Trig challenge problem area of a triangle Khan Academy.mp3", "Sentence": "And so what do we get? We get DM squared is equal to 225 minus 625 over 4. So let's put this over 4. So 225 with 4 as the denominator is the same thing as 900 over 4. In the last video, I mistakenly said 900 over 4 was 125. Obviously, boneheaded mistake. But the numbers show up again here."}, {"video_title": "Trig challenge problem area of a triangle Khan Academy.mp3", "Sentence": "So 225 with 4 as the denominator is the same thing as 900 over 4. In the last video, I mistakenly said 900 over 4 was 125. Obviously, boneheaded mistake. But the numbers show up again here. So 225 is obviously 900 over 4. And then we're going to subtract from that 625 over 4. And this is equal to, let's see, in our numerator, let's see, we have 900 minus 625 would be 300 minus 20."}, {"video_title": "Trig challenge problem area of a triangle Khan Academy.mp3", "Sentence": "But the numbers show up again here. So 225 is obviously 900 over 4. And then we're going to subtract from that 625 over 4. And this is equal to, let's see, in our numerator, let's see, we have 900 minus 625 would be 300 minus 20. It's going to be 200, 275 over 4. And so DM is going to be equal to the square root of this. So it's equal to the square root of 275 over 4."}, {"video_title": "Trig challenge problem area of a triangle Khan Academy.mp3", "Sentence": "And this is equal to, let's see, in our numerator, let's see, we have 900 minus 625 would be 300 minus 20. It's going to be 200, 275 over 4. And so DM is going to be equal to the square root of this. So it's equal to the square root of 275 over 4. 275 is 25 times 11, because 25 times 12 would be 300. So 25 times 11 over 4. So this is going to be equal to 5 square roots of 11 over 2."}, {"video_title": "Trig challenge problem area of a triangle Khan Academy.mp3", "Sentence": "So it's equal to the square root of 275 over 4. 275 is 25 times 11, because 25 times 12 would be 300. So 25 times 11 over 4. So this is going to be equal to 5 square roots of 11 over 2. So DM is 5 square roots of 11 over 2. Now, if we could just figure out, so what we need to do here is figure out the height of this triangle right over here. If we could figure out the height of that triangle, we're pretty much done."}, {"video_title": "Trig challenge problem area of a triangle Khan Academy.mp3", "Sentence": "So this is going to be equal to 5 square roots of 11 over 2. So DM is 5 square roots of 11 over 2. Now, if we could just figure out, so what we need to do here is figure out the height of this triangle right over here. If we could figure out the height of that triangle, we're pretty much done. 1 half the base times the height. But we don't know, let's see, we could figure that if we know the law of cosines, we could do something over there. If we know the sine of this angle right over here, if we know the sine of that angle, the sine of that angle is going to be this height over the side we just figured out."}, {"video_title": "Trig challenge problem area of a triangle Khan Academy.mp3", "Sentence": "If we could figure out the height of that triangle, we're pretty much done. 1 half the base times the height. But we don't know, let's see, we could figure that if we know the law of cosines, we could do something over there. If we know the sine of this angle right over here, if we know the sine of that angle, the sine of that angle is going to be this height over the side we just figured out. So if we could figure out the sine of that angle, then we'd be done, or we'll be very close to being done. But it's not any obvious way. But one thing we can do, if you look at this bigger triangle over here, so let me highlight it."}, {"video_title": "Trig challenge problem area of a triangle Khan Academy.mp3", "Sentence": "If we know the sine of this angle right over here, if we know the sine of that angle, the sine of that angle is going to be this height over the side we just figured out. So if we could figure out the sine of that angle, then we'd be done, or we'll be very close to being done. But it's not any obvious way. But one thing we can do, if you look at this bigger triangle over here, so let me highlight it. So if you look at triangle BMC, let me draw a triangle BMC. Well, I'll just draw it on this drawing right over here. I don't want it to get too overloaded."}, {"video_title": "Trig challenge problem area of a triangle Khan Academy.mp3", "Sentence": "But one thing we can do, if you look at this bigger triangle over here, so let me highlight it. So if you look at triangle BMC, let me draw a triangle BMC. Well, I'll just draw it on this drawing right over here. I don't want it to get too overloaded. But triangle BMC right over here. We know this side is 25 over 2. We know this side over here is 25 over 2."}, {"video_title": "Trig challenge problem area of a triangle Khan Academy.mp3", "Sentence": "I don't want it to get too overloaded. But triangle BMC right over here. We know this side is 25 over 2. We know this side over here is 25 over 2. We know this side over here is 24. And we know this, so what we want to do is figure out the sine of this angle right here, the sine of theta, of angle CMD. Now that's hard to do, but what we can do is use the law of cosines of theta plus 90."}, {"video_title": "Trig challenge problem area of a triangle Khan Academy.mp3", "Sentence": "We know this side over here is 25 over 2. We know this side over here is 24. And we know this, so what we want to do is figure out the sine of this angle right here, the sine of theta, of angle CMD. Now that's hard to do, but what we can do is use the law of cosines of theta plus 90. So let me redraw our triangle. So triangle BCM, I could redraw like this. I could redraw, it's actually an isosceles triangle."}, {"video_title": "Trig challenge problem area of a triangle Khan Academy.mp3", "Sentence": "Now that's hard to do, but what we can do is use the law of cosines of theta plus 90. So let me redraw our triangle. So triangle BCM, I could redraw like this. I could redraw, it's actually an isosceles triangle. So we have BCM, and what is this angle right over here? It's going to be our theta that we care about plus 90 degrees. So this angle right here is theta plus 90 degrees."}, {"video_title": "Trig challenge problem area of a triangle Khan Academy.mp3", "Sentence": "I could redraw, it's actually an isosceles triangle. So we have BCM, and what is this angle right over here? It's going to be our theta that we care about plus 90 degrees. So this angle right here is theta plus 90 degrees. And this over here is 24, this is 25 over 2, and this is 25 over 2. And so using this, we could use the law of cosines to figure out what theta actually is here. So let's do that."}, {"video_title": "Trig challenge problem area of a triangle Khan Academy.mp3", "Sentence": "So this angle right here is theta plus 90 degrees. And this over here is 24, this is 25 over 2, and this is 25 over 2. And so using this, we could use the law of cosines to figure out what theta actually is here. So let's do that. We're going to use a little bit of trig identities, but law of cosines. So we get the opposite to the angle squared, so we get 24 squared. 24 squared is equal to 25 over 2 squared, plus 25 over 2 squared, minus 2 times 25 over 2, times the cosine of this angle, times, let me scroll over, the cosine of theta plus 90 degrees."}, {"video_title": "Trig challenge problem area of a triangle Khan Academy.mp3", "Sentence": "So let's do that. We're going to use a little bit of trig identities, but law of cosines. So we get the opposite to the angle squared, so we get 24 squared. 24 squared is equal to 25 over 2 squared, plus 25 over 2 squared, minus 2 times 25 over 2, times the cosine of this angle, times, let me scroll over, the cosine of theta plus 90 degrees. Now you're saying, hey Sal, this has it in terms of cosine of theta plus 90 degrees, how can we figure out the sine of theta? That's what we actually care about to figure out the area of this triangle, actually to figure out the height of this triangle. And to do that, you just have to make the realization, we know the trig identity, that the cosine of theta is equal, I won't use theta because I don't want to overload theta, cosine of x is equal to sine of 90 minus x."}, {"video_title": "Trig challenge problem area of a triangle Khan Academy.mp3", "Sentence": "24 squared is equal to 25 over 2 squared, plus 25 over 2 squared, minus 2 times 25 over 2, times the cosine of this angle, times, let me scroll over, the cosine of theta plus 90 degrees. Now you're saying, hey Sal, this has it in terms of cosine of theta plus 90 degrees, how can we figure out the sine of theta? That's what we actually care about to figure out the area of this triangle, actually to figure out the height of this triangle. And to do that, you just have to make the realization, we know the trig identity, that the cosine of theta is equal, I won't use theta because I don't want to overload theta, cosine of x is equal to sine of 90 minus x. So the cosine of theta plus 90 degrees is going to be equal to the sine of 90 minus whatever is here, 90 minus theta, minus 90, which is equal to, the 90's cancel out, sine of negative theta, and we know sine of negative theta is equal to the negative sine of theta. So this over here simplifies to, this is the negative sine of theta, so we can write the sine of theta here, and then put the negative out here and this becomes a positive. So what does this simplify to?"}, {"video_title": "Trig challenge problem area of a triangle Khan Academy.mp3", "Sentence": "And to do that, you just have to make the realization, we know the trig identity, that the cosine of theta is equal, I won't use theta because I don't want to overload theta, cosine of x is equal to sine of 90 minus x. So the cosine of theta plus 90 degrees is going to be equal to the sine of 90 minus whatever is here, 90 minus theta, minus 90, which is equal to, the 90's cancel out, sine of negative theta, and we know sine of negative theta is equal to the negative sine of theta. So this over here simplifies to, this is the negative sine of theta, so we can write the sine of theta here, and then put the negative out here and this becomes a positive. So what does this simplify to? We have 24 squared, which is 576, 576 is equal to, let's see, we have, I won't skip any steps here, so we have 25 squared plus 25 squared, this is 2 times 25, this is 2 times 25 over 2 squared, plus 2 times, this is 25 over 2 squared again, 2 times 25 over 2 squared, times sine of theta. Now we just have to solve for sine of theta, so this is going to be equal to, well this is 576, is equal to 2 times 25 over 2 squared, I'm just factoring that out, 1 plus sine of theta, 1 plus sine of theta, or we can just divide both sides of the equation by this here, actually let me just simplify it, this thing over here is 625 over 4, but then we're going to multiply that by 2, so this thing over here is 625 over 2. So let's divide both sides of this by 625 over 2, and we get 576 times 2 over 625, just multiplying both sides by the inverse, is equal to, when you multiply both sides of the inverse by this, that actually cancels out, is equal to 1 plus sine of theta, or sine of theta, we just subtract 1 from both sides, we get sine of theta is equal to 576 times 2, let's see, 576 times 2 is 152, plus 1000, so it's 1152 over 625, that's that part there, minus 1, but instead of 1, let's say 625 over 625, so minus 625, and this is equal to, I'll have to do a little math on the side, so 1152 minus 625, you get a 12 there, this becomes a 4, 12 minus 5 is 7, 4 minus 2 is 2, 11 minus 6 is 5, so this is equal to 527 over 625."}, {"video_title": "Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy (2).mp3", "Sentence": "Determine the six trigonometric ratios for angle A in the right triangle below. So this right over here is angle A, it's at vertex A. And to help me remember the definitions of the trig ratios, and these are human constructed definitions that have ended up being very, very useful for analyzing a whole series of things in the world. But to help me remember them, I use the word SOH CAH TOA. So let me write that down. So SOH CAH TOA. Sometimes you can think of it as one word, but it's really the three parts that define at least three of the trig functions for you."}, {"video_title": "Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy (2).mp3", "Sentence": "But to help me remember them, I use the word SOH CAH TOA. So let me write that down. So SOH CAH TOA. Sometimes you can think of it as one word, but it's really the three parts that define at least three of the trig functions for you. And then we can get the other three by looking at the first three. So SOH tells us that sine of an angle, and in this case it's sine of A, so sine of A is equal to the opposite, opposite, that's the O, over the hypotenuse. Opposite over the hypotenuse."}, {"video_title": "Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy (2).mp3", "Sentence": "Sometimes you can think of it as one word, but it's really the three parts that define at least three of the trig functions for you. And then we can get the other three by looking at the first three. So SOH tells us that sine of an angle, and in this case it's sine of A, so sine of A is equal to the opposite, opposite, that's the O, over the hypotenuse. Opposite over the hypotenuse. Well in this context, what is the opposite side to angle A? Well we go across the triangle, it opens up onto side BC, it has length 12, so that is the opposite side. So this is going to be equal to 12."}, {"video_title": "Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy (2).mp3", "Sentence": "Opposite over the hypotenuse. Well in this context, what is the opposite side to angle A? Well we go across the triangle, it opens up onto side BC, it has length 12, so that is the opposite side. So this is going to be equal to 12. And what's the hypotenuse? Well the hypotenuse is the longest side of the triangle. It's opposite the 90 degree angle, and so we go opposite the 90 degree angle, longest side is side AB, it has length 13."}, {"video_title": "Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy (2).mp3", "Sentence": "So this is going to be equal to 12. And what's the hypotenuse? Well the hypotenuse is the longest side of the triangle. It's opposite the 90 degree angle, and so we go opposite the 90 degree angle, longest side is side AB, it has length 13. So this right over here is the hypotenuse. So the sine of A is 12 thirteens. Now let's go to CAH."}, {"video_title": "Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy (2).mp3", "Sentence": "It's opposite the 90 degree angle, and so we go opposite the 90 degree angle, longest side is side AB, it has length 13. So this right over here is the hypotenuse. So the sine of A is 12 thirteens. Now let's go to CAH. CAH defines cosine for us. It tells us that cosine of an angle, in this case cosine of A, is equal to the adjacent side, the adjacent side to the angle, over the hypotenuse. Over the hypotenuse."}, {"video_title": "Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy (2).mp3", "Sentence": "Now let's go to CAH. CAH defines cosine for us. It tells us that cosine of an angle, in this case cosine of A, is equal to the adjacent side, the adjacent side to the angle, over the hypotenuse. Over the hypotenuse. So what's the adjacent side to angle A? Well if we look at angle A, there's two sides that are next to it. One of them is the hypotenuse."}, {"video_title": "Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy (2).mp3", "Sentence": "Over the hypotenuse. So what's the adjacent side to angle A? Well if we look at angle A, there's two sides that are next to it. One of them is the hypotenuse. The other one has length five. The adjacent one is side CA. So it's five."}, {"video_title": "Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy (2).mp3", "Sentence": "One of them is the hypotenuse. The other one has length five. The adjacent one is side CA. So it's five. And what is the hypotenuse? Well we've already figured that out. The hypotenuse right over here, it's opposite the 90 degree angle, it's the longest side of the right triangle, it has length 13."}, {"video_title": "Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy (2).mp3", "Sentence": "So it's five. And what is the hypotenuse? Well we've already figured that out. The hypotenuse right over here, it's opposite the 90 degree angle, it's the longest side of the right triangle, it has length 13. So the cosine of A is 5 thirteens. And let me label this. This right over here is the adjacent side."}, {"video_title": "Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy (2).mp3", "Sentence": "The hypotenuse right over here, it's opposite the 90 degree angle, it's the longest side of the right triangle, it has length 13. So the cosine of A is 5 thirteens. And let me label this. This right over here is the adjacent side. And this is all specific to angle A. The hypotenuse would be the same regardless of what angle you pick. But the opposite and the adjacent is dependent on the angle that we choose in the right triangle."}, {"video_title": "Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy (2).mp3", "Sentence": "This right over here is the adjacent side. And this is all specific to angle A. The hypotenuse would be the same regardless of what angle you pick. But the opposite and the adjacent is dependent on the angle that we choose in the right triangle. Now let's go to TOA. TOA defines tangent for us. It tells us that the tangent, the tangent of an angle, is equal to the opposite, equal to the opposite side, over the adjacent side."}, {"video_title": "Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy (2).mp3", "Sentence": "But the opposite and the adjacent is dependent on the angle that we choose in the right triangle. Now let's go to TOA. TOA defines tangent for us. It tells us that the tangent, the tangent of an angle, is equal to the opposite, equal to the opposite side, over the adjacent side. So given this definition, what is the tangent of A? Well, the opposite side, we already figured out, has length 12, has length 12. And the adjacent side, we already figured out, has length five."}, {"video_title": "Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy (2).mp3", "Sentence": "It tells us that the tangent, the tangent of an angle, is equal to the opposite, equal to the opposite side, over the adjacent side. So given this definition, what is the tangent of A? Well, the opposite side, we already figured out, has length 12, has length 12. And the adjacent side, we already figured out, has length five. So the tangent of A, which is opposite over adjacent, is 12 fifths. Now we'll go to the other three trig ratios, which you could think of as the reciprocals of these right over here. But I'll define it."}, {"video_title": "Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy (2).mp3", "Sentence": "And the adjacent side, we already figured out, has length five. So the tangent of A, which is opposite over adjacent, is 12 fifths. Now we'll go to the other three trig ratios, which you could think of as the reciprocals of these right over here. But I'll define it. So first you have cosecant. And cosecant, it's always a little bit unintuitive why cosecant is the reciprocal of sine of A, even though it starts with a co, like cosine. But cosecant is the reciprocal of the sine of A."}, {"video_title": "Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy (2).mp3", "Sentence": "But I'll define it. So first you have cosecant. And cosecant, it's always a little bit unintuitive why cosecant is the reciprocal of sine of A, even though it starts with a co, like cosine. But cosecant is the reciprocal of the sine of A. So sine of A is opposite over hypotenuse. Cosecant of A is hypotenuse over opposite. And so what's the hypotenuse over the opposite?"}, {"video_title": "Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy (2).mp3", "Sentence": "But cosecant is the reciprocal of the sine of A. So sine of A is opposite over hypotenuse. Cosecant of A is hypotenuse over opposite. And so what's the hypotenuse over the opposite? Well, the hypotenuse is 13, and the opposite side is 12. And notice, 13 twelfths is the reciprocal of 12 thirteens. Now, secant of A is the reciprocal, so instead of being adjacent over hypotenuse, which we got from the co part of SOH CAH TOA, it's hypotenuse over adjacent."}, {"video_title": "Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy (2).mp3", "Sentence": "And so what's the hypotenuse over the opposite? Well, the hypotenuse is 13, and the opposite side is 12. And notice, 13 twelfths is the reciprocal of 12 thirteens. Now, secant of A is the reciprocal, so instead of being adjacent over hypotenuse, which we got from the co part of SOH CAH TOA, it's hypotenuse over adjacent. So what is the secant of A? Well, the hypotenuse, we've figured out multiple times already, is 13. And what is the adjacent side?"}, {"video_title": "Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy (2).mp3", "Sentence": "Now, secant of A is the reciprocal, so instead of being adjacent over hypotenuse, which we got from the co part of SOH CAH TOA, it's hypotenuse over adjacent. So what is the secant of A? Well, the hypotenuse, we've figured out multiple times already, is 13. And what is the adjacent side? It's five. So it's 13 fifths, which is once again the reciprocal of the cosine of A, five thirteens. Finally, let's get the cotangent."}, {"video_title": "Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy (2).mp3", "Sentence": "And what is the adjacent side? It's five. So it's 13 fifths, which is once again the reciprocal of the cosine of A, five thirteens. Finally, let's get the cotangent. And the cotangent is the reciprocal of tangent of A. Instead of being opposite over adjacent, it is adjacent over opposite. So what is the cotangent of A?"}, {"video_title": "Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy (2).mp3", "Sentence": "Finally, let's get the cotangent. And the cotangent is the reciprocal of tangent of A. Instead of being opposite over adjacent, it is adjacent over opposite. So what is the cotangent of A? Well, we figured out the adjacent side multiple times for angle A. It's length five. And the opposite side to angle A is 12."}, {"video_title": "Law of sines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And what I claim is that I can figure out everything else about this triangle just with this information. You give me two angles and a side, and I can figure out what the other two sides are going to be, and I can, of course, figure out the third angle. So let's try to figure that out. And the way that we're going to do it, we're going to use something called the law of sines. And in a future video, I will prove the law of sines. But here, I am just going to show you how we can actually apply it. And it's a fairly straightforward idea."}, {"video_title": "Law of sines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And the way that we're going to do it, we're going to use something called the law of sines. And in a future video, I will prove the law of sines. But here, I am just going to show you how we can actually apply it. And it's a fairly straightforward idea. The law of sines just tells us that the ratio between the sine of an angle and the side opposite to it is going to be constant for any of the angles in a triangle. So for example, for this triangle right over here, this is a 30-degree angle. This is a 45-degree angle."}, {"video_title": "Law of sines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And it's a fairly straightforward idea. The law of sines just tells us that the ratio between the sine of an angle and the side opposite to it is going to be constant for any of the angles in a triangle. So for example, for this triangle right over here, this is a 30-degree angle. This is a 45-degree angle. They have to add up to 180. So this right over here has to be a, let's see, so it's going to be 180 minus 45 minus 30. That's 180 minus 75."}, {"video_title": "Law of sines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "This is a 45-degree angle. They have to add up to 180. So this right over here has to be a, let's see, so it's going to be 180 minus 45 minus 30. That's 180 minus 75. So this is going to equal 105-degree angle right over here. 105-degree angle. And so applying the law of sines here, and actually, let me label the different sides."}, {"video_title": "Law of sines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "That's 180 minus 75. So this is going to equal 105-degree angle right over here. 105-degree angle. And so applying the law of sines here, and actually, let me label the different sides. Let's call this side right over here. Let's call this side right over here side a, or has length a. And let's call this side right over here."}, {"video_title": "Law of sines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And so applying the law of sines here, and actually, let me label the different sides. Let's call this side right over here. Let's call this side right over here side a, or has length a. And let's call this side right over here. Let's say this side right over here has length b. So the law of sines tells us that the ratio between the sine of an angle and the opposite side is going to be constant through this triangle. So it tells us that sine of this angle, sine of 30 degrees, over the length of the side opposite is going to be equal to sine of 105 degrees over the length of the side opposite to it, which is going to be equal to sine of 45 degrees equal to the length of the side opposite."}, {"video_title": "Law of sines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And let's call this side right over here. Let's say this side right over here has length b. So the law of sines tells us that the ratio between the sine of an angle and the opposite side is going to be constant through this triangle. So it tells us that sine of this angle, sine of 30 degrees, over the length of the side opposite is going to be equal to sine of 105 degrees over the length of the side opposite to it, which is going to be equal to sine of 45 degrees equal to the length of the side opposite. So sine of 45 degrees over b. And so if we wanted to figure out a, we could solve this equation right over here. And if we wanted to solve for b, we could just set this equal to that right over there."}, {"video_title": "Law of sines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So it tells us that sine of this angle, sine of 30 degrees, over the length of the side opposite is going to be equal to sine of 105 degrees over the length of the side opposite to it, which is going to be equal to sine of 45 degrees equal to the length of the side opposite. So sine of 45 degrees over b. And so if we wanted to figure out a, we could solve this equation right over here. And if we wanted to solve for b, we could just set this equal to that right over there. So let's solve each of these. So what is the sine of 30 degrees? Well, you just might remember it from your unit circles or from even 30, 60, 90 triangles that that's 1 half."}, {"video_title": "Law of sines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And if we wanted to solve for b, we could just set this equal to that right over there. So let's solve each of these. So what is the sine of 30 degrees? Well, you just might remember it from your unit circles or from even 30, 60, 90 triangles that that's 1 half. And if you don't remember it, you can use a calculator to verify that. I already verified that this is in degree mode. So it's 0.5."}, {"video_title": "Law of sines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Well, you just might remember it from your unit circles or from even 30, 60, 90 triangles that that's 1 half. And if you don't remember it, you can use a calculator to verify that. I already verified that this is in degree mode. So it's 0.5. So this is going to be equal to 1 half over 2. So another way of thinking about it, that's going to be equal to 1 fourth. This piece is equal to 1 fourth is equal to sine of 105 degrees over a."}, {"video_title": "Law of sines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So it's 0.5. So this is going to be equal to 1 half over 2. So another way of thinking about it, that's going to be equal to 1 fourth. This piece is equal to 1 fourth is equal to sine of 105 degrees over a. So let me write this. This is equal to sine of 105 degrees over a. And actually, we could also say, we could actually do both at the same time, that this is equal to that."}, {"video_title": "Law of sines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "This piece is equal to 1 fourth is equal to sine of 105 degrees over a. So let me write this. This is equal to sine of 105 degrees over a. And actually, we could also say, we could actually do both at the same time, that this is equal to that. That 1 fourth is equal to sine of 45 degrees over b. Actually, sine of 45 degrees is another one of those that it's easy to jump out of your unit circle. You might remember it's square root of 2 over 2."}, {"video_title": "Law of sines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And actually, we could also say, we could actually do both at the same time, that this is equal to that. That 1 fourth is equal to sine of 45 degrees over b. Actually, sine of 45 degrees is another one of those that it's easy to jump out of your unit circle. You might remember it's square root of 2 over 2. So let's just write that. That's square root of 2 over 2. And you could use a calculator, but you'll get some decimal value right over there."}, {"video_title": "Law of sines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "You might remember it's square root of 2 over 2. So let's just write that. That's square root of 2 over 2. And you could use a calculator, but you'll get some decimal value right over there. But in either case, in either of these equations, let's solve for a, and then let's solve for b. So one thing we could do is we could take the reciprocal of both sides of this equation. The reciprocal of 1 fourth is 4."}, {"video_title": "Law of sines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And you could use a calculator, but you'll get some decimal value right over there. But in either case, in either of these equations, let's solve for a, and then let's solve for b. So one thing we could do is we could take the reciprocal of both sides of this equation. The reciprocal of 1 fourth is 4. And the reciprocal of this right-hand side is a over the sine of 105 degrees. And now to solve for a, we could just multiply both sides times the sine of 105 degrees. So we get 4 times the sine of 105 degrees is equal to a."}, {"video_title": "Law of sines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "The reciprocal of 1 fourth is 4. And the reciprocal of this right-hand side is a over the sine of 105 degrees. And now to solve for a, we could just multiply both sides times the sine of 105 degrees. So we get 4 times the sine of 105 degrees is equal to a. So let's get our calculator out. So 4 times the sine of 105 gives us, let's see, it's approximately equal to, let's say, let's just round to the nearest hundreds, 3.86. So a is approximately equal to 3.86, which looks about right."}, {"video_title": "Law of sines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So we get 4 times the sine of 105 degrees is equal to a. So let's get our calculator out. So 4 times the sine of 105 gives us, let's see, it's approximately equal to, let's say, let's just round to the nearest hundreds, 3.86. So a is approximately equal to 3.86, which looks about right. If this is 2 and if I made my angles appropriately, that looks like about 3.86. And let's figure out what b is. We could once again take the reciprocal of both sides of this, and we get 4 is equal to b over square root of 2 over 2."}, {"video_title": "Law of sines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So a is approximately equal to 3.86, which looks about right. If this is 2 and if I made my angles appropriately, that looks like about 3.86. And let's figure out what b is. We could once again take the reciprocal of both sides of this, and we get 4 is equal to b over square root of 2 over 2. We can multiply both sides times square root of 2 over 2, and we would get b is equal to 4 times the square root of 2 over 2. Or we could think of it b as 4 times the sine of 45 degrees. But let's figure out what that is."}, {"video_title": "Law of sines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "We could once again take the reciprocal of both sides of this, and we get 4 is equal to b over square root of 2 over 2. We can multiply both sides times square root of 2 over 2, and we would get b is equal to 4 times the square root of 2 over 2. Or we could think of it b as 4 times the sine of 45 degrees. But let's figure out what that is. So if we want an actual numerical value, we could just write this as 2 square roots of 2. But let's actually figure out what that is. 2 square roots of 2 is equal to 2.83."}, {"video_title": "Law of sines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "But let's figure out what that is. So if we want an actual numerical value, we could just write this as 2 square roots of 2. But let's actually figure out what that is. 2 square roots of 2 is equal to 2.83. So b is approximately equal to 2.83. So let me be clear. This, 4 divided by 2 is 2 square roots of 2, which is approximately equal to 2.83 if we round to the nearest hundreds, which also seems pretty reasonable here."}, {"video_title": "Angle to aim to get alien Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "A tiny but horrible alien is standing at the top of the Eiffel Tower, so this is where the tiny but horrible alien is, which is 324 meters tall, and they label that the height of the Eiffel Tower, and threatening to destroy the city of Paris. A men in black, or I guess a men in black agent, I was about to say maybe it should be a man in black, a men in black agent is standing at ground level, 54 meters across the Eiffel Square, so 54 meters from, I guess you could say the center of the base of the Eiffel Tower, aiming his laser gun at the alien, so this is him aiming the laser gun. At what angle should the agent shoot his laser gun? Round your answer if necessary to two decimal places. So if we construct a right triangle here, and we can, so the height of this right triangle is 324 meters, this width right over here is 54 meters, it is a right triangle, what they're really asking us is what is the angle, what is this angle, what is this angle right over here? And they've given us two pieces of information, they gave us the side that is opposite the angle, and they've given us the side that is adjacent to the angle. So what trig function deals with opposite and adjacent?"}, {"video_title": "Angle to aim to get alien Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Round your answer if necessary to two decimal places. So if we construct a right triangle here, and we can, so the height of this right triangle is 324 meters, this width right over here is 54 meters, it is a right triangle, what they're really asking us is what is the angle, what is this angle, what is this angle right over here? And they've given us two pieces of information, they gave us the side that is opposite the angle, and they've given us the side that is adjacent to the angle. So what trig function deals with opposite and adjacent? And to remind ourselves, we can write like I always like to do, SOH CAH TOA. And these are really by definition, so you just kind of have to know this, and SOH CAH TOA helps us. Sine is opposite over hypotenuse."}, {"video_title": "Angle to aim to get alien Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So what trig function deals with opposite and adjacent? And to remind ourselves, we can write like I always like to do, SOH CAH TOA. And these are really by definition, so you just kind of have to know this, and SOH CAH TOA helps us. Sine is opposite over hypotenuse. Cosine is adjacent over hypotenuse. Tangent is opposite over adjacent. So we can say that the tangent, we can write that the tangent of theta, the tangent of theta is equal to the length of the opposite side, 324 meters, 324 meters over the length of the adjacent side, over 54 meters."}, {"video_title": "Angle to aim to get alien Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Sine is opposite over hypotenuse. Cosine is adjacent over hypotenuse. Tangent is opposite over adjacent. So we can say that the tangent, we can write that the tangent of theta, the tangent of theta is equal to the length of the opposite side, 324 meters, 324 meters over the length of the adjacent side, over 54 meters. Now you might say, well, okay, that's fine, so what angle gives, when I take its tangent, gives me 324 over 54? Well, for this, it will probably be useful to use a calculator. And the way that we'd use a calculator is we would use the inverse tan function."}, {"video_title": "Angle to aim to get alien Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So we can say that the tangent, we can write that the tangent of theta, the tangent of theta is equal to the length of the opposite side, 324 meters, 324 meters over the length of the adjacent side, over 54 meters. Now you might say, well, okay, that's fine, so what angle gives, when I take its tangent, gives me 324 over 54? Well, for this, it will probably be useful to use a calculator. And the way that we'd use a calculator is we would use the inverse tan function. So we could rewrite this as we're going to take the inverse tangent, and sometimes it's written as tangent, kind of this negative 1 superscript. So the inverse tangent of tan of theta is going to be equal to the inverse tangent of 324 over 54. And just to be clear, what is this inverse tangent?"}, {"video_title": "Angle to aim to get alien Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "And the way that we'd use a calculator is we would use the inverse tan function. So we could rewrite this as we're going to take the inverse tangent, and sometimes it's written as tangent, kind of this negative 1 superscript. So the inverse tangent of tan of theta is going to be equal to the inverse tangent of 324 over 54. And just to be clear, what is this inverse tangent? This literally says, this will return what is the angle that when I take the tangent of it, gives me 324 over 54. This says, what is the angle that when I take the tangent of it, gives me tangent of theta? So this right over here, this just simplifies to theta."}, {"video_title": "Angle to aim to get alien Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "And just to be clear, what is this inverse tangent? This literally says, this will return what is the angle that when I take the tangent of it, gives me 324 over 54. This says, what is the angle that when I take the tangent of it, gives me tangent of theta? So this right over here, this just simplifies to theta. Theta is the angle that when you get the tangent of it, gets you tangent of theta. And so we get theta is equal to tangent, inverse tangent of 324 over 54. Once again, this inverse tangent thing, you might find confusing, but all this is saying is, over here it's saying tangent of some angle is 324 over 54."}, {"video_title": "Angle to aim to get alien Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So this right over here, this just simplifies to theta. Theta is the angle that when you get the tangent of it, gets you tangent of theta. And so we get theta is equal to tangent, inverse tangent of 324 over 54. Once again, this inverse tangent thing, you might find confusing, but all this is saying is, over here it's saying tangent of some angle is 324 over 54. This is just saying is my angle is whatever angle I need, so that when I take the tangent of it, I get 324 over 54. It's how we will solve for theta. So let's get our calculator out, and let's say that we want our answer in degrees."}, {"video_title": "Angle to aim to get alien Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Once again, this inverse tangent thing, you might find confusing, but all this is saying is, over here it's saying tangent of some angle is 324 over 54. This is just saying is my angle is whatever angle I need, so that when I take the tangent of it, I get 324 over 54. It's how we will solve for theta. So let's get our calculator out, and let's say that we want our answer in degrees. And actually they should, well, I'm just going to assume that they want our answer in degrees. So let me make sure my calculator is actually in degree mode. So I'll go to the second mode right over here, and actually it's in radian mode right now."}, {"video_title": "Angle to aim to get alien Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So let's get our calculator out, and let's say that we want our answer in degrees. And actually they should, well, I'm just going to assume that they want our answer in degrees. So let me make sure my calculator is actually in degree mode. So I'll go to the second mode right over here, and actually it's in radian mode right now. So let me make sure I'm in degree mode to get my answer in degrees. Now let me exit out of here, and let me just type in the inverse tangent. So it's in this yellow color right here."}, {"video_title": "Angle to aim to get alien Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So I'll go to the second mode right over here, and actually it's in radian mode right now. So let me make sure I'm in degree mode to get my answer in degrees. Now let me exit out of here, and let me just type in the inverse tangent. So it's in this yellow color right here. Inverse tangent of 324 divided by 54 is going to be 80 point, and they told us to round to two decimal places, 80.54 degrees. So theta is equal to 80.54 degrees. That's the angle at which you should shoot the gun to help defeat this horrible alien."}, {"video_title": "Law of sines for missing angle Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And you measure the angle between the kite and the ground where you're standing. And you see that it's a 40 degree angle. And what you're curious about is whether you can use your powers of trigonometry to figure out the angle between the string and the ground. And I encourage you to pause the video now and figure out if you can do that using just the information that you have. So whenever I see a, I guess, a non-right triangle where I'm trying to figure out some lengths of sides or some lengths of angles, I immediately think maybe the law of cosine might be useful, or the law of cosines might be useful. And so let's think about which one could be useful in this case. Law of cosines, and I'll just rewrite them here."}, {"video_title": "Law of sines for missing angle Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And I encourage you to pause the video now and figure out if you can do that using just the information that you have. So whenever I see a, I guess, a non-right triangle where I'm trying to figure out some lengths of sides or some lengths of angles, I immediately think maybe the law of cosine might be useful, or the law of cosines might be useful. And so let's think about which one could be useful in this case. Law of cosines, and I'll just rewrite them here. The law of cosines is c squared is equal to a squared plus b squared minus 2ab cosine of theta. And so what it's doing is it's relating three sides of a triangle, so a, b, c, to an angle. So for example, if I knew two sides and the angle in between them, I could figure out the third side."}, {"video_title": "Law of sines for missing angle Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Law of cosines, and I'll just rewrite them here. The law of cosines is c squared is equal to a squared plus b squared minus 2ab cosine of theta. And so what it's doing is it's relating three sides of a triangle, so a, b, c, to an angle. So for example, if I knew two sides and the angle in between them, I could figure out the third side. Or if I know all three sides, then I could figure out this angle. But that's not the situation that we have over here. We're trying to figure out this question mark."}, {"video_title": "Law of sines for missing angle Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So for example, if I knew two sides and the angle in between them, I could figure out the third side. Or if I know all three sides, then I could figure out this angle. But that's not the situation that we have over here. We're trying to figure out this question mark. And we don't know three of the sides. We're trying to figure out an angle, but we don't know three of the sides. So the law of cosines doesn't seem, at least, in an obvious way, that's going to help me."}, {"video_title": "Law of sines for missing angle Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "We're trying to figure out this question mark. And we don't know three of the sides. We're trying to figure out an angle, but we don't know three of the sides. So the law of cosines doesn't seem, at least, in an obvious way, that's going to help me. I could also try to find this angle. But once again, we don't know all three sides to be able to solve for the angle. So maybe law of sines could be useful."}, {"video_title": "Law of sines for missing angle Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So the law of cosines doesn't seem, at least, in an obvious way, that's going to help me. I could also try to find this angle. But once again, we don't know all three sides to be able to solve for the angle. So maybe law of sines could be useful. So the law of sines, so let's say that the measure of this angle is a. Measure of this angle is lowercase b. Measure of this angle is lower case c. Length of this side is capital C. Length of this side is capital A."}, {"video_title": "Law of sines for missing angle Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So maybe law of sines could be useful. So the law of sines, so let's say that the measure of this angle is a. Measure of this angle is lowercase b. Measure of this angle is lower case c. Length of this side is capital C. Length of this side is capital A. Length of this side is capital B. Law of sines tells us that the ratio between the sine of each of these angles and the length of the opposite side is constant. So sine of lowercase a over capital A is the same as the sine of lowercase b over capital B, which is going to be the same as the sine of lowercase c over capital C. So let's see if we could leverage that somehow right over here."}, {"video_title": "Law of sines for missing angle Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Measure of this angle is lower case c. Length of this side is capital C. Length of this side is capital A. Length of this side is capital B. Law of sines tells us that the ratio between the sine of each of these angles and the length of the opposite side is constant. So sine of lowercase a over capital A is the same as the sine of lowercase b over capital B, which is going to be the same as the sine of lowercase c over capital C. So let's see if we could leverage that somehow right over here. So we know this angle and the opposite side. So we could write that ratio. Sine of 40 degrees over 30."}, {"video_title": "Law of sines for missing angle Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So sine of lowercase a over capital A is the same as the sine of lowercase b over capital B, which is going to be the same as the sine of lowercase c over capital C. So let's see if we could leverage that somehow right over here. So we know this angle and the opposite side. So we could write that ratio. Sine of 40 degrees over 30. Let's see. Can we say that that's going to be equal to the sine of this angle over that? Well, it would be, but we don't know either of these."}, {"video_title": "Law of sines for missing angle Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Sine of 40 degrees over 30. Let's see. Can we say that that's going to be equal to the sine of this angle over that? Well, it would be, but we don't know either of these. So that doesn't seem like it's going to help us. But we do know this side. And so maybe we could use the law of cosines to figure out this angle."}, {"video_title": "Law of sines for missing angle Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Well, it would be, but we don't know either of these. So that doesn't seem like it's going to help us. But we do know this side. And so maybe we could use the law of cosines to figure out this angle. Because if we know two angles of a triangle, then we could figure out a third angle. So let's do that. So let's say that this angle right over here is theta."}, {"video_title": "Law of sines for missing angle Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And so maybe we could use the law of cosines to figure out this angle. Because if we know two angles of a triangle, then we could figure out a third angle. So let's do that. So let's say that this angle right over here is theta. We know this distance right over here is 40 meters. So we could say that the sine of theta over 40. This ratio is going to be the same as the sine of 40 over 30."}, {"video_title": "Law of sines for missing angle Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So let's say that this angle right over here is theta. We know this distance right over here is 40 meters. So we could say that the sine of theta over 40. This ratio is going to be the same as the sine of 40 over 30. And now we can just solve for theta. Multiplying both sides times 40, you're going to get, let's see, 40 divided by 30 is 4 thirds. Sine of 40 degrees is equal to sine of theta."}, {"video_title": "Law of sines for missing angle Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "This ratio is going to be the same as the sine of 40 over 30. And now we can just solve for theta. Multiplying both sides times 40, you're going to get, let's see, 40 divided by 30 is 4 thirds. Sine of 40 degrees is equal to sine of theta. And now to solve for theta, we just have to take the inverse sine of both sides. So inverse sine of 4 over 3 sine of 40 degrees, I'll put some parentheses here, is equal to theta. And that will give us this angle."}, {"video_title": "Law of sines for missing angle Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Sine of 40 degrees is equal to sine of theta. And now to solve for theta, we just have to take the inverse sine of both sides. So inverse sine of 4 over 3 sine of 40 degrees, I'll put some parentheses here, is equal to theta. And that will give us this angle. And then we can use that information and this information to figure out the angle that we really care about. So let's get a calculator out and see if we can calculate it. So let me just verify I am in degree mode."}, {"video_title": "Law of sines for missing angle Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And that will give us this angle. And then we can use that information and this information to figure out the angle that we really care about. So let's get a calculator out and see if we can calculate it. So let me just verify I am in degree mode. Very important. All right. Now I'm going to take the inverse sine of 4 thirds times sine of 40 degrees."}, {"video_title": "Law of sines for missing angle Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So let me just verify I am in degree mode. Very important. All right. Now I'm going to take the inverse sine of 4 thirds times sine of 40 degrees. And that gets me, and I deserve a little bit of a drumroll now, 58, well, let's just round to the nearest. Let's just maintain our precision here. So 58.99 degrees, roughly."}, {"video_title": "Law of sines for missing angle Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Now I'm going to take the inverse sine of 4 thirds times sine of 40 degrees. And that gets me, and I deserve a little bit of a drumroll now, 58, well, let's just round to the nearest. Let's just maintain our precision here. So 58.99 degrees, roughly. So this is approximately equal to 58.99 degrees. So if that is 58.99 degrees, what is this one? Well, it's going to be 180 minus this angle's measure minus that angle's measure."}, {"video_title": "Law of sines for missing angle Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So 58.99 degrees, roughly. So this is approximately equal to 58.99 degrees. So if that is 58.99 degrees, what is this one? Well, it's going to be 180 minus this angle's measure minus that angle's measure. So let's calculate that. So it's going to be 180 degrees minus this angle, so minus 40, minus the angle we just figured out. And actually, I could get all of our precision by just typing in second answer."}, {"video_title": "Law of sines for missing angle Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Well, it's going to be 180 minus this angle's measure minus that angle's measure. So let's calculate that. So it's going to be 180 degrees minus this angle, so minus 40, minus the angle we just figured out. And actually, I could get all of our precision by just typing in second answer. So that just says our previous answer. So I get all that precision there. And so I get 81.01 degrees."}, {"video_title": "Law of sines for missing angle Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And actually, I could get all of our precision by just typing in second answer. So that just says our previous answer. So I get all that precision there. And so I get 81.01 degrees. If I want to round to the nearest, let's say I round to the nearest hundredth of a degree, then I'd say 81.01 degrees. So this right over here is approximately 81.01 degrees. And we're done."}, {"video_title": "Symmetry of trig values Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And for the sake of this video, we'll assume everything is in radians. So this angle right over here, we would call this theta. And now let's flip this, I guess we could say the terminal ray of this angle. Let's flip it over the x and y axes. So let's just make sure we've labeled our axes. So let's flip it over the positive x-axis. So if you flip it over the positive x-axis, you just go straight down."}, {"video_title": "Symmetry of trig values Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Let's flip it over the x and y axes. So let's just make sure we've labeled our axes. So let's flip it over the positive x-axis. So if you flip it over the positive x-axis, you just go straight down. Then you go the same distance on the other side. You get to that point right over there. And so you would get this ray that I'm attempting to draw in blue."}, {"video_title": "Symmetry of trig values Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So if you flip it over the positive x-axis, you just go straight down. Then you go the same distance on the other side. You get to that point right over there. And so you would get this ray that I'm attempting to draw in blue. You would get that ray right over there. Now what is the angle between this ray and the positive x-axis, if you start at the positive x-axis? Well, just using our conventions that counterclockwise from the x-axis is a positive angle, this is clockwise."}, {"video_title": "Symmetry of trig values Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And so you would get this ray that I'm attempting to draw in blue. You would get that ray right over there. Now what is the angle between this ray and the positive x-axis, if you start at the positive x-axis? Well, just using our conventions that counterclockwise from the x-axis is a positive angle, this is clockwise. So instead of going theta above the x-axis, we're going theta below. So we would call this, by our convention, an angle of negative theta. Now let's flip our original green ray."}, {"video_title": "Symmetry of trig values Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Well, just using our conventions that counterclockwise from the x-axis is a positive angle, this is clockwise. So instead of going theta above the x-axis, we're going theta below. So we would call this, by our convention, an angle of negative theta. Now let's flip our original green ray. Let's flip it over the positive y-axis. So if you flip it over the positive y-axis, we're going to go from there all the way to right over there. And we can draw ourselves a ray."}, {"video_title": "Symmetry of trig values Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Now let's flip our original green ray. Let's flip it over the positive y-axis. So if you flip it over the positive y-axis, we're going to go from there all the way to right over there. And we can draw ourselves a ray. So my best attempt at that is right over there. And what would be the measure of this angle right over here? What was the measure of that angle in radians?"}, {"video_title": "Symmetry of trig values Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And we can draw ourselves a ray. So my best attempt at that is right over there. And what would be the measure of this angle right over here? What was the measure of that angle in radians? Well, we know if we were to go all the way from the positive x-axis to the negative x-axis, that would be pi radians, because that's halfway around the circle. So this angle, since we know that that's theta, this is theta right over here, the angle that we want to figure out, this is going to be all the way around. It's going to be pi minus theta."}, {"video_title": "Symmetry of trig values Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "What was the measure of that angle in radians? Well, we know if we were to go all the way from the positive x-axis to the negative x-axis, that would be pi radians, because that's halfway around the circle. So this angle, since we know that that's theta, this is theta right over here, the angle that we want to figure out, this is going to be all the way around. It's going to be pi minus theta. Notice pi minus theta plus theta, these two are supplementary, and they add up to pi radians, or 180 degrees. Now let's flip this one over the negative x-axis. So if we flip this one over the negative x-axis, you're going to get right over there."}, {"video_title": "Symmetry of trig values Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "It's going to be pi minus theta. Notice pi minus theta plus theta, these two are supplementary, and they add up to pi radians, or 180 degrees. Now let's flip this one over the negative x-axis. So if we flip this one over the negative x-axis, you're going to get right over there. And so you're going to get an angle that looks like this. And now what is going to be the measure of this angle? So if we go all the way around like that, what is the measure of that angle?"}, {"video_title": "Symmetry of trig values Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So if we flip this one over the negative x-axis, you're going to get right over there. And so you're going to get an angle that looks like this. And now what is going to be the measure of this angle? So if we go all the way around like that, what is the measure of that angle? Well, to go this far is pi. And then you're going another theta. This angle right over here is theta."}, {"video_title": "Symmetry of trig values Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So if we go all the way around like that, what is the measure of that angle? Well, to go this far is pi. And then you're going another theta. This angle right over here is theta. So you're going pi plus another theta. So this whole angle right over here, this whole thing, is pi plus theta radians. Let me just write that down."}, {"video_title": "Symmetry of trig values Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "This angle right over here is theta. So you're going pi plus another theta. So this whole angle right over here, this whole thing, is pi plus theta radians. Let me just write that down. So this is pi plus theta. Now that we've figured out these have different symmetries about them, let's think about how the sines and cosines of these different angles relate to each other. So we already know that this coordinate right over here, that is sine of theta, sorry, the x-coordinate is cosine of theta."}, {"video_title": "Symmetry of trig values Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Let me just write that down. So this is pi plus theta. Now that we've figured out these have different symmetries about them, let's think about how the sines and cosines of these different angles relate to each other. So we already know that this coordinate right over here, that is sine of theta, sorry, the x-coordinate is cosine of theta. The x-coordinate is cosine of theta. And the y-coordinate is sine of theta. Or another way of thinking about it is this value on the x-axis is cosine of theta."}, {"video_title": "Symmetry of trig values Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So we already know that this coordinate right over here, that is sine of theta, sorry, the x-coordinate is cosine of theta. The x-coordinate is cosine of theta. And the y-coordinate is sine of theta. Or another way of thinking about it is this value on the x-axis is cosine of theta. And this value right over here on the y-axis is sine of theta. Now let's think about this one down over here. By the same convention, this point, this is really the unit circle definition of our trig functions."}, {"video_title": "Symmetry of trig values Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Or another way of thinking about it is this value on the x-axis is cosine of theta. And this value right over here on the y-axis is sine of theta. Now let's think about this one down over here. By the same convention, this point, this is really the unit circle definition of our trig functions. This point, since our angle is negative theta now, this point would be cosine of negative theta, comma, sine of negative theta. And we can apply the same thing over here. This point right over here, the x-coordinate, is cosine of pi minus theta."}, {"video_title": "Symmetry of trig values Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "By the same convention, this point, this is really the unit circle definition of our trig functions. This point, since our angle is negative theta now, this point would be cosine of negative theta, comma, sine of negative theta. And we can apply the same thing over here. This point right over here, the x-coordinate, is cosine of pi minus theta. That's what this angle is when we go from the positive x-axis. This is cosine of pi minus theta. And the y-coordinate is the sine of pi minus theta."}, {"video_title": "Symmetry of trig values Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "This point right over here, the x-coordinate, is cosine of pi minus theta. That's what this angle is when we go from the positive x-axis. This is cosine of pi minus theta. And the y-coordinate is the sine of pi minus theta. And then we could go all the way around to this point. I think you see where this is going. This is cosine of, I guess we could say, theta plus pi or pi plus theta."}, {"video_title": "Symmetry of trig values Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And the y-coordinate is the sine of pi minus theta. And then we could go all the way around to this point. I think you see where this is going. This is cosine of, I guess we could say, theta plus pi or pi plus theta. That's right, pi plus theta. And sine of pi plus theta. And now how do these all relate to each other?"}, {"video_title": "Symmetry of trig values Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "This is cosine of, I guess we could say, theta plus pi or pi plus theta. That's right, pi plus theta. And sine of pi plus theta. And now how do these all relate to each other? Well, notice over here, out here on the right-hand side, our x-coordinates are the exact same value. It's this value right over here. So we know that cosine of theta must be equal to the cosine of negative theta."}, {"video_title": "Symmetry of trig values Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And now how do these all relate to each other? Well, notice over here, out here on the right-hand side, our x-coordinates are the exact same value. It's this value right over here. So we know that cosine of theta must be equal to the cosine of negative theta. So that's pretty interesting. Let's write that down. Cosine of theta is equal to the cosine of negative theta."}, {"video_title": "Symmetry of trig values Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So we know that cosine of theta must be equal to the cosine of negative theta. So that's pretty interesting. Let's write that down. Cosine of theta is equal to the cosine of negative theta. That's a pretty interesting result. But what about their sines? Well, here the sine of theta is this distance above the x-axis."}, {"video_title": "Symmetry of trig values Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Cosine of theta is equal to the cosine of negative theta. That's a pretty interesting result. But what about their sines? Well, here the sine of theta is this distance above the x-axis. And here the sine of negative theta is the same distance below the x-axis. So they're going to be the negatives of each other. So we could say that sine of negative theta is equal to the negative sine of theta."}, {"video_title": "Symmetry of trig values Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Well, here the sine of theta is this distance above the x-axis. And here the sine of negative theta is the same distance below the x-axis. So they're going to be the negatives of each other. So we could say that sine of negative theta is equal to the negative sine of theta. It's the opposite. If you go the same amount above or below the x-axis, you're going to get the negative value for the sine. And we could do the same thing over here."}, {"video_title": "Symmetry of trig values Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So we could say that sine of negative theta is equal to the negative sine of theta. It's the opposite. If you go the same amount above or below the x-axis, you're going to get the negative value for the sine. And we could do the same thing over here. How does this one relate to that? Well, these two are going to have the same sine values. The sine of this, the y-coordinate, is the same as the sine of that."}, {"video_title": "Symmetry of trig values Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And we could do the same thing over here. How does this one relate to that? Well, these two are going to have the same sine values. The sine of this, the y-coordinate, is the same as the sine of that. So we see that this must be equal to that. Let's write that down. So we get sine of theta is equal to sine of pi minus theta."}, {"video_title": "Symmetry of trig values Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "The sine of this, the y-coordinate, is the same as the sine of that. So we see that this must be equal to that. Let's write that down. So we get sine of theta is equal to sine of pi minus theta. Now let's think about how do the cosines relate. Well, by the same argument, they're going to be the opposites of each other, where the x-coordinates are the same distance but on opposite sides of the origin. So we get cosine of theta is equal to the negative of the cosine of pi minus theta."}, {"video_title": "Symmetry of trig values Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So we get sine of theta is equal to sine of pi minus theta. Now let's think about how do the cosines relate. Well, by the same argument, they're going to be the opposites of each other, where the x-coordinates are the same distance but on opposite sides of the origin. So we get cosine of theta is equal to the negative of the cosine of pi minus theta. And now finally, let's think about how this one relates. Well, here, our cosine value, our x-coordinate, is the negative. And our sine value is also the negative."}, {"video_title": "Symmetry of trig values Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So we get cosine of theta is equal to the negative of the cosine of pi minus theta. And now finally, let's think about how this one relates. Well, here, our cosine value, our x-coordinate, is the negative. And our sine value is also the negative. We've kind of flipped over both axes. So let's write that down. Over here, we have sine of theta plus pi, which is the same thing as pi plus theta, is equal to the negative of the sine of theta."}, {"video_title": "Symmetry of trig values Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And our sine value is also the negative. We've kind of flipped over both axes. So let's write that down. Over here, we have sine of theta plus pi, which is the same thing as pi plus theta, is equal to the negative of the sine of theta. And we see that this is sine of theta. This is sine of pi plus theta, or sine of theta plus pi. And we get the cosine of theta plus pi is going to be the negative of cosine of theta."}, {"video_title": "Symmetry of trig values Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Over here, we have sine of theta plus pi, which is the same thing as pi plus theta, is equal to the negative of the sine of theta. And we see that this is sine of theta. This is sine of pi plus theta, or sine of theta plus pi. And we get the cosine of theta plus pi is going to be the negative of cosine of theta. So even here, and you could keep going. You could try to relate this one to that one, or that one to that one. You can get all sorts of interesting results."}, {"video_title": "Proof of the law of cosines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "But instead of just being able to do the Pythagorean theorem, and because it was a right triangle, it was just kind of a normal triangle. And it wasn't a right triangle. And we just kind of chugged through it using Sohcahtoa and just our very simple trig functions, and we got to the right answer. What I want to do now is introduce you to something called the law of cosines, which we essentially proved in the last video. But I want to kind of prove it in a more, you know, without the word problem getting in the way. And I want to show you, once you know the law of cosines, you can then apply it to a problem like we did in the past, and you'll do it faster. I have a bit of a mixed opinion about it, because I'm not a big fan of memorizing things."}, {"video_title": "Proof of the law of cosines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "What I want to do now is introduce you to something called the law of cosines, which we essentially proved in the last video. But I want to kind of prove it in a more, you know, without the word problem getting in the way. And I want to show you, once you know the law of cosines, you can then apply it to a problem like we did in the past, and you'll do it faster. I have a bit of a mixed opinion about it, because I'm not a big fan of memorizing things. You know, when you're 40 years old, you probably won't have the law of cosines still memorized. But if you have that ability to start with the trig functions and just move forward, then you'll always be set, and I'd be impressed if you're still doing trig at 40, but who knows. So let's go and let's see what this law of cosines is all about."}, {"video_title": "Proof of the law of cosines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "I have a bit of a mixed opinion about it, because I'm not a big fan of memorizing things. You know, when you're 40 years old, you probably won't have the law of cosines still memorized. But if you have that ability to start with the trig functions and just move forward, then you'll always be set, and I'd be impressed if you're still doing trig at 40, but who knows. So let's go and let's see what this law of cosines is all about. So let's say that I know this angle theta. And I know, let's call this side, I don't know, a, let's call this side b. I'm being a little arbitrary here. Let's call that, actually, let me stay in the colors of the side."}, {"video_title": "Proof of the law of cosines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So let's go and let's see what this law of cosines is all about. So let's say that I know this angle theta. And I know, let's call this side, I don't know, a, let's call this side b. I'm being a little arbitrary here. Let's call that, actually, let me stay in the colors of the side. So let's call that b, and let's call this c, and then let's call this side a. So if this was a right triangle, then we could have used the Pythagorean theorem somehow, but now we can't. So what do we do?"}, {"video_title": "Proof of the law of cosines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Let's call that, actually, let me stay in the colors of the side. So let's call that b, and let's call this c, and then let's call this side a. So if this was a right triangle, then we could have used the Pythagorean theorem somehow, but now we can't. So what do we do? So we know a, well, let's assume that we know b, we know c, we know theta, and that we want to solve for a. But in general, as long as you know three of these, you can solve for the fourth, once you know the law of cosines. So how can we do it?"}, {"video_title": "Proof of the law of cosines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So what do we do? So we know a, well, let's assume that we know b, we know c, we know theta, and that we want to solve for a. But in general, as long as you know three of these, you can solve for the fourth, once you know the law of cosines. So how can we do it? Well, we're going to do it the exact same way we did that last problem. We can drop a line here to make, oh my god, that's messy. I thought I was using the line tool."}, {"video_title": "Proof of the law of cosines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So how can we do it? Well, we're going to do it the exact same way we did that last problem. We can drop a line here to make, oh my god, that's messy. I thought I was using the line tool. Edit, undo. So I can drop a line like that. So I have two right angles."}, {"video_title": "Proof of the law of cosines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "I thought I was using the line tool. Edit, undo. So I can drop a line like that. So I have two right angles. And then once I have right triangles, then now I can start to use trig functions and Pythagorean theorem, et cetera, et cetera. So let's see, so this is a right angle, and this is a right angle. So what is this side here?"}, {"video_title": "Proof of the law of cosines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So I have two right angles. And then once I have right triangles, then now I can start to use trig functions and Pythagorean theorem, et cetera, et cetera. So let's see, so this is a right angle, and this is a right angle. So what is this side here? What is, let me pick another color. I'm probably going to get too involved with all of the colors, but it's for your improvement. So what is this side here?"}, {"video_title": "Proof of the law of cosines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So what is this side here? What is, let me pick another color. I'm probably going to get too involved with all of the colors, but it's for your improvement. So what is this side here? What is the length of that side, that purple side? Well, that purple side is just, you know, we use Soh-Kah-Toa. I always just get to write Soh-Kah-Toa up here."}, {"video_title": "Proof of the law of cosines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So what is this side here? What is the length of that side, that purple side? Well, that purple side is just, you know, we use Soh-Kah-Toa. I always just get to write Soh-Kah-Toa up here. Soh-Kah-Toa. So this purple side is adjacent to theta, and then this blue or mauve side, b, is a hypotenuse, right, of this right triangle. So we know that, actually I'm going to stick to one color because it'll take me forever if I keep switching colors."}, {"video_title": "Proof of the law of cosines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "I always just get to write Soh-Kah-Toa up here. Soh-Kah-Toa. So this purple side is adjacent to theta, and then this blue or mauve side, b, is a hypotenuse, right, of this right triangle. So we know that, actually I'm going to stick to one color because it'll take me forever if I keep switching colors. We know that cosine of theta, let's call this side, let's call this kind of sub-side, let's call this, I don't know, let's call this d, side d. We know that cosine of theta is equal to d over b, right? And we know b. Or that d is equal to what?"}, {"video_title": "Proof of the law of cosines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So we know that, actually I'm going to stick to one color because it'll take me forever if I keep switching colors. We know that cosine of theta, let's call this side, let's call this kind of sub-side, let's call this, I don't know, let's call this d, side d. We know that cosine of theta is equal to d over b, right? And we know b. Or that d is equal to what? It equals b cosine theta. Now let's call this side e, right here. e. Well, what's e?"}, {"video_title": "Proof of the law of cosines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Or that d is equal to what? It equals b cosine theta. Now let's call this side e, right here. e. Well, what's e? Well, e is this whole c side minus this d side, right? So e is equal to c minus d. We just solved for d. So side e is equal to c minus b cosine of theta. So that's e. We got e out of the way."}, {"video_title": "Proof of the law of cosines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "e. Well, what's e? Well, e is this whole c side minus this d side, right? So e is equal to c minus d. We just solved for d. So side e is equal to c minus b cosine of theta. So that's e. We got e out of the way. What's this magenta side going to be? Well, let's call this magenta. Let's call it m for magenta."}, {"video_title": "Proof of the law of cosines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So that's e. We got e out of the way. What's this magenta side going to be? Well, let's call this magenta. Let's call it m for magenta. Well, m is opposite to theta, and so what involves, now we know that we've solved for c as well, but we know b and b is simple, so what relationship gives us m over b, or involves the opposite in the hypotenuse? Well, that's sine, opposite over hypotenuse. So we know that m over b is equal to sine of theta."}, {"video_title": "Proof of the law of cosines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Let's call it m for magenta. Well, m is opposite to theta, and so what involves, now we know that we've solved for c as well, but we know b and b is simple, so what relationship gives us m over b, or involves the opposite in the hypotenuse? Well, that's sine, opposite over hypotenuse. So we know that m over b is equal to sine of theta. We know that, let me go here, m over b, right, because it's the hypotenuse, is equal to sine of theta, or that m is equal to b sine of theta. So we figured out m, we figured out e, and now we want to figure out a. And this should jump out at you."}, {"video_title": "Proof of the law of cosines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So we know that m over b is equal to sine of theta. We know that, let me go here, m over b, right, because it's the hypotenuse, is equal to sine of theta, or that m is equal to b sine of theta. So we figured out m, we figured out e, and now we want to figure out a. And this should jump out at you. We have two sides of a right triangle, and we want to figure out the hypotenuse. We can use the Pythagorean theorem. Pythagorean theorem tells us a squared is equal to m squared plus e squared, right?"}, {"video_title": "Proof of the law of cosines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And this should jump out at you. We have two sides of a right triangle, and we want to figure out the hypotenuse. We can use the Pythagorean theorem. Pythagorean theorem tells us a squared is equal to m squared plus e squared, right? Just the square of the other two sides. Well, what's m squared plus e squared? Let me switch to another color, just to be arbitrary."}, {"video_title": "Proof of the law of cosines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Pythagorean theorem tells us a squared is equal to m squared plus e squared, right? Just the square of the other two sides. Well, what's m squared plus e squared? Let me switch to another color, just to be arbitrary. So a squared is equal to m squared. m is b sine of theta. So it's b sine of theta squared plus e squared."}, {"video_title": "Proof of the law of cosines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Let me switch to another color, just to be arbitrary. So a squared is equal to m squared. m is b sine of theta. So it's b sine of theta squared plus e squared. Well, e we figured out is this. So it's plus c minus b cosine theta squared. Now let's just chug through some algebra."}, {"video_title": "Proof of the law of cosines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So it's b sine of theta squared plus e squared. Well, e we figured out is this. So it's plus c minus b cosine theta squared. Now let's just chug through some algebra. So that equals b squared sine squared theta. Sine squared theta just means sine of theta squared, right? Plus, and let me just FOIL this out."}, {"video_title": "Proof of the law of cosines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Now let's just chug through some algebra. So that equals b squared sine squared theta. Sine squared theta just means sine of theta squared, right? Plus, and let me just FOIL this out. Although I don't like using FOIL, I just multiply it out. But c squared minus 2 cb cosine theta plus b squared cosine theta. I just expanded this out by multiplying it out."}, {"video_title": "Proof of the law of cosines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Plus, and let me just FOIL this out. Although I don't like using FOIL, I just multiply it out. But c squared minus 2 cb cosine theta plus b squared cosine theta. I just expanded this out by multiplying it out. And now let's see if we can do anything interesting. Well, if we take this term and this term, we get that those two terms are b squared sine squared of theta plus b squared cosine. There should be a squared there, right?"}, {"video_title": "Proof of the law of cosines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "I just expanded this out by multiplying it out. And now let's see if we can do anything interesting. Well, if we take this term and this term, we get that those two terms are b squared sine squared of theta plus b squared cosine. There should be a squared there, right? Because we squared it. b squared cosine squared of theta. And then we have plus c squared minus 2 bc cosine theta."}, {"video_title": "Proof of the law of cosines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "There should be a squared there, right? Because we squared it. b squared cosine squared of theta. And then we have plus c squared minus 2 bc cosine theta. Well, what does this simplify to? Well, this is the same thing as this equals b squared times the sine squared theta plus cosine squared of theta. Something should be jumping out at you."}, {"video_title": "Proof of the law of cosines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And then we have plus c squared minus 2 bc cosine theta. Well, what does this simplify to? Well, this is the same thing as this equals b squared times the sine squared theta plus cosine squared of theta. Something should be jumping out at you. And that's plus c squared minus 2 bc cosine of theta. Well, this thing, sine squared plus cosine squared of any angle is 1. That's one of the earlier identities."}, {"video_title": "Proof of the law of cosines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Something should be jumping out at you. And that's plus c squared minus 2 bc cosine of theta. Well, this thing, sine squared plus cosine squared of any angle is 1. That's one of the earlier identities. That's a Pythagorean identity right there. So this equals 1. So then we're left with, going back to my original color, we're almost there."}, {"video_title": "Proof of the law of cosines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "That's one of the earlier identities. That's a Pythagorean identity right there. So this equals 1. So then we're left with, going back to my original color, we're almost there. a squared is equal to, this term just becomes 1, so b squared, we're just left with a b squared, right? Plus c squared minus 2 bc cosine of theta. That's pretty neat."}, {"video_title": "Proof of the law of cosines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So then we're left with, going back to my original color, we're almost there. a squared is equal to, this term just becomes 1, so b squared, we're just left with a b squared, right? Plus c squared minus 2 bc cosine of theta. That's pretty neat. And this is called the law of cosines. And it's useful because if you know an angle and 2 of the sides of any triangle, you can now solve for the other side. Or, really, if you want to, if you know 3 sides of a triangle, you can now solve for any angle."}, {"video_title": "Proof of the law of cosines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "That's pretty neat. And this is called the law of cosines. And it's useful because if you know an angle and 2 of the sides of any triangle, you can now solve for the other side. Or, really, if you want to, if you know 3 sides of a triangle, you can now solve for any angle. So that also is very useful. The only reason why I'm a little bit here, there, is if you are in trigonometry right now and you might have a test, you should memorize this because it'll make you faster and you'll get the answer right quicker. I'm not a big fan of just memorizing it without knowing where it came from because a year from now or two years from now, when you go to college and it's been four years since you took trigonometry, you probably won't have this memorized."}, {"video_title": "Proof of the law of cosines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Or, really, if you want to, if you know 3 sides of a triangle, you can now solve for any angle. So that also is very useful. The only reason why I'm a little bit here, there, is if you are in trigonometry right now and you might have a test, you should memorize this because it'll make you faster and you'll get the answer right quicker. I'm not a big fan of just memorizing it without knowing where it came from because a year from now or two years from now, when you go to college and it's been four years since you took trigonometry, you probably won't have this memorized. And if you face a trig problem all of a sudden, it's good to kind of get there from scratch. With that said, this is the law of cosines. And if you use the law of cosines, you could have done that problem we just did a lot faster because you just have to set up the triangle and then just substitute into this and you could have solved for a in that ship off-course problem."}, {"video_title": "Example Converting degrees to radians Trigonometry Khan Academy.mp3", "Sentence": "So let's think about the relationship between degrees and radians. And to do that, let me just draw a little circle here. So that's the center of the circle. And then do my best shot, best attempt to freehand draw a reasonable looking circle. Yeah, that's not, I've done worse. I've done worse than that. All right."}, {"video_title": "Example Converting degrees to radians Trigonometry Khan Academy.mp3", "Sentence": "And then do my best shot, best attempt to freehand draw a reasonable looking circle. Yeah, that's not, I've done worse. I've done worse than that. All right. Now, if we were to go in degrees, if we were to go one time around the circle like that, how many degrees is that? Well, we know that that would be 360 degrees. Well, if we did the same thing, how many radians is that?"}, {"video_title": "Example Converting degrees to radians Trigonometry Khan Academy.mp3", "Sentence": "All right. Now, if we were to go in degrees, if we were to go one time around the circle like that, how many degrees is that? Well, we know that that would be 360 degrees. Well, if we did the same thing, how many radians is that? If we were to go all the way around the circle? Well, we just have to remember, when we're measuring in terms of radians, we're really talking about the arc that subtends that angle. So if you go all the way around, you're really talking about the arc length of the entire circle, or essentially the circumference of the circle."}, {"video_title": "Example Converting degrees to radians Trigonometry Khan Academy.mp3", "Sentence": "Well, if we did the same thing, how many radians is that? If we were to go all the way around the circle? Well, we just have to remember, when we're measuring in terms of radians, we're really talking about the arc that subtends that angle. So if you go all the way around, you're really talking about the arc length of the entire circle, or essentially the circumference of the circle. And you're essentially saying how many radiuses this is, or radii, or how many radii is the circumference of the circle? You know a circumference of a circle is two pi, is two pi times the radius. Or you could say that the length of the circumference of the circle is two pi radii."}, {"video_title": "Example Converting degrees to radians Trigonometry Khan Academy.mp3", "Sentence": "So if you go all the way around, you're really talking about the arc length of the entire circle, or essentially the circumference of the circle. And you're essentially saying how many radiuses this is, or radii, or how many radii is the circumference of the circle? You know a circumference of a circle is two pi, is two pi times the radius. Or you could say that the length of the circumference of the circle is two pi radii. Two pi radii. If you wanna know the exact length, you just have to get the length of the radius and multiply it by two pi. That just comes from, really actually the definition of pi, but it comes from what we know as the formula for the circumference of a circle."}, {"video_title": "Example Converting degrees to radians Trigonometry Khan Academy.mp3", "Sentence": "Or you could say that the length of the circumference of the circle is two pi radii. Two pi radii. If you wanna know the exact length, you just have to get the length of the radius and multiply it by two pi. That just comes from, really actually the definition of pi, but it comes from what we know as the formula for the circumference of a circle. So if we were to go all the way around this, this is also two pi radians. This is also two pi, two pi radians. So that tells us that two pi radians, as an angle measure, is the exact same thing, and I'm gonna write it out as 360, 360 degrees."}, {"video_title": "Example Converting degrees to radians Trigonometry Khan Academy.mp3", "Sentence": "That just comes from, really actually the definition of pi, but it comes from what we know as the formula for the circumference of a circle. So if we were to go all the way around this, this is also two pi radians. This is also two pi, two pi radians. So that tells us that two pi radians, as an angle measure, is the exact same thing, and I'm gonna write it out as 360, 360 degrees. And then we can take all of this relationship and manipulate it in different ways. If we wanna simplify a little bit, we can divide both sides of this equation by two, in which case you are left with, if you divide both sides by two, you are left with pi radians, pi radians is equal to 180 degrees, is equal to 180 degrees, 180 degrees. So how can we use this relationship now to figure out what 150 degrees is?"}, {"video_title": "Example Converting degrees to radians Trigonometry Khan Academy.mp3", "Sentence": "So that tells us that two pi radians, as an angle measure, is the exact same thing, and I'm gonna write it out as 360, 360 degrees. And then we can take all of this relationship and manipulate it in different ways. If we wanna simplify a little bit, we can divide both sides of this equation by two, in which case you are left with, if you divide both sides by two, you are left with pi radians, pi radians is equal to 180 degrees, is equal to 180 degrees, 180 degrees. So how can we use this relationship now to figure out what 150 degrees is? Well, this relationship we could write in different ways. We could divide both sides by 180 degrees, and we could get pi radians, pi radians over 180 degrees, over 180 degrees is equal to one, which is just another way of saying that there are pi radians for every 180 degrees, or you could say pi over 180 radians per degree. The other option, you could divide both sides of this by pi radians."}, {"video_title": "Example Converting degrees to radians Trigonometry Khan Academy.mp3", "Sentence": "So how can we use this relationship now to figure out what 150 degrees is? Well, this relationship we could write in different ways. We could divide both sides by 180 degrees, and we could get pi radians, pi radians over 180 degrees, over 180 degrees is equal to one, which is just another way of saying that there are pi radians for every 180 degrees, or you could say pi over 180 radians per degree. The other option, you could divide both sides of this by pi radians. You could say, you would get, on the left-hand side, you get one, and you would also get, on the right-hand side, you would get 180 degrees for every pi radians, 180 degrees for every pi radians, for every pi radians, or you could interpret this as 180 over pi degrees per radian. So how would we figure out, how would we do what they asked us? So let's convert 150 degrees to radians."}, {"video_title": "Example Converting degrees to radians Trigonometry Khan Academy.mp3", "Sentence": "The other option, you could divide both sides of this by pi radians. You could say, you would get, on the left-hand side, you get one, and you would also get, on the right-hand side, you would get 180 degrees for every pi radians, 180 degrees for every pi radians, for every pi radians, or you could interpret this as 180 over pi degrees per radian. So how would we figure out, how would we do what they asked us? So let's convert 150 degrees to radians. Let me write the word out. So 150 degrees, 150 degrees. Well, we wanna convert this to radians, so we really care about how many radians there are, how many radians there are per degree."}, {"video_title": "Example Converting degrees to radians Trigonometry Khan Academy.mp3", "Sentence": "So let's convert 150 degrees to radians. Let me write the word out. So 150 degrees, 150 degrees. Well, we wanna convert this to radians, so we really care about how many radians there are, how many radians there are per degree. Actually, let me do that in that color. So we care about how many radians, radians there are per degree, per, let me do that same green color, per degree. Well, how many radians are there per degree?"}, {"video_title": "Example Converting degrees to radians Trigonometry Khan Academy.mp3", "Sentence": "Well, we wanna convert this to radians, so we really care about how many radians there are, how many radians there are per degree. Actually, let me do that in that color. So we care about how many radians, radians there are per degree, per, let me do that same green color, per degree. Well, how many radians are there per degree? We already know, there's pi radians for every 180 degrees, or there are pi, pi, let me do that yellow color, there are pi over 180, pi over 180 radians per degree. And so if we multiply, and this all works out, because you have degrees in the numerator, degrees in the denominator, these cancel out, and so you are left with 150 times pi divided by 180 radians. So what do we get?"}, {"video_title": "Example Converting degrees to radians Trigonometry Khan Academy.mp3", "Sentence": "Well, how many radians are there per degree? We already know, there's pi radians for every 180 degrees, or there are pi, pi, let me do that yellow color, there are pi over 180, pi over 180 radians per degree. And so if we multiply, and this all works out, because you have degrees in the numerator, degrees in the denominator, these cancel out, and so you are left with 150 times pi divided by 180 radians. So what do we get? Well, this becomes, let me just rewrite it, 150 times pi, times pi, all of that over 180, all of that over 180, so this is equal to, and we get it in radians, radians. And so if we simplify it, let's see, we can divide the numerator and the denominator both by, looks like 30. So if you divide the numerator by 30, you get five."}, {"video_title": "Example Converting degrees to radians Trigonometry Khan Academy.mp3", "Sentence": "So what do we get? Well, this becomes, let me just rewrite it, 150 times pi, times pi, all of that over 180, all of that over 180, so this is equal to, and we get it in radians, radians. And so if we simplify it, let's see, we can divide the numerator and the denominator both by, looks like 30. So if you divide the numerator by 30, you get five. You divide the denominator by 30, you get six. So you get five pi over six radians, or five six pi radians, depending how you want to do it. Now let's do the same thing for negative 45 degrees."}, {"video_title": "Example Converting degrees to radians Trigonometry Khan Academy.mp3", "Sentence": "So if you divide the numerator by 30, you get five. You divide the denominator by 30, you get six. So you get five pi over six radians, or five six pi radians, depending how you want to do it. Now let's do the same thing for negative 45 degrees. What do you get for negative 45 degrees if you were to convert that to radians? Well, same exact process. You have negative, and I'll do this one a little quicker, for negative 45 degrees, I'll write down the word, times pi radians, pi radians for every 180 degrees, for every 180 degrees, the degrees cancel out, and you are left with negative 45 pi over 180 radians."}, {"video_title": "Example Converting degrees to radians Trigonometry Khan Academy.mp3", "Sentence": "Now let's do the same thing for negative 45 degrees. What do you get for negative 45 degrees if you were to convert that to radians? Well, same exact process. You have negative, and I'll do this one a little quicker, for negative 45 degrees, I'll write down the word, times pi radians, pi radians for every 180 degrees, for every 180 degrees, the degrees cancel out, and you are left with negative 45 pi over 180 radians. So this is equal to negative 45 pi, pi over 180, over 180 radians, radians. So how can we simplify this? Well, it looks like they're both, at minimum, divisible by nine."}, {"video_title": "Example Converting degrees to radians Trigonometry Khan Academy.mp3", "Sentence": "You have negative, and I'll do this one a little quicker, for negative 45 degrees, I'll write down the word, times pi radians, pi radians for every 180 degrees, for every 180 degrees, the degrees cancel out, and you are left with negative 45 pi over 180 radians. So this is equal to negative 45 pi, pi over 180, over 180 radians, radians. So how can we simplify this? Well, it looks like they're both, at minimum, divisible by nine. Nine times five is 45, this is nine times 20, so actually it's gonna be divisible by more than just, let's see, nine, nine, they're both, well, I will just, actually, they're both divisible by 45, what am I doing? Okay, so if you divide the numerator by 45, you get one. You divide the denominator by 45, 45 goes into 180 four times, four times."}, {"video_title": "Interpreting solutions of trigonometric equations Trigonometry Precalculus Khan Academy.mp3", "Sentence": "Alvaro presses the treadle of a spinning wheel with his foot. It moves a bar up and down, making the wheel spin. So just to be clear what a treadle is, this is an old spinning wheel, and this little pedal, that is a treadle. And as this goes up and down, it's going to pull on this bar, which is then going to spin this wheel, which can then be used to essentially power the machine. So it says the function B of t models the height in centimeters of the top of the bar when Alvaro has pressed the treadle for t seconds. So it's telling us the height of, I can barely see where the top of the bar is, someplace over here. And this isn't exactly what they're probably talking about in this exercise here, but this is just to give you a visualization of what a treadle is and what the bar is, and then what the spinning wheel is."}, {"video_title": "Interpreting solutions of trigonometric equations Trigonometry Precalculus Khan Academy.mp3", "Sentence": "And as this goes up and down, it's going to pull on this bar, which is then going to spin this wheel, which can then be used to essentially power the machine. So it says the function B of t models the height in centimeters of the top of the bar when Alvaro has pressed the treadle for t seconds. So it's telling us the height of, I can barely see where the top of the bar is, someplace over here. And this isn't exactly what they're probably talking about in this exercise here, but this is just to give you a visualization of what a treadle is and what the bar is, and then what the spinning wheel is. Alvaro has pressed those treadle for t seconds. So they give us B of t right over here, 90 minus 12 times sine of 5t. The first question is, what does the solution set to y is equal to 90 minus 12 times sine of five times six represent?"}, {"video_title": "Interpreting solutions of trigonometric equations Trigonometry Precalculus Khan Academy.mp3", "Sentence": "And this isn't exactly what they're probably talking about in this exercise here, but this is just to give you a visualization of what a treadle is and what the bar is, and then what the spinning wheel is. Alvaro has pressed those treadle for t seconds. So they give us B of t right over here, 90 minus 12 times sine of 5t. The first question is, what does the solution set to y is equal to 90 minus 12 times sine of five times six represent? Pause this video and see if you can think about that. All right, so it looks like right over here, so we have the 90, 90, 12, 12, and we're subtracting 12 sine of five times t, five times t. So this right over here is t. So this is the solution, the solution set right over here tells us what is the height because that's what B of t is. So B of t is equal to y."}, {"video_title": "Interpreting solutions of trigonometric equations Trigonometry Precalculus Khan Academy.mp3", "Sentence": "The first question is, what does the solution set to y is equal to 90 minus 12 times sine of five times six represent? Pause this video and see if you can think about that. All right, so it looks like right over here, so we have the 90, 90, 12, 12, and we're subtracting 12 sine of five times t, five times t. So this right over here is t. So this is the solution, the solution set right over here tells us what is the height because that's what B of t is. So B of t is equal to y. What is the height when t is equal to six? And remember, t is in seconds. So this is height, height of top of bar, top of bar, top of bar at six seconds."}, {"video_title": "Interpreting solutions of trigonometric equations Trigonometry Precalculus Khan Academy.mp3", "Sentence": "So B of t is equal to y. What is the height when t is equal to six? And remember, t is in seconds. So this is height, height of top of bar, top of bar, top of bar at six seconds. All right, now we have more questions here. The next question asks us, what does the solution set to 95 equals 90 minus 12 sine of 5t represent? Pause the video and think about that."}, {"video_title": "Interpreting solutions of trigonometric equations Trigonometry Precalculus Khan Academy.mp3", "Sentence": "So this is height, height of top of bar, top of bar, top of bar at six seconds. All right, now we have more questions here. The next question asks us, what does the solution set to 95 equals 90 minus 12 sine of 5t represent? Pause the video and think about that. All right, so here they're saying that B of t is equal to 95. And so the solution set, you're really solving for t. So you're really solving for all of the times when our height is going to be 95 centimeters. So all times t when height of top of bar, of top of bar at 95 centimeters."}, {"video_title": "Interpreting solutions of trigonometric equations Trigonometry Precalculus Khan Academy.mp3", "Sentence": "Pause the video and think about that. All right, so here they're saying that B of t is equal to 95. And so the solution set, you're really solving for t. So you're really solving for all of the times when our height is going to be 95 centimeters. So all times t when height of top of bar, of top of bar at 95 centimeters. And that's going to keep happening over and over and over again as t goes forward in time. So you're going to have a very large, you're gonna have an infinite solution set over here. You're gonna have an infinite number of t's at which your solution, at which the top of the bar is at 95 centimeters."}, {"video_title": "Interpreting solutions of trigonometric equations Trigonometry Precalculus Khan Academy.mp3", "Sentence": "So all times t when height of top of bar, of top of bar at 95 centimeters. And that's going to keep happening over and over and over again as t goes forward in time. So you're going to have a very large, you're gonna have an infinite solution set over here. You're gonna have an infinite number of t's at which your solution, at which the top of the bar is at 95 centimeters. Now we have another question. This one is asking us, what does the solution set to y is equal to 90 minus 12 sine of pi over two represent? So pause the video and think about that."}, {"video_title": "Interpreting solutions of trigonometric equations Trigonometry Precalculus Khan Academy.mp3", "Sentence": "You're gonna have an infinite number of t's at which your solution, at which the top of the bar is at 95 centimeters. Now we have another question. This one is asking us, what does the solution set to y is equal to 90 minus 12 sine of pi over two represent? So pause the video and think about that. All right, now this is pretty interesting. We can actually evaluate what sine of pi over two is. So sine of pi over two radians or sine of 90 degrees, that is going to be equal to one."}, {"video_title": "Interpreting solutions of trigonometric equations Trigonometry Precalculus Khan Academy.mp3", "Sentence": "So pause the video and think about that. All right, now this is pretty interesting. We can actually evaluate what sine of pi over two is. So sine of pi over two radians or sine of 90 degrees, that is going to be equal to one. And so that's the maximum value that this sine over here can take on. Now we're going to subtract 12 times that. So this is taking on a max."}, {"video_title": "Interpreting solutions of trigonometric equations Trigonometry Precalculus Khan Academy.mp3", "Sentence": "So sine of pi over two radians or sine of 90 degrees, that is going to be equal to one. And so that's the maximum value that this sine over here can take on. Now we're going to subtract 12 times that. So this is taking on a max. Then when you subtract 12 times that, this is actually the minimum value that you can take on. You're gonna have, you can't get any lower than this. And so this is going to be the lowest, the lowest height for the top of the bar."}, {"video_title": "Using the Pythagorean trig identity Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "It's greater than negative 3 pi over 2, and it's less than negative pi. And we're also told that sine of theta is equal to 1 half. Just from this information, can we figure out what the tangent of theta is going to be equal to? And I encourage you to pause the video and try this on your own. In case you're stumped, I will give you a hint. You should use the Pythagorean identity, the fact that sine squared theta plus cosine squared theta is equal to 1. So let's do it."}, {"video_title": "Using the Pythagorean trig identity Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And I encourage you to pause the video and try this on your own. In case you're stumped, I will give you a hint. You should use the Pythagorean identity, the fact that sine squared theta plus cosine squared theta is equal to 1. So let's do it. So we know the Pythagorean identity. Sine squared theta plus cosine squared theta is equal to 1. We know what sine squared theta is."}, {"video_title": "Using the Pythagorean trig identity Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So let's do it. So we know the Pythagorean identity. Sine squared theta plus cosine squared theta is equal to 1. We know what sine squared theta is. Sine theta is 1 half, so this could be rewritten as 1 half squared plus cosine squared theta is equal to 1. Or we could write this as 1 fourth plus cosine squared theta is equal to 1. Or we could subtract 1 fourth from both sides, and we get cosine squared theta is equal to, see, you subtract 1 fourth from the left-hand side, and this 1 fourth goes away."}, {"video_title": "Using the Pythagorean trig identity Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "We know what sine squared theta is. Sine theta is 1 half, so this could be rewritten as 1 half squared plus cosine squared theta is equal to 1. Or we could write this as 1 fourth plus cosine squared theta is equal to 1. Or we could subtract 1 fourth from both sides, and we get cosine squared theta is equal to, see, you subtract 1 fourth from the left-hand side, and this 1 fourth goes away. That was the whole point. 1 minus 1 fourth is 3 fourths. So what could cosine of theta be?"}, {"video_title": "Using the Pythagorean trig identity Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Or we could subtract 1 fourth from both sides, and we get cosine squared theta is equal to, see, you subtract 1 fourth from the left-hand side, and this 1 fourth goes away. That was the whole point. 1 minus 1 fourth is 3 fourths. So what could cosine of theta be? Well, when I square it, I get positive 3 fourths. So it could be the positive or negative square root of 3 fourths. So cosine of theta could be equal to the positive or negative square root of 3 over 4, which is the same thing as the positive or negative square root of 3 over the square root of 4, which is 2."}, {"video_title": "Using the Pythagorean trig identity Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So what could cosine of theta be? Well, when I square it, I get positive 3 fourths. So it could be the positive or negative square root of 3 fourths. So cosine of theta could be equal to the positive or negative square root of 3 over 4, which is the same thing as the positive or negative square root of 3 over the square root of 4, which is 2. So it's the positive or negative square root of 3 over 2. But how do we know which one of these it actually is? Well, that's where this information becomes useful."}, {"video_title": "Using the Pythagorean trig identity Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So cosine of theta could be equal to the positive or negative square root of 3 over 4, which is the same thing as the positive or negative square root of 3 over the square root of 4, which is 2. So it's the positive or negative square root of 3 over 2. But how do we know which one of these it actually is? Well, that's where this information becomes useful. That's where this information actually becomes useful. Let's draw our unit circle. If you're saying, well, why am I even worried about cosine of theta?"}, {"video_title": "Using the Pythagorean trig identity Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Well, that's where this information becomes useful. That's where this information actually becomes useful. Let's draw our unit circle. If you're saying, well, why am I even worried about cosine of theta? Well, if you know sine of theta, you know cosine of theta. Tangent of theta is just sine of theta over cosine theta. So then you will know the tangent of theta."}, {"video_title": "Using the Pythagorean trig identity Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "If you're saying, well, why am I even worried about cosine of theta? Well, if you know sine of theta, you know cosine of theta. Tangent of theta is just sine of theta over cosine theta. So then you will know the tangent of theta. But let's look at the unit circle to figure out which value of cosine we should use. So let me draw it. So unit circle."}, {"video_title": "Using the Pythagorean trig identity Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So then you will know the tangent of theta. But let's look at the unit circle to figure out which value of cosine we should use. So let me draw it. So unit circle. That's my y-axis. That is my x-axis. And I will draw the unit circle in pink."}, {"video_title": "Using the Pythagorean trig identity Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So unit circle. That's my y-axis. That is my x-axis. And I will draw the unit circle in pink. So that's my best attempt at drawing a circle. Please forgive me for its lack of perfect roundness. And it says theta is greater than negative 3 pi over 2."}, {"video_title": "Using the Pythagorean trig identity Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And I will draw the unit circle in pink. So that's my best attempt at drawing a circle. Please forgive me for its lack of perfect roundness. And it says theta is greater than negative 3 pi over 2. So where's negative 3 pi over 2? So let's see. This is negative pi over 2."}, {"video_title": "Using the Pythagorean trig identity Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And it says theta is greater than negative 3 pi over 2. So where's negative 3 pi over 2? So let's see. This is negative pi over 2. So this is one side of the angle. Let me do this in a color. So let's see."}, {"video_title": "Using the Pythagorean trig identity Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "This is negative pi over 2. So this is one side of the angle. Let me do this in a color. So let's see. This one side of the angle is going to be along the positive x-axis. And we want to figure out what the other side is. So it's going to this right over here."}, {"video_title": "Using the Pythagorean trig identity Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So let's see. This one side of the angle is going to be along the positive x-axis. And we want to figure out what the other side is. So it's going to this right over here. That's negative pi over 2. This is negative pi. So it's between negative pi, which is right over here."}, {"video_title": "Using the Pythagorean trig identity Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So it's going to this right over here. That's negative pi over 2. This is negative pi. So it's between negative pi, which is right over here. So let me make that clear. Negative pi is right over here. It's between negative pi and negative 3 pi over 2."}, {"video_title": "Using the Pythagorean trig identity Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So it's between negative pi, which is right over here. So let me make that clear. Negative pi is right over here. It's between negative pi and negative 3 pi over 2. Negative 3 pi over 2 is right over here. So the angle, our angle theta, is going to put us someplace over here. And the whole reason I did this, so this whole arc right here, you could think of this as the measure of angle theta right over there."}, {"video_title": "Using the Pythagorean trig identity Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "It's between negative pi and negative 3 pi over 2. Negative 3 pi over 2 is right over here. So the angle, our angle theta, is going to put us someplace over here. And the whole reason I did this, so this whole arc right here, you could think of this as the measure of angle theta right over there. And the whole reason I did that is to think about whether the cosine of theta is going to be positive or negative. We clearly see it's in the second quadrant. The cosine of theta is the x-coordinate of this point where our angle intersects the unit circle."}, {"video_title": "Using the Pythagorean trig identity Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And the whole reason I did this, so this whole arc right here, you could think of this as the measure of angle theta right over there. And the whole reason I did that is to think about whether the cosine of theta is going to be positive or negative. We clearly see it's in the second quadrant. The cosine of theta is the x-coordinate of this point where our angle intersects the unit circle. So this point right over here, this right over here, actually let me do it in that orange color again, this right over here, that is the cosine of theta. Now is that a positive or negative value? Well, it's clearly a negative value."}, {"video_title": "Using the Pythagorean trig identity Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "The cosine of theta is the x-coordinate of this point where our angle intersects the unit circle. So this point right over here, this right over here, actually let me do it in that orange color again, this right over here, that is the cosine of theta. Now is that a positive or negative value? Well, it's clearly a negative value. So for the sake of this example, our cosine theta is not the positive one. It is the negative one. So we can write that cosine theta is equal to the negative square root of 3 over 2."}, {"video_title": "Using the Pythagorean trig identity Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Well, it's clearly a negative value. So for the sake of this example, our cosine theta is not the positive one. It is the negative one. So we can write that cosine theta is equal to the negative square root of 3 over 2. So we figured out cosine theta, but we still have to figure out tangent of theta. And we just have to remind ourselves that the tangent of theta is going to be equal to the sine of theta over the cosine of theta. Well, they told us the sine of theta is 1 half."}, {"video_title": "Using the Pythagorean trig identity Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So we can write that cosine theta is equal to the negative square root of 3 over 2. So we figured out cosine theta, but we still have to figure out tangent of theta. And we just have to remind ourselves that the tangent of theta is going to be equal to the sine of theta over the cosine of theta. Well, they told us the sine of theta is 1 half. So it's going to be 1 half over cosine of theta, which is negative square root of 3 over 2. And what does that equal? Well, that's the same thing as 1 half times the reciprocal of this."}, {"video_title": "Using the Pythagorean trig identity Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Well, they told us the sine of theta is 1 half. So it's going to be 1 half over cosine of theta, which is negative square root of 3 over 2. And what does that equal? Well, that's the same thing as 1 half times the reciprocal of this. So times negative 2 over the square root of 3. These 2's will cancel out. And we are left with negative 1 over the square root of 3."}, {"video_title": "Using the Pythagorean trig identity Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Well, that's the same thing as 1 half times the reciprocal of this. So times negative 2 over the square root of 3. These 2's will cancel out. And we are left with negative 1 over the square root of 3. Now some people don't like a radical in the denominator like this. They don't like an irrational denominator. So we could rationalize the denominator here by multiplying by square root of 3 over square root of 3."}, {"video_title": "Using the Pythagorean trig identity Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And we are left with negative 1 over the square root of 3. Now some people don't like a radical in the denominator like this. They don't like an irrational denominator. So we could rationalize the denominator here by multiplying by square root of 3 over square root of 3. And so that gets us, this will be equal to negative square root of 3 over 3 is the tangent of this angle right over here. And that actually makes sense, because the tangent of the angle is the slope of this line. And we see that it is indeed a negative slope."}, {"video_title": "Midline, amplitude and period of a function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "The midline is a line, a horizontal line, where half of the function is above it and half of the function is below it. Then I want you to think about the amplitude. How far does this function vary from that midline? Either how far above does it go or how far does it go below? And it should be the same amount because the midline should be between the highest and the lowest points. And then finally, think about what the period of this function is. How much do you have to have a change in x to get to the same point in the cycle of this periodic function?"}, {"video_title": "Midline, amplitude and period of a function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Either how far above does it go or how far does it go below? And it should be the same amount because the midline should be between the highest and the lowest points. And then finally, think about what the period of this function is. How much do you have to have a change in x to get to the same point in the cycle of this periodic function? So I encourage you to pause the video now and think about those questions. So let's tackle the midline first. So one way to think about it is, well, how high does this function go?"}, {"video_title": "Midline, amplitude and period of a function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "How much do you have to have a change in x to get to the same point in the cycle of this periodic function? So I encourage you to pause the video now and think about those questions. So let's tackle the midline first. So one way to think about it is, well, how high does this function go? Well, the highest y value for this function we see is four. It keeps hitting four on a fairly regular basis. And we'll talk about how regular that is when we talk about the period."}, {"video_title": "Midline, amplitude and period of a function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So one way to think about it is, well, how high does this function go? Well, the highest y value for this function we see is four. It keeps hitting four on a fairly regular basis. And we'll talk about how regular that is when we talk about the period. And what's the lowest value that this function gets to? Well, it gets to y equals negative two. So what's halfway between four and negative two?"}, {"video_title": "Midline, amplitude and period of a function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And we'll talk about how regular that is when we talk about the period. And what's the lowest value that this function gets to? Well, it gets to y equals negative two. So what's halfway between four and negative two? Well, you could eyeball it or you could count or you could literally just take the average between four and negative two. So four, so we could, the midline is going to be the horizontal line y is equal to four plus negative two over two, just literally the mean, the arithmetic mean between four and negative two, the average of four and negative two, which is just going to be equal to one. So the line y equals one is the midline."}, {"video_title": "Midline, amplitude and period of a function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So what's halfway between four and negative two? Well, you could eyeball it or you could count or you could literally just take the average between four and negative two. So four, so we could, the midline is going to be the horizontal line y is equal to four plus negative two over two, just literally the mean, the arithmetic mean between four and negative two, the average of four and negative two, which is just going to be equal to one. So the line y equals one is the midline. So that's the midline right over here. And you see that, you see that it's kind of cutting the function and it's where you have half of the function is above it and half of the function is below it. So that's the midline."}, {"video_title": "Midline, amplitude and period of a function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So the line y equals one is the midline. So that's the midline right over here. And you see that, you see that it's kind of cutting the function and it's where you have half of the function is above it and half of the function is below it. So that's the midline. Midline. Now let's think about the amplitude. Well, the amplitude is how much this function varies from the midline, either above the midline or below the midline."}, {"video_title": "Midline, amplitude and period of a function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So that's the midline. Midline. Now let's think about the amplitude. Well, the amplitude is how much this function varies from the midline, either above the midline or below the midline. And the midline's in the middle, so it's going to be the same amount whether you go above or below. So if you, one way to say it is, well, at this maximum point right over here, how far above the midline is this? How far above the midline is this?"}, {"video_title": "Midline, amplitude and period of a function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Well, the amplitude is how much this function varies from the midline, either above the midline or below the midline. And the midline's in the middle, so it's going to be the same amount whether you go above or below. So if you, one way to say it is, well, at this maximum point right over here, how far above the midline is this? How far above the midline is this? Well, to get from one to four, you have to go, you're three above the midline, or another way of thinking about it, this maximum point is y equals four minus y equals one. Well, you had a, your y can go as much as three above the midline. Or you could say your y value could be as much as three below the midline."}, {"video_title": "Midline, amplitude and period of a function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "How far above the midline is this? Well, to get from one to four, you have to go, you're three above the midline, or another way of thinking about it, this maximum point is y equals four minus y equals one. Well, you had a, your y can go as much as three above the midline. Or you could say your y value could be as much as three below the midline. That's this point right over here. One minus three is negative one. So your amplitude right over here is equal to three."}, {"video_title": "Midline, amplitude and period of a function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Or you could say your y value could be as much as three below the midline. That's this point right over here. One minus three is negative one. So your amplitude right over here is equal to three. You can vary as much as three, either above the midline or below the midline. Finally, the period. And when I think about the period, I try to look for a relatively convenient spot on the curve."}, {"video_title": "Midline, amplitude and period of a function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So your amplitude right over here is equal to three. You can vary as much as three, either above the midline or below the midline. Finally, the period. And when I think about the period, I try to look for a relatively convenient spot on the curve. And I'm calling this a convenient spot because it's at a nice, when x is at negative two, y is at one, it's at a nice integer value. And so what I want to do is keep traveling along this curve until I get to the same y value, but not just the same y value, but the y value, I get the same y value and I'm also traveling in the same direction. So for example, let's travel along this curve."}, {"video_title": "Midline, amplitude and period of a function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And when I think about the period, I try to look for a relatively convenient spot on the curve. And I'm calling this a convenient spot because it's at a nice, when x is at negative two, y is at one, it's at a nice integer value. And so what I want to do is keep traveling along this curve until I get to the same y value, but not just the same y value, but the y value, I get the same y value and I'm also traveling in the same direction. So for example, let's travel along this curve. So essentially, our x is increasing. Our x keeps increasing. Now you might say, hey, have I completed a cycle here?"}, {"video_title": "Midline, amplitude and period of a function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So for example, let's travel along this curve. So essentially, our x is increasing. Our x keeps increasing. Now you might say, hey, have I completed a cycle here? Because once again, y is equal to one. You haven't completed a cycle here because notice over here, our y is increasing as x increases, while here our y is decreasing as x increases. Our slope is positive here, our slope is negative here."}, {"video_title": "Midline, amplitude and period of a function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Now you might say, hey, have I completed a cycle here? Because once again, y is equal to one. You haven't completed a cycle here because notice over here, our y is increasing as x increases, while here our y is decreasing as x increases. Our slope is positive here, our slope is negative here. So this isn't the same point on the cycle. We need to get to the point where y once again equals one, or we could say, especially in this case, we're at the midline again, but our slope is increasing. So let's just keep going."}, {"video_title": "Midline, amplitude and period of a function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Our slope is positive here, our slope is negative here. So this isn't the same point on the cycle. We need to get to the point where y once again equals one, or we could say, especially in this case, we're at the midline again, but our slope is increasing. So let's just keep going. Let's just keep going. So that gets us to right over there. So notice, now we have completed one cycle."}, {"video_title": "Midline, amplitude and period of a function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So let's just keep going. Let's just keep going. So that gets us to right over there. So notice, now we have completed one cycle. So the change in x needed to complete one cycle, that is your period. So to go from negative two to zero, your period is two. So your period here is two, and you could do it again."}, {"video_title": "Midline, amplitude and period of a function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So notice, now we have completed one cycle. So the change in x needed to complete one cycle, that is your period. So to go from negative two to zero, your period is two. So your period here is two, and you could do it again. So we're at that point. Let's see, we wanna get back to a point where we're at the midline, and I just happened to start right over here at the midline. I could've started really at any point."}, {"video_title": "Midline, amplitude and period of a function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So your period here is two, and you could do it again. So we're at that point. Let's see, we wanna get back to a point where we're at the midline, and I just happened to start right over here at the midline. I could've started really at any point. You wanna get to the same point, but also where the slope is the same. We're at the same point in the cycle once again. So I could go, so if I travel one, I'm at the midline again, but I'm now going down."}, {"video_title": "Midline, amplitude and period of a function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "I could've started really at any point. You wanna get to the same point, but also where the slope is the same. We're at the same point in the cycle once again. So I could go, so if I travel one, I'm at the midline again, but I'm now going down. So I have to go further. Now I am back at that same point in the cycle. I'm at y equals one, and the slope is positive."}, {"video_title": "Proof of angle addition formula for sine Trigonometry Khan Academy.mp3", "Sentence": "What I hope to do in this video is prove the angle addition formula for sine, or in particular, prove that the sine of x plus y is equal to the sine of x times the cosine of y plus cosine of x times the sine of y. And the way I'm going to do it is with this diagram right over here that you can kind of view it as it has this red right triangle. So it has this right triangle here that has a hypotenuse of 1. You could say this triangle ADC. It has it stacked on top of its base is the hypotenuse of triangle ACD, which I could outline it in blue since I already labeled the measure of this angle as being y. So we have AC, which is the base of triangle ADC, is the hypotenuse of triangle ABC. They're just kind of stacked on top of each other like that."}, {"video_title": "Proof of angle addition formula for sine Trigonometry Khan Academy.mp3", "Sentence": "You could say this triangle ADC. It has it stacked on top of its base is the hypotenuse of triangle ACD, which I could outline it in blue since I already labeled the measure of this angle as being y. So we have AC, which is the base of triangle ADC, is the hypotenuse of triangle ABC. They're just kind of stacked on top of each other like that. And the way I'm going to think about it is first, if you just look at this, what is sine of x plus y going to be? Well, x plus y is this entire angle right over here. And if you look at this right triangle, right triangle ADF, we know that the sine of an angle is the opposite side over the hypotenuse."}, {"video_title": "Proof of angle addition formula for sine Trigonometry Khan Academy.mp3", "Sentence": "They're just kind of stacked on top of each other like that. And the way I'm going to think about it is first, if you just look at this, what is sine of x plus y going to be? Well, x plus y is this entire angle right over here. And if you look at this right triangle, right triangle ADF, we know that the sine of an angle is the opposite side over the hypotenuse. Well, the hypotenuse here is 1. So the sine of this is the opposite over 1. Or the sine of this angle, the sine of x plus y, is equal to the length of this opposite side."}, {"video_title": "Proof of angle addition formula for sine Trigonometry Khan Academy.mp3", "Sentence": "And if you look at this right triangle, right triangle ADF, we know that the sine of an angle is the opposite side over the hypotenuse. Well, the hypotenuse here is 1. So the sine of this is the opposite over 1. Or the sine of this angle, the sine of x plus y, is equal to the length of this opposite side. So sine of x plus y is going to be equal to the length of segment DF. And what I'm going to try to do is say, well, OK, length of segment DF is essentially what we're looking for. But we can decompose the length of segment DF into two segments."}, {"video_title": "Proof of angle addition formula for sine Trigonometry Khan Academy.mp3", "Sentence": "Or the sine of this angle, the sine of x plus y, is equal to the length of this opposite side. So sine of x plus y is going to be equal to the length of segment DF. And what I'm going to try to do is say, well, OK, length of segment DF is essentially what we're looking for. But we can decompose the length of segment DF into two segments. We can decompose it into length of segment DE and the length of segment EF. So we could say that DF, which is the same thing as sine of x plus y, the length of segment DF is the same thing, is equal to the length of segment DE plus the length of segment EF, and EF is, of course, the same thing as the length of segment CB. That ECBF, this right over here, is a rectangle."}, {"video_title": "Proof of angle addition formula for sine Trigonometry Khan Academy.mp3", "Sentence": "But we can decompose the length of segment DF into two segments. We can decompose it into length of segment DE and the length of segment EF. So we could say that DF, which is the same thing as sine of x plus y, the length of segment DF is the same thing, is equal to the length of segment DE plus the length of segment EF, and EF is, of course, the same thing as the length of segment CB. That ECBF, this right over here, is a rectangle. So EF is the same thing as CB. So this thing is going to be equal to DE right over here, length of segment DE, plus the length of segment CB. So once again, the way I'm going to address this, I'm saying, well, the sine of x plus y, which is the length of DF, and DF can be decomposed as the lengths of DE and CB."}, {"video_title": "Proof of angle addition formula for sine Trigonometry Khan Academy.mp3", "Sentence": "That ECBF, this right over here, is a rectangle. So EF is the same thing as CB. So this thing is going to be equal to DE right over here, length of segment DE, plus the length of segment CB. So once again, the way I'm going to address this, I'm saying, well, the sine of x plus y, which is the length of DF, and DF can be decomposed as the lengths of DE and CB. Now with that as a hint, I encourage you to figure out what the length of segment DE is in terms of x's and y's and sines and cosines, and also figure out what the length of segment CB is in terms of x's and y's and sines and cosines. So try to figure out as much as you can about this, and these two just might fall out of that. So I'm assuming you've given a go at it."}, {"video_title": "Proof of angle addition formula for sine Trigonometry Khan Academy.mp3", "Sentence": "So once again, the way I'm going to address this, I'm saying, well, the sine of x plus y, which is the length of DF, and DF can be decomposed as the lengths of DE and CB. Now with that as a hint, I encourage you to figure out what the length of segment DE is in terms of x's and y's and sines and cosines, and also figure out what the length of segment CB is in terms of x's and y's and sines and cosines. So try to figure out as much as you can about this, and these two just might fall out of that. So I'm assuming you've given a go at it. So now that we essentially know that sine of x plus y can be expressed this way, let's see if we can figure these things out. And I'm going to try to address it by just figuring out as many lengths and angles here as I can. So let's go to this top red triangle right over here."}, {"video_title": "Proof of angle addition formula for sine Trigonometry Khan Academy.mp3", "Sentence": "So I'm assuming you've given a go at it. So now that we essentially know that sine of x plus y can be expressed this way, let's see if we can figure these things out. And I'm going to try to address it by just figuring out as many lengths and angles here as I can. So let's go to this top red triangle right over here. Its hypotenuse has length 1. So what's going to be the length of segment DC? Well, that is the opposite side of our angle x."}, {"video_title": "Proof of angle addition formula for sine Trigonometry Khan Academy.mp3", "Sentence": "So let's go to this top red triangle right over here. Its hypotenuse has length 1. So what's going to be the length of segment DC? Well, that is the opposite side of our angle x. So we know sine of x is equal to DC over 1. Or DC over 1 is just DC. So this length right over here is sine of x."}, {"video_title": "Proof of angle addition formula for sine Trigonometry Khan Academy.mp3", "Sentence": "Well, that is the opposite side of our angle x. So we know sine of x is equal to DC over 1. Or DC over 1 is just DC. So this length right over here is sine of x. And segment AC, same exact logic. Cosine of x is A is the length of AC over 1, which is just the length of AC. So this length right over here, segment AC, its length is cosine of x."}, {"video_title": "Proof of angle addition formula for sine Trigonometry Khan Academy.mp3", "Sentence": "So this length right over here is sine of x. And segment AC, same exact logic. Cosine of x is A is the length of AC over 1, which is just the length of AC. So this length right over here, segment AC, its length is cosine of x. So that's kind of interesting. Now let's see what we could figure out about this triangle, triangle ACB right over here. So how could we figure out CB?"}, {"video_title": "Proof of angle addition formula for sine Trigonometry Khan Academy.mp3", "Sentence": "So this length right over here, segment AC, its length is cosine of x. So that's kind of interesting. Now let's see what we could figure out about this triangle, triangle ACB right over here. So how could we figure out CB? Well, we know that sine of y is equal to what? It's equal to the length of segment CB over the hypotenuse. The hypotenuse here is now cosine of x."}, {"video_title": "Proof of angle addition formula for sine Trigonometry Khan Academy.mp3", "Sentence": "So how could we figure out CB? Well, we know that sine of y is equal to what? It's equal to the length of segment CB over the hypotenuse. The hypotenuse here is now cosine of x. And I think you might see where all of this is leading. And at any point, if you get excited, pause the video and try to finish the proof on your own. So the length of segment CB, if we just multiply both sides by cosine of x, the length of segment CB is equal to cosine of x times sine of y, which is neat."}, {"video_title": "Proof of angle addition formula for sine Trigonometry Khan Academy.mp3", "Sentence": "The hypotenuse here is now cosine of x. And I think you might see where all of this is leading. And at any point, if you get excited, pause the video and try to finish the proof on your own. So the length of segment CB, if we just multiply both sides by cosine of x, the length of segment CB is equal to cosine of x times sine of y, which is neat. Because we just showed that this thing right over here is equal to this thing right over here. So to complete our proof, we just need to prove that this thing is equal to this thing right over there. If that's equal to that and that's equal to that, well, we already know that the sum of these is equal to the length of DF, which is sine of x plus y."}, {"video_title": "Proof of angle addition formula for sine Trigonometry Khan Academy.mp3", "Sentence": "So the length of segment CB, if we just multiply both sides by cosine of x, the length of segment CB is equal to cosine of x times sine of y, which is neat. Because we just showed that this thing right over here is equal to this thing right over here. So to complete our proof, we just need to prove that this thing is equal to this thing right over there. If that's equal to that and that's equal to that, well, we already know that the sum of these is equal to the length of DF, which is sine of x plus y. So let's see if we can express DE somehow. Now what angle would be useful? Well, somehow we could figure out this angle up here or maybe this angle."}, {"video_title": "Proof of angle addition formula for sine Trigonometry Khan Academy.mp3", "Sentence": "If that's equal to that and that's equal to that, well, we already know that the sum of these is equal to the length of DF, which is sine of x plus y. So let's see if we can express DE somehow. Now what angle would be useful? Well, somehow we could figure out this angle up here or maybe this angle. Well, let's see. If we could figure out this angle, then DE we could express in terms of this angle and sine of x. Let's see if we can figure out that angle."}, {"video_title": "Proof of angle addition formula for sine Trigonometry Khan Academy.mp3", "Sentence": "Well, somehow we could figure out this angle up here or maybe this angle. Well, let's see. If we could figure out this angle, then DE we could express in terms of this angle and sine of x. Let's see if we can figure out that angle. So we know this is angle y over here. And we also know that this is a right angle. So EC is parallel to AB."}, {"video_title": "Proof of angle addition formula for sine Trigonometry Khan Academy.mp3", "Sentence": "Let's see if we can figure out that angle. So we know this is angle y over here. And we also know that this is a right angle. So EC is parallel to AB. So you could view AC as a transversal. So if this is angle y right over here, then we know this is also angle y. These are, once again, we'll notice if AC is a transversal here and EC and AB are parallel, then if this is y, then that is y."}, {"video_title": "Proof of angle addition formula for sine Trigonometry Khan Academy.mp3", "Sentence": "So EC is parallel to AB. So you could view AC as a transversal. So if this is angle y right over here, then we know this is also angle y. These are, once again, we'll notice if AC is a transversal here and EC and AB are parallel, then if this is y, then that is y. Then if that's y, then this is 90 minus y. And if this is 90 degrees and this is 90 minus y, then these two angles combined add up to 180 minus y. And if all three of these add up to 180, then this thing up here must be equal to y. Validate that."}, {"video_title": "Proof of angle addition formula for sine Trigonometry Khan Academy.mp3", "Sentence": "These are, once again, we'll notice if AC is a transversal here and EC and AB are parallel, then if this is y, then that is y. Then if that's y, then this is 90 minus y. And if this is 90 degrees and this is 90 minus y, then these two angles combined add up to 180 minus y. And if all three of these add up to 180, then this thing up here must be equal to y. Validate that. y plus 90 minus y plus 90 is 180 degrees. And that is useful for us because now we can express segment DE in terms of y and sine of x. What is DE to y?"}, {"video_title": "Proof of angle addition formula for sine Trigonometry Khan Academy.mp3", "Sentence": "And if all three of these add up to 180, then this thing up here must be equal to y. Validate that. y plus 90 minus y plus 90 is 180 degrees. And that is useful for us because now we can express segment DE in terms of y and sine of x. What is DE to y? It's the adjacent angle. So we could think of cosine. We know that the cosine of angle y, if we look at triangle DEC right over here, we know that the cosine of y is equal to segment DE over its hypotenuse, over sine of x."}, {"video_title": "Proof of angle addition formula for sine Trigonometry Khan Academy.mp3", "Sentence": "What is DE to y? It's the adjacent angle. So we could think of cosine. We know that the cosine of angle y, if we look at triangle DEC right over here, we know that the cosine of y is equal to segment DE over its hypotenuse, over sine of x. And you should be getting excited right about now because we've just shown, if we multiply both sides by sine of x, we've just shown that DE is equal to sine of x times cosine of y. So we've now shown this is equal to this. We already showed that CB is equal to that."}, {"video_title": "Proof of angle addition formula for sine Trigonometry Khan Academy.mp3", "Sentence": "We know that the cosine of angle y, if we look at triangle DEC right over here, we know that the cosine of y is equal to segment DE over its hypotenuse, over sine of x. And you should be getting excited right about now because we've just shown, if we multiply both sides by sine of x, we've just shown that DE is equal to sine of x times cosine of y. So we've now shown this is equal to this. We already showed that CB is equal to that. So the sum of DE and CB, which is the same thing as the sum of DE and EF, is the sine of x plus y, which is that over there. So we are done. We have proven the angle addition formula for sine."}, {"video_title": "Using similarity to estimate ratio between side lengths High school geometry Khan Academy.mp3", "Sentence": "So they want us to figure out the ratio PN over MN. So pause this video and see if you can figure this out. All right, now let's work through this together. Now, given that they want us to figure out this ratio, and they want us to actually evaluate it or be able to approximate it, we are probably dealing with similarity. And so what I would wanna look for is, are one of these triangles similar to the triangle we have here? And we're dealing with similar triangles if we have two angles in common, because if we have two angles in common, then that means that we definitely have the third angle as well, because the third angle is completely determined by what the other two angles are. So we have a 35 degree angle here, and we have a 90 degree angle here."}, {"video_title": "Using similarity to estimate ratio between side lengths High school geometry Khan Academy.mp3", "Sentence": "Now, given that they want us to figure out this ratio, and they want us to actually evaluate it or be able to approximate it, we are probably dealing with similarity. And so what I would wanna look for is, are one of these triangles similar to the triangle we have here? And we're dealing with similar triangles if we have two angles in common, because if we have two angles in common, then that means that we definitely have the third angle as well, because the third angle is completely determined by what the other two angles are. So we have a 35 degree angle here, and we have a 90 degree angle here. And of all of these choices, this doesn't have a 35 degree angle, it has a 90. This doesn't have a 35, it has a 90. But triangle two here has a 35 degree angle, has a 90 degree angle, and has a 55 degree angle."}, {"video_title": "Using similarity to estimate ratio between side lengths High school geometry Khan Academy.mp3", "Sentence": "So we have a 35 degree angle here, and we have a 90 degree angle here. And of all of these choices, this doesn't have a 35 degree angle, it has a 90. This doesn't have a 35, it has a 90. But triangle two here has a 35 degree angle, has a 90 degree angle, and has a 55 degree angle. And if you did the math, knowing that 35 plus 90 plus this have to add up to 180 degrees, you would see that this too has a measure of 55 degrees. And so given that all of our angle measures are the same between triangle PNM and triangle number two right over here, we know that these two are similar triangles. And so the ratios between corresponding sides are going to be the same."}, {"video_title": "Using similarity to estimate ratio between side lengths High school geometry Khan Academy.mp3", "Sentence": "But triangle two here has a 35 degree angle, has a 90 degree angle, and has a 55 degree angle. And if you did the math, knowing that 35 plus 90 plus this have to add up to 180 degrees, you would see that this too has a measure of 55 degrees. And so given that all of our angle measures are the same between triangle PNM and triangle number two right over here, we know that these two are similar triangles. And so the ratios between corresponding sides are going to be the same. We could either take the ratio across triangles, or we could say the ratio within when we just look at one triangle. And so if you look at PN over MN, let me try to color code it. So PN right over here, that corresponds to the side that's opposite the 35 degree angle."}, {"video_title": "Using similarity to estimate ratio between side lengths High school geometry Khan Academy.mp3", "Sentence": "And so the ratios between corresponding sides are going to be the same. We could either take the ratio across triangles, or we could say the ratio within when we just look at one triangle. And so if you look at PN over MN, let me try to color code it. So PN right over here, that corresponds to the side that's opposite the 35 degree angle. So that would correspond to this side right over here on triangle two. And then MN, that's this that I'm coloring in this bluish color. Not so well."}, {"video_title": "Using similarity to estimate ratio between side lengths High school geometry Khan Academy.mp3", "Sentence": "So PN right over here, that corresponds to the side that's opposite the 35 degree angle. So that would correspond to this side right over here on triangle two. And then MN, that's this that I'm coloring in this bluish color. Not so well. Probably spent more time coloring. That's opposite the 55 degree angle. And so opposite the 55 degree angle would be right over there."}, {"video_title": "Using similarity to estimate ratio between side lengths High school geometry Khan Academy.mp3", "Sentence": "Not so well. Probably spent more time coloring. That's opposite the 55 degree angle. And so opposite the 55 degree angle would be right over there. Now, since these triangles are similar, the ratio of the length of the red side over the length of the blue side is going to be the same in either triangle. So PN, let me write it this way. The length of segment PN over the length of segment MN is going to be equivalent to 5.7 over 8.2."}, {"video_title": "Using similarity to estimate ratio between side lengths High school geometry Khan Academy.mp3", "Sentence": "And so opposite the 55 degree angle would be right over there. Now, since these triangles are similar, the ratio of the length of the red side over the length of the blue side is going to be the same in either triangle. So PN, let me write it this way. The length of segment PN over the length of segment MN is going to be equivalent to 5.7 over 8.2. Because this ratio is going to be the same for the corresponding sides, regardless of which triangle you look at. So if you take the side that's opposite 35 degrees, that's 5.7 over 8.2. Now, to be very clear, it doesn't mean that somehow the length of this side is 5.7 or that the length of this side is 8.2."}, {"video_title": "Using similarity to estimate ratio between side lengths High school geometry Khan Academy.mp3", "Sentence": "The length of segment PN over the length of segment MN is going to be equivalent to 5.7 over 8.2. Because this ratio is going to be the same for the corresponding sides, regardless of which triangle you look at. So if you take the side that's opposite 35 degrees, that's 5.7 over 8.2. Now, to be very clear, it doesn't mean that somehow the length of this side is 5.7 or that the length of this side is 8.2. We would only be able to make that conclusion if they were congruent. But with similarity, we know that the ratios, if we look at the ratio of the red side to the blue side on each of those triangles, that would be the same. And so this gives us that ratio."}, {"video_title": "Using similarity to estimate ratio between side lengths High school geometry Khan Academy.mp3", "Sentence": "Now, to be very clear, it doesn't mean that somehow the length of this side is 5.7 or that the length of this side is 8.2. We would only be able to make that conclusion if they were congruent. But with similarity, we know that the ratios, if we look at the ratio of the red side to the blue side on each of those triangles, that would be the same. And so this gives us that ratio. And let's see, 5.7 over 8.2. Which of these choices get close to that? Well, we could say that this is roughly, if I am approximating it, let's see, it's going to be larger than 0.57 because 8.2 is less than 10."}, {"video_title": "Using similarity to estimate ratio between side lengths High school geometry Khan Academy.mp3", "Sentence": "And so this gives us that ratio. And let's see, 5.7 over 8.2. Which of these choices get close to that? Well, we could say that this is roughly, if I am approximating it, let's see, it's going to be larger than 0.57 because 8.2 is less than 10. And so we are going to rule this choice out. And 5.7 is less than 8.2, so it can't be over one. And so we have to think between these two choices."}, {"video_title": "Using similarity to estimate ratio between side lengths High school geometry Khan Academy.mp3", "Sentence": "Well, we could say that this is roughly, if I am approximating it, let's see, it's going to be larger than 0.57 because 8.2 is less than 10. And so we are going to rule this choice out. And 5.7 is less than 8.2, so it can't be over one. And so we have to think between these two choices. Well, the simplest thing I can do is actually just try to start dividing it by hand. So 8.2 goes into 5.7 the same number of times as 82 goes into 57. And I'll add some decimals here."}, {"video_title": "Using similarity to estimate ratio between side lengths High school geometry Khan Academy.mp3", "Sentence": "And so we have to think between these two choices. Well, the simplest thing I can do is actually just try to start dividing it by hand. So 8.2 goes into 5.7 the same number of times as 82 goes into 57. And I'll add some decimals here. So it doesn't go into 57, but how many times does 82 go into 570? I would assume it's about six times, maybe seven times, looks like. So seven times two is 14, and then seven times eight is 56, I guess it's 57."}, {"video_title": "Using similarity to estimate ratio between side lengths High school geometry Khan Academy.mp3", "Sentence": "And I'll add some decimals here. So it doesn't go into 57, but how many times does 82 go into 570? I would assume it's about six times, maybe seven times, looks like. So seven times two is 14, and then seven times eight is 56, I guess it's 57. So it's actually a little less than 0.7. This got, made me go a little bit too high. So if I am approximating, it's gonna be 0.6 something."}, {"video_title": "Angle addition formula with cosine Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Write a formula for sine of theta in terms of C. And so I encourage you to pause the video and try to figure this out on your own before I work through it. So I'm assuming you've had a go, so let's see if we can work through it together, and I'm gonna get my scratch pad out. I have that copy and pasted the exact question right over here, and so let's think about it a little bit. They're telling us that cosine of two theta is equal to C, so let me write it this way. C is equal to cosine of two theta. Now, using my knowledge of angle addition formulas, we know, for example, that cosine of alpha plus beta, we know that cosine of alpha plus beta is equal to cosine alpha, is equal to cosine alpha cosine beta, cosine beta minus, minus sine alpha sine beta, minus sine alpha, sine alpha sine beta, sine beta. Now, why would this be useful here?"}, {"video_title": "Angle addition formula with cosine Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "They're telling us that cosine of two theta is equal to C, so let me write it this way. C is equal to cosine of two theta. Now, using my knowledge of angle addition formulas, we know, for example, that cosine of alpha plus beta, we know that cosine of alpha plus beta is equal to cosine alpha, is equal to cosine alpha cosine beta, cosine beta minus, minus sine alpha sine beta, minus sine alpha, sine alpha sine beta, sine beta. Now, why would this be useful here? Well, this is the sum of just theta plus theta, and so I can rewrite this, at least in terms of cosines and sines, and maybe I can rewrite the cosines in terms of sines, and then solve for the sines. So let's try to do that. So I can rewrite cosine of two theta, that's the same thing as cosine of theta, theta plus theta, cosine of theta plus theta is, of course, the same thing as two theta, and now I can use the angle addition formula for cosine."}, {"video_title": "Angle addition formula with cosine Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Now, why would this be useful here? Well, this is the sum of just theta plus theta, and so I can rewrite this, at least in terms of cosines and sines, and maybe I can rewrite the cosines in terms of sines, and then solve for the sines. So let's try to do that. So I can rewrite cosine of two theta, that's the same thing as cosine of theta, theta plus theta, cosine of theta plus theta is, of course, the same thing as two theta, and now I can use the angle addition formula for cosine. This is going to be equal to, this is going to be equal to cosine of theta, cosine of theta times cosine of theta, minus sine of theta, times sine of theta, which is, of course, equal to, this is equal to cosine squared theta, that's equal to cosine squared theta minus, And what we have right over here is sine squared theta. Sine squared theta. So let's see."}, {"video_title": "Angle addition formula with cosine Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So I can rewrite cosine of two theta, that's the same thing as cosine of theta, theta plus theta, cosine of theta plus theta is, of course, the same thing as two theta, and now I can use the angle addition formula for cosine. This is going to be equal to, this is going to be equal to cosine of theta, cosine of theta times cosine of theta, minus sine of theta, times sine of theta, which is, of course, equal to, this is equal to cosine squared theta, that's equal to cosine squared theta minus, And what we have right over here is sine squared theta. Sine squared theta. So let's see. We've been able to rewrite c in terms of cosine squared theta and sine squared theta. But ideally, we just want to write it in terms of sine theta so that we can solve for sine theta. So if we can re-express cosine theta in terms of sine, well, we already know from the Pythagorean identity that cosine squared theta plus sine squared theta is equal to 1."}, {"video_title": "Angle addition formula with cosine Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So let's see. We've been able to rewrite c in terms of cosine squared theta and sine squared theta. But ideally, we just want to write it in terms of sine theta so that we can solve for sine theta. So if we can re-express cosine theta in terms of sine, well, we already know from the Pythagorean identity that cosine squared theta plus sine squared theta is equal to 1. Or we could say that cosine squared theta is equal to, I'm just going to subtract sine squared from both sides, is equal to 1 minus sine squared theta. So let me rewrite this as 1 minus sine squared theta. And then, of course, we have minus this yellow sine squared theta."}, {"video_title": "Angle addition formula with cosine Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So if we can re-express cosine theta in terms of sine, well, we already know from the Pythagorean identity that cosine squared theta plus sine squared theta is equal to 1. Or we could say that cosine squared theta is equal to, I'm just going to subtract sine squared from both sides, is equal to 1 minus sine squared theta. So let me rewrite this as 1 minus sine squared theta. And then, of course, we have minus this yellow sine squared theta. And all of this is equal to c. All of this is equal to c. Or we could get that c is equal to 1 minus 2 sine squared theta. And what's useful about this is now we just have to solve for sine of theta. So let's see."}, {"video_title": "Angle addition formula with cosine Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And then, of course, we have minus this yellow sine squared theta. And all of this is equal to c. All of this is equal to c. Or we could get that c is equal to 1 minus 2 sine squared theta. And what's useful about this is now we just have to solve for sine of theta. So let's see. I could multiply both sides by a negative just so that I can switch the order over here. So I could write this as negative c is equal to 2 sine squared theta minus 1. I just multiplied both sides by a negative."}, {"video_title": "Angle addition formula with cosine Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So let's see. I could multiply both sides by a negative just so that I can switch the order over here. So I could write this as negative c is equal to 2 sine squared theta minus 1. I just multiplied both sides by a negative. And then, let's see. I could add 1 to both sides. And I'll go over here."}, {"video_title": "Angle addition formula with cosine Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "I just multiplied both sides by a negative. And then, let's see. I could add 1 to both sides. And I'll go over here. If I add 1 to both sides, I get 1 minus c is equal to 2 sine squared theta. I could divide both sides by 2. And then, so I get sine squared theta is equal to 1 minus c over 2."}, {"video_title": "Angle addition formula with cosine Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And I'll go over here. If I add 1 to both sides, I get 1 minus c is equal to 2 sine squared theta. I could divide both sides by 2. And then, so I get sine squared theta is equal to 1 minus c over 2. Or I could write that sine of theta is equal to the plus or minus square root of 1 minus c over 2. So that leads to a question. Is it both?"}, {"video_title": "Angle addition formula with cosine Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And then, so I get sine squared theta is equal to 1 minus c over 2. Or I could write that sine of theta is equal to the plus or minus square root of 1 minus c over 2. So that leads to a question. Is it both? Is it the plus and minus square root? Or is it just one of those? And I encourage you to pause the video again in case you haven't already figured it out and look at the information here and think about whether they give us the information of whether we should be looking at the positive or negative sine."}, {"video_title": "Angle addition formula with cosine Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Is it both? Is it the plus and minus square root? Or is it just one of those? And I encourage you to pause the video again in case you haven't already figured it out and look at the information here and think about whether they give us the information of whether we should be looking at the positive or negative sine. Well, they tell us that theta is between 0 and pi. So if I were to draw a unit circle here between 0 and pi radians, so this angle right over here is 0 radians. And pi is going all the way over here."}, {"video_title": "Angle addition formula with cosine Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And I encourage you to pause the video again in case you haven't already figured it out and look at the information here and think about whether they give us the information of whether we should be looking at the positive or negative sine. Well, they tell us that theta is between 0 and pi. So if I were to draw a unit circle here between 0 and pi radians, so this angle right over here is 0 radians. And pi is going all the way over here. So this angle places its terminal ray either in the first or second quadrants. So it could be an angle like this. It could be an angle like this."}, {"video_title": "Angle addition formula with cosine Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And pi is going all the way over here. So this angle places its terminal ray either in the first or second quadrants. So it could be an angle like this. It could be an angle like this. It cannot be an angle like this. And we know that the sine of an angle is the y-coordinate. And so we know if we're in the first or second quadrant, the y-coordinate is going to be non-negative."}, {"video_title": "Angle addition formula with cosine Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "It could be an angle like this. It cannot be an angle like this. And we know that the sine of an angle is the y-coordinate. And so we know if we're in the first or second quadrant, the y-coordinate is going to be non-negative. So we would want to take the positive square root right over here. So we would get sine of theta is equal to the principal root, or you could even think of it the positive square root, of 1 minus c over 2. So let's go back to our make sure that we can check our answer."}, {"video_title": "Matching ratios to trig functions Trigonometry Khan Academy.mp3", "Sentence": "On the right hand side, we have a bunch of expressions that are just ratios of different information given in these two diagrams. 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."}, {"video_title": "Matching ratios to trig functions Trigonometry Khan Academy.mp3", "Sentence": "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. 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."}, {"video_title": "Matching ratios to trig functions Trigonometry Khan Academy.mp3", "Sentence": "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. So sine is opposite over hypotenuse. Cosine is adjacent over hypotenuse."}, {"video_title": "Matching ratios to trig functions Trigonometry Khan Academy.mp3", "Sentence": "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. 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."}, {"video_title": "Matching ratios to trig functions Trigonometry Khan Academy.mp3", "Sentence": "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. So we have x, x is the x-coordinate. That's also the length of this side right over here."}, {"video_title": "Matching ratios to trig functions Trigonometry Khan Academy.mp3", "Sentence": "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. So x is equal to the adjacent side. What is one?"}, {"video_title": "Matching ratios to trig functions Trigonometry Khan Academy.mp3", "Sentence": "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. 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."}, {"video_title": "Matching ratios to trig functions Trigonometry Khan Academy.mp3", "Sentence": "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. 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."}, {"video_title": "Matching ratios to trig functions Trigonometry Khan Academy.mp3", "Sentence": "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. Now let's move over to y over one. Well, y is going to be the length of this side right over here."}, {"video_title": "Matching ratios to trig functions Trigonometry Khan Academy.mp3", "Sentence": "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. Y is going to be this length relative to angle theta. That is the opposite side."}, {"video_title": "Matching ratios to trig functions Trigonometry Khan Academy.mp3", "Sentence": "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. Now which trig function is opposite over hypotenuse? Opposite over hypotenuse, that's sine of theta."}, {"video_title": "Matching ratios to trig functions Trigonometry Khan Academy.mp3", "Sentence": "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. So sine of angle MKJ is the same thing as sine of theta. We see that they have the same measure."}, {"video_title": "Matching ratios to trig functions Trigonometry Khan Academy.mp3", "Sentence": "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. 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."}, {"video_title": "Matching ratios to trig functions Trigonometry Khan Academy.mp3", "Sentence": "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. 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."}, {"video_title": "Matching ratios to trig functions Trigonometry Khan Academy.mp3", "Sentence": "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. 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."}, {"video_title": "Matching ratios to trig functions Trigonometry Khan Academy.mp3", "Sentence": "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. We have adjacent over opposite. So this is equal to adjacent over opposite."}, {"video_title": "Matching ratios to trig functions Trigonometry Khan Academy.mp3", "Sentence": "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. 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."}, {"video_title": "Matching ratios to trig functions Trigonometry Khan Academy.mp3", "Sentence": "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. But it's not one of our choices. So we can rule this one out."}, {"video_title": "Matching ratios to trig functions Trigonometry Khan Academy.mp3", "Sentence": "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. Well, this is looking good. This is, y is opposite, opposite, x is adjacent relative to angle theta, adjacent."}, {"video_title": "Matching ratios to trig functions Trigonometry Khan Academy.mp3", "Sentence": "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. 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."}, {"video_title": "Matching ratios to trig functions Trigonometry Khan Academy.mp3", "Sentence": "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. 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."}, {"video_title": "Matching ratios to trig functions Trigonometry Khan Academy.mp3", "Sentence": "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. 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."}, {"video_title": "Matching ratios to trig functions Trigonometry Khan Academy.mp3", "Sentence": "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. Now k over j. Well now this is opposite over adjacent, opposite over adjacent."}, {"video_title": "Matching ratios to trig functions Trigonometry Khan Academy.mp3", "Sentence": "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. This is equal to tangent of theta, or tangent of angle mkj. So this is equal to k over j."}, {"video_title": "Matching ratios to trig functions Trigonometry Khan Academy.mp3", "Sentence": "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. This of course, this of course is equal to the hypotenuse. Hypotenuse over adjacent."}, {"video_title": "Matching ratios to trig functions Trigonometry Khan Academy.mp3", "Sentence": "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. 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."}, {"video_title": "Matching ratios to trig functions Trigonometry Khan Academy.mp3", "Sentence": "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. So this is equal to cosine of theta, or cosine of angle mkj. So we could write it down."}, {"video_title": "Matching ratios to trig functions Trigonometry Khan Academy.mp3", "Sentence": "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. Opposite over hypotenuse. That's going to be sine of theta."}, {"video_title": "Solution set to sin equation Trigonometry Khan Academy.mp3", "Sentence": "Which of these are contained in the solution set to sine of x is equal to 1 3rd? Answers should be rounded to the nearest hundredth. Select all that apply. I encourage you to pause the video right now and work on it on your own. I'm assuming you've given a go at it. Let's think about what this is asking. They're asking what are the x values?"}, {"video_title": "Solution set to sin equation Trigonometry Khan Academy.mp3", "Sentence": "I encourage you to pause the video right now and work on it on your own. I'm assuming you've given a go at it. Let's think about what this is asking. They're asking what are the x values? What is the solution set? What are the possible x values where sine of x is equal to 1 3rd? To help us visualize this, let's draw a unit circle."}, {"video_title": "Solution set to sin equation Trigonometry Khan Academy.mp3", "Sentence": "They're asking what are the x values? What is the solution set? What are the possible x values where sine of x is equal to 1 3rd? To help us visualize this, let's draw a unit circle. That's my y axis. This right over here is my x axis. This is x at 1."}, {"video_title": "Solution set to sin equation Trigonometry Khan Academy.mp3", "Sentence": "To help us visualize this, let's draw a unit circle. That's my y axis. This right over here is my x axis. This is x at 1. This is y at positive 1. Negative 1 along the x axis. Negative 1 on the y axis."}, {"video_title": "Solution set to sin equation Trigonometry Khan Academy.mp3", "Sentence": "This is x at 1. This is y at positive 1. Negative 1 along the x axis. Negative 1 on the y axis. Unit circle, I'm going to center it at 0. It's going to have a radius of 1. We just have to remind ourselves what the unit circle definition of the sine function is."}, {"video_title": "Solution set to sin equation Trigonometry Khan Academy.mp3", "Sentence": "Negative 1 on the y axis. Unit circle, I'm going to center it at 0. It's going to have a radius of 1. We just have to remind ourselves what the unit circle definition of the sine function is. If we have some angle, one side of the angle is going to be a ray along the positive x axis. We do this in a color you can see. Along the positive x axis."}, {"video_title": "Solution set to sin equation Trigonometry Khan Academy.mp3", "Sentence": "We just have to remind ourselves what the unit circle definition of the sine function is. If we have some angle, one side of the angle is going to be a ray along the positive x axis. We do this in a color you can see. Along the positive x axis. Then the other side, this is our angle right over here. Let's say that's some angle theta. The sine of this angle is going to be the y value of where this ray intersects the unit circle."}, {"video_title": "Solution set to sin equation Trigonometry Khan Academy.mp3", "Sentence": "Along the positive x axis. Then the other side, this is our angle right over here. Let's say that's some angle theta. The sine of this angle is going to be the y value of where this ray intersects the unit circle. This right over here, that is going to be sine of theta. With that review out of the way, let's think about what x values, and we're assuming we're dealing in radians, what x values, if I take the sine of it, are going to give me 1 third. When does y equal 1 third along the unit circle?"}, {"video_title": "Solution set to sin equation Trigonometry Khan Academy.mp3", "Sentence": "The sine of this angle is going to be the y value of where this ray intersects the unit circle. This right over here, that is going to be sine of theta. With that review out of the way, let's think about what x values, and we're assuming we're dealing in radians, what x values, if I take the sine of it, are going to give me 1 third. When does y equal 1 third along the unit circle? That's 2 thirds, 1 third right over here. We see it equals 1 third exactly two places. Here and here."}, {"video_title": "Solution set to sin equation Trigonometry Khan Academy.mp3", "Sentence": "When does y equal 1 third along the unit circle? That's 2 thirds, 1 third right over here. We see it equals 1 third exactly two places. Here and here. There's two angles, or at least two, if we just take one or two on each path of the unit circle, then we can keep adding multiples of 2 pi to get as many as we want. We see just on the unit circle, we could have this angle. We could have this angle right over here."}, {"video_title": "Solution set to sin equation Trigonometry Khan Academy.mp3", "Sentence": "Here and here. There's two angles, or at least two, if we just take one or two on each path of the unit circle, then we can keep adding multiples of 2 pi to get as many as we want. We see just on the unit circle, we could have this angle. We could have this angle right over here. Or we could go all the way around to that angle right over there. Then we could add any multiple of 2 pi to those angles to get other angles. That would also work, where if I took the sine of them, I would get 1 third."}, {"video_title": "Solution set to sin equation Trigonometry Khan Academy.mp3", "Sentence": "We could have this angle right over here. Or we could go all the way around to that angle right over there. Then we could add any multiple of 2 pi to those angles to get other angles. That would also work, where if I took the sine of them, I would get 1 third. Let's think about what these are. Here we can take our calculator out, and we could take the inverse sine of 1 third. Let's do that."}, {"video_title": "Solution set to sin equation Trigonometry Khan Academy.mp3", "Sentence": "That would also work, where if I took the sine of them, I would get 1 third. Let's think about what these are. Here we can take our calculator out, and we could take the inverse sine of 1 third. Let's do that. The inverse sine of 1 over 3. We have to remember what the range of the inverse sine function is. It's going to give us a value between negative pi over 2 and pi over 2, so a value that sticks us in either the first or the fourth quadrant, if we're thinking about the unit circle right over here."}, {"video_title": "Solution set to sin equation Trigonometry Khan Academy.mp3", "Sentence": "Let's do that. The inverse sine of 1 over 3. We have to remember what the range of the inverse sine function is. It's going to give us a value between negative pi over 2 and pi over 2, so a value that sticks us in either the first or the fourth quadrant, if we're thinking about the unit circle right over here. We see that gave us 0 point, if we round to the nearest hundredth, 3, 4. Essentially, they've given us this value. They've given us 0.34."}, {"video_title": "Solution set to sin equation Trigonometry Khan Academy.mp3", "Sentence": "It's going to give us a value between negative pi over 2 and pi over 2, so a value that sticks us in either the first or the fourth quadrant, if we're thinking about the unit circle right over here. We see that gave us 0 point, if we round to the nearest hundredth, 3, 4. Essentially, they've given us this value. They've given us 0.34. That's this angle right over here. How did I know that? It's a positive value."}, {"video_title": "Solution set to sin equation Trigonometry Khan Academy.mp3", "Sentence": "They've given us 0.34. That's this angle right over here. How did I know that? It's a positive value. It's greater than 0, but it's less than pi over 2. Pi is 3.14, so pi over 2 is going to be 1.57. We can go on and on and on."}, {"video_title": "Solution set to sin equation Trigonometry Khan Academy.mp3", "Sentence": "It's a positive value. It's greater than 0, but it's less than pi over 2. Pi is 3.14, so pi over 2 is going to be 1.57. We can go on and on and on. This right over here is 0.34 radians. What would this thing over here be? It's going to be whatever."}, {"video_title": "Solution set to sin equation Trigonometry Khan Academy.mp3", "Sentence": "We can go on and on and on. This right over here is 0.34 radians. What would this thing over here be? It's going to be whatever. If we go to the negative x-axis and we subtract 0.34, so this is 0.34, we're going to get to this angle. It's going to be, if we take pi minus the previous answer, it gets us, if we round to the nearest hundredth, 2.8 radians. This is 0.34 radians."}, {"video_title": "Solution set to sin equation Trigonometry Khan Academy.mp3", "Sentence": "It's going to be whatever. If we go to the negative x-axis and we subtract 0.34, so this is 0.34, we're going to get to this angle. It's going to be, if we take pi minus the previous answer, it gets us, if we round to the nearest hundredth, 2.8 radians. This is 0.34 radians. Then this one, let me do it in this purple color, this one right over here, if we were to go all the way around, it's pi minus 0.34, which is 2.80 radians, rounding to the nearest hundredth. That's not all of the values. We can add multiples of 2 pi to each of these."}, {"video_title": "Solution set to sin equation Trigonometry Khan Academy.mp3", "Sentence": "This is 0.34 radians. Then this one, let me do it in this purple color, this one right over here, if we were to go all the way around, it's pi minus 0.34, which is 2.80 radians, rounding to the nearest hundredth. That's not all of the values. We can add multiples of 2 pi to each of these. 2.80 plus any multiple of 2 pi, so 2 pi n where n is an integer, or we could take 0.34 and add any multiple of 2 pi. 2 pi n where n is an integer. Our solution set here, just to rewrite it outside of this messiness, is going to be 2.80 radians plus 2 pi n where n is an integer, and 0.34 plus 2 pi n where n is an integer."}, {"video_title": "Solution set to sin equation Trigonometry Khan Academy.mp3", "Sentence": "We can add multiples of 2 pi to each of these. 2.80 plus any multiple of 2 pi, so 2 pi n where n is an integer, or we could take 0.34 and add any multiple of 2 pi. 2 pi n where n is an integer. Our solution set here, just to rewrite it outside of this messiness, is going to be 2.80 radians plus 2 pi n where n is an integer, and 0.34 plus 2 pi n where n is an integer. Let's see which of these are at least a subset of this. We look at 0.34 plus 2 pi n where n is an integer. That's exactly what we wrote over here."}, {"video_title": "Solution set to sin equation Trigonometry Khan Academy.mp3", "Sentence": "Our solution set here, just to rewrite it outside of this messiness, is going to be 2.80 radians plus 2 pi n where n is an integer, and 0.34 plus 2 pi n where n is an integer. Let's see which of these are at least a subset of this. We look at 0.34 plus 2 pi n where n is an integer. That's exactly what we wrote over here. That's 0.34. If n is a positive integer, we'll go around this way, and we keep getting back to the same point. If it's a negative integer, we go around that way, we keep getting to the same point."}, {"video_title": "Solution set to sin equation Trigonometry Khan Academy.mp3", "Sentence": "That's exactly what we wrote over here. That's 0.34. If n is a positive integer, we'll go around this way, and we keep getting back to the same point. If it's a negative integer, we go around that way, we keep getting to the same point. That's definitely in the solution set. 0.34 plus pi n for n an integer. If we have 0.34, and if we were to not add 2 pi, but just pi, where would we get to?"}, {"video_title": "Solution set to sin equation Trigonometry Khan Academy.mp3", "Sentence": "If it's a negative integer, we go around that way, we keep getting to the same point. That's definitely in the solution set. 0.34 plus pi n for n an integer. If we have 0.34, and if we were to not add 2 pi, but just pi, where would we get to? We would get to right over there. The sine of this isn't going to be positive 1 third, it's going to be negative 1 third. We could rule that out."}, {"video_title": "Solution set to sin equation Trigonometry Khan Academy.mp3", "Sentence": "If we have 0.34, and if we were to not add 2 pi, but just pi, where would we get to? We would get to right over there. The sine of this isn't going to be positive 1 third, it's going to be negative 1 third. We could rule that out. Negative 0.34, that's this angle right over here. The sine of that's going to be negative 1 third. If you add a multiple of 2 pi to that, you're still going to get negative 1 third, so that doesn't work."}, {"video_title": "Solution set to sin equation Trigonometry Khan Academy.mp3", "Sentence": "We could rule that out. Negative 0.34, that's this angle right over here. The sine of that's going to be negative 1 third. If you add a multiple of 2 pi to that, you're still going to get negative 1 third, so that doesn't work. Same thing for this one right over here. 2.8 plus 2 pi n, that's what we wrote right over here. Going all the way to 2.8, and any multiple of it is going to get you back to that same point."}, {"video_title": "Solution set to sin equation Trigonometry Khan Academy.mp3", "Sentence": "If you add a multiple of 2 pi to that, you're still going to get negative 1 third, so that doesn't work. Same thing for this one right over here. 2.8 plus 2 pi n, that's what we wrote right over here. Going all the way to 2.8, and any multiple of it is going to get you back to that same point. That one works. 2.8 plus pi n, so if you're here, and if you added pi, you're going to get over here. The sine of that isn't going to be positive 1 third, it's going to be negative 1 third."}, {"video_title": "Solution set to sin equation Trigonometry Khan Academy.mp3", "Sentence": "Going all the way to 2.8, and any multiple of it is going to get you back to that same point. That one works. 2.8 plus pi n, so if you're here, and if you added pi, you're going to get over here. The sine of that isn't going to be positive 1 third, it's going to be negative 1 third. We can rule this one out as well. These are the only two applying. If you actually take them together, you have the entire solution set to this equation right over here."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy (2).mp3", "Sentence": "Use a trigonometric function to model the temperature in Santiago, Chile, using 365 days as the length of a year. Remember that January 7th is the summer in Santiago. How many days after January 7th is the first spring day when the temperature reaches 20 degrees Celsius? So let's do this in two parts. So first, let's try to figure out a trigonometric function that models the temperature in Santiago, Chile. So we'll have temperature as a function of days, where days are the number of days after January 7th. And once we have that trigonometric function to model that, then we can answer the second part, I guess the essential question, which is how many days after January 7th is the first spring day when the temperature reaches 20 degrees Celsius?"}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy (2).mp3", "Sentence": "So let's do this in two parts. So first, let's try to figure out a trigonometric function that models the temperature in Santiago, Chile. So we'll have temperature as a function of days, where days are the number of days after January 7th. And once we have that trigonometric function to model that, then we can answer the second part, I guess the essential question, which is how many days after January 7th is the first spring day when the temperature reaches 20 degrees Celsius? So to think about this, let's graph it. Let's graph it, and it should become pretty apparent why they are suggesting that we use a trigonometric function to model this, because our seasonal variations, they're cyclical, they go up and down. And actually, if you look at the average temperature for any city over the course of the year, it really does look like a trigonometric function."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy (2).mp3", "Sentence": "And once we have that trigonometric function to model that, then we can answer the second part, I guess the essential question, which is how many days after January 7th is the first spring day when the temperature reaches 20 degrees Celsius? So to think about this, let's graph it. Let's graph it, and it should become pretty apparent why they are suggesting that we use a trigonometric function to model this, because our seasonal variations, they're cyclical, they go up and down. And actually, if you look at the average temperature for any city over the course of the year, it really does look like a trigonometric function. So let's, so let's, this axis right over here, this is the passage of the days. Let's do D for days, and that's going to be in days after January 7th. So this right over here would be January 7th."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy (2).mp3", "Sentence": "And actually, if you look at the average temperature for any city over the course of the year, it really does look like a trigonometric function. So let's, so let's, this axis right over here, this is the passage of the days. Let's do D for days, and that's going to be in days after January 7th. So this right over here would be January 7th. And the vertical axis, this is the horizontal, the vertical axis is going to be in terms of Celsius degrees. So this, let's see, the high is 29, and I could write 29 degrees Celsius, and then, or the highest average day. And then if this is zero, then 14, which is the lowest average day, 14 degrees Celsius."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy (2).mp3", "Sentence": "So this right over here would be January 7th. And the vertical axis, this is the horizontal, the vertical axis is going to be in terms of Celsius degrees. So this, let's see, the high is 29, and I could write 29 degrees Celsius, and then, or the highest average day. And then if this is zero, then 14, which is the lowest average day, 14 degrees Celsius. And so our temperature will vary between these two extremes. Our temperature is going to vary, the highest average day, which they already told us is January 7th, we get to 29 degrees Celsius. And then the coldest day of the year, on average, you get to an average high of 14 degrees Celsius."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy (2).mp3", "Sentence": "And then if this is zero, then 14, which is the lowest average day, 14 degrees Celsius. And so our temperature will vary between these two extremes. Our temperature is going to vary, the highest average day, which they already told us is January 7th, we get to 29 degrees Celsius. And then the coldest day of the year, on average, you get to an average high of 14 degrees Celsius. So it looks like this. We're talking about average highs on a given day. And the reason why a trigonometric function is a good idea is because it's cyclical."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy (2).mp3", "Sentence": "And then the coldest day of the year, on average, you get to an average high of 14 degrees Celsius. So it looks like this. We're talking about average highs on a given day. And the reason why a trigonometric function is a good idea is because it's cyclical. So if this is January 7th, if you go 365 days in the future, you're back at January 7th. So if the average high temperature is 29 degrees Celsius on that day, the average high temperature is going to be 29 degrees Celsius on that day. Now, we're using a trigonometric function, so we're gonna hit our low point exactly halfway in between."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy (2).mp3", "Sentence": "And the reason why a trigonometric function is a good idea is because it's cyclical. So if this is January 7th, if you go 365 days in the future, you're back at January 7th. So if the average high temperature is 29 degrees Celsius on that day, the average high temperature is going to be 29 degrees Celsius on that day. Now, we're using a trigonometric function, so we're gonna hit our low point exactly halfway in between. So we're gonna hit our low point exactly halfway in between, something right like that. And so our function is going to look like this. Our function, so let me see, let me just draw the low point right over there."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy (2).mp3", "Sentence": "Now, we're using a trigonometric function, so we're gonna hit our low point exactly halfway in between. So we're gonna hit our low point exactly halfway in between, something right like that. And so our function is going to look like this. Our function, so let me see, let me just draw the low point right over there. And this is a high point. It's a high point right over there. That looks pretty good."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy (2).mp3", "Sentence": "Our function, so let me see, let me just draw the low point right over there. And this is a high point. It's a high point right over there. That looks pretty good. And then I have the high point right over here, and then I just need to connect them, and there you go. I've drawn one period of our trigonometric function. And our period is 365 days."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy (2).mp3", "Sentence": "That looks pretty good. And then I have the high point right over here, and then I just need to connect them, and there you go. I've drawn one period of our trigonometric function. And our period is 365 days. If we go 365 days later, we're at the same point in the cycle. We are at the same point in the year. We're at the same point in the year."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy (2).mp3", "Sentence": "And our period is 365 days. If we go 365 days later, we're at the same point in the cycle. We are at the same point in the year. We're at the same point in the year. So what I want you to do right now is given what I've just drawn, try to model this right. So this right over here is, let's write this, this is capital T as a function of D. Try to figure out an expression for capital T as a function of D. And remember, it's going to be some trigonometric function. So I'm assuming you've given a go at it, and you might say, well, this looks like a cosine curve."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy (2).mp3", "Sentence": "We're at the same point in the year. So what I want you to do right now is given what I've just drawn, try to model this right. So this right over here is, let's write this, this is capital T as a function of D. Try to figure out an expression for capital T as a function of D. And remember, it's going to be some trigonometric function. So I'm assuming you've given a go at it, and you might say, well, this looks like a cosine curve. Maybe it could be a sine curve. Which one should I use? And you could actually use either one, but I always like to go with the simpler one."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy (2).mp3", "Sentence": "So I'm assuming you've given a go at it, and you might say, well, this looks like a cosine curve. Maybe it could be a sine curve. Which one should I use? And you could actually use either one, but I always like to go with the simpler one. You just think about, well, if this were angles, either actually degrees or radians, which trigonometric function starts at your maximum point? Well, cosine of zero is one. The cosine starts at your maximum point."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy (2).mp3", "Sentence": "And you could actually use either one, but I always like to go with the simpler one. You just think about, well, if this were angles, either actually degrees or radians, which trigonometric function starts at your maximum point? Well, cosine of zero is one. The cosine starts at your maximum point. Sine of zero is zero. So I'm going to use cosine here. I'm going to use a cosine function."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy (2).mp3", "Sentence": "The cosine starts at your maximum point. Sine of zero is zero. So I'm going to use cosine here. I'm going to use a cosine function. So temperature as a function of days is going to be some amplitude times our cosine function. And we're going to have some argument to our cosine function. And then I'm probably going to have to shift it."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy (2).mp3", "Sentence": "I'm going to use a cosine function. So temperature as a function of days is going to be some amplitude times our cosine function. And we're going to have some argument to our cosine function. And then I'm probably going to have to shift it. So let's think about how we would do that. Well, what's the midline here? Well, the midline's the halfway point between our high and our low."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy (2).mp3", "Sentence": "And then I'm probably going to have to shift it. So let's think about how we would do that. Well, what's the midline here? Well, the midline's the halfway point between our high and our low. So our middle point, if we were to visualize it, looks just like that is our midline right over there. And what value is this? Well, what's the average of 29 and 14?"}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy (2).mp3", "Sentence": "Well, the midline's the halfway point between our high and our low. So our middle point, if we were to visualize it, looks just like that is our midline right over there. And what value is this? Well, what's the average of 29 and 14? 29 plus 14 is 43 divided by 2 is 21.5 degrees Celsius. So that's our midline. So essentially, we've shifted up our function by that amount."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy (2).mp3", "Sentence": "Well, what's the average of 29 and 14? 29 plus 14 is 43 divided by 2 is 21.5 degrees Celsius. So that's our midline. So essentially, we've shifted up our function by that amount. If we just had a regular cosine function, our midline would be at 0. But now we're at 21.5 degrees Celsius. So I'll just write plus 21.5."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy (2).mp3", "Sentence": "So essentially, we've shifted up our function by that amount. If we just had a regular cosine function, our midline would be at 0. But now we're at 21.5 degrees Celsius. So I'll just write plus 21.5. That's how much we've shifted it up. Now, what's the amplitude? Well, our amplitude is how much we diverge from the midline."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy (2).mp3", "Sentence": "So I'll just write plus 21.5. That's how much we've shifted it up. Now, what's the amplitude? Well, our amplitude is how much we diverge from the midline. So over here, we're 7.5 above the midline. So that's plus 7.5. Here, we're 7.5 below the midline."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy (2).mp3", "Sentence": "Well, our amplitude is how much we diverge from the midline. So over here, we're 7.5 above the midline. So that's plus 7.5. Here, we're 7.5 below the midline. So minus 7.5. So our amplitude is 7.5. So the maximum amount we go away from the midline is 7.5."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy (2).mp3", "Sentence": "Here, we're 7.5 below the midline. So minus 7.5. So our amplitude is 7.5. So the maximum amount we go away from the midline is 7.5. So that's our amplitude. And now let's think about our argument to the cosine function right over here. So it's going to be a function of the days."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy (2).mp3", "Sentence": "So the maximum amount we go away from the midline is 7.5. So that's our amplitude. And now let's think about our argument to the cosine function right over here. So it's going to be a function of the days. And what do we want? When 365 days have gone by, we want this entire argument to be 2 pi. So when d is 365, we want this whole thing to evaluate to 2 pi."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy (2).mp3", "Sentence": "So it's going to be a function of the days. And what do we want? When 365 days have gone by, we want this entire argument to be 2 pi. So when d is 365, we want this whole thing to evaluate to 2 pi. So we could put 2 pi over 365 in here. And you might remember your formulas. I always forget them."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy (2).mp3", "Sentence": "So when d is 365, we want this whole thing to evaluate to 2 pi. So we could put 2 pi over 365 in here. And you might remember your formulas. I always forget them. That's why I always try to reason through them again. The formula is, oh, you want 2 pi divided by your period and all the rest. But I just like to think, OK, look, after one period, which is 365 days, I want the whole argument over here to be 2 pi."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy (2).mp3", "Sentence": "I always forget them. That's why I always try to reason through them again. The formula is, oh, you want 2 pi divided by your period and all the rest. But I just like to think, OK, look, after one period, which is 365 days, I want the whole argument over here to be 2 pi. I want to go around the unit circle once. And so if this is 2 pi over 365, when you multiply by 365, your argument here is going to be 2 pi. So just like that, we've done the first part of this question."}, {"video_title": "IIT JEE trigonometric maximum Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So this is 1 over, so I have a sine squared of theta. I have a sine squared of theta. And then, at least in my brain, whenever I see a sine squared of theta, I always look for a cosine squared of theta, because I know that when I take the sum of them, it equals 1. And I don't have just 1 cosine squared theta here. I have 5 cosine squared theta. So let me just take one of them. So I have a plus cosine squared theta."}, {"video_title": "IIT JEE trigonometric maximum Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And I don't have just 1 cosine squared theta here. I have 5 cosine squared theta. So let me just take one of them. So I have a plus cosine squared theta. And since I took one of them, I only have 4 of these left. So plus 4 cosine squared theta. And then I have this stuff."}, {"video_title": "IIT JEE trigonometric maximum Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So I have a plus cosine squared theta. And since I took one of them, I only have 4 of these left. So plus 4 cosine squared theta. And then I have this stuff. Plus 3 sine of theta cosine of theta. So what this first step allowed me to do is just turn these two characters right over here, sine squared theta plus cosine squared theta. That is equal to 1."}, {"video_title": "IIT JEE trigonometric maximum Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And then I have this stuff. Plus 3 sine of theta cosine of theta. So what this first step allowed me to do is just turn these two characters right over here, sine squared theta plus cosine squared theta. That is equal to 1. So we've simplified it to 1 over 1 plus. Now let's think about how we can write cosine squared of theta. I'll write our identities here."}, {"video_title": "IIT JEE trigonometric maximum Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "That is equal to 1. So we've simplified it to 1 over 1 plus. Now let's think about how we can write cosine squared of theta. I'll write our identities here. Cosine squared of theta. And we've proved these in the trigonometry playlist. This is equal to 1 plus cosine of 2 theta."}, {"video_title": "IIT JEE trigonometric maximum Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "I'll write our identities here. Cosine squared of theta. And we've proved these in the trigonometry playlist. This is equal to 1 plus cosine of 2 theta. All of that over 2. And my goal here is I really just want to get everything. Well, I really just want to simplify it."}, {"video_title": "IIT JEE trigonometric maximum Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "This is equal to 1 plus cosine of 2 theta. All of that over 2. And my goal here is I really just want to get everything. Well, I really just want to simplify it. And maybe we'll do calculus. We'll use a little calculus to find the minimum value of the denominator, which would give us the maximum value of the numerator. So let's see."}, {"video_title": "IIT JEE trigonometric maximum Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Well, I really just want to simplify it. And maybe we'll do calculus. We'll use a little calculus to find the minimum value of the denominator, which would give us the maximum value of the numerator. So let's see. Cosine squared of theta is equal to this. So 4 times this is just going to be, the 4 divided by 2 is 2. So it's going to be 2 times this numerator."}, {"video_title": "IIT JEE trigonometric maximum Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So let's see. Cosine squared of theta is equal to this. So 4 times this is just going to be, the 4 divided by 2 is 2. So it's going to be 2 times this numerator. So it's going to be 2 plus 2 cosine of 2 theta. That's this term right here. And then this term right over here, we could use the trig identity that sine of 2 theta is equal to 2 sine of theta cosine of theta."}, {"video_title": "IIT JEE trigonometric maximum Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So it's going to be 2 times this numerator. So it's going to be 2 plus 2 cosine of 2 theta. That's this term right here. And then this term right over here, we could use the trig identity that sine of 2 theta is equal to 2 sine of theta cosine of theta. Or you divide both sides by 2, you get 1 half sine of 2 theta is equal to sine of theta cosine of theta. So this right over here, this part right over here is going to be 1 half sine of 2 theta, but we're multiplying it by 3. So it's going to be plus 3 halves sine of 2 theta."}, {"video_title": "IIT JEE trigonometric maximum Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And then this term right over here, we could use the trig identity that sine of 2 theta is equal to 2 sine of theta cosine of theta. Or you divide both sides by 2, you get 1 half sine of 2 theta is equal to sine of theta cosine of theta. So this right over here, this part right over here is going to be 1 half sine of 2 theta, but we're multiplying it by 3. So it's going to be plus 3 halves sine of 2 theta. And let's see. This part over here clearly simplifies. This is 3, so let me just rewrite it."}, {"video_title": "IIT JEE trigonometric maximum Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So it's going to be plus 3 halves sine of 2 theta. And let's see. This part over here clearly simplifies. This is 3, so let me just rewrite it. This is 1 over 3 plus 2 cosine of 2 theta plus 3 halves sine of 2 theta. And we're really just looking for the minimum value of the denominator, which would give us the same as the maximum value of the numerator. It would just be 1 over this minimum value."}, {"video_title": "IIT JEE trigonometric maximum Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "This is 3, so let me just rewrite it. This is 1 over 3 plus 2 cosine of 2 theta plus 3 halves sine of 2 theta. And we're really just looking for the minimum value of the denominator, which would give us the same as the maximum value of the numerator. It would just be 1 over this minimum value. So let's see how low we can get. Assuming we're above 0, let's see how low we can get for this denominator right here. We're just going to look for its minimum value."}, {"video_title": "IIT JEE trigonometric maximum Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "It would just be 1 over this minimum value. So let's see how low we can get. Assuming we're above 0, let's see how low we can get for this denominator right here. We're just going to look for its minimum value. So one thing we can do just to simplify things, the minimum value of this is going to be the same. The min of this thing, I don't want to write it there because it kind of confuses the problem. The minimum value of 3 plus 2 cosine 2 theta plus 3 halves sine of 2 theta is going to be the same thing as the minimum value of 3 plus."}, {"video_title": "IIT JEE trigonometric maximum Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "We're just going to look for its minimum value. So one thing we can do just to simplify things, the minimum value of this is going to be the same. The min of this thing, I don't want to write it there because it kind of confuses the problem. The minimum value of 3 plus 2 cosine 2 theta plus 3 halves sine of 2 theta is going to be the same thing as the minimum value of 3 plus. I'm just going to do the substitution that 2 theta is equal to x. That just simplifies things a little bit. You don't have to do that."}, {"video_title": "IIT JEE trigonometric maximum Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "The minimum value of 3 plus 2 cosine 2 theta plus 3 halves sine of 2 theta is going to be the same thing as the minimum value of 3 plus. I'm just going to do the substitution that 2 theta is equal to x. That just simplifies things a little bit. You don't have to do that. So 3 plus 2 cosine of x plus 3 halves sine of x. So this is a pretty simple expression. Let's see how we can figure out its minimum value."}, {"video_title": "IIT JEE trigonometric maximum Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "You don't have to do that. So 3 plus 2 cosine of x plus 3 halves sine of x. So this is a pretty simple expression. Let's see how we can figure out its minimum value. And my temptation is to take the derivative, find out where the derivative is equal to 0, and then that will either be a maximum or a minimum point. So let's take the derivative. The derivative of this expression right over here with respect to x."}, {"video_title": "IIT JEE trigonometric maximum Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Let's see how we can figure out its minimum value. And my temptation is to take the derivative, find out where the derivative is equal to 0, and then that will either be a maximum or a minimum point. So let's take the derivative. The derivative of this expression right over here with respect to x. Well, derivative of 3 with respect to x is 0. Derivative of 2 cosine of x is negative 2 sine of x. Derivative of 3 halves sine of x is going to be plus 3 halves cosine of x."}, {"video_title": "IIT JEE trigonometric maximum Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "The derivative of this expression right over here with respect to x. Well, derivative of 3 with respect to x is 0. Derivative of 2 cosine of x is negative 2 sine of x. Derivative of 3 halves sine of x is going to be plus 3 halves cosine of x. And then that is going to be equal to 0. We want to find where the slope is 0 because that's either going to be a maximum or a minimum point. And let's see, we could add 2 sine of x to both sides."}, {"video_title": "IIT JEE trigonometric maximum Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Derivative of 3 halves sine of x is going to be plus 3 halves cosine of x. And then that is going to be equal to 0. We want to find where the slope is 0 because that's either going to be a maximum or a minimum point. And let's see, we could add 2 sine of x to both sides. So we get 3 halves cosine of x is equal to 2 sine of x. And then we can divide both sides of this equation by 2 first. I don't want to skip too many steps."}, {"video_title": "IIT JEE trigonometric maximum Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And let's see, we could add 2 sine of x to both sides. So we get 3 halves cosine of x is equal to 2 sine of x. And then we can divide both sides of this equation by 2 first. I don't want to skip too many steps. So 3 fourths cosine of x is equal to sine of x. And we could divide both sides by cosine of x. So we get 3 over 4 is equal to sine of x over cosine of x, which is the same thing as the tangent of x."}, {"video_title": "IIT JEE trigonometric maximum Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "I don't want to skip too many steps. So 3 fourths cosine of x is equal to sine of x. And we could divide both sides by cosine of x. So we get 3 over 4 is equal to sine of x over cosine of x, which is the same thing as the tangent of x. So an x value that gives us 3 or the tangent of an x value gives us 3 fourths is going to give us either a maximum or a minimum point. So let's think about this a little bit. Let me draw my unit circle."}, {"video_title": "IIT JEE trigonometric maximum Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So we get 3 over 4 is equal to sine of x over cosine of x, which is the same thing as the tangent of x. So an x value that gives us 3 or the tangent of an x value gives us 3 fourths is going to give us either a maximum or a minimum point. So let's think about this a little bit. Let me draw my unit circle. So let's think about the 2 x values that will give us a tangent of 3 fourths. So let me draw my unit circle. This is always the hardest part."}, {"video_title": "IIT JEE trigonometric maximum Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Let me draw my unit circle. So let's think about the 2 x values that will give us a tangent of 3 fourths. So let me draw my unit circle. This is always the hardest part. So let me draw this. All right. There is my unit circle."}, {"video_title": "IIT JEE trigonometric maximum Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "This is always the hardest part. So let me draw this. All right. There is my unit circle. So how can I get a triangle where an angle is a tangent of 3 fourths? Remember, tangent is opposite over adjacent. So if this is my triangle right over here, if this is x, opposite over adjacent is equal to 3 fourths."}, {"video_title": "IIT JEE trigonometric maximum Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "There is my unit circle. So how can I get a triangle where an angle is a tangent of 3 fourths? Remember, tangent is opposite over adjacent. So if this is my triangle right over here, if this is x, opposite over adjacent is equal to 3 fourths. So opposite could be 3 and adjacent could be 4. And we hopefully immediately recognize this. This is a 3, 4, 5 triangle because it's a right triangle."}, {"video_title": "IIT JEE trigonometric maximum Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So if this is my triangle right over here, if this is x, opposite over adjacent is equal to 3 fourths. So opposite could be 3 and adjacent could be 4. And we hopefully immediately recognize this. This is a 3, 4, 5 triangle because it's a right triangle. 3 squared plus 4 squared is 25, which is 5 squared. So this is a 3, 4, 5 triangle. Now, there's 2 tangent values."}, {"video_title": "IIT JEE trigonometric maximum Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "This is a 3, 4, 5 triangle because it's a right triangle. 3 squared plus 4 squared is 25, which is 5 squared. So this is a 3, 4, 5 triangle. Now, there's 2 tangent values. So x could be like this. So this obviously isn't a unit hypotenuse over here, but we can divide everything by 5 and it would be. So we could have this situation over here where this is x, if this is a unit circle, the hypotenuse is 1."}, {"video_title": "IIT JEE trigonometric maximum Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Now, there's 2 tangent values. So x could be like this. So this obviously isn't a unit hypotenuse over here, but we can divide everything by 5 and it would be. So we could have this situation over here where this is x, if this is a unit circle, the hypotenuse is 1. This is 3 over 5 and this is 4 over 5. This would, tangent of x here, would give us 3 fourths. But is this going to give us a maximum value or minimum value?"}, {"video_title": "IIT JEE trigonometric maximum Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So we could have this situation over here where this is x, if this is a unit circle, the hypotenuse is 1. This is 3 over 5 and this is 4 over 5. This would, tangent of x here, would give us 3 fourths. But is this going to give us a maximum value or minimum value? Well, over here, both cosine of x and sine of x are going to be positive. So both of these are going to be positive values. So it's going to probably maximize this expression over here."}, {"video_title": "IIT JEE trigonometric maximum Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "But is this going to give us a maximum value or minimum value? Well, over here, both cosine of x and sine of x are going to be positive. So both of these are going to be positive values. So it's going to probably maximize this expression over here. Now, the other x that gives us the same tangent, and remember, the tangent is really just the slope of the radius of the unit circle, would be this angle. This has the same tangent value. So this x over here."}, {"video_title": "IIT JEE trigonometric maximum Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So it's going to probably maximize this expression over here. Now, the other x that gives us the same tangent, and remember, the tangent is really just the slope of the radius of the unit circle, would be this angle. This has the same tangent value. So this x over here. In this case, the tangent is still going to be 3 fourths, but over here, the sine and cosine are negative. So over here, the x-coordinate, or the cosine, is going to be negative 4 fifths. And the sine value, or the y value here, is going to be negative 3 fifths."}, {"video_title": "IIT JEE trigonometric maximum Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So this x over here. In this case, the tangent is still going to be 3 fourths, but over here, the sine and cosine are negative. So over here, the x-coordinate, or the cosine, is going to be negative 4 fifths. And the sine value, or the y value here, is going to be negative 3 fifths. And this will give us a minimum point, because here everything is, both the sine and the cosine are negative. So let's use this x right over here. And notice, we don't even have to figure out what the x is, because we know that if the tangent over here is 4 fifths, either both the sine is going to be 3 fifths and the cosine is going to be 4 fifths, which will give us a maximum point, or the tangent could be 3 fourths, and then the sine will be negative 3 fifths and the cosine will be negative 4 fifths."}, {"video_title": "IIT JEE trigonometric maximum Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And the sine value, or the y value here, is going to be negative 3 fifths. And this will give us a minimum point, because here everything is, both the sine and the cosine are negative. So let's use this x right over here. And notice, we don't even have to figure out what the x is, because we know that if the tangent over here is 4 fifths, either both the sine is going to be 3 fifths and the cosine is going to be 4 fifths, which will give us a maximum point, or the tangent could be 3 fourths, and then the sine will be negative 3 fifths and the cosine will be negative 4 fifths. So let's use these over here. So the minimum is going to be equal to 3 plus 2 times cosine of x. We're using this one over here."}, {"video_title": "IIT JEE trigonometric maximum Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And notice, we don't even have to figure out what the x is, because we know that if the tangent over here is 4 fifths, either both the sine is going to be 3 fifths and the cosine is going to be 4 fifths, which will give us a maximum point, or the tangent could be 3 fourths, and then the sine will be negative 3 fifths and the cosine will be negative 4 fifths. So let's use these over here. So the minimum is going to be equal to 3 plus 2 times cosine of x. We're using this one over here. So 2 times cosine of x. Cosine of x here is negative 4 fifths. And then plus 3 halves times the sine of x. Sine of x here is negative 3 fifths."}, {"video_title": "IIT JEE trigonometric maximum Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "We're using this one over here. So 2 times cosine of x. Cosine of x here is negative 4 fifths. And then plus 3 halves times the sine of x. Sine of x here is negative 3 fifths. And what is this going to be equal to? This is going to be equal to 3 plus, this is negative 8 fifths, maybe I should write 3 minus 8 fifths, minus 9 tenths, and so this is going to be equal to, we could put everything over 10, 30 over 10 minus 16 over 10, that's 8 fifths, minus 9 over 10, and this gives us what? This gives us 5 over 10, or 1 half."}, {"video_title": "IIT JEE trigonometric maximum Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Sine of x here is negative 3 fifths. And what is this going to be equal to? This is going to be equal to 3 plus, this is negative 8 fifths, maybe I should write 3 minus 8 fifths, minus 9 tenths, and so this is going to be equal to, we could put everything over 10, 30 over 10 minus 16 over 10, that's 8 fifths, minus 9 over 10, and this gives us what? This gives us 5 over 10, or 1 half. So the minimum value for our denominator, everything we've been dealing with so far has been our denominator, the minimum value of all of this business over here, the minimum value is 1 half. So the maximum value that this whole expression takes is when the minimum value is 1 half. So we get 1 over 1 half, which is equal to 2."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "So to think about that, let's actually draw the sine function out. And what I have here, on the left-hand side right over here, I've got a unit circle. And I can, let me truncate this a little bit. I don't need that space right there. So let me clear that out. So I have a unit circle on the left-hand side right over here. And I'm going to use that to figure out the values of sine of theta for a given theta."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "I don't need that space right there. So let me clear that out. So I have a unit circle on the left-hand side right over here. And I'm going to use that to figure out the values of sine of theta for a given theta. So in the unit circle, this is x and this is y. Or you could even view this as the, well, we could just use x or y. And so for a given theta, we can see where that angle entered, the terminal side of the angle intersects the unit circle."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "And I'm going to use that to figure out the values of sine of theta for a given theta. So in the unit circle, this is x and this is y. Or you could even view this as the, well, we could just use x or y. And so for a given theta, we can see where that angle entered, the terminal side of the angle intersects the unit circle. And then the y-coordinate of that point is going to be sine of theta. And over here, I'm going to graph still y in the vertical axis. But I'm going to graph the graph of y is equal to sine of theta."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "And so for a given theta, we can see where that angle entered, the terminal side of the angle intersects the unit circle. And then the y-coordinate of that point is going to be sine of theta. And over here, I'm going to graph still y in the vertical axis. But I'm going to graph the graph of y is equal to sine of theta. y is equal to sine of theta. And on the horizontal axis, I'm not going to graph x. But I'm going to graph theta."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "But I'm going to graph the graph of y is equal to sine of theta. y is equal to sine of theta. And on the horizontal axis, I'm not going to graph x. But I'm going to graph theta. You can view theta as the independent variable here. And it's going to be theta is going to be in radians. So we're essentially going to pick a bunch of thetas and then come up with what sine of theta is and then graph it."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "But I'm going to graph theta. You can view theta as the independent variable here. And it's going to be theta is going to be in radians. So we're essentially going to pick a bunch of thetas and then come up with what sine of theta is and then graph it. So let's set up a little bit of a table here. Let's set up a little bit of a table. And so over here, I have theta."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "So we're essentially going to pick a bunch of thetas and then come up with what sine of theta is and then graph it. So let's set up a little bit of a table here. Let's set up a little bit of a table. And so over here, I have theta. And over here, we're going to figure out what sine of theta is. And we could do a bunch of theta values. We could start, let's say we start at 0."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "And so over here, I have theta. And over here, we're going to figure out what sine of theta is. And we could do a bunch of theta values. We could start, let's say we start at 0. Let's say we start at theta is equal to 0. What is sine of theta going to be? Well, when the angle is 0, we intersect the unit circle right over there."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "We could start, let's say we start at 0. Let's say we start at theta is equal to 0. What is sine of theta going to be? Well, when the angle is 0, we intersect the unit circle right over there. The y-coordinate of this is still 0. This is the point 1, 0. The y-coordinate is 0."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "Well, when the angle is 0, we intersect the unit circle right over there. The y-coordinate of this is still 0. This is the point 1, 0. The y-coordinate is 0. So sine of theta is 0. Or we could say sine of 0 is equal to 0. Sine of 0 is equal to 0."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "The y-coordinate is 0. So sine of theta is 0. Or we could say sine of 0 is equal to 0. Sine of 0 is equal to 0. Now let's try theta is equal to pi over 2. Theta is equal to pi over 2. I'm just doing the ones that are really easy to figure out."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "Sine of 0 is equal to 0. Now let's try theta is equal to pi over 2. Theta is equal to pi over 2. I'm just doing the ones that are really easy to figure out. So if theta is equal to pi over 2, that's the same thing as a 90-degree angle. So the terminal side is going to be right along the y-axis, just like that. And where it intersects the unit circle is right over here."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "I'm just doing the ones that are really easy to figure out. So if theta is equal to pi over 2, that's the same thing as a 90-degree angle. So the terminal side is going to be right along the y-axis, just like that. And where it intersects the unit circle is right over here. And what point is that? Well, that's the point 0, 1. So what is sine of pi over 2?"}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "And where it intersects the unit circle is right over here. And what point is that? Well, that's the point 0, 1. So what is sine of pi over 2? Well, sine of pi over 2 is just the y-coordinate right over here. It is 1. Sine of pi over 2 is 1."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "So what is sine of pi over 2? Well, sine of pi over 2 is just the y-coordinate right over here. It is 1. Sine of pi over 2 is 1. Let's keep going. You might see a little pattern here. We're just going more and more around the circle."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "Sine of pi over 2 is 1. Let's keep going. You might see a little pattern here. We're just going more and more around the circle. So let's think about what happens when theta is equal to pi. What is sine of pi? Well, we intersect the unit circle right over there."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "We're just going more and more around the circle. So let's think about what happens when theta is equal to pi. What is sine of pi? Well, we intersect the unit circle right over there. That coordinate is negative 1, 0. Sine is the y-coordinate. So this right over here is sine of pi."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "Well, we intersect the unit circle right over there. That coordinate is negative 1, 0. Sine is the y-coordinate. So this right over here is sine of pi. Sine of pi is 0. Let's go to 3 pi over 2. 3 pi over 2."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "So this right over here is sine of pi. Sine of pi is 0. Let's go to 3 pi over 2. 3 pi over 2. Well, now we've gone 3 quarters of the way around the circle. We intersect. The terminal side of the angle intersects the unit circle right over here."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "3 pi over 2. Well, now we've gone 3 quarters of the way around the circle. We intersect. The terminal side of the angle intersects the unit circle right over here. And so based on that, what is sine of 3 pi over 2? Well, this point right over here is the point negative. Let me be careful."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "The terminal side of the angle intersects the unit circle right over here. And so based on that, what is sine of 3 pi over 2? Well, this point right over here is the point negative. Let me be careful. It is 0, negative 1. The sine of theta is the same thing as the y-coordinate. The y-coordinate is sine of theta."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "Let me be careful. It is 0, negative 1. The sine of theta is the same thing as the y-coordinate. The y-coordinate is sine of theta. So when theta is 3 pi over 2, sine of theta is equal to negative 1. Now let's come full circle here. So let's go all the way to theta equaling 2 pi."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "The y-coordinate is sine of theta. So when theta is 3 pi over 2, sine of theta is equal to negative 1. Now let's come full circle here. So let's go all the way to theta equaling 2 pi. Let me do a color. Yeah, I'll just use the yellow here. What happens when theta is equal to 2 pi?"}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "So let's go all the way to theta equaling 2 pi. Let me do a color. Yeah, I'll just use the yellow here. What happens when theta is equal to 2 pi? Well, then we've gone all the way around the circle. And we are back to where we started. And the y-coordinate is 0."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "What happens when theta is equal to 2 pi? Well, then we've gone all the way around the circle. And we are back to where we started. And the y-coordinate is 0. So sine of 2 pi is once again 0. And if we were to keep going around, we're going to start seeing, as we keep incrementing the angle, we're going to start seeing the same pattern emerge again. Well, let's try to graph this."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "And the y-coordinate is 0. So sine of 2 pi is once again 0. And if we were to keep going around, we're going to start seeing, as we keep incrementing the angle, we're going to start seeing the same pattern emerge again. Well, let's try to graph this. So when theta is equal to 0, sine of theta is 0. When theta is equal to pi over 2, sine of theta is 1. So we'll use the same scale."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "Well, let's try to graph this. So when theta is equal to 0, sine of theta is 0. When theta is equal to pi over 2, sine of theta is 1. So we'll use the same scale. So sine of theta is equal to 1. I'll make this 1 on this axis and on that axis. So we can maybe see a little bit of a parallel here."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "So we'll use the same scale. So sine of theta is equal to 1. I'll make this 1 on this axis and on that axis. So we can maybe see a little bit of a parallel here. When theta is equal to pi, sine of theta is 0. So when theta is equal to pi, sine of theta is 0. So we go back right over there."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "So we can maybe see a little bit of a parallel here. When theta is equal to pi, sine of theta is 0. So when theta is equal to pi, sine of theta is 0. So we go back right over there. When theta is equal to 3 pi over 2, so that would be right over here, sine of theta is negative 1. So this is negative 1 over here. I'll do the same scale over here."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "So we go back right over there. When theta is equal to 3 pi over 2, so that would be right over here, sine of theta is negative 1. So this is negative 1 over here. I'll do the same scale over here. I'll make this negative. I'll make this negative 1. And so sine of theta is negative 1."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "I'll do the same scale over here. I'll make this negative. I'll make this negative 1. And so sine of theta is negative 1. And then when theta is 2 pi, sine of theta is 0. So when theta is 2 pi, sine of theta is 0. And so we can connect the dots."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "And so sine of theta is negative 1. And then when theta is 2 pi, sine of theta is 0. So when theta is 2 pi, sine of theta is 0. And so we can connect the dots. You can try other points in between, and you get a graph that looks something like this. My best attempt at drawing it freehand. It looks something like this."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "And so we can connect the dots. You can try other points in between, and you get a graph that looks something like this. My best attempt at drawing it freehand. It looks something like this. There's a reason why curves that look like this are called sinusoids, because they're the graph of a sine function. So just like this. But that's not the entire graph."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "It looks something like this. There's a reason why curves that look like this are called sinusoids, because they're the graph of a sine function. So just like this. But that's not the entire graph. We could keep going. We could add another pi over 2. If you added another pi over 2, so if you go to 2 pi, and then you add another pi over 2, so you could view this as 2 and 1 half pi, or however you want to think about it, then you're going to go back over here."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "But that's not the entire graph. We could keep going. We could add another pi over 2. If you added another pi over 2, so if you go to 2 pi, and then you add another pi over 2, so you could view this as 2 and 1 half pi, or however you want to think about it, then you're going to go back over here. So then you're going to get back to sine of theta being equal to 1. So you're going to go back to this point right over here. And you could keep going."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "If you added another pi over 2, so if you go to 2 pi, and then you add another pi over 2, so you could view this as 2 and 1 half pi, or however you want to think about it, then you're going to go back over here. So then you're going to get back to sine of theta being equal to 1. So you're going to go back to this point right over here. And you could keep going. You go another pi over 2, you're going to go back to this point, and you're going to be over here. And so the curve, or the function sine of theta, is really defined for any theta value, any real theta value that you choose. So any theta value."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "And you could keep going. You go another pi over 2, you're going to go back to this point, and you're going to be over here. And so the curve, or the function sine of theta, is really defined for any theta value, any real theta value that you choose. So any theta value. You say, well, what about negatives? I mean, obviously I agree. As you keep increasing theta like this, we just keep going around and around the circle, and this pattern kind of emerges."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "So any theta value. You say, well, what about negatives? I mean, obviously I agree. As you keep increasing theta like this, we just keep going around and around the circle, and this pattern kind of emerges. But what happens when we go in the negative direction? Well, let's try it out. What happens if we were to take negative pi over 2?"}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "As you keep increasing theta like this, we just keep going around and around the circle, and this pattern kind of emerges. But what happens when we go in the negative direction? Well, let's try it out. What happens if we were to take negative pi over 2? So let me do that. So negative pi over 2, well, that's going right over here. And so we intersect the unit circle right over there."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "What happens if we were to take negative pi over 2? So let me do that. So negative pi over 2, well, that's going right over here. And so we intersect the unit circle right over there. The y-coordinate is negative 1. So sine of negative pi over 2 is negative 1. And we see that it just continues."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "And so we intersect the unit circle right over there. The y-coordinate is negative 1. So sine of negative pi over 2 is negative 1. And we see that it just continues. It just continues. So sine of theta is defined for any positive, negative, any theta, positive or negative, non-negative, 0, anything. So it's defined for anything."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "And we see that it just continues. It just continues. So sine of theta is defined for any positive, negative, any theta, positive or negative, non-negative, 0, anything. So it's defined for anything. So let's go back to the question. So I could just keep drawing this function on and on and on. So let's go back to the question."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "So it's defined for anything. So let's go back to the question. So I could just keep drawing this function on and on and on. So let's go back to the question. What is the domain of the sine function? And just as a reminder, the domain are all of the inputs over which the function is defined, or all of the valid inputs into the function that the function will actually spit out a valid answer. So what is the domain of the sine function?"}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "So let's go back to the question. What is the domain of the sine function? And just as a reminder, the domain are all of the inputs over which the function is defined, or all of the valid inputs into the function that the function will actually spit out a valid answer. So what is the domain of the sine function? Well, we already saw. We can put in any theta here. So you could say the domain is all real numbers."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "So what is the domain of the sine function? Well, we already saw. We can put in any theta here. So you could say the domain is all real numbers. Now, what about the range? What about the range? Well, just as a review, the range is sometimes in more technical math classes called the image."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "So you could say the domain is all real numbers. Now, what about the range? What about the range? Well, just as a review, the range is sometimes in more technical math classes called the image. It's the set of all the values that the function can actually take on. Well, what is that set? What is the range here?"}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "Well, just as a review, the range is sometimes in more technical math classes called the image. It's the set of all the values that the function can actually take on. Well, what is that set? What is the range here? What is all the values that y equals sine of theta could actually take on? Well, we see that it keeps going between positive 1. It keeps going between positive 1 and then to negative 1, and then back to positive 1, and then negative 1."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "What is the range here? What is all the values that y equals sine of theta could actually take on? Well, we see that it keeps going between positive 1. It keeps going between positive 1 and then to negative 1, and then back to positive 1, and then negative 1. It takes on all the values in between. So you see that sine of theta is always going to be less than or equal to 1, and it's always going to be greater than or equal to negative 1. So you could say that the range of sine of theta is the set of all numbers between negative 1 and positive 1, and it includes negative 1 and 1."}, {"video_title": "IIT JEE trigonometry problem 1 Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So let's see if we can work our way through this. So let's just draw the triangle, just so we have a visualization of what all of the letters represent. So we have the angles A, B, and C. Let me just draw them like this. So we have the angles A, B, and C. And then the sides opposite them are the lowercase versions. So the side opposite to capital A is lowercase a. The side opposite to capital B is lowercase b. And the side opposite to the angle capital C is lowercase c. Now the first piece of information they tell us is that the angles capital A, capital B, and capital C of the triangle are in arithmetic progression."}, {"video_title": "IIT JEE trigonometry problem 1 Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So we have the angles A, B, and C. And then the sides opposite them are the lowercase versions. So the side opposite to capital A is lowercase a. The side opposite to capital B is lowercase b. And the side opposite to the angle capital C is lowercase c. Now the first piece of information they tell us is that the angles capital A, capital B, and capital C of the triangle are in arithmetic progression. Arithmetic progression. Very fancy word, but all an arithmetic progression is is a series of numbers that are separated by the same amount. So let me give you some examples."}, {"video_title": "IIT JEE trigonometry problem 1 Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And the side opposite to the angle capital C is lowercase c. Now the first piece of information they tell us is that the angles capital A, capital B, and capital C of the triangle are in arithmetic progression. Arithmetic progression. Very fancy word, but all an arithmetic progression is is a series of numbers that are separated by the same amount. So let me give you some examples. So 1, 2, 3, that's an arithmetic progression. 2, 4, 6, arithmetic progression. We're separated by 2 every time."}, {"video_title": "IIT JEE trigonometry problem 1 Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So let me give you some examples. So 1, 2, 3, that's an arithmetic progression. 2, 4, 6, arithmetic progression. We're separated by 2 every time. I could do 10, 20, 30. Also an arithmetic progression. These are all arithmetic progressions."}, {"video_title": "IIT JEE trigonometry problem 1 Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "We're separated by 2 every time. I could do 10, 20, 30. Also an arithmetic progression. These are all arithmetic progressions. So all they're saying is to go from angle A to angle B, however much that is, it's the same amount to go from angle B to angle C. So let's see what that tells us about, or maybe it doesn't tell us anything about those angles. So if we could say we have angle A, and then we have the notion angle B. So we could say that B is equal to A plus some constant."}, {"video_title": "IIT JEE trigonometry problem 1 Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "These are all arithmetic progressions. So all they're saying is to go from angle A to angle B, however much that is, it's the same amount to go from angle B to angle C. So let's see what that tells us about, or maybe it doesn't tell us anything about those angles. So if we could say we have angle A, and then we have the notion angle B. So we could say that B is equal to A plus some constant. We don't know what that is. It could go up by 1, it could go up by 2, it could go up by 10. We don't know what it is."}, {"video_title": "IIT JEE trigonometry problem 1 Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So we could say that B is equal to A plus some constant. We don't know what that is. It could go up by 1, it could go up by 2, it could go up by 10. We don't know what it is. So A plus N. And then C would be equal to B plus N, which is the same thing as B is A plus N. So this is A plus N plus N, which is equal to A plus 2N. So what does that do us? Well, the other thing we know about 3 angles in a triangle is that they have to add up to 180 degrees."}, {"video_title": "IIT JEE trigonometry problem 1 Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "We don't know what it is. So A plus N. And then C would be equal to B plus N, which is the same thing as B is A plus N. So this is A plus N plus N, which is equal to A plus 2N. So what does that do us? Well, the other thing we know about 3 angles in a triangle is that they have to add up to 180 degrees. So this, this, and this have to add up to 180 degrees. So let's try it out. So we have A plus A plus N plus A plus 2N is going to be equal to 180 degrees."}, {"video_title": "IIT JEE trigonometry problem 1 Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Well, the other thing we know about 3 angles in a triangle is that they have to add up to 180 degrees. So this, this, and this have to add up to 180 degrees. So let's try it out. So we have A plus A plus N plus A plus 2N is going to be equal to 180 degrees. We have 1, 2, 3 As here. So we get 3A plus, we have 1N and then another 2N. 3A plus 3N is equal to 180 degrees."}, {"video_title": "IIT JEE trigonometry problem 1 Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So we have A plus A plus N plus A plus 2N is going to be equal to 180 degrees. We have 1, 2, 3 As here. So we get 3A plus, we have 1N and then another 2N. 3A plus 3N is equal to 180 degrees. Or you can divide both sides by 3 and you get A plus N is equal to 60 degrees. So what does that tell us? Well, A could still be anything, because if N is 1, then A is 59."}, {"video_title": "IIT JEE trigonometry problem 1 Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "3A plus 3N is equal to 180 degrees. Or you can divide both sides by 3 and you get A plus N is equal to 60 degrees. So what does that tell us? Well, A could still be anything, because if N is 1, then A is 59. If N is 10, then A is going to be 50. So it doesn't give us much information about the angle A. But if you look up here, do you see an A plus N anywhere?"}, {"video_title": "IIT JEE trigonometry problem 1 Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Well, A could still be anything, because if N is 1, then A is 59. If N is 10, then A is going to be 50. So it doesn't give us much information about the angle A. But if you look up here, do you see an A plus N anywhere? Well, you see it right over here. B is equal to A plus N. And we just figured out that A plus N has to be equal to 60 degrees. So using this first piece of information, we were able to come up with something pretty tangible."}, {"video_title": "IIT JEE trigonometry problem 1 Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "But if you look up here, do you see an A plus N anywhere? Well, you see it right over here. B is equal to A plus N. And we just figured out that A plus N has to be equal to 60 degrees. So using this first piece of information, we were able to come up with something pretty tangible. B must be equal to 60 degrees. And you could try it out with a bunch of numbers. These could be 59, 60, and 61."}, {"video_title": "IIT JEE trigonometry problem 1 Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So using this first piece of information, we were able to come up with something pretty tangible. B must be equal to 60 degrees. And you could try it out with a bunch of numbers. These could be 59, 60, and 61. That's an arithmetic progression. And once again, B is the middle one right over here. These could be 50, 60, and 70."}, {"video_title": "IIT JEE trigonometry problem 1 Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "These could be 59, 60, and 61. That's an arithmetic progression. And once again, B is the middle one right over here. These could be 50, 60, and 70. It could be 40, 60, and 80. But no matter what the arithmetic progression is, in order for these three angles to add up to 180, the middle one has to be equal to 60 degrees. So we're doing pretty well so far."}, {"video_title": "IIT JEE trigonometry problem 1 Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "These could be 50, 60, and 70. It could be 40, 60, and 80. But no matter what the arithmetic progression is, in order for these three angles to add up to 180, the middle one has to be equal to 60 degrees. So we're doing pretty well so far. So let's see what we can do with the next part of the problem. I'm trying to save some screen real estate right over here. OK, so they want us to figure out the value of the expression A over C sine of 2C, capital C, plus C over A sine of 2A."}, {"video_title": "IIT JEE trigonometry problem 1 Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So we're doing pretty well so far. So let's see what we can do with the next part of the problem. I'm trying to save some screen real estate right over here. OK, so they want us to figure out the value of the expression A over C sine of 2C, capital C, plus C over A sine of 2A. So let me just write it down. So we have, I'll do it in blue, A over C sine of 2 times capital C plus C over A sine of 2 times capital A. What's that going to be equal to?"}, {"video_title": "IIT JEE trigonometry problem 1 Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "OK, so they want us to figure out the value of the expression A over C sine of 2C, capital C, plus C over A sine of 2A. So let me just write it down. So we have, I'll do it in blue, A over C sine of 2 times capital C plus C over A sine of 2 times capital A. What's that going to be equal to? So whenever you see stuff like this, you've got a 2 here, a 2 here, frankly, the best thing you should do is just experiment with your trigonometric identities and see if anything pops out at you that might be useful. And a little bit of a clue here. The first part of the problem helped us figure out what B is."}, {"video_title": "IIT JEE trigonometry problem 1 Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "What's that going to be equal to? So whenever you see stuff like this, you've got a 2 here, a 2 here, frankly, the best thing you should do is just experiment with your trigonometric identities and see if anything pops out at you that might be useful. And a little bit of a clue here. The first part of the problem helped us figure out what B is. It helped us figure out what B is. But right now, the expression has no B in it. So right now, this information seems kind of useless."}, {"video_title": "IIT JEE trigonometry problem 1 Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "The first part of the problem helped us figure out what B is. It helped us figure out what B is. But right now, the expression has no B in it. So right now, this information seems kind of useless. But if we could put this somehow in terms of B, then we'll be making progress because we know information about angle B. So let's see what we can do. So the first thing I would use is sine of 2A."}, {"video_title": "IIT JEE trigonometry problem 1 Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So right now, this information seems kind of useless. But if we could put this somehow in terms of B, then we'll be making progress because we know information about angle B. So let's see what we can do. So the first thing I would use is sine of 2A. Let me just rewrite each of these. So sine of 2 times anything, that's just the same thing as, I think this is called the double angle formula. So this is, although I might be wrong, I always forget the actual names of them, but sine of 2 times something is 2 sine of that something times the cosine of that times the cosine of that something."}, {"video_title": "IIT JEE trigonometry problem 1 Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So the first thing I would use is sine of 2A. Let me just rewrite each of these. So sine of 2 times anything, that's just the same thing as, I think this is called the double angle formula. So this is, although I might be wrong, I always forget the actual names of them, but sine of 2 times something is 2 sine of that something times the cosine of that times the cosine of that something. And you'll see that in any trigonometric book on the inside cover, even a lot of calculus books. Let's do that for the same thing right over here. So sine of 2A over here is going to be 2 sine of A cosine of A."}, {"video_title": "IIT JEE trigonometry problem 1 Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So this is, although I might be wrong, I always forget the actual names of them, but sine of 2 times something is 2 sine of that something times the cosine of that times the cosine of that something. And you'll see that in any trigonometric book on the inside cover, even a lot of calculus books. Let's do that for the same thing right over here. So sine of 2A over here is going to be 2 sine of A cosine of A. That's just a standard trigonometric identity. And in the trigonometric playlist, we prove that identity. I think we do it multiple times."}, {"video_title": "IIT JEE trigonometry problem 1 Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So sine of 2A over here is going to be 2 sine of A cosine of A. That's just a standard trigonometric identity. And in the trigonometric playlist, we prove that identity. I think we do it multiple times. And then out in front we have our coefficient still. We have A over C times this plus C over A times this. Now is there anything we can do?"}, {"video_title": "IIT JEE trigonometry problem 1 Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "I think we do it multiple times. And then out in front we have our coefficient still. We have A over C times this plus C over A times this. Now is there anything we can do? And remember, in the back of our mind, we should be thinking of how can we use this information that B is equal to 60. So if we can somehow put this in the form, get a B here. And when I think about how do you get a B here, I think, well, you know, we have a triangle here."}, {"video_title": "IIT JEE trigonometry problem 1 Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Now is there anything we can do? And remember, in the back of our mind, we should be thinking of how can we use this information that B is equal to 60. So if we can somehow put this in the form, get a B here. And when I think about how do you get a B here, I think, well, you know, we have a triangle here. So the things that relate the sides of a triangle, especially when it's not a right triangle, is we're either going to deal with the law of sines or the law of cosines. And the law of sines, let me just rewrite it over here just for our reference. So the law of sines would say sine of A over A is equal to sine of B over B, which is equal to sine of C over C. And it looks like we might be able to use that."}, {"video_title": "IIT JEE trigonometry problem 1 Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And when I think about how do you get a B here, I think, well, you know, we have a triangle here. So the things that relate the sides of a triangle, especially when it's not a right triangle, is we're either going to deal with the law of sines or the law of cosines. And the law of sines, let me just rewrite it over here just for our reference. So the law of sines would say sine of A over A is equal to sine of B over B, which is equal to sine of C over C. And it looks like we might be able to use that. And let me just write the law of cosines here just in case it's useful in the future. So the law of cosines, C squared, it's really the Pythagorean theorem with a little adjustment for the fact that it's not a right triangle. So C squared is equal to A squared plus B squared minus 2AB cosine, cosine of C, of capital C. So it's the law of sines and the law of cosines."}, {"video_title": "IIT JEE trigonometry problem 1 Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So the law of sines would say sine of A over A is equal to sine of B over B, which is equal to sine of C over C. And it looks like we might be able to use that. And let me just write the law of cosines here just in case it's useful in the future. So the law of cosines, C squared, it's really the Pythagorean theorem with a little adjustment for the fact that it's not a right triangle. So C squared is equal to A squared plus B squared minus 2AB cosine, cosine of C, of capital C. So it's the law of sines and the law of cosines. So let's see if we can somehow use both of these to put these in terms of B, which we have information about. Well, the first thing is I could rewrite this. So this says sine of C over C, and this is sine of A over A, so let me do that."}, {"video_title": "IIT JEE trigonometry problem 1 Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So C squared is equal to A squared plus B squared minus 2AB cosine, cosine of C, of capital C. So it's the law of sines and the law of cosines. So let's see if we can somehow use both of these to put these in terms of B, which we have information about. Well, the first thing is I could rewrite this. So this says sine of C over C, and this is sine of A over A, so let me do that. So I have the 2A, I have 2A cosine of C. Let me write that separately. So I have 2A cosine, cosine of capital C, and then times sine of C over C. Times, I'll do that in white, sine of C, that's a capital C, sine of capital C over lowercase c. That's that term and that term right over there. And then to that I'm adding, to that I'm adding, I'm going to do the same thing over here."}, {"video_title": "IIT JEE trigonometry problem 1 Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So this says sine of C over C, and this is sine of A over A, so let me do that. So I have the 2A, I have 2A cosine of C. Let me write that separately. So I have 2A cosine, cosine of capital C, and then times sine of C over C. Times, I'll do that in white, sine of C, that's a capital C, sine of capital C over lowercase c. That's that term and that term right over there. And then to that I'm adding, to that I'm adding, I'm going to do the same thing over here. I have 2 times, I'm going to separate these guys out. Actually, no, I want to do the sines. So let me separate, I'm going to separate this guy and this guy out."}, {"video_title": "IIT JEE trigonometry problem 1 Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And then to that I'm adding, to that I'm adding, I'm going to do the same thing over here. I have 2 times, I'm going to separate these guys out. Actually, no, I want to do the sines. So let me separate, I'm going to separate this guy and this guy out. And so I get plus 2C cosine of A times sine of capital A over lowercase a. Times sine of capital A over lowercase a. Now what did this do for me?"}, {"video_title": "IIT JEE trigonometry problem 1 Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So let me separate, I'm going to separate this guy and this guy out. And so I get plus 2C cosine of A times sine of capital A over lowercase a. Times sine of capital A over lowercase a. Now what did this do for me? Well look at the law of sines right over there. I have sine of C over C, that's that over there. And then I have sine of A over A, that's that over there, capital A over lowercase a."}, {"video_title": "IIT JEE trigonometry problem 1 Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Now what did this do for me? Well look at the law of sines right over there. I have sine of C over C, that's that over there. And then I have sine of A over A, that's that over there, capital A over lowercase a. They're both equal to sine of B over B, so we're making progress. We've started to introduce B into the equation, or the expression, and that's what we actually have information about. So this can be rewritten as sine of B over B."}, {"video_title": "IIT JEE trigonometry problem 1 Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And then I have sine of A over A, that's that over there, capital A over lowercase a. They're both equal to sine of B over B, so we're making progress. We've started to introduce B into the equation, or the expression, and that's what we actually have information about. So this can be rewritten as sine of B over B. So this is the same thing as sine of capital B over lowercase b. And this is the same thing as sine of capital B over lowercase b. And they're both being multiplied, or both of these terms are multiplying, are being multiplied by that."}, {"video_title": "IIT JEE trigonometry problem 1 Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So this can be rewritten as sine of B over B. So this is the same thing as sine of capital B over lowercase b. And this is the same thing as sine of capital B over lowercase b. And they're both being multiplied, or both of these terms are multiplying, are being multiplied by that. 2A cosine of capital C times that, and then plus 2C, it's a lowercase c, cosine of capital A times that. So we can factor out the sine of B over B. So let's do that, let's factor it out."}, {"video_title": "IIT JEE trigonometry problem 1 Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And they're both being multiplied, or both of these terms are multiplying, are being multiplied by that. 2A cosine of capital C times that, and then plus 2C, it's a lowercase c, cosine of capital A times that. So we can factor out the sine of B over B. So let's do that, let's factor it out. So this is the same thing as 2A, and I already have a sense of what the next step is, so I'm going to leave a little space here, 2A times cosine of C, plus, and these are being multiplied, I left some space there, plus 2C, 2 lowercase c, times the cosine of A, and all of this times the sine of B over B. And we already know that B is 60 degrees, so we can evaluate this pretty easily. But let's just continue, see if we can somehow put this right over here in terms of B."}, {"video_title": "IIT JEE trigonometry problem 1 Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So let's do that, let's factor it out. So this is the same thing as 2A, and I already have a sense of what the next step is, so I'm going to leave a little space here, 2A times cosine of C, plus, and these are being multiplied, I left some space there, plus 2C, 2 lowercase c, times the cosine of A, and all of this times the sine of B over B. And we already know that B is 60 degrees, so we can evaluate this pretty easily. But let's just continue, see if we can somehow put this right over here in terms of B. Well if you look over here, we have 2A cosine of C, 2C cosine of A, it's starting to look pretty darn close, each of these terms look pretty darn close to this part of the law of cosines over there. And actually let's solve for that part of the law of cosines to see what we can do. So if you add 2AB cosine C to both sides, you get 2AB cosine of capital C, plus C squared is equal to A squared plus B squared, or if you subtract C squared from both sides, you get 2AB cosine of capital C is equal to A squared plus B squared minus C squared."}, {"video_title": "IIT JEE trigonometry problem 1 Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "But let's just continue, see if we can somehow put this right over here in terms of B. Well if you look over here, we have 2A cosine of C, 2C cosine of A, it's starting to look pretty darn close, each of these terms look pretty darn close to this part of the law of cosines over there. And actually let's solve for that part of the law of cosines to see what we can do. So if you add 2AB cosine C to both sides, you get 2AB cosine of capital C, plus C squared is equal to A squared plus B squared, or if you subtract C squared from both sides, you get 2AB cosine of capital C is equal to A squared plus B squared minus C squared. And this is interesting, and we can switch around the letters later on, but this looks pretty darn close to this. So what if, and this looks pretty darn close to this, except here we're dealing with an A instead of a C, we've just switched the letters around. And we could rewrite this."}, {"video_title": "IIT JEE trigonometry problem 1 Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So if you add 2AB cosine C to both sides, you get 2AB cosine of capital C, plus C squared is equal to A squared plus B squared, or if you subtract C squared from both sides, you get 2AB cosine of capital C is equal to A squared plus B squared minus C squared. And this is interesting, and we can switch around the letters later on, but this looks pretty darn close to this. So what if, and this looks pretty darn close to this, except here we're dealing with an A instead of a C, we've just switched the letters around. And we could rewrite this. Actually let me rewrite it just for fun. I could rewrite this over here as 2CB, not rewrite it, I can swap the letters, times the cosine of A, here I'm swapping the As and the Cs, is equal to C squared plus B squared minus A squared. There's nothing unique about the side C, I can do this with all of the sides."}, {"video_title": "IIT JEE trigonometry problem 1 Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And we could rewrite this. Actually let me rewrite it just for fun. I could rewrite this over here as 2CB, not rewrite it, I can swap the letters, times the cosine of A, here I'm swapping the As and the Cs, is equal to C squared plus B squared minus A squared. There's nothing unique about the side C, I can do this with all of the sides. So here, when you have a big C here, you have an A and a B out front, and then you have the A squared plus B squared minus the small c squared. If you have a big A, then you're going to have the CB in the front, and then you're subtracting the A squared right over here. Now this is useful because this term right over here, this term right over here, looks almost like this term over here if we could just multiply this by B."}, {"video_title": "IIT JEE trigonometry problem 1 Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "There's nothing unique about the side C, I can do this with all of the sides. So here, when you have a big C here, you have an A and a B out front, and then you have the A squared plus B squared minus the small c squared. If you have a big A, then you're going to have the CB in the front, and then you're subtracting the A squared right over here. Now this is useful because this term right over here, this term right over here, looks almost like this term over here if we could just multiply this by B. So let's do that. We can multiply that by B. In fact, let's multiply this whole numerator, this whole term by B."}, {"video_title": "IIT JEE trigonometry problem 1 Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Now this is useful because this term right over here, this term right over here, looks almost like this term over here if we could just multiply this by B. So let's do that. We can multiply that by B. In fact, let's multiply this whole numerator, this whole term by B. So if we multiply this whole term by B, what do we get? We get a B there, we get a B there, and of course you can't just arbitrarily multiply an expression by B, that'll change its value. So what we could do is multiply the expression by B, which we just did, we distributed the B across here, but then we'll also divide by B."}, {"video_title": "IIT JEE trigonometry problem 1 Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "In fact, let's multiply this whole numerator, this whole term by B. So if we multiply this whole term by B, what do we get? We get a B there, we get a B there, and of course you can't just arbitrarily multiply an expression by B, that'll change its value. So what we could do is multiply the expression by B, which we just did, we distributed the B across here, but then we'll also divide by B. So I'll divide by B, that's the equivalent of multiplying the denominator there, not B squared, that's the equivalent of multiplying the denominator there by B, that's the same thing as dividing by B. We've multiplied by B, divided by B, or that's the same thing as just turning this into B squared. Now, what does this give us?"}, {"video_title": "IIT JEE trigonometry problem 1 Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So what we could do is multiply the expression by B, which we just did, we distributed the B across here, but then we'll also divide by B. So I'll divide by B, that's the equivalent of multiplying the denominator there, not B squared, that's the equivalent of multiplying the denominator there by B, that's the same thing as dividing by B. We've multiplied by B, divided by B, or that's the same thing as just turning this into B squared. Now, what does this give us? Well, we have this term right over here, this term right over here is now the exact same thing as that over there. So it is now A squared plus B squared minus C squared, and then this term right over here is now the exact same thing as this thing over here, which is the same thing as that. We're using the law of cosines."}, {"video_title": "IIT JEE trigonometry problem 1 Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Now, what does this give us? Well, we have this term right over here, this term right over here is now the exact same thing as that over there. So it is now A squared plus B squared minus C squared, and then this term right over here is now the exact same thing as this thing over here, which is the same thing as that. We're using the law of cosines. So this is plus C squared plus B squared minus A squared, and then all of that times this, sine of B, sine of capital B over B squared. Now, what does this give us? We have an A squared and a negative A squared."}, {"video_title": "IIT JEE trigonometry problem 1 Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "We're using the law of cosines. So this is plus C squared plus B squared minus A squared, and then all of that times this, sine of B, sine of capital B over B squared. Now, what does this give us? We have an A squared and a negative A squared. Things are starting to simplify. A squared, negative A squared. We have a negative C squared and a positive C squared."}, {"video_title": "IIT JEE trigonometry problem 1 Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "We have an A squared and a negative A squared. Things are starting to simplify. A squared, negative A squared. We have a negative C squared and a positive C squared. So what are we left with? We're just left with a 2B squared. So our whole expression has simplified to 2B squared sine of B, sine of capital B, over lowercase b squared."}, {"video_title": "IIT JEE trigonometry problem 1 Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "We have a negative C squared and a positive C squared. So what are we left with? We're just left with a 2B squared. So our whole expression has simplified to 2B squared sine of B, sine of capital B, over lowercase b squared. These cancel out, so our whole expression simplifies to 2 sine of B. And from the get-go, we knew what B was. We know it's 60 degrees."}, {"video_title": "IIT JEE trigonometry problem 1 Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So our whole expression has simplified to 2B squared sine of B, sine of capital B, over lowercase b squared. These cancel out, so our whole expression simplifies to 2 sine of B. And from the get-go, we knew what B was. We know it's 60 degrees. So this is equal to 2 times the sine of 60 degrees. And if you don't have the sine of 60 degrees memorized, you can always just break out a 30-60-90 triangle. So let me draw."}, {"video_title": "IIT JEE trigonometry problem 1 Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "We know it's 60 degrees. So this is equal to 2 times the sine of 60 degrees. And if you don't have the sine of 60 degrees memorized, you can always just break out a 30-60-90 triangle. So let me draw. This is a right triangle right over here. This is 60 degrees. Hypotenuse has length 1."}, {"video_title": "IIT JEE trigonometry problem 1 Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So let me draw. This is a right triangle right over here. This is 60 degrees. Hypotenuse has length 1. We're dealing with the unit circle. This side is 30 degrees. The side opposite the 30 degrees is 1 half."}, {"video_title": "IIT JEE trigonometry problem 1 Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Hypotenuse has length 1. We're dealing with the unit circle. This side is 30 degrees. The side opposite the 30 degrees is 1 half. The side opposite the 60 degrees is square root of 3 times that, so it's square root of 3 over 2. You could even use the Pythagorean theorem to figure out, once you know one of them, you could figure out the other one. So the sine is opposite over hypotenuse, so square root of 3 over 2 over 1, or it's just square root of 3 over 2."}, {"video_title": "IIT JEE trigonometry problem 1 Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "The side opposite the 30 degrees is 1 half. The side opposite the 60 degrees is square root of 3 times that, so it's square root of 3 over 2. You could even use the Pythagorean theorem to figure out, once you know one of them, you could figure out the other one. So the sine is opposite over hypotenuse, so square root of 3 over 2 over 1, or it's just square root of 3 over 2. So this is equal to 2 times the whole stretch. It's very exciting. Square root of 3 over 2."}, {"video_title": "IIT JEE trigonometry problem 1 Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So the sine is opposite over hypotenuse, so square root of 3 over 2 over 1, or it's just square root of 3 over 2. So this is equal to 2 times the whole stretch. It's very exciting. Square root of 3 over 2. These cancel out, so we are left with the square root of 3. That's a pretty neat problem. And just in case you're curious, this came from the 2010 IIT."}, {"video_title": "Tau versus pi Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "What I want to do in this video is revisit a little bit of what we know about pi and really how we measure angles and radians and then think about whether pi is necessarily the best number to be paying attention to. So let's think a little bit about what I just said. So pi, we know, is defined, and I'll write defined as a triple equal sign, I guess you could call it that way. Pi is defined as the ratio of the circumference of a circle to its diameter, which is the same thing as the ratio of the circumference of the circle to 2 times the radius. And from that we get all these interesting formulas that you get in geometry class that, hey, if you have the radius and you want to calculate the circumference, multiply both sides of this definition or this equation by 2 times the radius and you get 2 times the radius times pi is equal to the circumference, or more familiarly, it would be circumference is equal to 2 pi r. This is one of those fundamental things that you learn early on in your career and you use it to find circumferences usually or figure out radiuses if you know circumference. And from that comes how we measure our angles in radians once we get to trigonometry class. And just as a review here, let me draw myself a circle."}, {"video_title": "Tau versus pi Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Pi is defined as the ratio of the circumference of a circle to its diameter, which is the same thing as the ratio of the circumference of the circle to 2 times the radius. And from that we get all these interesting formulas that you get in geometry class that, hey, if you have the radius and you want to calculate the circumference, multiply both sides of this definition or this equation by 2 times the radius and you get 2 times the radius times pi is equal to the circumference, or more familiarly, it would be circumference is equal to 2 pi r. This is one of those fundamental things that you learn early on in your career and you use it to find circumferences usually or figure out radiuses if you know circumference. And from that comes how we measure our angles in radians once we get to trigonometry class. And just as a review here, let me draw myself a circle. Let me draw myself a better circle. So there is my, well, it'll do the job. And here is the positive x-axis."}, {"video_title": "Tau versus pi Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And just as a review here, let me draw myself a circle. Let me draw myself a better circle. So there is my, well, it'll do the job. And here is the positive x-axis. And let me make some angle here. I'll make the angle kind of obvious just so that it, so let me make this angle. And the way that we measure angles when we talk about radians, we're really talking about the angle subtended by something of a certain arc length."}, {"video_title": "Tau versus pi Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And here is the positive x-axis. And let me make some angle here. I'll make the angle kind of obvious just so that it, so let me make this angle. And the way that we measure angles when we talk about radians, we're really talking about the angle subtended by something of a certain arc length. And we measure the arc length in, well, the way I like to think about it is the angle is in radians and the arc length itself is in radiuses, which isn't really a word, but that's how I think about it. How many radiuses is this arc length that subtends the angle in radians? So let me tell you, let me show you what I'm talking about."}, {"video_title": "Tau versus pi Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And the way that we measure angles when we talk about radians, we're really talking about the angle subtended by something of a certain arc length. And we measure the arc length in, well, the way I like to think about it is the angle is in radians and the arc length itself is in radiuses, which isn't really a word, but that's how I think about it. How many radiuses is this arc length that subtends the angle in radians? So let me tell you, let me show you what I'm talking about. So this arc length right here, if the radius is r, what is the length of this arc length? Well, we know from basic geometry the entire circumference over here is going to be 2 pi r, right? This entire circumference, that's really by definition."}, {"video_title": "Tau versus pi Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So let me tell you, let me show you what I'm talking about. So this arc length right here, if the radius is r, what is the length of this arc length? Well, we know from basic geometry the entire circumference over here is going to be 2 pi r, right? This entire circumference, that's really by definition. This entire circumference is going to be 2 pi r. So what is just this arc length here? And I'm assuming this is a, that this is a fourth of the circle. So it's going to be 2 pi r over 4."}, {"video_title": "Tau versus pi Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "This entire circumference, that's really by definition. This entire circumference is going to be 2 pi r. So what is just this arc length here? And I'm assuming this is a, that this is a fourth of the circle. So it's going to be 2 pi r over 4. So this arc length over here, this arc length is going to be 2 pi r over 4, which is the same thing as pi over 2 r. Or you could say this is the same thing as pi over 2 radiuses. One of those, you know, not a real word, but that's how I like to think about it. Or you could say it subtends an angle of pi over 2 radians."}, {"video_title": "Tau versus pi Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So it's going to be 2 pi r over 4. So this arc length over here, this arc length is going to be 2 pi r over 4, which is the same thing as pi over 2 r. Or you could say this is the same thing as pi over 2 radiuses. One of those, you know, not a real word, but that's how I like to think about it. Or you could say it subtends an angle of pi over 2 radians. So over here, theta is pi over 2 radians. And so really when you're measuring angles in radians, it's really you're saying, okay, that angle is subtended by an arc of, that has a length of how many radiusi, or I don't even know what the plural of radius is. Actually, I think it's radii, but it's fun to try to say radiuses is radii."}, {"video_title": "Tau versus pi Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Or you could say it subtends an angle of pi over 2 radians. So over here, theta is pi over 2 radians. And so really when you're measuring angles in radians, it's really you're saying, okay, that angle is subtended by an arc of, that has a length of how many radiusi, or I don't even know what the plural of radius is. Actually, I think it's radii, but it's fun to try to say radiuses is radii. Actually, let me do that just so no one says that, Sal, you're teaching people the wrong plural form of radius, radii. So this arc length is pi over 2 radii, and it subtends an angle of pi over 2 radians. We could do another one just for the sake of making the point clear."}, {"video_title": "Tau versus pi Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Actually, I think it's radii, but it's fun to try to say radiuses is radii. Actually, let me do that just so no one says that, Sal, you're teaching people the wrong plural form of radius, radii. So this arc length is pi over 2 radii, and it subtends an angle of pi over 2 radians. We could do another one just for the sake of making the point clear. If you went all the way around the circle, and you got back to the positive x-axis here, what is the arc length? Well, now all of a sudden, the arc length is the entire circumference of the circle. It would be 2 pi r, which is the same thing as 2 pi radii."}, {"video_title": "Tau versus pi Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "We could do another one just for the sake of making the point clear. If you went all the way around the circle, and you got back to the positive x-axis here, what is the arc length? Well, now all of a sudden, the arc length is the entire circumference of the circle. It would be 2 pi r, which is the same thing as 2 pi radii. We would say that the angle subtended by this arc length, the angle that we care about going all the way around the circle, is 2 pi radians. Out of this comes all of the things that we know about how to graph trigonometric functions, or at least how we measure the graph on the x-axis. I'll also touch on Euler's formula, which is the most beautiful formula, I think, in all of mathematics."}, {"video_title": "Tau versus pi Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "It would be 2 pi r, which is the same thing as 2 pi radii. We would say that the angle subtended by this arc length, the angle that we care about going all the way around the circle, is 2 pi radians. Out of this comes all of the things that we know about how to graph trigonometric functions, or at least how we measure the graph on the x-axis. I'll also touch on Euler's formula, which is the most beautiful formula, I think, in all of mathematics. Let's visit those right now, just to remind ourselves of how pi fits into all of that. If I think about our trigonometric functions, remember, on the trigonometric functions, we assume we have a unit circle here. On the trig functions, this is the unit circle definition of the trig functions, so this is a nicer view of all of that."}, {"video_title": "Tau versus pi Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "I'll also touch on Euler's formula, which is the most beautiful formula, I think, in all of mathematics. Let's visit those right now, just to remind ourselves of how pi fits into all of that. If I think about our trigonometric functions, remember, on the trigonometric functions, we assume we have a unit circle here. On the trig functions, this is the unit circle definition of the trig functions, so this is a nicer view of all of that. You assume you have a unit circle, a circle of radius 1. The trig functions are defined as, for any angle you have here, for any angle theta, cosine of theta is how far you have to move in the x-coordinate of the point along the arc that subtends this angle, so that's cosine of theta. Then sine of theta is the y-value of that point."}, {"video_title": "Tau versus pi Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "On the trig functions, this is the unit circle definition of the trig functions, so this is a nicer view of all of that. You assume you have a unit circle, a circle of radius 1. The trig functions are defined as, for any angle you have here, for any angle theta, cosine of theta is how far you have to move in the x-coordinate of the point along the arc that subtends this angle, so that's cosine of theta. Then sine of theta is the y-value of that point. Cosine of theta is the x-value, sine of theta is the y-value. If you were to graph one of these functions, and I'll just do sine of theta for convenience, but you could try it with cosine of theta. Let's graph sine of theta."}, {"video_title": "Tau versus pi Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Then sine of theta is the y-value of that point. Cosine of theta is the x-value, sine of theta is the y-value. If you were to graph one of these functions, and I'll just do sine of theta for convenience, but you could try it with cosine of theta. Let's graph sine of theta. Let's do one revolution of sine of theta. We tend to label it, so let's do sine. When the angle is 0, sine of theta is 0."}, {"video_title": "Tau versus pi Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Let's graph sine of theta. Let's do one revolution of sine of theta. We tend to label it, so let's do sine. When the angle is 0, sine of theta is 0. Let me draw the x- and y-axis, just so you remember this. This is the y-axis, and this is the x-axis. This right here is the x-axis."}, {"video_title": "Tau versus pi Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "When the angle is 0, sine of theta is 0. Let me draw the x- and y-axis, just so you remember this. This is the y-axis, and this is the x-axis. This right here is the x-axis. When the angle is 0, we're right here on the unit circle. The y-value there is 0, so sine of theta is going to be right like that. Let me draw it like this."}, {"video_title": "Tau versus pi Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "This right here is the x-axis. When the angle is 0, we're right here on the unit circle. The y-value there is 0, so sine of theta is going to be right like that. Let me draw it like this. This is our theta, and this is, I'm going to graph sine of theta along the y-axis. We'll say y is equal to sine of theta in this graph that I'm drawing right over here. Then we could do, I'll just do the simple points here."}, {"video_title": "Tau versus pi Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Let me draw it like this. This is our theta, and this is, I'm going to graph sine of theta along the y-axis. We'll say y is equal to sine of theta in this graph that I'm drawing right over here. Then we could do, I'll just do the simple points here. Then if we make the angle go, if we did it in degrees, 90 degrees, or if we do it in radians, pi over 2 radians, what is sine of theta? Now it is 1. This is a unit circle."}, {"video_title": "Tau versus pi Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Then we could do, I'll just do the simple points here. Then if we make the angle go, if we did it in degrees, 90 degrees, or if we do it in radians, pi over 2 radians, what is sine of theta? Now it is 1. This is a unit circle. It has a radius 1. When we get to pi over 2, so when theta is equal to pi over 2, then sine of theta is equal to 1. This is 1 right here."}, {"video_title": "Tau versus pi Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "This is a unit circle. It has a radius 1. When we get to pi over 2, so when theta is equal to pi over 2, then sine of theta is equal to 1. This is 1 right here. Sine of theta is equal to 1. If theta, and then if we go 180 degrees, or halfway around the circle, theta is now equal to pi. Theta is, let me do this in a color, I'll do it in orange."}, {"video_title": "Tau versus pi Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "This is 1 right here. Sine of theta is equal to 1. If theta, and then if we go 180 degrees, or halfway around the circle, theta is now equal to pi. Theta is, let me do this in a color, I'll do it in orange. I already used orange. Theta is now equal to pi. When theta is equal to pi, the y value of this point right here is once again 0."}, {"video_title": "Tau versus pi Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Theta is, let me do this in a color, I'll do it in orange. I already used orange. Theta is now equal to pi. When theta is equal to pi, the y value of this point right here is once again 0. We go back to 0. Remember, we're talking about sine of theta. Then we can go all the way down here where you could view this either as 270 degrees, or you could view this as 3 pi over 2 radians."}, {"video_title": "Tau versus pi Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "When theta is equal to pi, the y value of this point right here is once again 0. We go back to 0. Remember, we're talking about sine of theta. Then we can go all the way down here where you could view this either as 270 degrees, or you could view this as 3 pi over 2 radians. This is in radians, this axis. 3 pi over 2 radians, sine of theta is the y coordinate on the unit circle right over here, so it's going to be negative 1. This is negative 1."}, {"video_title": "Tau versus pi Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Then we can go all the way down here where you could view this either as 270 degrees, or you could view this as 3 pi over 2 radians. This is in radians, this axis. 3 pi over 2 radians, sine of theta is the y coordinate on the unit circle right over here, so it's going to be negative 1. This is negative 1. Then finally, when you go all the way around the circle, you've gone 2 pi radians, and you're back where you began, and the sine of theta, or the y coordinate, is now 0 once again. If you connect the dots, or if you plotted more points, you would see a sine curve over just the part that we've graphed right over here. That's another application."}, {"video_title": "Tau versus pi Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "This is negative 1. Then finally, when you go all the way around the circle, you've gone 2 pi radians, and you're back where you began, and the sine of theta, or the y coordinate, is now 0 once again. If you connect the dots, or if you plotted more points, you would see a sine curve over just the part that we've graphed right over here. That's another application. You say, hey, Sal, where is this going? I'm showing you, I'm reminding you of all of these things, because we're going to revisit it with a different number other than pi. I want to do one last visit with pi."}, {"video_title": "Tau versus pi Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "That's another application. You say, hey, Sal, where is this going? I'm showing you, I'm reminding you of all of these things, because we're going to revisit it with a different number other than pi. I want to do one last visit with pi. We say, look, pi is powerful because, or one of the reasons why pi seems to have some type of mystical power, and we've shown this in the calculus playlist, is Euler's formula that e to the i theta is equal to cosine of theta plus i sine of theta. This by itself is just one of those mind-boggling formulas, but it sometimes looks even more mind-boggling when you put pi in for theta, because then from Euler's formula, you would get e to the i pi is equal to, well, what's cosine of pi? Cosine of pi is negative 1, and then sine of pi is 0, so 0 times i."}, {"video_title": "Tau versus pi Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "I want to do one last visit with pi. We say, look, pi is powerful because, or one of the reasons why pi seems to have some type of mystical power, and we've shown this in the calculus playlist, is Euler's formula that e to the i theta is equal to cosine of theta plus i sine of theta. This by itself is just one of those mind-boggling formulas, but it sometimes looks even more mind-boggling when you put pi in for theta, because then from Euler's formula, you would get e to the i pi is equal to, well, what's cosine of pi? Cosine of pi is negative 1, and then sine of pi is 0, so 0 times i. You get this formula, which is pretty profound. Then you say, okay, if I want to put all of the fundamental numbers together in one, in one formula, I can add 1 to both sides of this. You get e to the i pi plus 1 is equal to 0."}, {"video_title": "Tau versus pi Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Cosine of pi is negative 1, and then sine of pi is 0, so 0 times i. You get this formula, which is pretty profound. Then you say, okay, if I want to put all of the fundamental numbers together in one, in one formula, I can add 1 to both sides of this. You get e to the i pi plus 1 is equal to 0. Sometimes this is called Euler's identity, the most beautiful formula or equation in all of mathematics. It is pretty profound. You have all of the fundamental numbers in one equation, e, i, pi, 1, 0."}, {"video_title": "Tau versus pi Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "You get e to the i pi plus 1 is equal to 0. Sometimes this is called Euler's identity, the most beautiful formula or equation in all of mathematics. It is pretty profound. You have all of the fundamental numbers in one equation, e, i, pi, 1, 0. Although for my aesthetic taste, it would have been even more powerful if this was a 1 right over here, because then this would have said, look, e to the i pi, this bizarre thing, this bizarre thing would have equaled unity. That would have been super-duper profound to me. It seems a little bit of a hack to add 1 to both sides and say, oh, look, now I have 0 here."}, {"video_title": "Tau versus pi Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "You have all of the fundamental numbers in one equation, e, i, pi, 1, 0. Although for my aesthetic taste, it would have been even more powerful if this was a 1 right over here, because then this would have said, look, e to the i pi, this bizarre thing, this bizarre thing would have equaled unity. That would have been super-duper profound to me. It seems a little bit of a hack to add 1 to both sides and say, oh, look, now I have 0 here. But this is pretty darn good. With that, I'm going to make, well, I'm not going to argue for it. I'm going to show an argument for another number, a number different than pi."}, {"video_title": "Tau versus pi Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "It seems a little bit of a hack to add 1 to both sides and say, oh, look, now I have 0 here. But this is pretty darn good. With that, I'm going to make, well, I'm not going to argue for it. I'm going to show an argument for another number, a number different than pi. I want to make it clear that these ideas are not my own. It comes from, well, it's inspired by, many people are on this movement now, the Tau movement, but these are kind of the people that gave me the thinking on this. The first is Robert Pallet on pi is wrong."}, {"video_title": "Tau versus pi Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "I'm going to show an argument for another number, a number different than pi. I want to make it clear that these ideas are not my own. It comes from, well, it's inspired by, many people are on this movement now, the Tau movement, but these are kind of the people that gave me the thinking on this. The first is Robert Pallet on pi is wrong. He doesn't argue that pi is calculated wrong. He still agrees that it is the ratio of the circumference to the diameter of the circle, that is 3.14159. But what he's saying is that we're paying attention to the wrong number."}, {"video_title": "Tau versus pi Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "The first is Robert Pallet on pi is wrong. He doesn't argue that pi is calculated wrong. He still agrees that it is the ratio of the circumference to the diameter of the circle, that is 3.14159. But what he's saying is that we're paying attention to the wrong number. Also, you have Michael Hartel, the Tau Manifesto. All of this is available online. What they argue for is a number called tau, or what they call tau."}, {"video_title": "Tau versus pi Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "But what he's saying is that we're paying attention to the wrong number. Also, you have Michael Hartel, the Tau Manifesto. All of this is available online. What they argue for is a number called tau, or what they call tau. They define tau, and it's a very simple change from pi. They define tau not as the ratio of the circumference to the diameter, or the ratio of the circumference to 2 times the radius. They say, hey, wouldn't it be natural to define some number that's the ratio of the circumference to the radius?"}, {"video_title": "Tau versus pi Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "What they argue for is a number called tau, or what they call tau. They define tau, and it's a very simple change from pi. They define tau not as the ratio of the circumference to the diameter, or the ratio of the circumference to 2 times the radius. They say, hey, wouldn't it be natural to define some number that's the ratio of the circumference to the radius? As you see here, this pi is just 1 half times this over here, right? Circumference over 2r, this is the same thing as 1 half times circumference over r. So pi is just half of tau. Or another way to think about it is that tau is just 2 times pi."}, {"video_title": "Tau versus pi Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "They say, hey, wouldn't it be natural to define some number that's the ratio of the circumference to the radius? As you see here, this pi is just 1 half times this over here, right? Circumference over 2r, this is the same thing as 1 half times circumference over r. So pi is just half of tau. Or another way to think about it is that tau is just 2 times pi. Or, and I'm sure you probably don't have this memorized, because you're just like, wait, I spent all my life memorizing pi. But it's 6.283185 and keeps going on and on and on, never repeating, just like pi. It's 2 times pi."}, {"video_title": "Tau versus pi Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Or another way to think about it is that tau is just 2 times pi. Or, and I'm sure you probably don't have this memorized, because you're just like, wait, I spent all my life memorizing pi. But it's 6.283185 and keeps going on and on and on, never repeating, just like pi. It's 2 times pi. So you're saying, hey, Sal, pi has been around for millennia, really. Why mess with such a fundamental number, especially when you just spent all this time showing how profound it is? And the argument that they'd make, and it seems like a pretty good argument, is that actually things seem a little bit more elegant when you pay attention to this number instead of half of this number, when you pay attention to tau."}, {"video_title": "Tau versus pi Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "It's 2 times pi. So you're saying, hey, Sal, pi has been around for millennia, really. Why mess with such a fundamental number, especially when you just spent all this time showing how profound it is? And the argument that they'd make, and it seems like a pretty good argument, is that actually things seem a little bit more elegant when you pay attention to this number instead of half of this number, when you pay attention to tau. And to see that, let's revisit everything that we did here. Now all of a sudden, if you pay attention to 2 pi, as opposed to pi, or we should call it, if you pay attention to tau instead of tau over 2, what is this angle that we did in magenta? What is this angle we did in magenta?"}, {"video_title": "Tau versus pi Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And the argument that they'd make, and it seems like a pretty good argument, is that actually things seem a little bit more elegant when you pay attention to this number instead of half of this number, when you pay attention to tau. And to see that, let's revisit everything that we did here. Now all of a sudden, if you pay attention to 2 pi, as opposed to pi, or we should call it, if you pay attention to tau instead of tau over 2, what is this angle that we did in magenta? What is this angle we did in magenta? Well, first of all, let's think about this formula right over here. What is the circumference in terms of the radius? Well, now we can say the circumference is equal to tau times the radius, because tau is the same thing as 2 pi."}, {"video_title": "Tau versus pi Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "What is this angle we did in magenta? Well, first of all, let's think about this formula right over here. What is the circumference in terms of the radius? Well, now we can say the circumference is equal to tau times the radius, because tau is the same thing as 2 pi. So it makes that formula a little bit neater, although it does make the pi r squared a little bit messier, so you could argue both sides of that. But it makes the measure of radians much more intuitive, because you could say that this is pi over 2 radians, or you could say that this is pi over 2 radians is the same thing as tau over 4 radians. And where did I get that from?"}, {"video_title": "Tau versus pi Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Well, now we can say the circumference is equal to tau times the radius, because tau is the same thing as 2 pi. So it makes that formula a little bit neater, although it does make the pi r squared a little bit messier, so you could argue both sides of that. But it makes the measure of radians much more intuitive, because you could say that this is pi over 2 radians, or you could say that this is pi over 2 radians is the same thing as tau over 4 radians. And where did I get that from? Remember, if you go all the way around the circle, that is the circumference. The arc length would be the circumference. It would be tau radii, or it would be tau radians, would be the angle subtended by that arc length."}, {"video_title": "Tau versus pi Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And where did I get that from? Remember, if you go all the way around the circle, that is the circumference. The arc length would be the circumference. It would be tau radii, or it would be tau radians, would be the angle subtended by that arc length. It would be tau radians. All the way around is tau radians. So that by itself is intuitive."}, {"video_title": "Tau versus pi Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "It would be tau radii, or it would be tau radians, would be the angle subtended by that arc length. It would be tau radians. All the way around is tau radians. So that by itself is intuitive. One revolution is one tau radians. If you go only one-fourth of that, it's going to be tau over 4 radians. So the reason why tau is more intuitive here is because it immediately, you don't have to do this weird conversion where you're saying, oh, you know, divide by 2, multiply by 2, all that."}, {"video_title": "Tau versus pi Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So that by itself is intuitive. One revolution is one tau radians. If you go only one-fourth of that, it's going to be tau over 4 radians. So the reason why tau is more intuitive here is because it immediately, you don't have to do this weird conversion where you're saying, oh, you know, divide by 2, multiply by 2, all that. You're just like, look, however many radians in terms of tau, that's really how many revolutions you've gone around the circle. And so if you've gone one-fourth around, that's tau over 4 radians. If you've gone halfway around, that would be tau over 2 radians."}, {"video_title": "Tau versus pi Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So the reason why tau is more intuitive here is because it immediately, you don't have to do this weird conversion where you're saying, oh, you know, divide by 2, multiply by 2, all that. You're just like, look, however many radians in terms of tau, that's really how many revolutions you've gone around the circle. And so if you've gone one-fourth around, that's tau over 4 radians. If you've gone halfway around, that would be tau over 2 radians. If you go three-fourths around, that would be 3 tau over 4 radians. If you go all the way around, that would be tau radians. If someone tells you that they have an angle of 10 tau radians, you'd go around exactly 10 times."}, {"video_title": "Tau versus pi Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "If you've gone halfway around, that would be tau over 2 radians. If you go three-fourths around, that would be 3 tau over 4 radians. If you go all the way around, that would be tau radians. If someone tells you that they have an angle of 10 tau radians, you'd go around exactly 10 times. It would be much more intuitive. You wouldn't have to do this little mental math, converting, you know, saying do I multiply or divide by 2 when I convert to radians in terms of pi? No, when you do it in terms of tau radians, it's just natural."}, {"video_title": "Tau versus pi Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "If someone tells you that they have an angle of 10 tau radians, you'd go around exactly 10 times. It would be much more intuitive. You wouldn't have to do this little mental math, converting, you know, saying do I multiply or divide by 2 when I convert to radians in terms of pi? No, when you do it in terms of tau radians, it's just natural. One revolution is 1 tau radian. And it makes a sine function over here. Instead of writing pi over 2, well, pi over 2, you know, when you look at a graph like this, you're like, where was this on the unit circle?"}, {"video_title": "Tau versus pi Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "No, when you do it in terms of tau radians, it's just natural. One revolution is 1 tau radian. And it makes a sine function over here. Instead of writing pi over 2, well, pi over 2, you know, when you look at a graph like this, you're like, where was this on the unit circle? Was this one-fourth around the circle? Was this one-half? And this is actually one-fourth around the circle."}, {"video_title": "Tau versus pi Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Instead of writing pi over 2, well, pi over 2, you know, when you look at a graph like this, you're like, where was this on the unit circle? Was this one-fourth around the circle? Was this one-half? And this is actually one-fourth around the circle. You're right over here. But now it becomes obvious if you write it in tau. Pi over 2 is the same thing as tau over 4."}, {"video_title": "Tau versus pi Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And this is actually one-fourth around the circle. You're right over here. But now it becomes obvious if you write it in tau. Pi over 2 is the same thing as tau over 4. Pi is the same thing as tau over 2. 3 pi over 2 is 3 tau over 4, 3 fourths tau. And then one revolution is tau."}, {"video_title": "Tau versus pi Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Pi over 2 is the same thing as tau over 4. Pi is the same thing as tau over 2. 3 pi over 2 is 3 tau over 4, 3 fourths tau. And then one revolution is tau. And then immediately now when you look at it this way, you know exactly where you are in the unit circle. You're one-fourth around the unit circle. You're halfway around the unit circle."}, {"video_title": "Tau versus pi Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And then one revolution is tau. And then immediately now when you look at it this way, you know exactly where you are in the unit circle. You're one-fourth around the unit circle. You're halfway around the unit circle. You're three-fourths the way around the unit circle. And then you're all the way around the unit circle. And so the last thing that I think the strong pi defenders would say is, well, look, Sal, you just pointed out one of the most beautiful identities or formulas in mathematics."}, {"video_title": "Tau versus pi Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "You're halfway around the unit circle. You're three-fourths the way around the unit circle. And then you're all the way around the unit circle. And so the last thing that I think the strong pi defenders would say is, well, look, Sal, you just pointed out one of the most beautiful identities or formulas in mathematics. How does tau hold up to this? Well, let's just try it out and see what happens. So if we take e to the i tau, that will give us cosine of tau plus i sine of tau."}, {"video_title": "Tau versus pi Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And so the last thing that I think the strong pi defenders would say is, well, look, Sal, you just pointed out one of the most beautiful identities or formulas in mathematics. How does tau hold up to this? Well, let's just try it out and see what happens. So if we take e to the i tau, that will give us cosine of tau plus i sine of tau. And once again, let's just think about what this is. Tau radians means we've gone all the way around the unit circle. So cosine of tau, remember, we're back at the beginning of the unit circle right over here."}, {"video_title": "Tau versus pi Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So if we take e to the i tau, that will give us cosine of tau plus i sine of tau. And once again, let's just think about what this is. Tau radians means we've gone all the way around the unit circle. So cosine of tau, remember, we're back at the beginning of the unit circle right over here. So cosine of tau is going to be equal to 1. And then sine of tau is equal to 0. So e to the i tau is equal to 1."}, {"video_title": "Example Converting radians to degrees Trigonometry Khan Academy.mp3", "Sentence": "So the first question I'll ask you, if you do one revolution, if you have an angle that went all the way around once, how many radians is that? Well, we know that that is 2 pi radians. Now, that exact same angle, if we were to measure it in degrees, how many degrees is that? Well, you've heard of people doing a 360, doing one full revolution. That is equal to 360 degrees. Now, can we simplify this? It's important to write this little superscript circle."}, {"video_title": "Example Converting radians to degrees Trigonometry Khan Academy.mp3", "Sentence": "Well, you've heard of people doing a 360, doing one full revolution. That is equal to 360 degrees. Now, can we simplify this? It's important to write this little superscript circle. That's literally the units under question. Sometimes it doesn't look like a unit, but it is a unit. You could literally write degrees instead of that little symbol."}, {"video_title": "Example Converting radians to degrees Trigonometry Khan Academy.mp3", "Sentence": "It's important to write this little superscript circle. That's literally the units under question. Sometimes it doesn't look like a unit, but it is a unit. You could literally write degrees instead of that little symbol. And the units right here, of course, are the word radians. Now, can we simplify this a little bit? Well, sure."}, {"video_title": "Example Converting radians to degrees Trigonometry Khan Academy.mp3", "Sentence": "You could literally write degrees instead of that little symbol. And the units right here, of course, are the word radians. Now, can we simplify this a little bit? Well, sure. Both 2 pi and 360 are divisible by 2, so let's divide things by 2. And if we do that, what do we get for what pi radians are equal to? Well, on the left-hand side here, we're just left with pi radians."}, {"video_title": "Example Converting radians to degrees Trigonometry Khan Academy.mp3", "Sentence": "Well, sure. Both 2 pi and 360 are divisible by 2, so let's divide things by 2. And if we do that, what do we get for what pi radians are equal to? Well, on the left-hand side here, we're just left with pi radians. And on the right-hand side here, 360 divided by 2 is 180, and we have still the units, which are degrees. So we get pi radians are equal to 180 degrees, which actually answered the first part of our question. We wanted to convert pi radians."}, {"video_title": "Example Converting radians to degrees Trigonometry Khan Academy.mp3", "Sentence": "Well, on the left-hand side here, we're just left with pi radians. And on the right-hand side here, 360 divided by 2 is 180, and we have still the units, which are degrees. So we get pi radians are equal to 180 degrees, which actually answered the first part of our question. We wanted to convert pi radians. Well, we just figured out pi radians are equal to 100, 180 degrees. Pi radians are equal to 180 degrees. And if you want to think about it, we know pi radians are halfway around a circle like that, and that's the same thing as 180 degrees."}, {"video_title": "Example Converting radians to degrees Trigonometry Khan Academy.mp3", "Sentence": "We wanted to convert pi radians. Well, we just figured out pi radians are equal to 100, 180 degrees. Pi radians are equal to 180 degrees. And if you want to think about it, we know pi radians are halfway around a circle like that, and that's the same thing as 180 degrees. So now let's think about the second part of it. We want to convert negative pi over 3 radians. Let me do this in a new color."}, {"video_title": "Example Converting radians to degrees Trigonometry Khan Academy.mp3", "Sentence": "And if you want to think about it, we know pi radians are halfway around a circle like that, and that's the same thing as 180 degrees. So now let's think about the second part of it. We want to convert negative pi over 3 radians. Let me do this in a new color. Negative pi over 3. So negative pi over 3 radians. How can we convert that to degrees?"}, {"video_title": "Example Converting radians to degrees Trigonometry Khan Academy.mp3", "Sentence": "Let me do this in a new color. Negative pi over 3. So negative pi over 3 radians. How can we convert that to degrees? What do we get based on this information right over here? Well, to figure this out, we need to know how many degrees there are per radian. If we need to multiply this times degrees, and I'm going to write the word out because if I just wrote a little circle here, it would be hard to visualize that as a unit."}, {"video_title": "Example Converting radians to degrees Trigonometry Khan Academy.mp3", "Sentence": "How can we convert that to degrees? What do we get based on this information right over here? Well, to figure this out, we need to know how many degrees there are per radian. If we need to multiply this times degrees, and I'm going to write the word out because if I just wrote a little circle here, it would be hard to visualize that as a unit. Degrees per radian. So how many degrees are there per radian? Well, we know that for every 180 degrees, we have pi radians."}, {"video_title": "Example Converting radians to degrees Trigonometry Khan Academy.mp3", "Sentence": "If we need to multiply this times degrees, and I'm going to write the word out because if I just wrote a little circle here, it would be hard to visualize that as a unit. Degrees per radian. So how many degrees are there per radian? Well, we know that for every 180 degrees, we have pi radians. Or you could say that there are 180 over pi degrees per radian. And this is going to work out. We have however many radians we have times the number of degrees per radian."}, {"video_title": "Example Converting radians to degrees Trigonometry Khan Academy.mp3", "Sentence": "Well, we know that for every 180 degrees, we have pi radians. Or you could say that there are 180 over pi degrees per radian. And this is going to work out. We have however many radians we have times the number of degrees per radian. So of course, the units are going to work out. Radians cancel out. The pi also cancels out."}, {"video_title": "Example Converting radians to degrees Trigonometry Khan Academy.mp3", "Sentence": "We have however many radians we have times the number of degrees per radian. So of course, the units are going to work out. Radians cancel out. The pi also cancels out. So you're left with negative 180 divided by 3, leaving us with negative 60. And we don't want to forget the units. We could write them out."}, {"video_title": "Example Converting radians to degrees Trigonometry Khan Academy.mp3", "Sentence": "The pi also cancels out. So you're left with negative 180 divided by 3, leaving us with negative 60. And we don't want to forget the units. We could write them out. That's the only units that's left. Degrees, which we could write out. We could write out the word degrees or we could just put that little symbol there."}, {"video_title": "Inverse trig functions arcsin Trigonometry Khan Academy.mp3", "Sentence": "You either have that memorized or you would draw the unit circle right there. That's not the best looking unit circle, but you get the idea. You'd go to pi over 4 radians, which is the same thing as 45 degrees. You would draw that unit radius out. And the sine is defined as a y-coordinate on the unit circle, so you would just want to know this value right here. And you would immediately see, OK, this is a 45 degrees. Let me draw the triangle a little bit larger."}, {"video_title": "Inverse trig functions arcsin Trigonometry Khan Academy.mp3", "Sentence": "You would draw that unit radius out. And the sine is defined as a y-coordinate on the unit circle, so you would just want to know this value right here. And you would immediately see, OK, this is a 45 degrees. Let me draw the triangle a little bit larger. The triangle looks like this. This is 45. That's 45 degrees."}, {"video_title": "Inverse trig functions arcsin Trigonometry Khan Academy.mp3", "Sentence": "Let me draw the triangle a little bit larger. The triangle looks like this. This is 45. That's 45 degrees. This is 90. And you can solve a 45-45-90 triangle. The hypotenuse is 1."}, {"video_title": "Inverse trig functions arcsin Trigonometry Khan Academy.mp3", "Sentence": "That's 45 degrees. This is 90. And you can solve a 45-45-90 triangle. The hypotenuse is 1. This is x. This is x. They're going to be the same values."}, {"video_title": "Inverse trig functions arcsin Trigonometry Khan Academy.mp3", "Sentence": "The hypotenuse is 1. This is x. This is x. They're going to be the same values. This is an isosceles triangle. Their base angles are the same. So you say, look, x squared plus x squared is equal to 1 squared, which is just 1."}, {"video_title": "Inverse trig functions arcsin Trigonometry Khan Academy.mp3", "Sentence": "They're going to be the same values. This is an isosceles triangle. Their base angles are the same. So you say, look, x squared plus x squared is equal to 1 squared, which is just 1. 2x squared is equal to 1. x squared is equal to 1 half. x is equal to the square root of 1 half, which is 1 over the square root of 2. I can put that in rational form by multiplying that by the square root of 2 over 2."}, {"video_title": "Inverse trig functions arcsin Trigonometry Khan Academy.mp3", "Sentence": "So you say, look, x squared plus x squared is equal to 1 squared, which is just 1. 2x squared is equal to 1. x squared is equal to 1 half. x is equal to the square root of 1 half, which is 1 over the square root of 2. I can put that in rational form by multiplying that by the square root of 2 over 2. And I get x is equal to the square root of 2 over 2. So the height here is square root of 2 over 2. And if you wanted to know this distance, too, it would also be the same thing."}, {"video_title": "Inverse trig functions arcsin Trigonometry Khan Academy.mp3", "Sentence": "I can put that in rational form by multiplying that by the square root of 2 over 2. And I get x is equal to the square root of 2 over 2. So the height here is square root of 2 over 2. And if you wanted to know this distance, too, it would also be the same thing. But we just cared about the height, because the sine of this is just this height right here, the y-coordinate. And we got that as the square root of 2 over 2. This is all review."}, {"video_title": "Inverse trig functions arcsin Trigonometry Khan Academy.mp3", "Sentence": "And if you wanted to know this distance, too, it would also be the same thing. But we just cared about the height, because the sine of this is just this height right here, the y-coordinate. And we got that as the square root of 2 over 2. This is all review. We learned this in the unit circle video. But what if someone else, let's say on another day, I come up to you and I say, you, please tell me what the arc sine of the square root of 2 over 2 is. What is the arc sine?"}, {"video_title": "Inverse trig functions arcsin Trigonometry Khan Academy.mp3", "Sentence": "This is all review. We learned this in the unit circle video. But what if someone else, let's say on another day, I come up to you and I say, you, please tell me what the arc sine of the square root of 2 over 2 is. What is the arc sine? And you're stumped. You're like, I know what the sine of an angle is, but this is some new trigonometric function that Sal has devised. And all you have to realize when they have this word arc in front of it, this is also sometimes referred to as the inverse sine, this could have just as easily been written as what is the inverse sine of the square root of 2 over 2."}, {"video_title": "Inverse trig functions arcsin Trigonometry Khan Academy.mp3", "Sentence": "What is the arc sine? And you're stumped. You're like, I know what the sine of an angle is, but this is some new trigonometric function that Sal has devised. And all you have to realize when they have this word arc in front of it, this is also sometimes referred to as the inverse sine, this could have just as easily been written as what is the inverse sine of the square root of 2 over 2. All this is asking is, what angle would I have to take the sine of in order to get the value square root of 2 over 2? This is also asking, what angle would I have to take the sine of in order to get square root of 2 over 2? I could rewrite either of these statements as saying sine of what is equal to the square root of 2 over 2?"}, {"video_title": "Inverse trig functions arcsin Trigonometry Khan Academy.mp3", "Sentence": "And all you have to realize when they have this word arc in front of it, this is also sometimes referred to as the inverse sine, this could have just as easily been written as what is the inverse sine of the square root of 2 over 2. All this is asking is, what angle would I have to take the sine of in order to get the value square root of 2 over 2? This is also asking, what angle would I have to take the sine of in order to get square root of 2 over 2? I could rewrite either of these statements as saying sine of what is equal to the square root of 2 over 2? And this, I think, is a much easier question for you to answer. Sine of what is square root of 2 over 2? Well, I just figured out that the sine of pi over 4 is square root of 2 over 2."}, {"video_title": "Inverse trig functions arcsin Trigonometry Khan Academy.mp3", "Sentence": "I could rewrite either of these statements as saying sine of what is equal to the square root of 2 over 2? And this, I think, is a much easier question for you to answer. Sine of what is square root of 2 over 2? Well, I just figured out that the sine of pi over 4 is square root of 2 over 2. So in this case, I know that the sine of pi over 4 is equal to square root of 2 over 2. So my question mark is equal to pi over 4. Or I could have rewritten this as the arc sine of square root of 2 over 2 is equal to pi over 4."}, {"video_title": "Inverse trig functions arcsin Trigonometry Khan Academy.mp3", "Sentence": "Well, I just figured out that the sine of pi over 4 is square root of 2 over 2. So in this case, I know that the sine of pi over 4 is equal to square root of 2 over 2. So my question mark is equal to pi over 4. Or I could have rewritten this as the arc sine of square root of 2 over 2 is equal to pi over 4. Now you might say, so just as a review, I'm giving you a value, and I'm saying give me an angle that gives me, when I take the sine of that angle, that gives me that value. But you're like, hey, Sal, look. Let me go over here."}, {"video_title": "Inverse trig functions arcsin Trigonometry Khan Academy.mp3", "Sentence": "Or I could have rewritten this as the arc sine of square root of 2 over 2 is equal to pi over 4. Now you might say, so just as a review, I'm giving you a value, and I'm saying give me an angle that gives me, when I take the sine of that angle, that gives me that value. But you're like, hey, Sal, look. Let me go over here. You're like, look, pi over 2 worked. 45 degrees worked. But I could just keep adding 360 degrees, or I could keep just adding 2 pi, and all of those would work."}, {"video_title": "Inverse trig functions arcsin Trigonometry Khan Academy.mp3", "Sentence": "Let me go over here. You're like, look, pi over 2 worked. 45 degrees worked. But I could just keep adding 360 degrees, or I could keep just adding 2 pi, and all of those would work. Because those would all get me to that same point on the unit circle. And you'd be correct. And so all of those values, you would think, would be valid answers for this."}, {"video_title": "Inverse trig functions arcsin Trigonometry Khan Academy.mp3", "Sentence": "But I could just keep adding 360 degrees, or I could keep just adding 2 pi, and all of those would work. Because those would all get me to that same point on the unit circle. And you'd be correct. And so all of those values, you would think, would be valid answers for this. Because if you take the sine of any of those angles, you could just keep adding 360 degrees. If you take the sine of any of them, you would get square root of 2 over 2. And that's a problem."}, {"video_title": "Inverse trig functions arcsin Trigonometry Khan Academy.mp3", "Sentence": "And so all of those values, you would think, would be valid answers for this. Because if you take the sine of any of those angles, you could just keep adding 360 degrees. If you take the sine of any of them, you would get square root of 2 over 2. And that's a problem. You can't have a function where if I take the function f of x, where it maps to multiple values, where it maps to pi over 4, or it maps to pi over 4 plus 2 pi, or pi over 4 plus 4 pi. So in order for this to be a valid function, in order for the inverse sine function to be valid, I have to restrict its range. And the way that we'll just restrict its range to the most natural place."}, {"video_title": "Inverse trig functions arcsin Trigonometry Khan Academy.mp3", "Sentence": "And that's a problem. You can't have a function where if I take the function f of x, where it maps to multiple values, where it maps to pi over 4, or it maps to pi over 4 plus 2 pi, or pi over 4 plus 4 pi. So in order for this to be a valid function, in order for the inverse sine function to be valid, I have to restrict its range. And the way that we'll just restrict its range to the most natural place. So let's restrict its range. And actually, just as a side note, what's its domain restricted to? So if I'm taking the arc sine of something, so if I'm taking the arc sine of x, and I'm saying that that is equal to theta, what's the domain restricted to?"}, {"video_title": "Inverse trig functions arcsin Trigonometry Khan Academy.mp3", "Sentence": "And the way that we'll just restrict its range to the most natural place. So let's restrict its range. And actually, just as a side note, what's its domain restricted to? So if I'm taking the arc sine of something, so if I'm taking the arc sine of x, and I'm saying that that is equal to theta, what's the domain restricted to? What are the valid values of x? x could be equal to what? Well, if I take the sine of any angle, I can only get values between 1 and negative 1, right?"}, {"video_title": "Inverse trig functions arcsin Trigonometry Khan Academy.mp3", "Sentence": "So if I'm taking the arc sine of something, so if I'm taking the arc sine of x, and I'm saying that that is equal to theta, what's the domain restricted to? What are the valid values of x? x could be equal to what? Well, if I take the sine of any angle, I can only get values between 1 and negative 1, right? So x is going to be greater than or equal to negative 1, and then less than or equal to 1. That's the domain. Now, in order to make this a valid function, I have to restrict the range, the possible values."}, {"video_title": "Inverse trig functions arcsin Trigonometry Khan Academy.mp3", "Sentence": "Well, if I take the sine of any angle, I can only get values between 1 and negative 1, right? So x is going to be greater than or equal to negative 1, and then less than or equal to 1. That's the domain. Now, in order to make this a valid function, I have to restrict the range, the possible values. I have to restrict the range. And for arc sine, the convention is to restrict it to the first and fourth quadrants, to restrict the possible angles to this area right here, along the unit circle. So theta is restricted to being less than or equal to pi over 2, and then greater than or equal to minus pi over 2."}, {"video_title": "Inverse trig functions arcsin Trigonometry Khan Academy.mp3", "Sentence": "Now, in order to make this a valid function, I have to restrict the range, the possible values. I have to restrict the range. And for arc sine, the convention is to restrict it to the first and fourth quadrants, to restrict the possible angles to this area right here, along the unit circle. So theta is restricted to being less than or equal to pi over 2, and then greater than or equal to minus pi over 2. So given that, we now understand what arc sine is. Let's do another problem. Let me clear out some space here."}, {"video_title": "Inverse trig functions arcsin Trigonometry Khan Academy.mp3", "Sentence": "So theta is restricted to being less than or equal to pi over 2, and then greater than or equal to minus pi over 2. So given that, we now understand what arc sine is. Let's do another problem. Let me clear out some space here. Let me do another arc sine. So let's say I were to ask you what the arc sine of minus square root of 3 over 2 is. Minus square root of 3 over 2."}, {"video_title": "Inverse trig functions arcsin Trigonometry Khan Academy.mp3", "Sentence": "Let me clear out some space here. Let me do another arc sine. So let's say I were to ask you what the arc sine of minus square root of 3 over 2 is. Minus square root of 3 over 2. Now, you might have that memorized. You say, oh, I immediately know that sine of x, or sine of theta, is square root of 3 over 2, and you'd be done. But I don't have that memorized, so let me just draw my unit circle."}, {"video_title": "Inverse trig functions arcsin Trigonometry Khan Academy.mp3", "Sentence": "Minus square root of 3 over 2. Now, you might have that memorized. You say, oh, I immediately know that sine of x, or sine of theta, is square root of 3 over 2, and you'd be done. But I don't have that memorized, so let me just draw my unit circle. And when I'm dealing with arc sine, I just have to draw the first and fourth quadrants of my unit circle. That's the y-axis. That's my x-axis."}, {"video_title": "Inverse trig functions arcsin Trigonometry Khan Academy.mp3", "Sentence": "But I don't have that memorized, so let me just draw my unit circle. And when I'm dealing with arc sine, I just have to draw the first and fourth quadrants of my unit circle. That's the y-axis. That's my x-axis. x and y. And where am I? If the sine of something is minus square root of 3 over 2, that means the y-coordinate on the unit circle is minus square root of 3 over 2."}, {"video_title": "Inverse trig functions arcsin Trigonometry Khan Academy.mp3", "Sentence": "That's my x-axis. x and y. And where am I? If the sine of something is minus square root of 3 over 2, that means the y-coordinate on the unit circle is minus square root of 3 over 2. So it means we are right about there. So this is minus square root of 3 over 2. This is where we are."}, {"video_title": "Inverse trig functions arcsin Trigonometry Khan Academy.mp3", "Sentence": "If the sine of something is minus square root of 3 over 2, that means the y-coordinate on the unit circle is minus square root of 3 over 2. So it means we are right about there. So this is minus square root of 3 over 2. This is where we are. Now, what angle gives me that? Let's think about it a little bit. My y-coordinate is minus square root of 3 over 2."}, {"video_title": "Inverse trig functions arcsin Trigonometry Khan Academy.mp3", "Sentence": "This is where we are. Now, what angle gives me that? Let's think about it a little bit. My y-coordinate is minus square root of 3 over 2. This is the angle. It's going to be a negative angle, because we're going below the x-axis in the clockwise direction. And to figure out, let me just draw a little triangle here."}, {"video_title": "Inverse trig functions arcsin Trigonometry Khan Academy.mp3", "Sentence": "My y-coordinate is minus square root of 3 over 2. This is the angle. It's going to be a negative angle, because we're going below the x-axis in the clockwise direction. And to figure out, let me just draw a little triangle here. Let me pick a better color than that. So that's a triangle. Let me do it in this blue color."}, {"video_title": "Inverse trig functions arcsin Trigonometry Khan Academy.mp3", "Sentence": "And to figure out, let me just draw a little triangle here. Let me pick a better color than that. So that's a triangle. Let me do it in this blue color. So let me zoom up that triangle like that. This is theta. That's theta."}, {"video_title": "Inverse trig functions arcsin Trigonometry Khan Academy.mp3", "Sentence": "Let me do it in this blue color. So let me zoom up that triangle like that. This is theta. That's theta. And what's this length right here? Well, that's the same as the y-height, I guess we could call it, which is square root of 3 over 2. It's a minus, because we're going down."}, {"video_title": "Inverse trig functions arcsin Trigonometry Khan Academy.mp3", "Sentence": "That's theta. And what's this length right here? Well, that's the same as the y-height, I guess we could call it, which is square root of 3 over 2. It's a minus, because we're going down. But let's just figure out this angle, and we know it's a negative angle. So when you see square root of 3 over 2, hopefully you recognize, hey, this is a 30-60-90 triangle. The square root of 3 over 2, this side is 1 half, and then of course this side is 1, because this is a unit circle."}, {"video_title": "Inverse trig functions arcsin Trigonometry Khan Academy.mp3", "Sentence": "It's a minus, because we're going down. But let's just figure out this angle, and we know it's a negative angle. So when you see square root of 3 over 2, hopefully you recognize, hey, this is a 30-60-90 triangle. The square root of 3 over 2, this side is 1 half, and then of course this side is 1, because this is a unit circle. So its radius is 1. So in a 30-60-90 triangle, the side opposite to the square root of 3 over 2 is 60 degrees. The side over here is 30 degrees."}, {"video_title": "Inverse trig functions arcsin Trigonometry Khan Academy.mp3", "Sentence": "The square root of 3 over 2, this side is 1 half, and then of course this side is 1, because this is a unit circle. So its radius is 1. So in a 30-60-90 triangle, the side opposite to the square root of 3 over 2 is 60 degrees. The side over here is 30 degrees. So we know that our theta is, this is 60 degrees. That's its magnitude, but it's going downward. So it's minus 60 degrees."}, {"video_title": "Inverse trig functions arcsin Trigonometry Khan Academy.mp3", "Sentence": "The side over here is 30 degrees. So we know that our theta is, this is 60 degrees. That's its magnitude, but it's going downward. So it's minus 60 degrees. So theta is equal to minus 60 degrees. But if we're dealing in radians, that's not good enough, so we can multiply that times pi radians for every 180 degrees. Degrees cancel out, and we're left with theta is equal to minus pi over 3 radians."}, {"video_title": "Inverse trig functions arcsin Trigonometry Khan Academy.mp3", "Sentence": "So it's minus 60 degrees. So theta is equal to minus 60 degrees. But if we're dealing in radians, that's not good enough, so we can multiply that times pi radians for every 180 degrees. Degrees cancel out, and we're left with theta is equal to minus pi over 3 radians. And so we can now make the statements that the arc sine of minus square root of 3 over 2 is equal to minus pi over 3 radians, or we could say the inverse sine of minus square root of 3 over 2 is equal to minus pi over 3 radians. And to confirm this, let me get a little calculator out. I put this in radian mode already."}, {"video_title": "Inverse trig functions arcsin Trigonometry Khan Academy.mp3", "Sentence": "Degrees cancel out, and we're left with theta is equal to minus pi over 3 radians. And so we can now make the statements that the arc sine of minus square root of 3 over 2 is equal to minus pi over 3 radians, or we could say the inverse sine of minus square root of 3 over 2 is equal to minus pi over 3 radians. And to confirm this, let me get a little calculator out. I put this in radian mode already. You can just check that per second mode. I'm in radian mode, so I know I'm going to get, hopefully, the right answer. And I want to figure out the inverse sine."}, {"video_title": "Inverse trig functions arcsin Trigonometry Khan Academy.mp3", "Sentence": "I put this in radian mode already. You can just check that per second mode. I'm in radian mode, so I know I'm going to get, hopefully, the right answer. And I want to figure out the inverse sine. So the inverse sine, the second in the sine button, of the minus square root of 3 over 2, it equals minus 1.04. So it's telling me that this is equal to minus 1.04 radians. So pi over 3 must be equal to 1.04."}, {"video_title": "Inverse trig functions arcsin Trigonometry Khan Academy.mp3", "Sentence": "And I want to figure out the inverse sine. So the inverse sine, the second in the sine button, of the minus square root of 3 over 2, it equals minus 1.04. So it's telling me that this is equal to minus 1.04 radians. So pi over 3 must be equal to 1.04. Let's see if I can confirm that. So if I were to write minus pi divided by 3, what do I get? I get the exact same value."}, {"video_title": "Periodicity of tan example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So we're saying that the tangent right over here is, so the tangent, so let me write this down. So we're saying that the tangent of 0.46 radians is equal to half. And another way of thinking about the tangent of an angle is that's the slope of that angle's terminal ray. So it's the slope of this ray right over here. That makes sense that that slope is about a half. Now what other angles have a tangent of 1 half? So let's look at these choices."}, {"video_title": "Periodicity of tan example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So it's the slope of this ray right over here. That makes sense that that slope is about a half. Now what other angles have a tangent of 1 half? So let's look at these choices. So this is our original angle, 0.46 radians plus pi over 2. If you think in degrees, pi is 180, pi over 2 is 90 degrees. So this one, this one, actually let me do it in a color you're more likely to see."}, {"video_title": "Periodicity of tan example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So let's look at these choices. So this is our original angle, 0.46 radians plus pi over 2. If you think in degrees, pi is 180, pi over 2 is 90 degrees. So this one, this one, actually let me do it in a color you're more likely to see. This one is going to look like this. It's going to look like this. Where this is an angle of pi over 2."}, {"video_title": "Periodicity of tan example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So this one, this one, actually let me do it in a color you're more likely to see. This one is going to look like this. It's going to look like this. Where this is an angle of pi over 2. And just eyeballing it, you immediately see that the slope of this ray is very different than the slope of this ray right over here. In fact, they look like, in fact they are. They are perpendicular because they have an angle of pi over 2 between them."}, {"video_title": "Periodicity of tan example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Where this is an angle of pi over 2. And just eyeballing it, you immediately see that the slope of this ray is very different than the slope of this ray right over here. In fact, they look like, in fact they are. They are perpendicular because they have an angle of pi over 2 between them. But they're definitely not going to have the same tangent. So they're not going to have the same tangent. They don't have the same slope."}, {"video_title": "Periodicity of tan example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "They are perpendicular because they have an angle of pi over 2 between them. But they're definitely not going to have the same tangent. So they're not going to have the same tangent. They don't have the same slope. Now let's think about pi minus 0.46. So that's essentially, pi is going along the positive x-axis. If you go all the way around, or halfway around, you're at pi radians."}, {"video_title": "Periodicity of tan example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "They don't have the same slope. Now let's think about pi minus 0.46. So that's essentially, pi is going along the positive x-axis. If you go all the way around, or halfway around, you're at pi radians. But then we're going to subtract 0.46. So it's going to look something like this. It's going to look something like that, where this is 0.46 that we have subtracted."}, {"video_title": "Periodicity of tan example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "If you go all the way around, or halfway around, you're at pi radians. But then we're going to subtract 0.46. So it's going to look something like this. It's going to look something like that, where this is 0.46 that we have subtracted. Another way to think about it, if we take our original terminal ray and we flip it over the y-axis, we get to this terminal ray right over here. And you can immediately see that the slope of the terminal ray is not the same as the slope of this one, of our first one, of our original. In fact, they look like the negatives of each other."}, {"video_title": "Periodicity of tan example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "It's going to look something like that, where this is 0.46 that we have subtracted. Another way to think about it, if we take our original terminal ray and we flip it over the y-axis, we get to this terminal ray right over here. And you can immediately see that the slope of the terminal ray is not the same as the slope of this one, of our first one, of our original. In fact, they look like the negatives of each other. So we could rule that one out as well. 0.46 plus pi, or pi plus 0.46. So that's going to take us, if you add pi to this, you're essentially going halfway around the unit circle and you're getting to a point where you're forming a ray that is collinear with the original ray."}, {"video_title": "Periodicity of tan example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "In fact, they look like the negatives of each other. So we could rule that one out as well. 0.46 plus pi, or pi plus 0.46. So that's going to take us, if you add pi to this, you're essentially going halfway around the unit circle and you're getting to a point where you're forming a ray that is collinear with the original ray. So that's that angle right over here. So pi plus 0.46 is this entire angle right over there. And when you just look at this ray, you see it's collinear."}, {"video_title": "Periodicity of tan example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So that's going to take us, if you add pi to this, you're essentially going halfway around the unit circle and you're getting to a point where you're forming a ray that is collinear with the original ray. So that's that angle right over here. So pi plus 0.46 is this entire angle right over there. And when you just look at this ray, you see it's collinear. It's going to have the exact same slope as the terminal ray for 0.46 radians. So just that tells you that the tangent is going to be the same. So I could check that there."}, {"video_title": "Periodicity of tan example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And when you just look at this ray, you see it's collinear. It's going to have the exact same slope as the terminal ray for 0.46 radians. So just that tells you that the tangent is going to be the same. So I could check that there. And in previous videos, when we explored the symmetries of the tangent function, we in fact saw that, that if you took an angle and you add pi to it, you're going to have the same tangent. And if you want to dig a little bit deeper, I encourage you to look at that video on the symmetries of unit circle symmetries for the tangent function. So let's look at these other choices."}, {"video_title": "Periodicity of tan example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So I could check that there. And in previous videos, when we explored the symmetries of the tangent function, we in fact saw that, that if you took an angle and you add pi to it, you're going to have the same tangent. And if you want to dig a little bit deeper, I encourage you to look at that video on the symmetries of unit circle symmetries for the tangent function. So let's look at these other choices. 2 pi minus 0.46. So 2 pi, if this is 0 degrees, 2 pi gets you back to the positive x-axis. And then you're going to subtract 0.46."}, {"video_title": "Periodicity of tan example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So let's look at these other choices. 2 pi minus 0.46. So 2 pi, if this is 0 degrees, 2 pi gets you back to the positive x-axis. And then you're going to subtract 0.46. So that's going to be this angle right over here. And that looks like it has the negative slope of this original ray right up here. So these aren't going to have the same tangent."}, {"video_title": "Periodicity of tan example Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And then you're going to subtract 0.46. So that's going to be this angle right over here. And that looks like it has the negative slope of this original ray right up here. So these aren't going to have the same tangent. Now this one, you're taking 0.46 and you're adding 2 pi. So you're taking 0.46 and then you're adding 2 pi, which is essentially just going around the unit circle once, and you get to the exact same point. So you add 2 pi to any angle measure."}, {"video_title": "Example Graphing y=3\u22c5sin(\u00bd\u22c5x)-2 Trigonometry Algebra 2 Khan Academy.mp3", "Sentence": "So we're asked to graph y is equal to 3 times sine of 1 half x minus 2 in the interactive widget. And this is the interactive widget that you would find on Khan Academy. And it first bears mentioning how this widget works. So this point right over here, it helps you define the midline, the thing that you could imagine your sine or cosine function oscillates around. And then you also define a neighboring extreme point, either a maximum or a minimum point, to graph your function. So let's think about how we would do this. And like always, I encourage you to pause this video and think about how you would do it yourself."}, {"video_title": "Example Graphing y=3\u22c5sin(\u00bd\u22c5x)-2 Trigonometry Algebra 2 Khan Academy.mp3", "Sentence": "So this point right over here, it helps you define the midline, the thing that you could imagine your sine or cosine function oscillates around. And then you also define a neighboring extreme point, either a maximum or a minimum point, to graph your function. So let's think about how we would do this. And like always, I encourage you to pause this video and think about how you would do it yourself. But the first way I like to think about it is what would a regular just... If this just said y is equal to sine of x, how would I graph that? Well, sine of 0 is 0."}, {"video_title": "Example Graphing y=3\u22c5sin(\u00bd\u22c5x)-2 Trigonometry Algebra 2 Khan Academy.mp3", "Sentence": "And like always, I encourage you to pause this video and think about how you would do it yourself. But the first way I like to think about it is what would a regular just... If this just said y is equal to sine of x, how would I graph that? Well, sine of 0 is 0. Sine of pi over 2 is 1. And then sine of pi is 0 again. And so this is what just regular sine of x would look like."}, {"video_title": "Example Graphing y=3\u22c5sin(\u00bd\u22c5x)-2 Trigonometry Algebra 2 Khan Academy.mp3", "Sentence": "Well, sine of 0 is 0. Sine of pi over 2 is 1. And then sine of pi is 0 again. And so this is what just regular sine of x would look like. But let's think about how this is different. Well, first of all, it's not just sine of x. It's sine of 1 half x."}, {"video_title": "Example Graphing y=3\u22c5sin(\u00bd\u22c5x)-2 Trigonometry Algebra 2 Khan Academy.mp3", "Sentence": "And so this is what just regular sine of x would look like. But let's think about how this is different. Well, first of all, it's not just sine of x. It's sine of 1 half x. So what would be the graph of just sine of 1 half x? Well, one way to think about it, there's actually two ways to think about it, is a coefficient right over here on your x term that tells you how fast the thing that's being inputted into sine is growing. And now it's going to grow half as fast."}, {"video_title": "Example Graphing y=3\u22c5sin(\u00bd\u22c5x)-2 Trigonometry Algebra 2 Khan Academy.mp3", "Sentence": "It's sine of 1 half x. So what would be the graph of just sine of 1 half x? Well, one way to think about it, there's actually two ways to think about it, is a coefficient right over here on your x term that tells you how fast the thing that's being inputted into sine is growing. And now it's going to grow half as fast. And so one way to think about it is your period is now going to be twice as long. So one way to think about it is instead of getting to this next maximum point at pi over 2, you're going to get there at pi. And you could test that."}, {"video_title": "Example Graphing y=3\u22c5sin(\u00bd\u22c5x)-2 Trigonometry Algebra 2 Khan Academy.mp3", "Sentence": "And now it's going to grow half as fast. And so one way to think about it is your period is now going to be twice as long. So one way to think about it is instead of getting to this next maximum point at pi over 2, you're going to get there at pi. And you could test that. If you...when x is equal to pi, this will be 1 half pi. Sine of 1 half pi is indeed equal to 1. Another way to think about it is you might be familiar with the formula, although I always like you to think about where these formulas come from, that to figure out the period of a sine or cosine function, you take 2 pi and you divide it by whatever this coefficient is."}, {"video_title": "Example Graphing y=3\u22c5sin(\u00bd\u22c5x)-2 Trigonometry Algebra 2 Khan Academy.mp3", "Sentence": "And you could test that. If you...when x is equal to pi, this will be 1 half pi. Sine of 1 half pi is indeed equal to 1. Another way to think about it is you might be familiar with the formula, although I always like you to think about where these formulas come from, that to figure out the period of a sine or cosine function, you take 2 pi and you divide it by whatever this coefficient is. So 2 pi divided by 1 half is going to be 4 pi. And you can see the period here, we go up, down, and back to where we were over 4 pi. And that makes sense because if you just had a 1 coefficient here, your period would be 2 pi."}, {"video_title": "Example Graphing y=3\u22c5sin(\u00bd\u22c5x)-2 Trigonometry Algebra 2 Khan Academy.mp3", "Sentence": "Another way to think about it is you might be familiar with the formula, although I always like you to think about where these formulas come from, that to figure out the period of a sine or cosine function, you take 2 pi and you divide it by whatever this coefficient is. So 2 pi divided by 1 half is going to be 4 pi. And you can see the period here, we go up, down, and back to where we were over 4 pi. And that makes sense because if you just had a 1 coefficient here, your period would be 2 pi. 2 pi radians, you make one circle around the unit circle is one way to think about it. So right here we have the graph of sine of 1 half x. Now what if we wanted to instead think about 3 times the graph of sine of 1 half x, or 3 sine 1 half x?"}, {"video_title": "Example Graphing y=3\u22c5sin(\u00bd\u22c5x)-2 Trigonometry Algebra 2 Khan Academy.mp3", "Sentence": "And that makes sense because if you just had a 1 coefficient here, your period would be 2 pi. 2 pi radians, you make one circle around the unit circle is one way to think about it. So right here we have the graph of sine of 1 half x. Now what if we wanted to instead think about 3 times the graph of sine of 1 half x, or 3 sine 1 half x? Well, then our amplitude is just going to be 3 times as much. And so instead of our maximum point going from... instead of our maximum point being at 1, it will now be at 3. Or another way to think about it is we're going 3 above the midline and 3 below the midline."}, {"video_title": "Example Graphing y=3\u22c5sin(\u00bd\u22c5x)-2 Trigonometry Algebra 2 Khan Academy.mp3", "Sentence": "Now what if we wanted to instead think about 3 times the graph of sine of 1 half x, or 3 sine 1 half x? Well, then our amplitude is just going to be 3 times as much. And so instead of our maximum point going from... instead of our maximum point being at 1, it will now be at 3. Or another way to think about it is we're going 3 above the midline and 3 below the midline. So this right over here is the graph of 3 sine of 1 half x. Now we have one thing left to do, and this is this minus 2. So this minus 2 is just going to shift everything down by 2."}, {"video_title": "Example Graphing y=3\u22c5sin(\u00bd\u22c5x)-2 Trigonometry Algebra 2 Khan Academy.mp3", "Sentence": "Or another way to think about it is we're going 3 above the midline and 3 below the midline. So this right over here is the graph of 3 sine of 1 half x. Now we have one thing left to do, and this is this minus 2. So this minus 2 is just going to shift everything down by 2. So we just have to shift everything down. So let me shift this one down by 2, and let me shift this one down by 2. And so there you have it."}, {"video_title": "Example Graphing y=3\u22c5sin(\u00bd\u22c5x)-2 Trigonometry Algebra 2 Khan Academy.mp3", "Sentence": "So this minus 2 is just going to shift everything down by 2. So we just have to shift everything down. So let me shift this one down by 2, and let me shift this one down by 2. And so there you have it. Notice our period is still 4 pi. Our amplitude, how much we oscillate above or below the midline, is still 3. And now we have this minus 2."}, {"video_title": "Cosine, sine and tangent of \u03c0 6 and \u03c0 3.mp3", "Sentence": "In this video, we're going to figure out what the sine, cosine, and tangent of two very important angles are, angles that you'll see a lot in your trigonometric and just in general in your regular life. So these are the angles pi over three radians and pi over six radians. And sometimes it's useful to visualize them as degrees. Pi over three, you might remember pi radians is 180 degrees. So you divide that by three. This is equivalent to 60 degrees. And once again, 180 degrees, which is the same thing as pi radians divided by six is the same thing as 30 degrees."}, {"video_title": "Cosine, sine and tangent of \u03c0 6 and \u03c0 3.mp3", "Sentence": "Pi over three, you might remember pi radians is 180 degrees. So you divide that by three. This is equivalent to 60 degrees. And once again, 180 degrees, which is the same thing as pi radians divided by six is the same thing as 30 degrees. Now I'm going to do it using the unit circle definition of trig functions, but to help us there, I'm gonna give us a little bit of a reminder of what some of you might be familiar with as 30, 60, 90 triangles, which I guess we could also call pi over six, pi over three, pi over two triangles. And so let me just draw one because this is going to be really helpful in establishing these trig functions using the unit circle definition. So let me draw a triangle here."}, {"video_title": "Cosine, sine and tangent of \u03c0 6 and \u03c0 3.mp3", "Sentence": "And once again, 180 degrees, which is the same thing as pi radians divided by six is the same thing as 30 degrees. Now I'm going to do it using the unit circle definition of trig functions, but to help us there, I'm gonna give us a little bit of a reminder of what some of you might be familiar with as 30, 60, 90 triangles, which I guess we could also call pi over six, pi over three, pi over two triangles. And so let me just draw one because this is going to be really helpful in establishing these trig functions using the unit circle definition. So let me draw a triangle here. It's hand drawn, so it's not as neat as it could be. So this right over here is a right angle. And let's say that this one is pi over three radians, which is the same thing as 60 degrees."}, {"video_title": "Cosine, sine and tangent of \u03c0 6 and \u03c0 3.mp3", "Sentence": "So let me draw a triangle here. It's hand drawn, so it's not as neat as it could be. So this right over here is a right angle. And let's say that this one is pi over three radians, which is the same thing as 60 degrees. And this one over here is pi over six radians, which is the same thing as 30 degrees. Now let's also say that the longest side, the hypotenuse here has length one. Now to help us think about what the other two sides are, what I'm gonna do is flip this triangle over this side right over here and essentially construct a mirror image."}, {"video_title": "Cosine, sine and tangent of \u03c0 6 and \u03c0 3.mp3", "Sentence": "And let's say that this one is pi over three radians, which is the same thing as 60 degrees. And this one over here is pi over six radians, which is the same thing as 30 degrees. Now let's also say that the longest side, the hypotenuse here has length one. Now to help us think about what the other two sides are, what I'm gonna do is flip this triangle over this side right over here and essentially construct a mirror image. So because this right over here is a mirror image, we immediately know a few things. We know that this length right over here is going to be congruent to this length over here. And let me actually finish drawing the entire triangle."}, {"video_title": "Cosine, sine and tangent of \u03c0 6 and \u03c0 3.mp3", "Sentence": "Now to help us think about what the other two sides are, what I'm gonna do is flip this triangle over this side right over here and essentially construct a mirror image. So because this right over here is a mirror image, we immediately know a few things. We know that this length right over here is going to be congruent to this length over here. And let me actually finish drawing the entire triangle. It's gonna look something like this. And since it's, once again, it's a reflection, this length over here is gonna have length one. This is going to be pi over six radians."}, {"video_title": "Cosine, sine and tangent of \u03c0 6 and \u03c0 3.mp3", "Sentence": "And let me actually finish drawing the entire triangle. It's gonna look something like this. And since it's, once again, it's a reflection, this length over here is gonna have length one. This is going to be pi over six radians. This is going to be pi over three radians. So what else do we know about this larger triangle now? Well, we know it's an equilateral triangle."}, {"video_title": "Cosine, sine and tangent of \u03c0 6 and \u03c0 3.mp3", "Sentence": "This is going to be pi over six radians. This is going to be pi over three radians. So what else do we know about this larger triangle now? Well, we know it's an equilateral triangle. All the angles, pi over three radians, pi over three radians. And if you add two pi over sixes together, you're going to get pi over three as well. So it's a 60 degree, 60 degree, 60 degree triangle."}, {"video_title": "Cosine, sine and tangent of \u03c0 6 and \u03c0 3.mp3", "Sentence": "Well, we know it's an equilateral triangle. All the angles, pi over three radians, pi over three radians. And if you add two pi over sixes together, you're going to get pi over three as well. So it's a 60 degree, 60 degree, 60 degree triangle. And so all the sides are gonna have the same length. There's gonna be one, one, and one. And if these two sides are congruent of the smaller triangles, of the smaller right triangles, well, then this right over here must be 1 1\u20442."}, {"video_title": "Cosine, sine and tangent of \u03c0 6 and \u03c0 3.mp3", "Sentence": "So it's a 60 degree, 60 degree, 60 degree triangle. And so all the sides are gonna have the same length. There's gonna be one, one, and one. And if these two sides are congruent of the smaller triangles, of the smaller right triangles, well, then this right over here must be 1 1\u20442. And then this right over here must be 1 1\u20442 as well. Now that's going to be useful for figuring out what this length right over here is going to be. Because we have two right triangles, we could use either one."}, {"video_title": "Cosine, sine and tangent of \u03c0 6 and \u03c0 3.mp3", "Sentence": "And if these two sides are congruent of the smaller triangles, of the smaller right triangles, well, then this right over here must be 1 1\u20442. And then this right over here must be 1 1\u20442 as well. Now that's going to be useful for figuring out what this length right over here is going to be. Because we have two right triangles, we could use either one. But if we just use this bottom right triangle here, the Pythagorean theorem tells us that 1 1\u20442 squared, let's call this b, so plus b squared, I'm just pattern matching a squared plus b squared is equal to c squared, where c is the length of the hypotenuse, is equal to one squared. And so we get that 1\u20444 plus b squared is equal to one, or subtracting 1\u20444 from both sides, b squared is equal to 3\u20444. And then taking the principal root of both sides, we get b is equal to the square root of three over two."}, {"video_title": "Cosine, sine and tangent of \u03c0 6 and \u03c0 3.mp3", "Sentence": "Because we have two right triangles, we could use either one. But if we just use this bottom right triangle here, the Pythagorean theorem tells us that 1 1\u20442 squared, let's call this b, so plus b squared, I'm just pattern matching a squared plus b squared is equal to c squared, where c is the length of the hypotenuse, is equal to one squared. And so we get that 1\u20444 plus b squared is equal to one, or subtracting 1\u20444 from both sides, b squared is equal to 3\u20444. And then taking the principal root of both sides, we get b is equal to the square root of three over two. So just like that, we have figured out what all the lengths of this 30, 60, 90 triangle are. So b here is equal to square root of three over two. Now I said this would be useful as we go into the unit circle definitions of sine, cosine, and tangent, and we're about to see why."}, {"video_title": "Cosine, sine and tangent of \u03c0 6 and \u03c0 3.mp3", "Sentence": "And then taking the principal root of both sides, we get b is equal to the square root of three over two. So just like that, we have figured out what all the lengths of this 30, 60, 90 triangle are. So b here is equal to square root of three over two. Now I said this would be useful as we go into the unit circle definitions of sine, cosine, and tangent, and we're about to see why. So here I have two different unit circles. I'm going to use one for each of these angles. So first let's think about pi over three radians."}, {"video_title": "Cosine, sine and tangent of \u03c0 6 and \u03c0 3.mp3", "Sentence": "Now I said this would be useful as we go into the unit circle definitions of sine, cosine, and tangent, and we're about to see why. So here I have two different unit circles. I'm going to use one for each of these angles. So first let's think about pi over three radians. And so pi over three would look something like this. So this is pi over three radians. And the cosine and sine can be determined by the x and y coordinates of this point where this radius intersects the actual unit circle."}, {"video_title": "Cosine, sine and tangent of \u03c0 6 and \u03c0 3.mp3", "Sentence": "So first let's think about pi over three radians. And so pi over three would look something like this. So this is pi over three radians. And the cosine and sine can be determined by the x and y coordinates of this point where this radius intersects the actual unit circle. The coordinates here are going to be cosine of pi over three radians, and sine of pi over three radians. Or another way to think about it is, I can set up a 30, 60, 90 triangle here. So I'm gonna drop a perpendicular."}, {"video_title": "Cosine, sine and tangent of \u03c0 6 and \u03c0 3.mp3", "Sentence": "And the cosine and sine can be determined by the x and y coordinates of this point where this radius intersects the actual unit circle. The coordinates here are going to be cosine of pi over three radians, and sine of pi over three radians. Or another way to think about it is, I can set up a 30, 60, 90 triangle here. So I'm gonna drop a perpendicular. This would be 90 degrees or pi over two radians. And then this angle over here, if this is 60, this is 90, this is going to be 30. Or another way of thinking about it, it's going to be pi over six radians."}, {"video_title": "Cosine, sine and tangent of \u03c0 6 and \u03c0 3.mp3", "Sentence": "So I'm gonna drop a perpendicular. This would be 90 degrees or pi over two radians. And then this angle over here, if this is 60, this is 90, this is going to be 30. Or another way of thinking about it, it's going to be pi over six radians. It's going to be just like one of these triangles here. And so the x coordinate, which is going to be the same thing as the cosine of pi over three, is going to be the length of this side right over here. Well, what's that going to be?"}, {"video_title": "Cosine, sine and tangent of \u03c0 6 and \u03c0 3.mp3", "Sentence": "Or another way of thinking about it, it's going to be pi over six radians. It's going to be just like one of these triangles here. And so the x coordinate, which is going to be the same thing as the cosine of pi over three, is going to be the length of this side right over here. Well, what's that going to be? Well, when your hypotenuse is one, we know that the shorter side, the side opposite the pi over six radians is 1 1\u20442. So just like that, we have been able to establish that cosine of pi over three radians is equal to 1 1\u20442. This right over here is 1 1\u20442."}, {"video_title": "Cosine, sine and tangent of \u03c0 6 and \u03c0 3.mp3", "Sentence": "Well, what's that going to be? Well, when your hypotenuse is one, we know that the shorter side, the side opposite the pi over six radians is 1 1\u20442. So just like that, we have been able to establish that cosine of pi over three radians is equal to 1 1\u20442. This right over here is 1 1\u20442. That is the x coordinate where this radius intersects the unit circle. Now, what about the y coordinate? What is sine of pi over three going to be?"}, {"video_title": "Cosine, sine and tangent of \u03c0 6 and \u03c0 3.mp3", "Sentence": "This right over here is 1 1\u20442. That is the x coordinate where this radius intersects the unit circle. Now, what about the y coordinate? What is sine of pi over three going to be? Well, the y coordinate is the same thing as the length of this side. And once again, it goes back to being this triangle. If this is one, this is 1 1\u20442, this is one, this is 1 1\u20442, this other side is going to be square root of three over two."}, {"video_title": "Cosine, sine and tangent of \u03c0 6 and \u03c0 3.mp3", "Sentence": "What is sine of pi over three going to be? Well, the y coordinate is the same thing as the length of this side. And once again, it goes back to being this triangle. If this is one, this is 1 1\u20442, this is one, this is 1 1\u20442, this other side is going to be square root of three over two. So sine of pi over three is going to be square root of three over two. So let me write that down. Sine of pi over three is equal to square root of three over two."}, {"video_title": "Cosine, sine and tangent of \u03c0 6 and \u03c0 3.mp3", "Sentence": "If this is one, this is 1 1\u20442, this is one, this is 1 1\u20442, this other side is going to be square root of three over two. So sine of pi over three is going to be square root of three over two. So let me write that down. Sine of pi over three is equal to square root of three over two. And these are good ones to know. I never say really memorize things. It's always good to know how to derive things in case you forget."}, {"video_title": "Cosine, sine and tangent of \u03c0 6 and \u03c0 3.mp3", "Sentence": "Sine of pi over three is equal to square root of three over two. And these are good ones to know. I never say really memorize things. It's always good to know how to derive things in case you forget. But if you have to memorize some, I would highly recommend memorizing these. And then of course, from these, we can figure out the tangent. The tangent is just going to be the sine over the cosine."}, {"video_title": "Cosine, sine and tangent of \u03c0 6 and \u03c0 3.mp3", "Sentence": "It's always good to know how to derive things in case you forget. But if you have to memorize some, I would highly recommend memorizing these. And then of course, from these, we can figure out the tangent. The tangent is just going to be the sine over the cosine. So let me write it down here. The tangent of pi over three is going to be the sine, which is square root of three over two over the cosine, which is 1 1\u20442. Got a little squinchy down there."}, {"video_title": "Cosine, sine and tangent of \u03c0 6 and \u03c0 3.mp3", "Sentence": "The tangent is just going to be the sine over the cosine. So let me write it down here. The tangent of pi over three is going to be the sine, which is square root of three over two over the cosine, which is 1 1\u20442. Got a little squinchy down there. And so this is just going to be square root of three over two times two is just going to be square root of three. So now let's just use that same logic for pi over six. And in fact, I encourage you to pause this video and see if you can do that on your own."}, {"video_title": "Cosine, sine and tangent of \u03c0 6 and \u03c0 3.mp3", "Sentence": "Got a little squinchy down there. And so this is just going to be square root of three over two times two is just going to be square root of three. So now let's just use that same logic for pi over six. And in fact, I encourage you to pause this video and see if you can do that on your own. All right, now let's draw a radius that forms a pi over six radian angle with a positive x-axis. Might look like that. So if that's going to be pi over six radians, you might imagine it's interesting to drop a perpendicular here and see what type of triangle we've constructed."}, {"video_title": "Cosine, sine and tangent of \u03c0 6 and \u03c0 3.mp3", "Sentence": "And in fact, I encourage you to pause this video and see if you can do that on your own. All right, now let's draw a radius that forms a pi over six radian angle with a positive x-axis. Might look like that. So if that's going to be pi over six radians, you might imagine it's interesting to drop a perpendicular here and see what type of triangle we've constructed. So this has length one. This is pi over six radians. This is a right angle."}, {"video_title": "Cosine, sine and tangent of \u03c0 6 and \u03c0 3.mp3", "Sentence": "So if that's going to be pi over six radians, you might imagine it's interesting to drop a perpendicular here and see what type of triangle we've constructed. So this has length one. This is pi over six radians. This is a right angle. So this again is going to follow the same pattern. This will be pi over three radians. And so the sides are exactly the exact same as this top blue triangle here."}, {"video_title": "Cosine, sine and tangent of \u03c0 6 and \u03c0 3.mp3", "Sentence": "This is a right angle. So this again is going to follow the same pattern. This will be pi over three radians. And so the sides are exactly the exact same as this top blue triangle here. So we know that this length over here is going to be 1 1\u20442. We know that this length over here is going to be square root of three over two. And that's useful because that tells us the coordinates here."}, {"video_title": "Cosine, sine and tangent of \u03c0 6 and \u03c0 3.mp3", "Sentence": "And so the sides are exactly the exact same as this top blue triangle here. So we know that this length over here is going to be 1 1\u20442. We know that this length over here is going to be square root of three over two. And that's useful because that tells us the coordinates here. The coordinates here, the x-coordinate of this point where the radius intersects the unit circle is square root of three over two. And then the y-coordinate is 1 1\u20442. And that immediately tells us the cosine and the sine of pi over six."}, {"video_title": "Cosine, sine and tangent of \u03c0 6 and \u03c0 3.mp3", "Sentence": "And that's useful because that tells us the coordinates here. The coordinates here, the x-coordinate of this point where the radius intersects the unit circle is square root of three over two. And then the y-coordinate is 1 1\u20442. And that immediately tells us the cosine and the sine of pi over six. Let's just write it down. So this tells us that cosine of pi over six is equal to square root of three over two. And sine of pi over six is equal to 1 1\u20442."}, {"video_title": "Cosine, sine and tangent of \u03c0 6 and \u03c0 3.mp3", "Sentence": "And that immediately tells us the cosine and the sine of pi over six. Let's just write it down. So this tells us that cosine of pi over six is equal to square root of three over two. And sine of pi over six is equal to 1 1\u20442. Notice, we just actually just swap these two things around because now the angle that we're taking the sine or cosine of is a different angle on a 30-60-90 triangle. But we're essentially utilizing the same side measures is one way to think about it. And then what's the tangent going to be?"}, {"video_title": "Cosine, sine and tangent of \u03c0 6 and \u03c0 3.mp3", "Sentence": "And sine of pi over six is equal to 1 1\u20442. Notice, we just actually just swap these two things around because now the angle that we're taking the sine or cosine of is a different angle on a 30-60-90 triangle. But we're essentially utilizing the same side measures is one way to think about it. And then what's the tangent going to be? I'll write it down here. The tangent of pi over six is going to be the sine over the cosine, square root of three over two. And so that's going to be equal to 1 1\u20442 times two over the square root of three, which is equal to one over the square root of three."}, {"video_title": "Cosine, sine and tangent of \u03c0 6 and \u03c0 3.mp3", "Sentence": "And then what's the tangent going to be? I'll write it down here. The tangent of pi over six is going to be the sine over the cosine, square root of three over two. And so that's going to be equal to 1 1\u20442 times two over the square root of three, which is equal to one over the square root of three. Now, some people sometimes don't like radicals in the denominator. And so you can multiply the numerator and the denominator by square root of three if you like to get something like this. You multiply the numerator and the denominator by square root of three, you get square root of three over three, which is another way of writing tangent of pi over six."}, {"video_title": "Cosine, sine and tangent of \u03c0 6 and \u03c0 3.mp3", "Sentence": "And so that's going to be equal to 1 1\u20442 times two over the square root of three, which is equal to one over the square root of three. Now, some people sometimes don't like radicals in the denominator. And so you can multiply the numerator and the denominator by square root of three if you like to get something like this. You multiply the numerator and the denominator by square root of three, you get square root of three over three, which is another way of writing tangent of pi over six. But either way, we're done. It's very useful to know the cosine, sine, and tangent of both pi over three and pi over six. And now you also know how to derive it."}, {"video_title": "Using the tangent angle addition identity Trigonometry Precalculus Khan Academy.mp3", "Sentence": "First of all, you can rewrite tangent of 13 pi over 12 as tangent of, instead of 13 pi over 12, we can express that in terms of angles where we might be able to figure out the tangents just based on other things we know about the unit circle. 13 pi over 12 is the same thing as 15 pi over 12 minus two pi over 12, which is the same thing as the tangent of five pi over four minus pi over six. Or we could even view it as plus negative pi over six. So that's my hint right over there. So pause this video and see if you can keep going with this train of reasoning to evaluate what tangent of 13 pi over 12 is without using a calculator. All right, now let's keep on going together. Well, we already know what the tangent of the sum of two angles are."}, {"video_title": "Using the tangent angle addition identity Trigonometry Precalculus Khan Academy.mp3", "Sentence": "So that's my hint right over there. So pause this video and see if you can keep going with this train of reasoning to evaluate what tangent of 13 pi over 12 is without using a calculator. All right, now let's keep on going together. Well, we already know what the tangent of the sum of two angles are. We've proven that in another video. We know that this is going to be equal to the tangent of the first of these angles, five pi over four, plus the tangent of the second angle, tangent of negative pi over six. All of that is going to be over one minus tangent of the first angle, five pi over four, times the tangent of the second angle, negative pi over six."}, {"video_title": "Using the tangent angle addition identity Trigonometry Precalculus Khan Academy.mp3", "Sentence": "Well, we already know what the tangent of the sum of two angles are. We've proven that in another video. We know that this is going to be equal to the tangent of the first of these angles, five pi over four, plus the tangent of the second angle, tangent of negative pi over six. All of that is going to be over one minus tangent of the first angle, five pi over four, times the tangent of the second angle, negative pi over six. And so now we can break out our unit circles to figure out what these things are. So I have pre-put some unit circles here. And so let's first think about what five pi over four looks like."}, {"video_title": "Using the tangent angle addition identity Trigonometry Precalculus Khan Academy.mp3", "Sentence": "All of that is going to be over one minus tangent of the first angle, five pi over four, times the tangent of the second angle, negative pi over six. And so now we can break out our unit circles to figure out what these things are. So I have pre-put some unit circles here. And so let's first think about what five pi over four looks like. Pi over four, you might already associate it with 45 degrees, that's pi over four right over there. Two pi over fours would get you here. Three pi over fours would get you there."}, {"video_title": "Using the tangent angle addition identity Trigonometry Precalculus Khan Academy.mp3", "Sentence": "And so let's first think about what five pi over four looks like. Pi over four, you might already associate it with 45 degrees, that's pi over four right over there. Two pi over fours would get you here. Three pi over fours would get you there. Four pis over four, which is the same thing as pi, gets us over there. Five pi over four would get us right about there. Now, you might already recognize the tangent of an angle as the slope of the radius."}, {"video_title": "Using the tangent angle addition identity Trigonometry Precalculus Khan Academy.mp3", "Sentence": "Three pi over fours would get you there. Four pis over four, which is the same thing as pi, gets us over there. Five pi over four would get us right about there. Now, you might already recognize the tangent of an angle as the slope of the radius. And so you might already be able to intuit that the tangent here is going to be one. But we can also break out our knowledge of triangles and the unit circle to figure this out if you didn't realize that. So what we need to do is figure out the coordinates of that point right over there."}, {"video_title": "Using the tangent angle addition identity Trigonometry Precalculus Khan Academy.mp3", "Sentence": "Now, you might already recognize the tangent of an angle as the slope of the radius. And so you might already be able to intuit that the tangent here is going to be one. But we can also break out our knowledge of triangles and the unit circle to figure this out if you didn't realize that. So what we need to do is figure out the coordinates of that point right over there. And to help us do that, we can set up a little bit of a right triangle, which you might immediately recognize is a 45, 45, 90 triangle. How do I know that? Well, five pi over four, remember we go four pi over four to get here, and then we have one more pi over four to go down here."}, {"video_title": "Using the tangent angle addition identity Trigonometry Precalculus Khan Academy.mp3", "Sentence": "So what we need to do is figure out the coordinates of that point right over there. And to help us do that, we can set up a little bit of a right triangle, which you might immediately recognize is a 45, 45, 90 triangle. How do I know that? Well, five pi over four, remember we go four pi over four to get here, and then we have one more pi over four to go down here. So this angle right over here is pi over four, or you could view it as 45 degrees. And of course, if that's 45 degrees, that's 90, then this has to be 45 degrees because they all add up to 180 degrees. And we know a triangle like this by the Pythagorean theorem."}, {"video_title": "Using the tangent angle addition identity Trigonometry Precalculus Khan Academy.mp3", "Sentence": "Well, five pi over four, remember we go four pi over four to get here, and then we have one more pi over four to go down here. So this angle right over here is pi over four, or you could view it as 45 degrees. And of course, if that's 45 degrees, that's 90, then this has to be 45 degrees because they all add up to 180 degrees. And we know a triangle like this by the Pythagorean theorem. If our hypotenuse is one, each of the other two sides is square root of two over two times the hypotenuse. So this is square root of two over two, and then this is square root of two over two. Now, if we think about the coordinates, our x-coordinate is square root of two over two in the negative direction."}, {"video_title": "Using the tangent angle addition identity Trigonometry Precalculus Khan Academy.mp3", "Sentence": "And we know a triangle like this by the Pythagorean theorem. If our hypotenuse is one, each of the other two sides is square root of two over two times the hypotenuse. So this is square root of two over two, and then this is square root of two over two. Now, if we think about the coordinates, our x-coordinate is square root of two over two in the negative direction. So our x-coordinate is negative square root of two over two, and our y-coordinate is square root of two over two going down. That's also negative square root of two over two. And the tangent is just the y-coordinate over the x-coordinate here."}, {"video_title": "Using the tangent angle addition identity Trigonometry Precalculus Khan Academy.mp3", "Sentence": "Now, if we think about the coordinates, our x-coordinate is square root of two over two in the negative direction. So our x-coordinate is negative square root of two over two, and our y-coordinate is square root of two over two going down. That's also negative square root of two over two. And the tangent is just the y-coordinate over the x-coordinate here. So the tangent is just going to be negative square root of two over two over negative square root of two over two, which is once again, one, which was our intuition. So we can write the tangent of five pi over four is equal to one. And then what about negative pi over six?"}, {"video_title": "Using the tangent angle addition identity Trigonometry Precalculus Khan Academy.mp3", "Sentence": "And the tangent is just the y-coordinate over the x-coordinate here. So the tangent is just going to be negative square root of two over two over negative square root of two over two, which is once again, one, which was our intuition. So we can write the tangent of five pi over four is equal to one. And then what about negative pi over six? Well, negative pi over six, you might recognize pi over six as being a 30-degree angle. Pi is 180 degrees, so divided by six is 30 degrees. And negative pi over six would be going 30 degrees below the positive x-axis."}, {"video_title": "Using the tangent angle addition identity Trigonometry Precalculus Khan Academy.mp3", "Sentence": "And then what about negative pi over six? Well, negative pi over six, you might recognize pi over six as being a 30-degree angle. Pi is 180 degrees, so divided by six is 30 degrees. And negative pi over six would be going 30 degrees below the positive x-axis. So it would look just like that. And as we said, this angle right over here is pi over six, which you can also view as a 30-degree angle. If we were to drop a perpendicular right over here, you might immediately recognize this as a 30, 60, 90 triangle."}, {"video_title": "Using the tangent angle addition identity Trigonometry Precalculus Khan Academy.mp3", "Sentence": "And negative pi over six would be going 30 degrees below the positive x-axis. So it would look just like that. And as we said, this angle right over here is pi over six, which you can also view as a 30-degree angle. If we were to drop a perpendicular right over here, you might immediately recognize this as a 30, 60, 90 triangle. We know if the hypotenuse is of length one, the side opposite the 30-degree side is 1 1\u20442 the hypotenuse. And then the longer non-hypotenuse side is going to be square root of three times the shorter side. So square root of three over two."}, {"video_title": "Using the tangent angle addition identity Trigonometry Precalculus Khan Academy.mp3", "Sentence": "If we were to drop a perpendicular right over here, you might immediately recognize this as a 30, 60, 90 triangle. We know if the hypotenuse is of length one, the side opposite the 30-degree side is 1 1\u20442 the hypotenuse. And then the longer non-hypotenuse side is going to be square root of three times the shorter side. So square root of three over two. And so our coordinates right over here, we are moving square root of three over two in the positive x-direction, square root of three over two. And then we are going negative 1 1\u20442 in the y-direction. So we put negative 1 1\u20442 right over there."}, {"video_title": "Using the tangent angle addition identity Trigonometry Precalculus Khan Academy.mp3", "Sentence": "So square root of three over two. And so our coordinates right over here, we are moving square root of three over two in the positive x-direction, square root of three over two. And then we are going negative 1 1\u20442 in the y-direction. So we put negative 1 1\u20442 right over there. And so now we know that the tangent of negative pi over six is going to be equal to negative 1 1\u20442 over square root of three over two, which is the same thing as negative 1 1\u20442 times, let me write it this way, negative 1 1\u20442 times two over the square root of three, which is equal to negative one over the square root of three. This right over here is one. We saw that there."}, {"video_title": "Using the tangent angle addition identity Trigonometry Precalculus Khan Academy.mp3", "Sentence": "So we put negative 1 1\u20442 right over there. And so now we know that the tangent of negative pi over six is going to be equal to negative 1 1\u20442 over square root of three over two, which is the same thing as negative 1 1\u20442 times, let me write it this way, negative 1 1\u20442 times two over the square root of three, which is equal to negative one over the square root of three. This right over here is one. We saw that there. This right over here is one. And then this right over here is negative one over the square root of three. And then this is negative one over the square root of three."}, {"video_title": "Using the tangent angle addition identity Trigonometry Precalculus Khan Academy.mp3", "Sentence": "We saw that there. This right over here is one. And then this right over here is negative one over the square root of three. And then this is negative one over the square root of three. And so I can rewrite this entire expression as being equal to one minus one over the square root of three over the square root of three, all of that over one, and then I have a negative here, negative here, so then that becomes a negative times a negative, so positive, one plus one over the square root of three. And if we multiply both the numerator and the denominator by square root of three, what we're going to get in the numerator is square root of three minus one, and then our denominator is going to be square root of three plus one. And we are done, that's tangent of 13 pi over 12 without using a calculator."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Write a trigonometric function that models the temperature, capital T, in Johannesburg, lowercase t, hours after midnight. So let's see if we can start to think about what a graph might look like of all of this. So this, let's say this is our temperature axis in Celsius degrees, so that is temperature, temperature, and I'm actually gonna first, I'm gonna do two different functions. So that's my temperature axis, and then this right over here is my time in hours. So that's lowercase t, time in hours, and let's think about the range of temperatures. So the daily low temperature's around three degrees Celsius, so let's actually, and the high is 18, so let's make this right over here 18, this right over here is three, and we can also think about the midpoint between 18 and three that we hit at both 10 a.m. and 10 p.m. 18 plus three is 21, divided by two is 10.5. So the midpoint, or we could say the midline of our trigonometric function is going to be 10.5 degrees Celsius, so let's, let me draw the midline."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So that's my temperature axis, and then this right over here is my time in hours. So that's lowercase t, time in hours, and let's think about the range of temperatures. So the daily low temperature's around three degrees Celsius, so let's actually, and the high is 18, so let's make this right over here 18, this right over here is three, and we can also think about the midpoint between 18 and three that we hit at both 10 a.m. and 10 p.m. 18 plus three is 21, divided by two is 10.5. So the midpoint, or we could say the midline of our trigonometric function is going to be 10.5 degrees Celsius, so let's, let me draw the midline. So we're gonna essentially oscillate around this right over here. We're gonna oscillate around this. The daily high is around 18 degrees Celsius."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So the midpoint, or we could say the midline of our trigonometric function is going to be 10.5 degrees Celsius, so let's, let me draw the midline. So we're gonna essentially oscillate around this right over here. We're gonna oscillate around this. The daily high is around 18 degrees Celsius. The daily high is around 18 degrees Celsius, and the daily low is around three degrees. The three degrees Celsius, just like that. So we're gonna oscillate around this midline."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "The daily high is around 18 degrees Celsius. The daily high is around 18 degrees Celsius, and the daily low is around three degrees. The three degrees Celsius, just like that. So we're gonna oscillate around this midline. We're gonna hit the lows and the highs. Now to simplify things, because we hit this 10.5 degrees at 10 a.m. and 10 p.m., to simplify this, I'm not going to tackle their question that they want immediately, the hour in terms of t hours after midnight. I'm gonna define a new function, f of t, f of lowercase t, which is equal to the temperature, temperature in Johannesburg, where we're assuming everything is in Johannesburg, temperature t hours after, I'm gonna say t hours after 10 a.m."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So we're gonna oscillate around this midline. We're gonna hit the lows and the highs. Now to simplify things, because we hit this 10.5 degrees at 10 a.m. and 10 p.m., to simplify this, I'm not going to tackle their question that they want immediately, the hour in terms of t hours after midnight. I'm gonna define a new function, f of t, f of lowercase t, which is equal to the temperature, temperature in Johannesburg, where we're assuming everything is in Johannesburg, temperature t hours after, I'm gonna say t hours after 10 a.m. The reason why I'm picking 10 a.m. is because we know that the temperature is right at the midline at 10 a.m., t hours after 10 a.m. Because if I want to graph f of t at t equals zero, that means we're at 10 a.m., that means that we're halfway between, they tell us, we're halfway between the daily low and the daily high. Now what is the period of this trigonometric function going to be? Well, after 24 hours, we're back to, we're going to be back to 10 a.m."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "I'm gonna define a new function, f of t, f of lowercase t, which is equal to the temperature, temperature in Johannesburg, where we're assuming everything is in Johannesburg, temperature t hours after, I'm gonna say t hours after 10 a.m. The reason why I'm picking 10 a.m. is because we know that the temperature is right at the midline at 10 a.m., t hours after 10 a.m. Because if I want to graph f of t at t equals zero, that means we're at 10 a.m., that means that we're halfway between, they tell us, we're halfway between the daily low and the daily high. Now what is the period of this trigonometric function going to be? Well, after 24 hours, we're back to, we're going to be back to 10 a.m. So our period is going to be 24 hours. So let me put 24 hours there, and then this is, halfway is 12 hours. So what happens after 12 hours?"}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Well, after 24 hours, we're back to, we're going to be back to 10 a.m. So our period is going to be 24 hours. So let me put 24 hours there, and then this is, halfway is 12 hours. So what happens after 12 hours? After 12 hours, we're back at 10 p.m., where we're back at the midway between our lows and our highs. And then, after 24 hours, we're back at 10 a.m. again. So those are going to be points on f of t. And now let's think about what'll happen as we go beyond, as we start at 10 a.m. and go forward."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So what happens after 12 hours? After 12 hours, we're back at 10 p.m., where we're back at the midway between our lows and our highs. And then, after 24 hours, we're back at 10 a.m. again. So those are going to be points on f of t. And now let's think about what'll happen as we go beyond, as we start at 10 a.m. and go forward. So as we go, start at 10 a.m. and go forward, they tell us that the hottest part, the hottest part, the highest temperatures are in the afternoon. The afternoon is going to be around here. So we should be going up in temperature, and the highest point is actually going to be halfway between these two."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So those are going to be points on f of t. And now let's think about what'll happen as we go beyond, as we start at 10 a.m. and go forward. So as we go, start at 10 a.m. and go forward, they tell us that the hottest part, the hottest part, the highest temperatures are in the afternoon. The afternoon is going to be around here. So we should be going up in temperature, and the highest point is actually going to be halfway between these two. So it's going to be six hours after 10 a.m., which is 4 p.m. So that's going to be the high at 4 p.m. So let me draw a curve, draw our curve like this."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So we should be going up in temperature, and the highest point is actually going to be halfway between these two. So it's going to be six hours after 10 a.m., which is 4 p.m. So that's going to be the high at 4 p.m. So let me draw a curve, draw our curve like this. So it'll look like this. And then our low, so now we're at 10 p.m. And then you go six hours after 10 p.m., you're now at 4 a.m., which is gonna be the low. This is 18 hours after 10 a.m. After 10 a.m., you're gonna be at your low temperature, roughly right over there."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So let me draw a curve, draw our curve like this. So it'll look like this. And then our low, so now we're at 10 p.m. And then you go six hours after 10 p.m., you're now at 4 a.m., which is gonna be the low. This is 18 hours after 10 a.m. After 10 a.m., you're gonna be at your low temperature, roughly right over there. And your curve will look something like this. So what would be, before we even try to model T of t, what would be an expression, and obviously we keep going, we keep going like that, and we could even go hours before 10 a.m. This is obviously, this keeps on cycling on and on and on forever."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "This is 18 hours after 10 a.m. After 10 a.m., you're gonna be at your low temperature, roughly right over there. And your curve will look something like this. So what would be, before we even try to model T of t, what would be an expression, and obviously we keep going, we keep going like that, and we could even go hours before 10 a.m. This is obviously, this keeps on cycling on and on and on forever. Now, what would be an expression for F of t? And I encourage you once again to pause the video and try to think about that. Well, one thing that you could say, well, you say this could be a sine or a cosine function."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "This is obviously, this keeps on cycling on and on and on forever. Now, what would be an expression for F of t? And I encourage you once again to pause the video and try to think about that. Well, one thing that you could say, well, you say this could be a sine or a cosine function. Actually, you could model it with either of them, but it's always easiest to do the simplest one. Which function is essentially at its midline, at its midline, when the argument to the function is zero? Well, the sine of zero is zero, and if we didn't shift this function up or down, the midline of just a sine function, without it being shifted, is zero."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Well, one thing that you could say, well, you say this could be a sine or a cosine function. Actually, you could model it with either of them, but it's always easiest to do the simplest one. Which function is essentially at its midline, at its midline, when the argument to the function is zero? Well, the sine of zero is zero, and if we didn't shift this function up or down, the midline of just a sine function, without it being shifted, is zero. So sine of zero is zero, and then sine begins to increase and oscillate like this. So it feels like sine is a good candidate to model it with. Once again, you could model it with either, but I have a feeling this is gonna be a little bit simpler."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Well, the sine of zero is zero, and if we didn't shift this function up or down, the midline of just a sine function, without it being shifted, is zero. So sine of zero is zero, and then sine begins to increase and oscillate like this. So it feels like sine is a good candidate to model it with. Once again, you could model it with either, but I have a feeling this is gonna be a little bit simpler. Now, let's think about the amplitude. Well, how much do we vary? What's our maximum variance from our midline?"}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Once again, you could model it with either, but I have a feeling this is gonna be a little bit simpler. Now, let's think about the amplitude. Well, how much do we vary? What's our maximum variance from our midline? So here we are 7.5 above our midline. Here we are 7.5 below our midline. So our amplitude is 7.5."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "What's our maximum variance from our midline? So here we are 7.5 above our midline. Here we are 7.5 below our midline. So our amplitude is 7.5. And actually, let me just do that in a different color, just so you see where things are coming from. So this is 7.5, this is 7.5. So our amplitude looks like it's 7.5."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So our amplitude is 7.5. And actually, let me just do that in a different color, just so you see where things are coming from. So this is 7.5, this is 7.5. So our amplitude looks like it's 7.5. And now, what is our period? Well, we've already talked about it. Our period is 24, 24 hours."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So our amplitude looks like it's 7.5. And now, what is our period? Well, we've already talked about it. Our period is 24, 24 hours. This distance right over here is 24 hours, which makes complete sense. After 24 hours, you're at the same point in the day. So we would divide two pi by the period, divided by 24 times t. And if you forget, hey, divide two pi by the period here, you could just remind yourself that what t value will make us go from, so when t is equal to zero, the whole argument to the sine function is going to be zero."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Our period is 24, 24 hours. This distance right over here is 24 hours, which makes complete sense. After 24 hours, you're at the same point in the day. So we would divide two pi by the period, divided by 24 times t. And if you forget, hey, divide two pi by the period here, you could just remind yourself that what t value will make us go from, so when t is equal to zero, the whole argument to the sine function is going to be zero. That's when we're over here. And then when t is equal to 24, the whole argument's going to be two pi. So we would have made one rotation around the unit circle if we think about the input into the sine function."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So we would divide two pi by the period, divided by 24 times t. And if you forget, hey, divide two pi by the period here, you could just remind yourself that what t value will make us go from, so when t is equal to zero, the whole argument to the sine function is going to be zero. That's when we're over here. And then when t is equal to 24, the whole argument's going to be two pi. So we would have made one rotation around the unit circle if we think about the input into the sine function. Now, we're almost done. If I were to just graph this, this would have a midline around zero, but we see that we've shifted everything up by 10.5. So we have to shift everything up by 10.5."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So we would have made one rotation around the unit circle if we think about the input into the sine function. Now, we're almost done. If I were to just graph this, this would have a midline around zero, but we see that we've shifted everything up by 10.5. So we have to shift everything up by 10.5. Now, this is, we've just successfully modeled it, and we could simplify a little bit. We could write this as pi over 12 instead of two pi over 24. But this right over here models the temperature in Johannesburg T hours after 10 a.m. After 10 a.m. That's not what they wanted."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So we have to shift everything up by 10.5. Now, this is, we've just successfully modeled it, and we could simplify a little bit. We could write this as pi over 12 instead of two pi over 24. But this right over here models the temperature in Johannesburg T hours after 10 a.m. After 10 a.m. That's not what they wanted. They want us to model, they want us to model the temperature T hours after midnight. So what would T of T be? We're gonna have to shift this a little bit."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "But this right over here models the temperature in Johannesburg T hours after 10 a.m. After 10 a.m. That's not what they wanted. They want us to model, they want us to model the temperature T hours after midnight. So what would T of T be? We're gonna have to shift this a little bit. So let's just think about it a little. Let me just write it out. So T of T, so T of T, this is now we're modeling T hours after midnight."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "We're gonna have to shift this a little bit. So let's just think about it a little. Let me just write it out. So T of T, so T of T, this is now we're modeling T hours after midnight. So we're gonna have the same amplitude. We're just gonna have the same variance from the midline. So it's gonna be 7.5 times sine of, actually, we do the same color so you see what I'm changing and not changing, times the sine of, instead of two pi over 24, I'll just write pi over 12."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So T of T, so T of T, this is now we're modeling T hours after midnight. So we're gonna have the same amplitude. We're just gonna have the same variance from the midline. So it's gonna be 7.5 times sine of, actually, we do the same color so you see what I'm changing and not changing, times the sine of, instead of two pi over 24, I'll just write pi over 12. Instead of writing T, I'm gonna shift T either to the right or the left. And actually, you could shift in either direction because this is a periodic function. We're gonna have to think about how much we're shifting it."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So it's gonna be 7.5 times sine of, actually, we do the same color so you see what I'm changing and not changing, times the sine of, instead of two pi over 24, I'll just write pi over 12. Instead of writing T, I'm gonna shift T either to the right or the left. And actually, you could shift in either direction because this is a periodic function. We're gonna have to think about how much we're shifting it. So T is gonna be plus or minus something right over here. I'm gonna shift it plus 10.5, plus 10.5. Now this is always a little bit, at least in my brain, I have to think about this in a lot of different ways so that I make sure that I'm shifting it in the right direction."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "We're gonna have to think about how much we're shifting it. So T is gonna be plus or minus something right over here. I'm gonna shift it plus 10.5, plus 10.5. Now this is always a little bit, at least in my brain, I have to think about this in a lot of different ways so that I make sure that I'm shifting it in the right direction. So here at 10 a.m., we were at this point. When T is equal to zero, this is zero hours after 10 a.m. But in this function, when is 10 a.m.?"}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Now this is always a little bit, at least in my brain, I have to think about this in a lot of different ways so that I make sure that I'm shifting it in the right direction. So here at 10 a.m., we were at this point. When T is equal to zero, this is zero hours after 10 a.m. But in this function, when is 10 a.m.? Well, in this function, 10 a.m., let me write it this way. 10 a.m. is 10 hours after midnight. So T, capital T of 10, this is 10 hours after midnight, should be equal to, should be equal to f of zero because here, the argument is hours after 10 a.m."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "But in this function, when is 10 a.m.? Well, in this function, 10 a.m., let me write it this way. 10 a.m. is 10 hours after midnight. So T, capital T of 10, this is 10 hours after midnight, should be equal to, should be equal to f of zero because here, the argument is hours after 10 a.m. So this is 10 a.m., this right over here represents temperature at 10 a.m. And this over here, because capital, this capital T function, this is hours after midnight. This is also temperature at 10 a.m. So we want T of 10 to be the same thing as f of zero."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So T, capital T of 10, this is 10 hours after midnight, should be equal to, should be equal to f of zero because here, the argument is hours after 10 a.m. So this is 10 a.m., this right over here represents temperature at 10 a.m. And this over here, because capital, this capital T function, this is hours after midnight. This is also temperature at 10 a.m. So we want T of 10 to be the same thing as f of zero. Or a same, another way of thinking about it, when f of zero, this whole argument is zero. So we want this whole argument to be zero when T is equal to 10. So how would we do that?"}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So we want T of 10 to be the same thing as f of zero. Or a same, another way of thinking about it, when f of zero, this whole argument is zero. So we want this whole argument to be zero when T is equal to 10. So how would we do that? Well, if this is T minus 10, notice, T of 10, you put a 10 here, this whole thing becomes zero. This whole thing becomes zero, and you're left with 10.5. And over here, f of zero, well, the same thing."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So how would we do that? Well, if this is T minus 10, notice, T of 10, you put a 10 here, this whole thing becomes zero. This whole thing becomes zero, and you're left with 10.5. And over here, f of zero, well, the same thing. This whole thing becomes zero, and all you're left with over here is 10.5. So T of 10 should be f of zero. So if we wanted to graph it, we've already answered their question."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And over here, f of zero, well, the same thing. This whole thing becomes zero, and all you're left with over here is 10.5. So T of 10 should be f of zero. So if we wanted to graph it, we've already answered their question. If we put a 10 here, the argument to the sign becomes zero. These two things are going to be equivalent. But let's actually graph this."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So if we wanted to graph it, we've already answered their question. If we put a 10 here, the argument to the sign becomes zero. These two things are going to be equivalent. But let's actually graph this. So T of 10, so if we're graphing capital T, T of 10, so this is six, 12, let's see. So this is, let me be, so 10 is going to be someplace around here. So T of 10 is going to be the same thing as f of zero."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "But let's actually graph this. So T of 10, so if we're graphing capital T, T of 10, so this is six, 12, let's see. So this is, let me be, so 10 is going to be someplace around here. So T of 10 is going to be the same thing as f of zero. So it's going to be like that. And then it's just going to, and then we've essentially just shifted everything to the right, everything to the right by 10. And that makes sense."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So T of 10 is going to be the same thing as f of zero. So it's going to be like that. And then it's just going to, and then we've essentially just shifted everything to the right, everything to the right by 10. And that makes sense. Because zero after, whatever hours you are after 10 a.m., it's going to be 10 more hours to get to that same point after midnight. So your curve is going to look, so this is going to be shifted by 10, this is going to be shifted by 10, and your curve's going to look something like, let me see, this is going to be shifted by 10, so you're going to get, this is going to be at 16 hours, so let's see, it's going to look something like that. And of course, it'll keep oscillating like that."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "The longest day of the year in Juneau, Alaska is June 21st. It's 1,096.5 minutes long. Half a year later, when the days are at their shortest, the days are about 382.5 minutes long. If it's not a leap year, the year is 365 days long. And June 21st is the 172nd day of the year. Write a trigonometric function that models the length, L, of the t-th day of the year. So it's going to be L as a function of t, assuming it's not a leap year."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "If it's not a leap year, the year is 365 days long. And June 21st is the 172nd day of the year. Write a trigonometric function that models the length, L, of the t-th day of the year. So it's going to be L as a function of t, assuming it's not a leap year. So I encourage you to pause this video and try to do this on your own before I try to work through it. So let me give a go at it. So instead of first starting at L of t, I'm going to actually start with L as a function of u, where u is another variable."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "So it's going to be L as a function of t, assuming it's not a leap year. So I encourage you to pause this video and try to do this on your own before I try to work through it. So let me give a go at it. So instead of first starting at L of t, I'm going to actually start with L as a function of u, where u is another variable. I'll just use this kind of an intermediary variable that'll help set it up in a little bit of a simpler way. Where u is days after June 21st. So let's just think about this a little bit."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "So instead of first starting at L of t, I'm going to actually start with L as a function of u, where u is another variable. I'll just use this kind of an intermediary variable that'll help set it up in a little bit of a simpler way. Where u is days after June 21st. So let's just think about this a little bit. June 21st, if we're thinking about in terms of u, u is going to be equal to 0, because it's 0 days after June 21st. But if we're thinking about in terms of t, June 21st is the 172nd day of the year. So 172."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "So let's just think about this a little bit. June 21st, if we're thinking about in terms of u, u is going to be equal to 0, because it's 0 days after June 21st. But if we're thinking about in terms of t, June 21st is the 172nd day of the year. So 172. So what's the relationship between u and t? Well, it's shifted by 172 days. u is going to be equal to t minus 172."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "So 172. So what's the relationship between u and t? Well, it's shifted by 172 days. u is going to be equal to t minus 172. Notice when t is 172, u is equal to 0. So let's figure out L of u first, and then later we can just substitute u with t minus 172. So first of all, what's happening when u is equal to 0?"}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "u is going to be equal to t minus 172. Notice when t is 172, u is equal to 0. So let's figure out L of u first, and then later we can just substitute u with t minus 172. So first of all, what's happening when u is equal to 0? Let me write all this down. So what is happening when u is equal to 0? Well, u equals 0 is June 21st, and that's the maximum point."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "So first of all, what's happening when u is equal to 0? Let me write all this down. So what is happening when u is equal to 0? Well, u equals 0 is June 21st, and that's the maximum point. So what trig function hits its maximum point when the input into the trig function is 0? Well, sine of 0 is 0, while cosine of 0 is 1. Cosine hits its maximum point."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "Well, u equals 0 is June 21st, and that's the maximum point. So what trig function hits its maximum point when the input into the trig function is 0? Well, sine of 0 is 0, while cosine of 0 is 1. Cosine hits its maximum point. So it seems a little bit easier to model this with a cosine. So it's going to be some amplitude times cosine of, and let's say I'll write some coefficient c right over here. Actually, let me just use a b, since I already used an a."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "Cosine hits its maximum point. So it seems a little bit easier to model this with a cosine. So it's going to be some amplitude times cosine of, and let's say I'll write some coefficient c right over here. Actually, let me just use a b, since I already used an a. Some coefficient over here times our u plus some constant. That'll shift the entire function up or down. So this is the form that our function of u is going to take."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "Actually, let me just use a b, since I already used an a. Some coefficient over here times our u plus some constant. That'll shift the entire function up or down. So this is the form that our function of u is going to take. And now we just have to figure out what each of these parts are. So first let's think about the amplitude and what the midline is going to be. The midline is essentially how much we're shifting the function up."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "So this is the form that our function of u is going to take. And now we just have to figure out what each of these parts are. So first let's think about the amplitude and what the midline is going to be. The midline is essentially how much we're shifting the function up. So let's get our calculator out. So the midline is going to be halfway between these two numbers, so we could say 1,096.5 plus 382.5 divided by 2 gets us to 739.5. So that's what c is equal to."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "The midline is essentially how much we're shifting the function up. So let's get our calculator out. So the midline is going to be halfway between these two numbers, so we could say 1,096.5 plus 382.5 divided by 2 gets us to 739.5. So that's what c is equal to. c is equal to 739.5. Now the amplitude is how much do we vary from that midline? So we could take 1,096 minus this, or we could take this minus 382.5."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "So that's what c is equal to. c is equal to 739.5. Now the amplitude is how much do we vary from that midline? So we could take 1,096 minus this, or we could take this minus 382.5. So let's do that. So let's take 1,096.5 minus what we just got, 739.5. And we get 357."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "So we could take 1,096 minus this, or we could take this minus 382.5. So let's do that. So let's take 1,096.5 minus what we just got, 739.5. And we get 357. So this is how much we vary from that midline. So a is equal to 357. So this right over here is equal to 357."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "And we get 357. So this is how much we vary from that midline. So a is equal to 357. So this right over here is equal to 357. So what's b going to be equal to? And for that, I always think about, well, what's the behavior of the function? What's the period of the function going to be?"}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "So this right over here is equal to 357. So what's b going to be equal to? And for that, I always think about, well, what's the behavior of the function? What's the period of the function going to be? Let me make a little table here. So when u, let's put some different inputs from u. When u is 0, we're 0 days after June 21."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "What's the period of the function going to be? Let me make a little table here. So when u, let's put some different inputs from u. When u is 0, we're 0 days after June 21. We're at our maximum point. And we already said that what we want the cosine function to evaluate to at that point is essentially we want it to evaluate as 357 times cosine of 0 plus 739.5. Now what happens?"}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "When u is 0, we're 0 days after June 21. We're at our maximum point. And we already said that what we want the cosine function to evaluate to at that point is essentially we want it to evaluate as 357 times cosine of 0 plus 739.5. Now what happens? What's a full period? Well, a full period is a year. At a year, we get to the same point in the year, which is, I guess, a little bit of common sense."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "Now what happens? What's a full period? Well, a full period is a year. At a year, we get to the same point in the year, which is, I guess, a little bit of common sense. So you go all the way to 365. When u is 365, we should have completed a period. We should be back to that maximum point."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "At a year, we get to the same point in the year, which is, I guess, a little bit of common sense. So you go all the way to 365. When u is 365, we should have completed a period. We should be back to that maximum point. So this should essentially be 357 times cosine of 2 pi. If we were just thinking in terms of a traditional trig function, if we just had a theta in here, you complete a period every 2 pi. So this should be equivalent to what I'm writing out right over here, plus 739.5."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "We should be back to that maximum point. So this should essentially be 357 times cosine of 2 pi. If we were just thinking in terms of a traditional trig function, if we just had a theta in here, you complete a period every 2 pi. So this should be equivalent to what I'm writing out right over here, plus 739.5. So one way to think about it is b times 365 should be equal to 2 pi. Notice, this is going to be b times 365. So let's write that down."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "So this should be equivalent to what I'm writing out right over here, plus 739.5. So one way to think about it is b times 365 should be equal to 2 pi. Notice, this is going to be b times 365. So let's write that down. b times 365, that's the input into the cosine function, needs to be equal to 2 pi. Or b is equal to 2 pi over 365. b is equal to 2 pi over 365. And we are almost done."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "So let's write that down. b times 365, that's the input into the cosine function, needs to be equal to 2 pi. Or b is equal to 2 pi over 365. b is equal to 2 pi over 365. And we are almost done. We figured out what a, b, and c are. Now we just have to substitute u with t minus 172 to get our function of t. So let's just do that. So we get, we deserve a little bit of a drum roll now, L of t is equal to a, which is 357, times cosine of b, 2 pi over 365, times not u, but now we're going to write it in terms of t. We want to think about day of the year, not days after June 21st."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "And we are almost done. We figured out what a, b, and c are. Now we just have to substitute u with t minus 172 to get our function of t. So let's just do that. So we get, we deserve a little bit of a drum roll now, L of t is equal to a, which is 357, times cosine of b, 2 pi over 365, times not u, but now we're going to write it in terms of t. We want to think about day of the year, not days after June 21st. So times t minus 172, and then finally plus our midline, plus 739.5. And we are done. So it seems like a very complicated expression, but if you just break it down and kind of think about it, make the point that we're talking about, the extreme point, either the minimum or the maximum, make that something, make that when the input into our function is 0 or 2 pi."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "So we get, we deserve a little bit of a drum roll now, L of t is equal to a, which is 357, times cosine of b, 2 pi over 365, times not u, but now we're going to write it in terms of t. We want to think about day of the year, not days after June 21st. So times t minus 172, and then finally plus our midline, plus 739.5. And we are done. So it seems like a very complicated expression, but if you just break it down and kind of think about it, make the point that we're talking about, the extreme point, either the minimum or the maximum, make that something, make that when the input into our function is 0 or 2 pi. 0 is actually the easiest one. And then later you can worry about the shift. Hopefully you found that helpful."}, {"video_title": "Inverse trig functions arctan Trigonometry Khan Academy.mp3", "Sentence": "This is just the same thing as saying that, look, the sine of some angle is equal to x. And we solved it in a couple of cases in the last example. So using the same pattern, let me say this, I could have also rewritten this as the inverse sine of x is equal to what? These are equivalent statements, two ways of writing the inverse sine function. This is the inverse sine function. You're not taking this to the negative one power. You're just saying the sine of what question mark, what angle is equal to x?"}, {"video_title": "Inverse trig functions arctan Trigonometry Khan Academy.mp3", "Sentence": "These are equivalent statements, two ways of writing the inverse sine function. This is the inverse sine function. You're not taking this to the negative one power. You're just saying the sine of what question mark, what angle is equal to x? We did this in the last video. So by the same pattern, if I were to walk up to you on the street and I were to say, the inverse tangent of x is equal to what? You should immediately in your head say, oh, he's just asking me, he's just saying, look, the tangent of some angle is equal to x."}, {"video_title": "Inverse trig functions arctan Trigonometry Khan Academy.mp3", "Sentence": "You're just saying the sine of what question mark, what angle is equal to x? We did this in the last video. So by the same pattern, if I were to walk up to you on the street and I were to say, the inverse tangent of x is equal to what? You should immediately in your head say, oh, he's just asking me, he's just saying, look, the tangent of some angle is equal to x. And I just need to figure out what that angle is. So let's do an example. Let's say I were to walk up to you on the street."}, {"video_title": "Inverse trig functions arctan Trigonometry Khan Academy.mp3", "Sentence": "You should immediately in your head say, oh, he's just asking me, he's just saying, look, the tangent of some angle is equal to x. And I just need to figure out what that angle is. So let's do an example. Let's say I were to walk up to you on the street. There's a lot of walking up on a lot of streets. I would write, and I would say, what is the arc tangent of minus 1? Or I could have equivalently asked you, what is the inverse tangent of minus 1?"}, {"video_title": "Inverse trig functions arctan Trigonometry Khan Academy.mp3", "Sentence": "Let's say I were to walk up to you on the street. There's a lot of walking up on a lot of streets. I would write, and I would say, what is the arc tangent of minus 1? Or I could have equivalently asked you, what is the inverse tangent of minus 1? These are equivalent questions. And what you should do is, in your head, if you don't have this memorized, you should draw the unit circle. And actually, let me just do a refresher of what tangent is even asking us."}, {"video_title": "Inverse trig functions arctan Trigonometry Khan Academy.mp3", "Sentence": "Or I could have equivalently asked you, what is the inverse tangent of minus 1? These are equivalent questions. And what you should do is, in your head, if you don't have this memorized, you should draw the unit circle. And actually, let me just do a refresher of what tangent is even asking us. The tangent of theta, this is just a straight up vanilla non-inverse function tangent, that's equal to the sine of theta over the cosine of theta. And the sine of theta is the y value on the unit circle, and the cosine of theta is the x value. And so if you draw a line, let me draw a little unit circle here."}, {"video_title": "Inverse trig functions arctan Trigonometry Khan Academy.mp3", "Sentence": "And actually, let me just do a refresher of what tangent is even asking us. The tangent of theta, this is just a straight up vanilla non-inverse function tangent, that's equal to the sine of theta over the cosine of theta. And the sine of theta is the y value on the unit circle, and the cosine of theta is the x value. And so if you draw a line, let me draw a little unit circle here. So if I have a unit circle like that, and let's say I'm at some angle, let's say that's my angle theta, and this is my y, my coordinate, xy. We know already that the y value, this is the sine of theta. Let me scroll over here."}, {"video_title": "Inverse trig functions arctan Trigonometry Khan Academy.mp3", "Sentence": "And so if you draw a line, let me draw a little unit circle here. So if I have a unit circle like that, and let's say I'm at some angle, let's say that's my angle theta, and this is my y, my coordinate, xy. We know already that the y value, this is the sine of theta. Let me scroll over here. Sine of theta. And we already know that this x value is the cosine of theta. So what's the tangent going to be?"}, {"video_title": "Inverse trig functions arctan Trigonometry Khan Academy.mp3", "Sentence": "Let me scroll over here. Sine of theta. And we already know that this x value is the cosine of theta. So what's the tangent going to be? It's going to be this distance, it's going to be this distance divided by this distance. Or from your algebra one, this might ring a bell, because we're starting at the origin from the point 0,0. This is our change in y over our change in x."}, {"video_title": "Inverse trig functions arctan Trigonometry Khan Academy.mp3", "Sentence": "So what's the tangent going to be? It's going to be this distance, it's going to be this distance divided by this distance. Or from your algebra one, this might ring a bell, because we're starting at the origin from the point 0,0. This is our change in y over our change in x. Or it's our rise over run. Or you can kind of view the tangent of theta, or it really is, as the slope of this line. The slope."}, {"video_title": "Inverse trig functions arctan Trigonometry Khan Academy.mp3", "Sentence": "This is our change in y over our change in x. Or it's our rise over run. Or you can kind of view the tangent of theta, or it really is, as the slope of this line. The slope. So you could write slope is equal to the tangent of theta. So let's just bear that in mind when we go to our example. If I'm asking you, and I'll rewrite it here, what is the inverse tangent of minus 1, and I'll keep rewriting it, or the arc tangent of minus 1."}, {"video_title": "Inverse trig functions arctan Trigonometry Khan Academy.mp3", "Sentence": "The slope. So you could write slope is equal to the tangent of theta. So let's just bear that in mind when we go to our example. If I'm asking you, and I'll rewrite it here, what is the inverse tangent of minus 1, and I'll keep rewriting it, or the arc tangent of minus 1. I'm saying, what angle gives me a slope of minus 1 on the unit circle? So let's draw the unit circle like that. And then I have my axes like that."}, {"video_title": "Inverse trig functions arctan Trigonometry Khan Academy.mp3", "Sentence": "If I'm asking you, and I'll rewrite it here, what is the inverse tangent of minus 1, and I'll keep rewriting it, or the arc tangent of minus 1. I'm saying, what angle gives me a slope of minus 1 on the unit circle? So let's draw the unit circle like that. And then I have my axes like that. And I want a slope of minus 1. A slope of minus 1 looks like this. It looks like that."}, {"video_title": "Inverse trig functions arctan Trigonometry Khan Academy.mp3", "Sentence": "And then I have my axes like that. And I want a slope of minus 1. A slope of minus 1 looks like this. It looks like that. If it was like that, it would be a slope of plus 1. So what angle is this? So in order to have a slope of minus 1, this distance is the same as this distance."}, {"video_title": "Inverse trig functions arctan Trigonometry Khan Academy.mp3", "Sentence": "It looks like that. If it was like that, it would be a slope of plus 1. So what angle is this? So in order to have a slope of minus 1, this distance is the same as this distance. And you might already recognize that this is a right angle. So these angles have to be the same. And so this has to be a 45-45-90 triangle."}, {"video_title": "Inverse trig functions arctan Trigonometry Khan Academy.mp3", "Sentence": "So in order to have a slope of minus 1, this distance is the same as this distance. And you might already recognize that this is a right angle. So these angles have to be the same. And so this has to be a 45-45-90 triangle. This is an isosceles triangle. These two have to add up to 90 and have to be the same. So this is 45-45-90."}, {"video_title": "Inverse trig functions arctan Trigonometry Khan Academy.mp3", "Sentence": "And so this has to be a 45-45-90 triangle. This is an isosceles triangle. These two have to add up to 90 and have to be the same. So this is 45-45-90. And if you know you're 45-45-90, actually you don't even have to know the sides of it. In the previous video we saw that this is going to be right here. This distance is going to be square root of 2 over 2."}, {"video_title": "Inverse trig functions arctan Trigonometry Khan Academy.mp3", "Sentence": "So this is 45-45-90. And if you know you're 45-45-90, actually you don't even have to know the sides of it. In the previous video we saw that this is going to be right here. This distance is going to be square root of 2 over 2. So this coordinate in the y direction is minus square root of 2 over 2. And this coordinate right here on the x direction is square root of 2 over 2. Because this length right there is that."}, {"video_title": "Inverse trig functions arctan Trigonometry Khan Academy.mp3", "Sentence": "This distance is going to be square root of 2 over 2. So this coordinate in the y direction is minus square root of 2 over 2. And this coordinate right here on the x direction is square root of 2 over 2. Because this length right there is that. So square root of 2 over 2 squared plus the square root of 2 over 2 squared is equal to 1 squared. But the important thing to realize is this is a 45-45-90 triangle. So this angle right here is, well if you're just looking at the triangle by itself, you would say that this is a 45 degree angle."}, {"video_title": "Inverse trig functions arctan Trigonometry Khan Academy.mp3", "Sentence": "Because this length right there is that. So square root of 2 over 2 squared plus the square root of 2 over 2 squared is equal to 1 squared. But the important thing to realize is this is a 45-45-90 triangle. So this angle right here is, well if you're just looking at the triangle by itself, you would say that this is a 45 degree angle. But since we're going clockwise below the x axis, we'll call this a minus 45 degree angle. So if the tangent of minus 45 degrees, let me write that down. So if I'm in degrees, and that tends to be how I think, so I could write the tangent of minus 45 degrees equals this negative value minus square root of 2 over 2 over square root of 2 over 2, which is equal to minus 1."}, {"video_title": "Inverse trig functions arctan Trigonometry Khan Academy.mp3", "Sentence": "So this angle right here is, well if you're just looking at the triangle by itself, you would say that this is a 45 degree angle. But since we're going clockwise below the x axis, we'll call this a minus 45 degree angle. So if the tangent of minus 45 degrees, let me write that down. So if I'm in degrees, and that tends to be how I think, so I could write the tangent of minus 45 degrees equals this negative value minus square root of 2 over 2 over square root of 2 over 2, which is equal to minus 1. Or I could write the arc tangent of minus 1 is equal to minus 45 degrees. Now if we're dealing with radians, we just have to convert this to radians. So we multiply that times, we get pi radians for every 180 degrees."}, {"video_title": "Inverse trig functions arctan Trigonometry Khan Academy.mp3", "Sentence": "So if I'm in degrees, and that tends to be how I think, so I could write the tangent of minus 45 degrees equals this negative value minus square root of 2 over 2 over square root of 2 over 2, which is equal to minus 1. Or I could write the arc tangent of minus 1 is equal to minus 45 degrees. Now if we're dealing with radians, we just have to convert this to radians. So we multiply that times, we get pi radians for every 180 degrees. The degrees cancel out. You have 45 over 180, this goes 4 times, so this is equal to, you have the minus sign, minus pi over 4 radians. So the arc tangent of minus 1 is equal to minus pi over 4."}, {"video_title": "Inverse trig functions arctan Trigonometry Khan Academy.mp3", "Sentence": "So we multiply that times, we get pi radians for every 180 degrees. The degrees cancel out. You have 45 over 180, this goes 4 times, so this is equal to, you have the minus sign, minus pi over 4 radians. So the arc tangent of minus 1 is equal to minus pi over 4. Or the inverse tangent of minus 1 is also equal to minus pi over 4. Now, you could say, look, if I'm at minus pi over 4, that's there, that's fine. This gives me a value of minus 1, because the slope of this line is minus 1."}, {"video_title": "Inverse trig functions arctan Trigonometry Khan Academy.mp3", "Sentence": "So the arc tangent of minus 1 is equal to minus pi over 4. Or the inverse tangent of minus 1 is also equal to minus pi over 4. Now, you could say, look, if I'm at minus pi over 4, that's there, that's fine. This gives me a value of minus 1, because the slope of this line is minus 1. But I could keep going around the unit circle. I could add 2 pi to this. Maybe I could add 2 pi to this, and that would also give me, if I took the tangent of that angle, it would also give me minus 1."}, {"video_title": "Inverse trig functions arctan Trigonometry Khan Academy.mp3", "Sentence": "This gives me a value of minus 1, because the slope of this line is minus 1. But I could keep going around the unit circle. I could add 2 pi to this. Maybe I could add 2 pi to this, and that would also give me, if I took the tangent of that angle, it would also give me minus 1. Or I could add 2 pi again, and it would again give me minus 1. In fact, I could go to this point right here, and the tangent would also give me minus 1, because the slope is right there. Like I said in the inverse sine video, you can't have a function that has a one-to-many mapping."}, {"video_title": "Inverse trig functions arctan Trigonometry Khan Academy.mp3", "Sentence": "Maybe I could add 2 pi to this, and that would also give me, if I took the tangent of that angle, it would also give me minus 1. Or I could add 2 pi again, and it would again give me minus 1. In fact, I could go to this point right here, and the tangent would also give me minus 1, because the slope is right there. Like I said in the inverse sine video, you can't have a function that has a one-to-many mapping. You can't, you know, tangent inverse of x can't map to a bunch of different values. It can't map to minus pi over 4. It can't map to 3 pi over 4."}, {"video_title": "Inverse trig functions arctan Trigonometry Khan Academy.mp3", "Sentence": "Like I said in the inverse sine video, you can't have a function that has a one-to-many mapping. You can't, you know, tangent inverse of x can't map to a bunch of different values. It can't map to minus pi over 4. It can't map to 3 pi over 4. I don't know. It would be, well, I'll just say 2 pi minus pi over 4, or 4 pi minus pi. It can't map to all of these different things."}, {"video_title": "Inverse trig functions arctan Trigonometry Khan Academy.mp3", "Sentence": "It can't map to 3 pi over 4. I don't know. It would be, well, I'll just say 2 pi minus pi over 4, or 4 pi minus pi. It can't map to all of these different things. So I have to constrict the range on the inverse tan function, and we'll restrict it very similarly to the way we restricted the inverse sine range. We're going to restrict it to the first and fourth quadrants. So the answer to your inverse tangent is always going to be something in these quadrants, but it can't be this point and that point, because a tangent function becomes undefined at pi over 2 and at minus pi over 2, because your slope goes vertical."}, {"video_title": "Inverse trig functions arctan Trigonometry Khan Academy.mp3", "Sentence": "It can't map to all of these different things. So I have to constrict the range on the inverse tan function, and we'll restrict it very similarly to the way we restricted the inverse sine range. We're going to restrict it to the first and fourth quadrants. So the answer to your inverse tangent is always going to be something in these quadrants, but it can't be this point and that point, because a tangent function becomes undefined at pi over 2 and at minus pi over 2, because your slope goes vertical. You start dividing. Your change in x is 0. You're dividing."}, {"video_title": "Inverse trig functions arctan Trigonometry Khan Academy.mp3", "Sentence": "So the answer to your inverse tangent is always going to be something in these quadrants, but it can't be this point and that point, because a tangent function becomes undefined at pi over 2 and at minus pi over 2, because your slope goes vertical. You start dividing. Your change in x is 0. You're dividing. Your cosine of theta goes to 0, so if you divide by that, it's undefined. So your range, so let me write this down. So if I have an inverse tangent of x, I'm going to, well, what are all the values that the tangent can take on?"}, {"video_title": "Inverse trig functions arctan Trigonometry Khan Academy.mp3", "Sentence": "You're dividing. Your cosine of theta goes to 0, so if you divide by that, it's undefined. So your range, so let me write this down. So if I have an inverse tangent of x, I'm going to, well, what are all the values that the tangent can take on? So if I have the tangent of theta is equal to x, what are all the different values that x could take on? These are all the possible values for the slope, and that slope can take on anything. So x could be anywhere between minus infinity and positive infinity."}, {"video_title": "Inverse trig functions arctan Trigonometry Khan Academy.mp3", "Sentence": "So if I have an inverse tangent of x, I'm going to, well, what are all the values that the tangent can take on? So if I have the tangent of theta is equal to x, what are all the different values that x could take on? These are all the possible values for the slope, and that slope can take on anything. So x could be anywhere between minus infinity and positive infinity. x can pretty much take on any value. But what about theta? Well, I just said, theta you can only go from minus pi over 2 all the way to pi over 2, and you can't even include pi over 2 or minus pi over 2, because then you'd be vertical."}, {"video_title": "Inverse trig functions arctan Trigonometry Khan Academy.mp3", "Sentence": "So x could be anywhere between minus infinity and positive infinity. x can pretty much take on any value. But what about theta? Well, I just said, theta you can only go from minus pi over 2 all the way to pi over 2, and you can't even include pi over 2 or minus pi over 2, because then you'd be vertical. So then you'd say you're, so if I'm just dealing with vanilla tangent, not the inverse, the domain, well, the domain of tangent can go multiple times around, so let me not make that statement. But if I want to do inverse tangent, so I don't have a one-to-many mapping, I want to cross out all of these, I'm going to restrict theta, or my range, to be greater than minus pi over 2 and less than positive pi over 2. And so if I restrict my range to this right here, and I exclude that point and that point, then I can only get one answer."}, {"video_title": "Inverse trig functions arctan Trigonometry Khan Academy.mp3", "Sentence": "Well, I just said, theta you can only go from minus pi over 2 all the way to pi over 2, and you can't even include pi over 2 or minus pi over 2, because then you'd be vertical. So then you'd say you're, so if I'm just dealing with vanilla tangent, not the inverse, the domain, well, the domain of tangent can go multiple times around, so let me not make that statement. But if I want to do inverse tangent, so I don't have a one-to-many mapping, I want to cross out all of these, I'm going to restrict theta, or my range, to be greater than minus pi over 2 and less than positive pi over 2. And so if I restrict my range to this right here, and I exclude that point and that point, then I can only get one answer. When I say tangent of what gives me a slope of minus 1, and that's the question I'm asking right there, there's only one answer, because if I keep, this one falls out of it, and obviously as I go around and around, those fall out of that valid range for theta that I was giving you. So just to kind of make sure we did it right, our answer was pi over 4. Let's see if we get that when we use our calculator."}, {"video_title": "Inverse trig functions arctan Trigonometry Khan Academy.mp3", "Sentence": "And so if I restrict my range to this right here, and I exclude that point and that point, then I can only get one answer. When I say tangent of what gives me a slope of minus 1, and that's the question I'm asking right there, there's only one answer, because if I keep, this one falls out of it, and obviously as I go around and around, those fall out of that valid range for theta that I was giving you. So just to kind of make sure we did it right, our answer was pi over 4. Let's see if we get that when we use our calculator. So the inverse tangent of minus 1 is equal to that. Let's see if that's the same thing as minus pi over 4. Minus pi over 4 is equal to that."}, {"video_title": "Introduction to the unit circle Trigonometry Khan Academy.mp3", "Sentence": "And the fact that I'm calling it a unit circle means it has a radius of one. So this length from the center, and I centered it at the origin, this length from the center to any point on the circle is of length one. So what would this coordinate be right over there, right where it intersects along the x-axis? Well, it would be, x would be one, y would be zero. What would this coordinate be up here? Well, we've gone one above the origin, but we haven't moved to the left or the right, so our x value is zero, our y value is one. What about back here?"}, {"video_title": "Introduction to the unit circle Trigonometry Khan Academy.mp3", "Sentence": "Well, it would be, x would be one, y would be zero. What would this coordinate be up here? Well, we've gone one above the origin, but we haven't moved to the left or the right, so our x value is zero, our y value is one. What about back here? Well, here, our x value is negative one, we've moved one to the left, and we haven't moved up or down, so our y value is zero. And what about down here? Well, we've gone a unit down, or one below the origin, but we haven't moved in the xy direction, so our x is zero, and our y is negative one."}, {"video_title": "Introduction to the unit circle Trigonometry Khan Academy.mp3", "Sentence": "What about back here? Well, here, our x value is negative one, we've moved one to the left, and we haven't moved up or down, so our y value is zero. And what about down here? Well, we've gone a unit down, or one below the origin, but we haven't moved in the xy direction, so our x is zero, and our y is negative one. Now, with that out of the way, I'm going to draw an angle. And then this, the way I'm gonna draw this angle, I'm gonna define a convention for positive angles. I'm gonna say a positive angle, well, the terminal, sorry, the initial side of an angle, we're always gonna do along the positive x-axis."}, {"video_title": "Introduction to the unit circle Trigonometry Khan Academy.mp3", "Sentence": "Well, we've gone a unit down, or one below the origin, but we haven't moved in the xy direction, so our x is zero, and our y is negative one. Now, with that out of the way, I'm going to draw an angle. And then this, the way I'm gonna draw this angle, I'm gonna define a convention for positive angles. I'm gonna say a positive angle, well, the terminal, sorry, the initial side of an angle, we're always gonna do along the positive x-axis. So this is the, you can kind of view it as the starting side of the angle, the initial side of an angle. And then, to draw an angle, a positive angle, the terminal side, we're going to move in a counterclockwise direction. So positive angle means we're going counterclockwise."}, {"video_title": "Introduction to the unit circle Trigonometry Khan Academy.mp3", "Sentence": "I'm gonna say a positive angle, well, the terminal, sorry, the initial side of an angle, we're always gonna do along the positive x-axis. So this is the, you can kind of view it as the starting side of the angle, the initial side of an angle. And then, to draw an angle, a positive angle, the terminal side, we're going to move in a counterclockwise direction. So positive angle means we're going counterclockwise. Counterclockwise. And this is just the convention I'm going to use, and it's also the convention that is typically used. And so you can imagine a negative angle would move in a clockwise, clockwise direction."}, {"video_title": "Introduction to the unit circle Trigonometry Khan Academy.mp3", "Sentence": "So positive angle means we're going counterclockwise. Counterclockwise. And this is just the convention I'm going to use, and it's also the convention that is typically used. And so you can imagine a negative angle would move in a clockwise, clockwise direction. So let me draw a positive angle. So a positive angle might look, might look something like this. This is the initial side."}, {"video_title": "Introduction to the unit circle Trigonometry Khan Academy.mp3", "Sentence": "And so you can imagine a negative angle would move in a clockwise, clockwise direction. So let me draw a positive angle. So a positive angle might look, might look something like this. This is the initial side. And then from that, I go in a counterclockwise direction until I get, until I measure out the angle, and then this is the terminal side. This is a positive angle theta. And what I wanna do is think about this point of intersection between the terminal side of this angle and my unit circle."}, {"video_title": "Introduction to the unit circle Trigonometry Khan Academy.mp3", "Sentence": "This is the initial side. And then from that, I go in a counterclockwise direction until I get, until I measure out the angle, and then this is the terminal side. This is a positive angle theta. And what I wanna do is think about this point of intersection between the terminal side of this angle and my unit circle. And let's just say it has the coordinates A, B. The X value where it intersects is A, the Y value where it intersects is B. And I'm also, the whole point of what I'm doing here is I'm gonna see how this unit circle might be able to help us extend our traditional definitions of trig functions."}, {"video_title": "Introduction to the unit circle Trigonometry Khan Academy.mp3", "Sentence": "And what I wanna do is think about this point of intersection between the terminal side of this angle and my unit circle. And let's just say it has the coordinates A, B. The X value where it intersects is A, the Y value where it intersects is B. And I'm also, the whole point of what I'm doing here is I'm gonna see how this unit circle might be able to help us extend our traditional definitions of trig functions. And so what I wanna do is I wanna make this theta part of a right triangle. So to make it part of a right triangle, let me drop an altitude right over here, and let me make it clear that this is a 90 degree angle. So this theta is part of this right triangle."}, {"video_title": "Introduction to the unit circle Trigonometry Khan Academy.mp3", "Sentence": "And I'm also, the whole point of what I'm doing here is I'm gonna see how this unit circle might be able to help us extend our traditional definitions of trig functions. And so what I wanna do is I wanna make this theta part of a right triangle. So to make it part of a right triangle, let me drop an altitude right over here, and let me make it clear that this is a 90 degree angle. So this theta is part of this right triangle. So let's see what we can figure out about the sides of this right triangle. So the first question I have to ask you is what is the length of the hypotenuse? What is the length of the hypotenuse of this right triangle that I have just constructed?"}, {"video_title": "Introduction to the unit circle Trigonometry Khan Academy.mp3", "Sentence": "So this theta is part of this right triangle. So let's see what we can figure out about the sides of this right triangle. So the first question I have to ask you is what is the length of the hypotenuse? What is the length of the hypotenuse of this right triangle that I have just constructed? Well, this hypotenuse is just a radius of a unit circle. The unit circle has a radius of one, so the hypotenuse has length one. Now, what is the length of this blue side right over here, this blue side?"}, {"video_title": "Introduction to the unit circle Trigonometry Khan Academy.mp3", "Sentence": "What is the length of the hypotenuse of this right triangle that I have just constructed? Well, this hypotenuse is just a radius of a unit circle. The unit circle has a radius of one, so the hypotenuse has length one. Now, what is the length of this blue side right over here, this blue side? You could view this as the opposite side to the angle. Well, this height is the exact same thing, is the exact same thing as the Y coordinate of this point of intersection. So this height right over here is going to be equal to b."}, {"video_title": "Introduction to the unit circle Trigonometry Khan Academy.mp3", "Sentence": "Now, what is the length of this blue side right over here, this blue side? You could view this as the opposite side to the angle. Well, this height is the exact same thing, is the exact same thing as the Y coordinate of this point of intersection. So this height right over here is going to be equal to b. The Y coordinate right over here is b. This height is equal to b. Now, exact same logic."}, {"video_title": "Introduction to the unit circle Trigonometry Khan Academy.mp3", "Sentence": "So this height right over here is going to be equal to b. The Y coordinate right over here is b. This height is equal to b. Now, exact same logic. What is the length of this base going to be, the base just of the right triangle? Well, this is going to be the X coordinate of this point of intersection. If you were to drop this down, this is the point X is equal to a, or this whole length between the origin and that is of length a."}, {"video_title": "Introduction to the unit circle Trigonometry Khan Academy.mp3", "Sentence": "Now, exact same logic. What is the length of this base going to be, the base just of the right triangle? Well, this is going to be the X coordinate of this point of intersection. If you were to drop this down, this is the point X is equal to a, or this whole length between the origin and that is of length a. Now that we have set that up, what is the cosine, and what is the cosine, let me use the same green, what is the cosine of my angle going to be in terms of a's and b's and any other numbers that might show up? Well, to think about that, we just need our Soh-Cah-Toa definition. That's the only one we have now."}, {"video_title": "Introduction to the unit circle Trigonometry Khan Academy.mp3", "Sentence": "If you were to drop this down, this is the point X is equal to a, or this whole length between the origin and that is of length a. Now that we have set that up, what is the cosine, and what is the cosine, let me use the same green, what is the cosine of my angle going to be in terms of a's and b's and any other numbers that might show up? Well, to think about that, we just need our Soh-Cah-Toa definition. That's the only one we have now. We are actually in the process of extending it. Soh-Cah-Toa definition of trig functions, and the Cah part is what helps us with cosine. It tells us that the cosine of an angle is equal to the length of the adjacent side over the hypotenuse."}, {"video_title": "Introduction to the unit circle Trigonometry Khan Academy.mp3", "Sentence": "That's the only one we have now. We are actually in the process of extending it. Soh-Cah-Toa definition of trig functions, and the Cah part is what helps us with cosine. It tells us that the cosine of an angle is equal to the length of the adjacent side over the hypotenuse. So what's this going to be? The length of the adjacent side, for this angle, the adjacent side has length a, so it's going to be equal to a, over what's the length of the hypotenuse? Well, that's just one."}, {"video_title": "Introduction to the unit circle Trigonometry Khan Academy.mp3", "Sentence": "It tells us that the cosine of an angle is equal to the length of the adjacent side over the hypotenuse. So what's this going to be? The length of the adjacent side, for this angle, the adjacent side has length a, so it's going to be equal to a, over what's the length of the hypotenuse? Well, that's just one. So the cosine of theta is just equal to a. Let me write this down again. So the cosine of theta is just equal to a."}, {"video_title": "Introduction to the unit circle Trigonometry Khan Academy.mp3", "Sentence": "Well, that's just one. So the cosine of theta is just equal to a. Let me write this down again. So the cosine of theta is just equal to a. It's equal to the x-coordinate of where this terminal side of the angle intersected the unit circle. Now let's think about the sine of theta. Sine of theta, and I'm going to do it in, well, let me see, I'll do it in orange."}, {"video_title": "Introduction to the unit circle Trigonometry Khan Academy.mp3", "Sentence": "So the cosine of theta is just equal to a. It's equal to the x-coordinate of where this terminal side of the angle intersected the unit circle. Now let's think about the sine of theta. Sine of theta, and I'm going to do it in, well, let me see, I'll do it in orange. So what's the sine of theta going to be? Well, we just have to look at the Soh part of our Soh-Cah-Toa definition. It tells us that sine is opposite over hypotenuse."}, {"video_title": "Introduction to the unit circle Trigonometry Khan Academy.mp3", "Sentence": "Sine of theta, and I'm going to do it in, well, let me see, I'll do it in orange. So what's the sine of theta going to be? Well, we just have to look at the Soh part of our Soh-Cah-Toa definition. It tells us that sine is opposite over hypotenuse. Well, the opposite side here has length b, and the hypotenuse has length one. So our sine of theta is equal to b. So an interesting thing, this coordinate, this point where our terminal side of our angle intersected the unit circle, that point a, b, we could also view this as a is the same thing as cosine of theta."}, {"video_title": "Introduction to the unit circle Trigonometry Khan Academy.mp3", "Sentence": "It tells us that sine is opposite over hypotenuse. Well, the opposite side here has length b, and the hypotenuse has length one. So our sine of theta is equal to b. So an interesting thing, this coordinate, this point where our terminal side of our angle intersected the unit circle, that point a, b, we could also view this as a is the same thing as cosine of theta. A is the same thing as cosine of theta, and b is the same thing as sine of theta. Well, that's interesting. That was just, we just used our Soh-Cah-Toa definition."}, {"video_title": "Introduction to the unit circle Trigonometry Khan Academy.mp3", "Sentence": "So an interesting thing, this coordinate, this point where our terminal side of our angle intersected the unit circle, that point a, b, we could also view this as a is the same thing as cosine of theta. A is the same thing as cosine of theta, and b is the same thing as sine of theta. Well, that's interesting. That was just, we just used our Soh-Cah-Toa definition. Now, can we in some way use this to extend Soh-Cah-Toa? Because Soh-Cah-Toa has a problem. It works out fine if our angle is greater than zero degrees, if we're dealing with degrees, and if it's less than 90 degrees."}, {"video_title": "Introduction to the unit circle Trigonometry Khan Academy.mp3", "Sentence": "That was just, we just used our Soh-Cah-Toa definition. Now, can we in some way use this to extend Soh-Cah-Toa? Because Soh-Cah-Toa has a problem. It works out fine if our angle is greater than zero degrees, if we're dealing with degrees, and if it's less than 90 degrees. We can always make it part of a right triangle. But Soh-Cah-Toa starts to break down as our angle is either zero, or maybe even becomes negative, or as our angle is 90 degrees or more. You can't have a right triangle with two 90 degree angles in it."}, {"video_title": "Introduction to the unit circle Trigonometry Khan Academy.mp3", "Sentence": "It works out fine if our angle is greater than zero degrees, if we're dealing with degrees, and if it's less than 90 degrees. We can always make it part of a right triangle. But Soh-Cah-Toa starts to break down as our angle is either zero, or maybe even becomes negative, or as our angle is 90 degrees or more. You can't have a right triangle with two 90 degree angles in it. It starts to break down. Let me make this clear. So sure, that's, so this is a right triangle, so the angle is pretty large."}, {"video_title": "Introduction to the unit circle Trigonometry Khan Academy.mp3", "Sentence": "You can't have a right triangle with two 90 degree angles in it. It starts to break down. Let me make this clear. So sure, that's, so this is a right triangle, so the angle is pretty large. I can make the angle even larger, and still have a right triangle even larger. But I can never get quite to 90 degrees. At 90 degrees, it's not clear that I have a right triangle anymore."}, {"video_title": "Introduction to the unit circle Trigonometry Khan Academy.mp3", "Sentence": "So sure, that's, so this is a right triangle, so the angle is pretty large. I can make the angle even larger, and still have a right triangle even larger. But I can never get quite to 90 degrees. At 90 degrees, it's not clear that I have a right triangle anymore. It all seems to break down, and especially the case what happens when I go beyond 90 degrees. So let's go, let's see if we can use what we set up here. Let's set up a new definition of our trig functions, which is really an extension of Soh-Cah-Toa, and it's consistent with Soh-Cah-Toa."}, {"video_title": "Introduction to the unit circle Trigonometry Khan Academy.mp3", "Sentence": "At 90 degrees, it's not clear that I have a right triangle anymore. It all seems to break down, and especially the case what happens when I go beyond 90 degrees. So let's go, let's see if we can use what we set up here. Let's set up a new definition of our trig functions, which is really an extension of Soh-Cah-Toa, and it's consistent with Soh-Cah-Toa. Instead of defining cosine as, oh, if I have a right triangle, and saying, okay, it's the adjacent over the hypotenuse, sine is the opposite over the hypotenuse, tangent is opposite over adjacent, why don't I just say, for any angle, I can draw it in the unit circle using this convention that I just set up, and let's just say that the cosine of our angle, the cosine of our angle is equal to the x-coordinate where we intersect, is equal to the x-coordinate, coordinate where the terminal side of our angle intersects the unit circle, where angle, I'll write the terminal side, terminal side of angle, side of angle intersects, intersects the unit, the unit circle, and why don't we define sine of theta, sine of theta to be equal to the y-coordinate, where the terminal side of the angle intersects the unit circle. So essentially, for any angle, this point is going to define cosine of theta and sine of theta, and so what would be the reasonable definition for tangent of theta? Well, tangent of theta, even with Sohcahtoa, could be defined as sine of theta over cosine of theta, which in this case is just going to be the y-coordinate where we intersect the unit circle over the x-coordinate."}, {"video_title": "Cosine equation algebraic solution set.mp3", "Sentence": "And like always, I encourage you to pause this video and see if you can have a go at this before we do it together. And a reminder, we want the entire solution set, not just one solution. All right, now let's work through this together. Some of you might recognize that it would be valuable to isolate the cosine of eight x and a good way of doing that would be first to subtract four from both sides. And then that would get us negative six times cosine of eight x. I subtracted four from the left, so that four is going to be gone. And then if I subtract four from the five, I am going to get a one there. And now I can multiply both sides of this equation by negative 1 6th."}, {"video_title": "Cosine equation algebraic solution set.mp3", "Sentence": "Some of you might recognize that it would be valuable to isolate the cosine of eight x and a good way of doing that would be first to subtract four from both sides. And then that would get us negative six times cosine of eight x. I subtracted four from the left, so that four is going to be gone. And then if I subtract four from the five, I am going to get a one there. And now I can multiply both sides of this equation by negative 1 6th. I just wanna have a one in front of the cosine, so negative 1 6th. And so this is going to be one, so I'm just gonna have cosine of eight x is equal to negative 1 6th. Now, if I just keep going, I could take the inverse cosine of negative 1 6th and whatever that is, divide by eight, I would get a solution."}, {"video_title": "Cosine equation algebraic solution set.mp3", "Sentence": "And now I can multiply both sides of this equation by negative 1 6th. I just wanna have a one in front of the cosine, so negative 1 6th. And so this is going to be one, so I'm just gonna have cosine of eight x is equal to negative 1 6th. Now, if I just keep going, I could take the inverse cosine of negative 1 6th and whatever that is, divide by eight, I would get a solution. But this is a good time to pause and to make sure that we were capturing all of the solutions. And I'll give us or I'll refresh our memories with some identities. So, and to help with these identities, I like to draw a quick unit circle."}, {"video_title": "Cosine equation algebraic solution set.mp3", "Sentence": "Now, if I just keep going, I could take the inverse cosine of negative 1 6th and whatever that is, divide by eight, I would get a solution. But this is a good time to pause and to make sure that we were capturing all of the solutions. And I'll give us or I'll refresh our memories with some identities. So, and to help with these identities, I like to draw a quick unit circle. So this is our x-axis, this is our y-axis. And so my quick hand drawn unit circle is like, might look something like this. It's not that nice looking."}, {"video_title": "Cosine equation algebraic solution set.mp3", "Sentence": "So, and to help with these identities, I like to draw a quick unit circle. So this is our x-axis, this is our y-axis. And so my quick hand drawn unit circle is like, might look something like this. It's not that nice looking. But we wanna think about all of the angles that when I take the cosine, I get to negative 1 6th. So negative 1 6th might be something like right over here. And so you can see that there might be an angle like this that would get us there."}, {"video_title": "Cosine equation algebraic solution set.mp3", "Sentence": "It's not that nice looking. But we wanna think about all of the angles that when I take the cosine, I get to negative 1 6th. So negative 1 6th might be something like right over here. And so you can see that there might be an angle like this that would get us there. So let me draw that, draw the radius. We know the cosine of an angle is the x-coordinate of where that radius that's defined by that angle, where that radius intersects the unit circle. But we also see there's another place."}, {"video_title": "Cosine equation algebraic solution set.mp3", "Sentence": "And so you can see that there might be an angle like this that would get us there. So let me draw that, draw the radius. We know the cosine of an angle is the x-coordinate of where that radius that's defined by that angle, where that radius intersects the unit circle. But we also see there's another place. If we essentially take the negative of that angle, we could go right over here, and we would also get the same cosine. So we could go to the negative of the angle, go that way. And that's where we get the identity that cosine of negative theta is equal to cosine of theta."}, {"video_title": "Cosine equation algebraic solution set.mp3", "Sentence": "But we also see there's another place. If we essentially take the negative of that angle, we could go right over here, and we would also get the same cosine. So we could go to the negative of the angle, go that way. And that's where we get the identity that cosine of negative theta is equal to cosine of theta. And so if cosine of eight x is equal to negative 1 6th, using this identity, we also know that cosine of the negative of this will also be equal to negative 1 6th. So let me write that down, cosine of negative eight x is also going to be equal to negative 1 6th. Now, already we have expanded our solution set because this is going to give us another x value that's going to get us the result that we want."}, {"video_title": "Cosine equation algebraic solution set.mp3", "Sentence": "And that's where we get the identity that cosine of negative theta is equal to cosine of theta. And so if cosine of eight x is equal to negative 1 6th, using this identity, we also know that cosine of the negative of this will also be equal to negative 1 6th. So let me write that down, cosine of negative eight x is also going to be equal to negative 1 6th. Now, already we have expanded our solution set because this is going to give us another x value that's going to get us the result that we want. But are we done? Well, the other thing to realize is, let's say I have some angle here, where if I take the cosine, I get to negative 1 6th. But then if I add two pi again, I'm gonna get to the same place."}, {"video_title": "Cosine equation algebraic solution set.mp3", "Sentence": "Now, already we have expanded our solution set because this is going to give us another x value that's going to get us the result that we want. But are we done? Well, the other thing to realize is, let's say I have some angle here, where if I take the cosine, I get to negative 1 6th. But then if I add two pi again, I'm gonna get to the same place. And the cosine is once again going to be negative 1 6th. And I could add two pi again. I can essentially add two pi an arbitrary integer number of times."}, {"video_title": "Cosine equation algebraic solution set.mp3", "Sentence": "But then if I add two pi again, I'm gonna get to the same place. And the cosine is once again going to be negative 1 6th. And I could add two pi again. I can essentially add two pi an arbitrary integer number of times. So I could rewrite this right over here as cosine, instead of just eight x, it's eight x plus an integer multiple of two pi. That's also going to be equal to negative 1 6th. And similarly for negative eight x, I could say cosine of negative eight x plus an integer multiple of two pi."}, {"video_title": "Cosine equation algebraic solution set.mp3", "Sentence": "I can essentially add two pi an arbitrary integer number of times. So I could rewrite this right over here as cosine, instead of just eight x, it's eight x plus an integer multiple of two pi. That's also going to be equal to negative 1 6th. And similarly for negative eight x, I could say cosine of negative eight x plus an integer multiple of two pi. And is going to be some integer in both of these situations. That's also going to get us to negative 1 6th. And so now we can feel pretty good that we're capturing all of the solutions when we solve for x."}, {"video_title": "Cosine equation algebraic solution set.mp3", "Sentence": "And similarly for negative eight x, I could say cosine of negative eight x plus an integer multiple of two pi. And is going to be some integer in both of these situations. That's also going to get us to negative 1 6th. And so now we can feel pretty good that we're capturing all of the solutions when we solve for x. So in both of these, now let's take the inverse cosine of negative 1 6th in order to solve for x here. So if we were to take the inverse cosine of both sides, we could get that eight x plus two pi times some arbitrary integer n is equal to the inverse cosine of negative 1 6th. And then now let's solve for x."}, {"video_title": "Cosine equation algebraic solution set.mp3", "Sentence": "And so now we can feel pretty good that we're capturing all of the solutions when we solve for x. So in both of these, now let's take the inverse cosine of negative 1 6th in order to solve for x here. So if we were to take the inverse cosine of both sides, we could get that eight x plus two pi times some arbitrary integer n is equal to the inverse cosine of negative 1 6th. And then now let's solve for x. We can subtract two pi n from both sides. So we could get eight x is equal to the inverse cosine of negative 1 6th minus two pi n. Now it's interesting to note that the sign on this two pi n term actually doesn't matter so much because n could be a negative integer, but I'll just stick with this negative two pi n. And so if we wanted to solve for x, we just divide both sides by eight. We get x is equal to 1 8th times the inverse cosine of negative 1 6th minus pi over four n. And now we can do the exact same thing in the other scenario."}, {"video_title": "Cosine equation algebraic solution set.mp3", "Sentence": "And then now let's solve for x. We can subtract two pi n from both sides. So we could get eight x is equal to the inverse cosine of negative 1 6th minus two pi n. Now it's interesting to note that the sign on this two pi n term actually doesn't matter so much because n could be a negative integer, but I'll just stick with this negative two pi n. And so if we wanted to solve for x, we just divide both sides by eight. We get x is equal to 1 8th times the inverse cosine of negative 1 6th minus pi over four n. And now we can do the exact same thing in the other scenario. I'll call this the yellow scenario, where if I take the inverse cosine, I get negative eight x plus two pi n is equal to the inverse cosine of negative 1 6th. And now I can subtract two pi n from both sides. So I get negative eight x is equal to inverse cosine of negative 1 6th minus two pi n. Now I can multiply both sides by negative 1 8th or divide both sides by negative eight, and I get x is equal to negative 1 8th times the inverse cosine of negative 1 6th plus pi over four n. So I will stop here for this video, where at least algebraically, we know the solution set."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "We're asked to graph the function y is equal to 2 sine of negative x on the interval, the closed interval, so it includes the endpoints, negative 2 pi to 2 pi. So to do this, I'm first going to graph the function y is equal to sine of x and then think about how it's changed by the 2 and this negative in front of the x right over here. So let's do y equal to sine of x first. So let me draw our x-axis. Let me draw the y-axis. Pretty straightforward. And we care about it between negative 2 pi and 2 pi."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "So let me draw our x-axis. Let me draw the y-axis. Pretty straightforward. And we care about it between negative 2 pi and 2 pi. So let's say that this is negative 2 pi, and then this right over here would be negative pi. This, of course, is 0. Then this is positive pi, and then this right over here is 2 pi again."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "And we care about it between negative 2 pi and 2 pi. So let's say that this is negative 2 pi, and then this right over here would be negative pi. This, of course, is 0. Then this is positive pi, and then this right over here is 2 pi again. This right over here is 2 pi, and then this could be 1, this could be 2, this could be negative 1, and this could be negative 2. And let me copy this thing so I can use it for later when I adjust this graph. So let me copy."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "Then this is positive pi, and then this right over here is 2 pi again. This right over here is 2 pi, and then this could be 1, this could be 2, this could be negative 1, and this could be negative 2. And let me copy this thing so I can use it for later when I adjust this graph. So let me copy. All right. So let's think about sine of x. So what happens when sine is 0?"}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "So let me copy. All right. So let's think about sine of x. So what happens when sine is 0? When sine is 0, or sorry, when x is 0, sine of 0 is 0. And I'll draw a little unit circle here for reference. This is what I like to do in my head as I'm trying to figure out the value of the trigonometric functions."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "So what happens when sine is 0? When sine is 0, or sorry, when x is 0, sine of 0 is 0. And I'll draw a little unit circle here for reference. This is what I like to do in my head as I'm trying to figure out the value of the trigonometric functions. So this is x, this is y. Draw a unit circle. And remember, over here, x is referring to the angle."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "This is what I like to do in my head as I'm trying to figure out the value of the trigonometric functions. So this is x, this is y. Draw a unit circle. And remember, over here, x is referring to the angle. So that's my unit circle, radius 1. So when the angle is 0, sine is going to be the y-coordinate here. So sine of 0 is 0."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "And remember, over here, x is referring to the angle. So that's my unit circle, radius 1. So when the angle is 0, sine is going to be the y-coordinate here. So sine of 0 is 0. When, as sine increases, we go up all the way to sine of pi over 2 is 1. So sine of pi over 2 is going to get you to 1. Then you go sine of pi gets you to 0."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "So sine of 0 is 0. When, as sine increases, we go up all the way to sine of pi over 2 is 1. So sine of pi over 2 is going to get you to 1. Then you go sine of pi gets you to 0. Sine of 3 pi over 2 gets you to negative 1. And then sine of 2 pi gets you back to 0. So if I were to graph it, it looks something like this."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "Then you go sine of pi gets you to 0. Sine of 3 pi over 2 gets you to negative 1. And then sine of 2 pi gets you back to 0. So if I were to graph it, it looks something like this. So this is between 0 and 2 pi. It looks something like that. And we also want to go in the negative direction."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "So if I were to graph it, it looks something like this. So this is between 0 and 2 pi. It looks something like that. And we also want to go in the negative direction. So as we go in the negative direction, so sine of negative pi over 2 is negative 1. Negative pi over 2 is negative 1. Then you go back to negative pi, go back to 0."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "And we also want to go in the negative direction. So as we go in the negative direction, so sine of negative pi over 2 is negative 1. Negative pi over 2 is negative 1. Then you go back to negative pi, go back to 0. Negative 3 pi over 2, you're going all the way around like that. That gets you back to sine is equal to 1. So sine is equal to 1."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "Then you go back to negative pi, go back to 0. Negative 3 pi over 2, you're going all the way around like that. That gets you back to sine is equal to 1. So sine is equal to 1. And then you go 2 pi, sine is back, is equaling 0. So the curve will look something like this in the negative. So as we go from between 0 and negative 2 pi."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "So sine is equal to 1. And then you go 2 pi, sine is back, is equaling 0. So the curve will look something like this in the negative. So as we go from between 0 and negative 2 pi. This is consistent with everything else that we know about sine. The period of sine of x, well you see here you have a coefficient of 1 here. So the period is just going to be 2 pi over the absolute value of 1, which is a little bit obvious."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "So as we go from between 0 and negative 2 pi. This is consistent with everything else that we know about sine. The period of sine of x, well you see here you have a coefficient of 1 here. So the period is just going to be 2 pi over the absolute value of 1, which is a little bit obvious. It's just 2 pi. Or you just see here that the period was 2 pi. It took 2 pi length to do one of our smallest repeating pattern right over here."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "So the period is just going to be 2 pi over the absolute value of 1, which is a little bit obvious. It's just 2 pi. Or you just see here that the period was 2 pi. It took 2 pi length to do one of our smallest repeating pattern right over here. And what is the amplitude? Well, we vary between 1 and negative 1. The total difference between the minimum and the maximum is 2."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "It took 2 pi length to do one of our smallest repeating pattern right over here. And what is the amplitude? Well, we vary between 1 and negative 1. The total difference between the minimum and the maximum is 2. Half of that is 1. Or another way of thinking about it, we vary 1 from our middle point. So that was pretty straightforward."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "The total difference between the minimum and the maximum is 2. Half of that is 1. Or another way of thinking about it, we vary 1 from our middle point. So that was pretty straightforward. Let's change it up a little bit. Now let's graph y is equal to 2 sine of x. So let me put my little axes there."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "So that was pretty straightforward. Let's change it up a little bit. Now let's graph y is equal to 2 sine of x. So let me put my little axes there. I want to do it right under it. And so what is going to happen now that we have y is equal to 2 sine of x? How is the graph going to change?"}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "So let me put my little axes there. I want to do it right under it. And so what is going to happen now that we have y is equal to 2 sine of x? How is the graph going to change? Well, all we did is we multiplied this function by 2. So whatever values it takes on, it's going to be twice as high now. So 2 times 0 is 0."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "How is the graph going to change? Well, all we did is we multiplied this function by 2. So whatever values it takes on, it's going to be twice as high now. So 2 times 0 is 0. 2 times 1 is now 2. 2 times 1 is 2. 2 times 0 is 2."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "So 2 times 0 is 0. 2 times 1 is now 2. 2 times 1 is 2. 2 times 0 is 2. That's at pi over 2. 2 times 0 is 0. 2 times negative 1 is negative 2."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "2 times 0 is 2. That's at pi over 2. 2 times 0 is 0. 2 times negative 1 is negative 2. 2 times 0 is 0. So it looks something like this between 0 and 2 pi. It looks something like that."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "2 times negative 1 is negative 2. 2 times 0 is 0. So it looks something like this between 0 and 2 pi. It looks something like that. And we could keep going in the negative direction. 2 times negative 1 is negative 2. 2 times 0 is 0."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "It looks something like that. And we could keep going in the negative direction. 2 times negative 1 is negative 2. 2 times 0 is 0. 2 times 1 is positive 2. 2 times 0 is 0. So in the negative direction, it looks something like that."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "2 times 0 is 0. 2 times 1 is positive 2. 2 times 0 is 0. So in the negative direction, it looks something like that. My best attempt to draw a relatively smooth curve. Hopefully you get the idea. So it will look something like that."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "So in the negative direction, it looks something like that. My best attempt to draw a relatively smooth curve. Hopefully you get the idea. So it will look something like that. So what just happened? Well, the difference between the minimum value and the maximum value just increased by a factor of 2. The total difference is 4."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "So it will look something like that. So what just happened? Well, the difference between the minimum value and the maximum value just increased by a factor of 2. The total difference is 4. Half of that difference is now 2. So what is the amplitude here? Well, the amplitude is 2."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "The total difference is 4. Half of that difference is now 2. So what is the amplitude here? Well, the amplitude is 2. So the amplitude, you could view it as the absolute value of this thing. The absolute value of 2 is now equal to 2. And it's common sense."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "Well, the amplitude is 2. So the amplitude, you could view it as the absolute value of this thing. The absolute value of 2 is now equal to 2. And it's common sense. The amplitude here was 1. And now you're swaying from that middle position twice as far because you're multiplying by 2. Now let's go back to sine of x."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "And it's common sense. The amplitude here was 1. And now you're swaying from that middle position twice as far because you're multiplying by 2. Now let's go back to sine of x. And let's change it in a different way. Let's graph sine of negative x. So let me once again put some graph paper here."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "Now let's go back to sine of x. And let's change it in a different way. Let's graph sine of negative x. So let me once again put some graph paper here. And now my goal is to graph y is equal to sine of negative x. So for at least the time being, I got rid of that 2 there. And I'm just going straight from sine of x to sine of negative x."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "So let me once again put some graph paper here. And now my goal is to graph y is equal to sine of negative x. So for at least the time being, I got rid of that 2 there. And I'm just going straight from sine of x to sine of negative x. So let's think about how the values are going to work out. So when x is 0, this is still going to be sine of 0, which is equal to 0. But then what happens as x increases?"}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "And I'm just going straight from sine of x to sine of negative x. So let's think about how the values are going to work out. So when x is 0, this is still going to be sine of 0, which is equal to 0. But then what happens as x increases? What happens when x is pi over 2? When x is pi over 2, the angle that we're inputting into sine, we're going to have to multiply it by this negative. So when x is pi over 2, we're really taking sine of negative pi over 2."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "But then what happens as x increases? What happens when x is pi over 2? When x is pi over 2, the angle that we're inputting into sine, we're going to have to multiply it by this negative. So when x is pi over 2, we're really taking sine of negative pi over 2. Well, what's sine of negative pi over 2? Well, we could see that right over here. It's negative 1."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "So when x is pi over 2, we're really taking sine of negative pi over 2. Well, what's sine of negative pi over 2? Well, we could see that right over here. It's negative 1. And then when x is equal to pi, well, sine of negative pi we already see is 0. When x is 3 pi over 2, well, it's going to be sine of negative 3 pi over 2, which is 1. And once again, when x is 2 pi, it's going to be sine of negative 2 pi."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "It's negative 1. And then when x is equal to pi, well, sine of negative pi we already see is 0. When x is 3 pi over 2, well, it's going to be sine of negative 3 pi over 2, which is 1. And once again, when x is 2 pi, it's going to be sine of negative 2 pi. Sine of negative 2 pi is 0. So notice what was happening as I was trying to graph between 0 and 2 pi. I kept referring to the points in the negative direction."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "And once again, when x is 2 pi, it's going to be sine of negative 2 pi. Sine of negative 2 pi is 0. So notice what was happening as I was trying to graph between 0 and 2 pi. I kept referring to the points in the negative direction. So you can imagine taking this negative side right over here, between 0 and negative 2 pi, and then flipping it over to get this one right over here. That's what that negative x seemed to do. And by that same logic, when we go in the negative direction, you say when x is equal to negative pi over 2, well, you have that negative in front of it, so it's going to be sine of pi over 2."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "I kept referring to the points in the negative direction. So you can imagine taking this negative side right over here, between 0 and negative 2 pi, and then flipping it over to get this one right over here. That's what that negative x seemed to do. And by that same logic, when we go in the negative direction, you say when x is equal to negative pi over 2, well, you have that negative in front of it, so it's going to be sine of pi over 2. Well, it's going to be equal to 1. And then you can flip this over the y-axis. So essentially what we have done is we have flipped it."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "And by that same logic, when we go in the negative direction, you say when x is equal to negative pi over 2, well, you have that negative in front of it, so it's going to be sine of pi over 2. Well, it's going to be equal to 1. And then you can flip this over the y-axis. So essentially what we have done is we have flipped it. We have reflected the graph of sine of x over the y-axis. This is the y-axis here, of course. So we have reflected it over the y-axis."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "So essentially what we have done is we have flipped it. We have reflected the graph of sine of x over the y-axis. This is the y-axis here, of course. So we have reflected it over the y-axis. So let me make sure I'm so it looks something like this. This is the y-axis. Hopefully you see that reflection right over there."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "So we have reflected it over the y-axis. So let me make sure I'm so it looks something like this. This is the y-axis. Hopefully you see that reflection right over there. That's what that negative x has done. So now let's think about kind of the combo, having the 2 out the front and the negative x right over there. So let me put our graph, my little axes there one more time."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "Hopefully you see that reflection right over there. That's what that negative x has done. So now let's think about kind of the combo, having the 2 out the front and the negative x right over there. So let me put our graph, my little axes there one more time. And now let's try to do what was asked of us. So I'll do it in a new color. I'll do it in blue."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "So let me put our graph, my little axes there one more time. And now let's try to do what was asked of us. So I'll do it in a new color. I'll do it in blue. Now let's graph. This is our y-axis. This is a y-axis."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "I'll do it in blue. Now let's graph. This is our y-axis. This is a y-axis. Let's graph y is equal to 2 times sine of negative x. So based on everything we've done, how will this look? What are the transformations we would do if we're going from original sine of x to y is equal to 2 sine of negative x?"}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "This is a y-axis. Let's graph y is equal to 2 times sine of negative x. So based on everything we've done, how will this look? What are the transformations we would do if we're going from original sine of x to y is equal to 2 sine of negative x? Well, there's two ways you could think about it. You could either take 2 sine of x. So here we multiplied by 2 to get double the amplitude."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "What are the transformations we would do if we're going from original sine of x to y is equal to 2 sine of negative x? Well, there's two ways you could think about it. You could either take 2 sine of x. So here we multiplied by 2 to get double the amplitude. And you could say, well, I'm going to now flip it over to get the negative sine of x. And so if you did that, you would get, so let me make it clear what I'm flipping. So if I took between 0 and negative 2 pi and I were to flip it over, what used to be here, you flip it over, you reflect it over the y-axis, and you now have, so it'll go negative first."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "So here we multiplied by 2 to get double the amplitude. And you could say, well, I'm going to now flip it over to get the negative sine of x. And so if you did that, you would get, so let me make it clear what I'm flipping. So if I took between 0 and negative 2 pi and I were to flip it over, what used to be here, you flip it over, you reflect it over the y-axis, and you now have, so it'll go negative first. Then it'll go back to 0. Then it'll go positive. And then you get right over there."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "So if I took between 0 and negative 2 pi and I were to flip it over, what used to be here, you flip it over, you reflect it over the y-axis, and you now have, so it'll go negative first. Then it'll go back to 0. Then it'll go positive. And then you get right over there. So all I did to go from 2 sine of x to 2 sine of negative x is I just reflected over the y-axis. And then, of course, what is between 0 and negative 2 pi? You just have to look between 0 and 2 pi."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "And then you get right over there. So all I did to go from 2 sine of x to 2 sine of negative x is I just reflected over the y-axis. And then, of course, what is between 0 and negative 2 pi? You just have to look between 0 and 2 pi. So now it's going to go up and down. Let me make it a little bit better, draw it a little bit neater. And then down and up."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "You just have to look between 0 and 2 pi. So now it's going to go up and down. Let me make it a little bit better, draw it a little bit neater. And then down and up. So it was a reflection of what was between 0 and 2 pi. So you see that right over here. Or if you start with sine of negative x and you go to 2 sine of negative x, notice what happens between sine of negative x and 2 sine of negative x."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "And then down and up. So it was a reflection of what was between 0 and 2 pi. So you see that right over here. Or if you start with sine of negative x and you go to 2 sine of negative x, notice what happens between sine of negative x and 2 sine of negative x. What's the difference between this graph and this graph? Well, we just have twice the amplitude. We're multiplying this one by 2, and so you get twice the amplitude."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "Or if you start with sine of negative x and you go to 2 sine of negative x, notice what happens between sine of negative x and 2 sine of negative x. What's the difference between this graph and this graph? Well, we just have twice the amplitude. We're multiplying this one by 2, and so you get twice the amplitude. So the last thought or question I have for you is how does the period of 2 sine of negative x relate to the period of sine of x? Well, there's two ways to think about it. Actually, I'll let you think about that for a second."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "We're multiplying this one by 2, and so you get twice the amplitude. So the last thought or question I have for you is how does the period of 2 sine of negative x relate to the period of sine of x? Well, there's two ways to think about it. Actually, I'll let you think about that for a second. Well, there's two ways to think about it. You could refer to the graphs right over here, or you could think about the formula, which might be a little bit intuitive right now. If you wanted to refer to the kind of classical formula, you would say the period is just going to be 2 pi, and you divide by the absolute value of the coefficient to figure out how much faster are you going to get to 2 pi right over here."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "Actually, I'll let you think about that for a second. Well, there's two ways to think about it. You could refer to the graphs right over here, or you could think about the formula, which might be a little bit intuitive right now. If you wanted to refer to the kind of classical formula, you would say the period is just going to be 2 pi, and you divide by the absolute value of the coefficient to figure out how much faster are you going to get to 2 pi right over here. So the absolute value of negative x, or the absolute value of the negative 1, is just 1. So you get 2 pi. So you have the exact same period as the period of sine of x."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "If you wanted to refer to the kind of classical formula, you would say the period is just going to be 2 pi, and you divide by the absolute value of the coefficient to figure out how much faster are you going to get to 2 pi right over here. So the absolute value of negative x, or the absolute value of the negative 1, is just 1. So you get 2 pi. So you have the exact same period as the period of sine of x. And if you see that, you complete one cycle every 2 pi. Now, what is the difference? Well, the period's the same, but remember, this negative x, it's not like you can just completely ignore it."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "So you have the exact same period as the period of sine of x. And if you see that, you complete one cycle every 2 pi. Now, what is the difference? Well, the period's the same, but remember, this negative x, it's not like you can just completely ignore it. It doesn't change the period, but it does change how the graph looks. When you start getting increased x's, instead of sine being positive, as it would be in the case of the traditional sine function, as x grows, you're taking the sine of negative x. You're taking the sine of a negative angle, and so that's why you start off having negative values of sine."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "Well, the period's the same, but remember, this negative x, it's not like you can just completely ignore it. It doesn't change the period, but it does change how the graph looks. When you start getting increased x's, instead of sine being positive, as it would be in the case of the traditional sine function, as x grows, you're taking the sine of negative x. You're taking the sine of a negative angle, and so that's why you start off having negative values of sine. And that's also another way, if you want to think about it, that it's the reflection over the y-axis of just sine of x. These two are reflections, and these two are reflections. This one is 2 times the amplitude as this one."}, {"video_title": "Relating trig function through angle rotations Trigonometry Khan Academy.mp3", "Sentence": "So the cosine of theta is the x, is the, let me do this in a color I haven't used before. The cosine of theta is the x-coordinate of where this terminal ray intersects the unit circle, or another way of thinking about it is, the cosine of theta is the length of what I'm drawing in purple right over here. It's this length. That length right over there is cosine of theta, and the sine of theta is the y-coordinate, or another way of thinking about it, the sine of, the sine of theta is the length of this line right over here. The how high you are above the x-axis, that is essentially the y-coordinate, and so the length of that is sine theta. And this makes sense. This is actually, this actually shows why the unit circle definition is an extension of the SOH CAH TOA definition."}, {"video_title": "Relating trig function through angle rotations Trigonometry Khan Academy.mp3", "Sentence": "That length right over there is cosine of theta, and the sine of theta is the y-coordinate, or another way of thinking about it, the sine of, the sine of theta is the length of this line right over here. The how high you are above the x-axis, that is essentially the y-coordinate, and so the length of that is sine theta. And this makes sense. This is actually, this actually shows why the unit circle definition is an extension of the SOH CAH TOA definition. Remember, SOH CAH TOA, let me write it down. SOH CAH TOA. SOH CAH, SOH CAH TOA, TOA, tells sine is opposite over hypotenuse."}, {"video_title": "Relating trig function through angle rotations Trigonometry Khan Academy.mp3", "Sentence": "This is actually, this actually shows why the unit circle definition is an extension of the SOH CAH TOA definition. Remember, SOH CAH TOA, let me write it down. SOH CAH TOA. SOH CAH, SOH CAH TOA, TOA, tells sine is opposite over hypotenuse. So if I want to do the sine of theta, what's it going to be? So if I think about the sine of theta, sine of theta by the SOH CAH TOA definition, it's going to be equal to the length of the opposite side, well we're saying that that's sine of theta, it's sine of theta over the hypotenuse. Well the hypotenuse here, this is a unit circle, so it's going to be one."}, {"video_title": "Relating trig function through angle rotations Trigonometry Khan Academy.mp3", "Sentence": "SOH CAH, SOH CAH TOA, TOA, tells sine is opposite over hypotenuse. So if I want to do the sine of theta, what's it going to be? So if I think about the sine of theta, sine of theta by the SOH CAH TOA definition, it's going to be equal to the length of the opposite side, well we're saying that that's sine of theta, it's sine of theta over the hypotenuse. Well the hypotenuse here, this is a unit circle, so it's going to be one. So this shows that this is consistent. Or another way of thinking about it, it is, or another way of thinking about it, sine of theta is equal to the opposite side over the hypotenuse. In this case, it's going to be equal to the opposite side, and what's the hypotenuse?"}, {"video_title": "Relating trig function through angle rotations Trigonometry Khan Academy.mp3", "Sentence": "Well the hypotenuse here, this is a unit circle, so it's going to be one. So this shows that this is consistent. Or another way of thinking about it, it is, or another way of thinking about it, sine of theta is equal to the opposite side over the hypotenuse. In this case, it's going to be equal to the opposite side, and what's the hypotenuse? This is a unit circle, so it's going to be one. So in this case, sine of theta is equal to the length of the opposite side. The length of the opposite side is equal to sine theta."}, {"video_title": "Relating trig function through angle rotations Trigonometry Khan Academy.mp3", "Sentence": "In this case, it's going to be equal to the opposite side, and what's the hypotenuse? This is a unit circle, so it's going to be one. So in this case, sine of theta is equal to the length of the opposite side. The length of the opposite side is equal to sine theta. And same exact logic. The cosine of theta, the cosine of theta is equal to adjacent over hypotenuse, is equal to adjacent over hypotenuse. And so that's, since the hypotenuse is equal to one, that's just the length of the adjacent side."}, {"video_title": "Relating trig function through angle rotations Trigonometry Khan Academy.mp3", "Sentence": "The length of the opposite side is equal to sine theta. And same exact logic. The cosine of theta, the cosine of theta is equal to adjacent over hypotenuse, is equal to adjacent over hypotenuse. And so that's, since the hypotenuse is equal to one, that's just the length of the adjacent side. So cosine of theta is the length of the adjacent side. So this is all a little bit of review, just showing how the unit circle definition is an extension of the Sohcahtoa definition. But now let's do something interesting."}, {"video_title": "Relating trig function through angle rotations Trigonometry Khan Academy.mp3", "Sentence": "And so that's, since the hypotenuse is equal to one, that's just the length of the adjacent side. So cosine of theta is the length of the adjacent side. So this is all a little bit of review, just showing how the unit circle definition is an extension of the Sohcahtoa definition. But now let's do something interesting. This is the angle theta. Let's think about the angle theta plus pi over two. So the angle theta plus pi over two."}, {"video_title": "Relating trig function through angle rotations Trigonometry Khan Academy.mp3", "Sentence": "But now let's do something interesting. This is the angle theta. Let's think about the angle theta plus pi over two. So the angle theta plus pi over two. So if I were to essentially add pi over two to this, I'm going to get a ray that is perpendicular to the first ray, pi over two, if we think in degrees, pi over two radians. So when I say theta plus pi over two, I'm talking in radians. Pi over two radians is equivalent to 90 degrees."}, {"video_title": "Relating trig function through angle rotations Trigonometry Khan Academy.mp3", "Sentence": "So the angle theta plus pi over two. So if I were to essentially add pi over two to this, I'm going to get a ray that is perpendicular to the first ray, pi over two, if we think in degrees, pi over two radians. So when I say theta plus pi over two, I'm talking in radians. Pi over two radians is equivalent to 90 degrees. So we're essentially adding 90 degrees to it. So this angle right over here, that angle right over here is theta plus pi over two. Now, what I want to explore in this video, and I guess this is the interesting part of the video, is can we relate sine of theta plus pi over two to somehow sine of theta or cosine of theta?"}, {"video_title": "Relating trig function through angle rotations Trigonometry Khan Academy.mp3", "Sentence": "Pi over two radians is equivalent to 90 degrees. So we're essentially adding 90 degrees to it. So this angle right over here, that angle right over here is theta plus pi over two. Now, what I want to explore in this video, and I guess this is the interesting part of the video, is can we relate sine of theta plus pi over two to somehow sine of theta or cosine of theta? And I encourage you to pause this video and try to think this through on your own before I work it out. Well, let's think about what sine of theta plus pi over two is. We know from the unit circle definition, the sine of this angle, which is theta plus pi over two, is the y-coordinate."}, {"video_title": "Relating trig function through angle rotations Trigonometry Khan Academy.mp3", "Sentence": "Now, what I want to explore in this video, and I guess this is the interesting part of the video, is can we relate sine of theta plus pi over two to somehow sine of theta or cosine of theta? And I encourage you to pause this video and try to think this through on your own before I work it out. Well, let's think about what sine of theta plus pi over two is. We know from the unit circle definition, the sine of this angle, which is theta plus pi over two, is the y-coordinate. It's that. It's this value right over here. Or another way of thinking about it, it's the length of this line in magenta."}, {"video_title": "Relating trig function through angle rotations Trigonometry Khan Academy.mp3", "Sentence": "We know from the unit circle definition, the sine of this angle, which is theta plus pi over two, is the y-coordinate. It's that. It's this value right over here. Or another way of thinking about it, it's the length of this line in magenta. This right over here is the sine of theta plus pi over two. So that right over there. Now, how does that relate to what we have over here?"}, {"video_title": "Relating trig function through angle rotations Trigonometry Khan Academy.mp3", "Sentence": "Or another way of thinking about it, it's the length of this line in magenta. This right over here is the sine of theta plus pi over two. So that right over there. Now, how does that relate to what we have over here? Well, when you look at it, it looks like we just took this triangle and we just kind of, we rotated it. We rotated it counterclockwise by 90 degrees, which is essentially what we did do, because we took this terminal side and we added 90 degrees to it, or pi over two radians. And if you want to get a little bit more rigorous about it, if this whole white angle here is theta plus pi over two, and the part that's in the first quadrant is pi over two, then this part right over here, that must be equal to theta."}, {"video_title": "Relating trig function through angle rotations Trigonometry Khan Academy.mp3", "Sentence": "Now, how does that relate to what we have over here? Well, when you look at it, it looks like we just took this triangle and we just kind of, we rotated it. We rotated it counterclockwise by 90 degrees, which is essentially what we did do, because we took this terminal side and we added 90 degrees to it, or pi over two radians. And if you want to get a little bit more rigorous about it, if this whole white angle here is theta plus pi over two, and the part that's in the first quadrant is pi over two, then this part right over here, that must be equal to theta. And if we think about it, if we try to relate this side, this side that I put in magenta, relative to this angle theta using the Sohcahtoa definition, here, relative to this angle theta in yellow, this is the adjacent side. So let's think about it a little bit. So what deals with the adjacent and the hypotenuse?"}, {"video_title": "Relating trig function through angle rotations Trigonometry Khan Academy.mp3", "Sentence": "And if you want to get a little bit more rigorous about it, if this whole white angle here is theta plus pi over two, and the part that's in the first quadrant is pi over two, then this part right over here, that must be equal to theta. And if we think about it, if we try to relate this side, this side that I put in magenta, relative to this angle theta using the Sohcahtoa definition, here, relative to this angle theta in yellow, this is the adjacent side. So let's think about it a little bit. So what deals with the adjacent and the hypotenuse? And in this case, of course, our hypotenuse has length one. This is a unit circle. Well, cosine deals with adjacent and hypotenuse."}, {"video_title": "Relating trig function through angle rotations Trigonometry Khan Academy.mp3", "Sentence": "So what deals with the adjacent and the hypotenuse? And in this case, of course, our hypotenuse has length one. This is a unit circle. Well, cosine deals with adjacent and hypotenuse. So we could say that the cosine of this theta, so the cosine of that theta, is equal to the adjacent side, the length of the adjacent side, which we already know is sine of theta plus pi over two. Let me write it this way. Sine of theta plus pi over two over the hypotenuse, over the hypotenuse, which is just one, so that doesn't change its value."}, {"video_title": "Relating trig function through angle rotations Trigonometry Khan Academy.mp3", "Sentence": "Well, cosine deals with adjacent and hypotenuse. So we could say that the cosine of this theta, so the cosine of that theta, is equal to the adjacent side, the length of the adjacent side, which we already know is sine of theta plus pi over two. Let me write it this way. Sine of theta plus pi over two over the hypotenuse, over the hypotenuse, which is just one, so that doesn't change its value. So that was pretty neat. Just like that, we were able to come up with a pretty neat relationship between cosine and sine. The cosine of theta is equal to sine of theta plus pi over two, or you could say sine of theta plus pi over two is equal to cosine of theta."}, {"video_title": "Relating trig function through angle rotations Trigonometry Khan Academy.mp3", "Sentence": "Sine of theta plus pi over two over the hypotenuse, over the hypotenuse, which is just one, so that doesn't change its value. So that was pretty neat. Just like that, we were able to come up with a pretty neat relationship between cosine and sine. The cosine of theta is equal to sine of theta plus pi over two, or you could say sine of theta plus pi over two is equal to cosine of theta. Now, what I encourage you to do is, after this video, see if you can come up with other results. Think about what happens to, how what sine of theta relates to, or what cosine of theta plus pi over two might relate to. So I encourage you to explore that on your own."}, {"video_title": "Proof Law of sines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Let's see, that's one side right there. And then I get another side here. And I'll try to make it look a little strange so you realize it can apply to any triangle. And let's say we know the following information. We know this angle. Well, actually, I'm not going to say what we know or don't know, but the law of sines is just a relationship between different angles and different sides. So let's say that this angle right here is alpha."}, {"video_title": "Proof Law of sines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And let's say we know the following information. We know this angle. Well, actually, I'm not going to say what we know or don't know, but the law of sines is just a relationship between different angles and different sides. So let's say that this angle right here is alpha. This side here is a. The length here is a. Let's say that this side here is beta."}, {"video_title": "Proof Law of sines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So let's say that this angle right here is alpha. This side here is a. The length here is a. Let's say that this side here is beta. And that the length here is b. Beta is just b with a long n there. So let's see if we can find a relationship that connects a and b and alpha and beta."}, {"video_title": "Proof Law of sines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Let's say that this side here is beta. And that the length here is b. Beta is just b with a long n there. So let's see if we can find a relationship that connects a and b and alpha and beta. So what can we do? So let me draw, and hopefully that relationship we find will be the law of sines. Otherwise, I would have to rename this video."}, {"video_title": "Proof Law of sines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So let's see if we can find a relationship that connects a and b and alpha and beta. So what can we do? So let me draw, and hopefully that relationship we find will be the law of sines. Otherwise, I would have to rename this video. So let me draw an altitude here. So I think that's the proper term. So if I just draw a line from this side coming straight down and it's going to be perpendicular to this bottom side, which I haven't labeled."}, {"video_title": "Proof Law of sines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Otherwise, I would have to rename this video. So let me draw an altitude here. So I think that's the proper term. So if I just draw a line from this side coming straight down and it's going to be perpendicular to this bottom side, which I haven't labeled. But if I have to label it, probably label it c because that's a and b. And this is going to be a 90 degree angle. And I don't know the length of that."}, {"video_title": "Proof Law of sines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So if I just draw a line from this side coming straight down and it's going to be perpendicular to this bottom side, which I haven't labeled. But if I have to label it, probably label it c because that's a and b. And this is going to be a 90 degree angle. And I don't know the length of that. I don't know anything about it. All I know is I went from this vertex and I dropped a line that's perpendicular to this other side. So what can we do with this line?"}, {"video_title": "Proof Law of sines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And I don't know the length of that. I don't know anything about it. All I know is I went from this vertex and I dropped a line that's perpendicular to this other side. So what can we do with this line? Well, let me just say that it has length x. The length of this line is x. So can we find a relationship between a, the length of this line, x, and beta?"}, {"video_title": "Proof Law of sines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So what can we do with this line? Well, let me just say that it has length x. The length of this line is x. So can we find a relationship between a, the length of this line, x, and beta? Well, sure. Let's see. Let me find an appropriate color."}, {"video_title": "Proof Law of sines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So can we find a relationship between a, the length of this line, x, and beta? Well, sure. Let's see. Let me find an appropriate color. OK, that's I think a good color. So what's a relationship? If we look at this angle right here, beta, x is opposite to it, and a is a hypotenuse if we look at this right triangle right here, right?"}, {"video_title": "Proof Law of sines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Let me find an appropriate color. OK, that's I think a good color. So what's a relationship? If we look at this angle right here, beta, x is opposite to it, and a is a hypotenuse if we look at this right triangle right here, right? So what deals with opposite and hypotenuse? Whenever we do trigonometry, we should always just write Sohcahtoa at the top of the page. So what deals with opposite and hypotenuse?"}, {"video_title": "Proof Law of sines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "If we look at this angle right here, beta, x is opposite to it, and a is a hypotenuse if we look at this right triangle right here, right? So what deals with opposite and hypotenuse? Whenever we do trigonometry, we should always just write Sohcahtoa at the top of the page. So what deals with opposite and hypotenuse? Sine, right? So, and you should probably guess that because I'm proving the law of sines. So let's see."}, {"video_title": "Proof Law of sines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So what deals with opposite and hypotenuse? Sine, right? So, and you should probably guess that because I'm proving the law of sines. So let's see. The sine of beta is equal to the opposite over the hypotenuse. So it's equal to this opposite, which is x, over the hypotenuse, which is a in this case. And if we wanted to solve for x, and I'll just do that because it'll be convenient later, we can multiply both sides of this equation by a, and you get a sine of beta is equal to x."}, {"video_title": "Proof Law of sines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So let's see. The sine of beta is equal to the opposite over the hypotenuse. So it's equal to this opposite, which is x, over the hypotenuse, which is a in this case. And if we wanted to solve for x, and I'll just do that because it'll be convenient later, we can multiply both sides of this equation by a, and you get a sine of beta is equal to x. Fair enough. That got us someplace. Well, let's see if we can find a relationship between alpha, b, and x."}, {"video_title": "Proof Law of sines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And if we wanted to solve for x, and I'll just do that because it'll be convenient later, we can multiply both sides of this equation by a, and you get a sine of beta is equal to x. Fair enough. That got us someplace. Well, let's see if we can find a relationship between alpha, b, and x. Well, similarly, if we look at this right triangle, because this is also a right triangle, of course, what deals, x here relative to alpha is also the opposite side, and b now is a hypotenuse. So we can also write that sine of alpha, let me do it in a different color, sine of alpha is equal to opposite over hypotenuse. The opposite is x, and the hypotenuse is b."}, {"video_title": "Proof Law of sines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Well, let's see if we can find a relationship between alpha, b, and x. Well, similarly, if we look at this right triangle, because this is also a right triangle, of course, what deals, x here relative to alpha is also the opposite side, and b now is a hypotenuse. So we can also write that sine of alpha, let me do it in a different color, sine of alpha is equal to opposite over hypotenuse. The opposite is x, and the hypotenuse is b. And let's solve for x again, just to do it. Multiply both sides by b, and you get b sine of alpha is equal to x. So now what do we have?"}, {"video_title": "Proof Law of sines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "The opposite is x, and the hypotenuse is b. And let's solve for x again, just to do it. Multiply both sides by b, and you get b sine of alpha is equal to x. So now what do we have? We have two different ways that we solve for this thing that I dropped down from this side, this x. We have a sine of beta is equal to x, and that b sine of alpha is equal to x. Well, if they're both equal to x, then they're both equal to each other."}, {"video_title": "Proof Law of sines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So now what do we have? We have two different ways that we solve for this thing that I dropped down from this side, this x. We have a sine of beta is equal to x, and that b sine of alpha is equal to x. Well, if they're both equal to x, then they're both equal to each other. So let me write that down. Let me write that down in a soothing color. So we know that a sine of beta is equal to x, which is also equal to b sine of alpha."}, {"video_title": "Proof Law of sines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Well, if they're both equal to x, then they're both equal to each other. So let me write that down. Let me write that down in a soothing color. So we know that a sine of beta is equal to x, which is also equal to b sine of alpha. If we divide both sides of this equation by a, what do we get? We get sine of beta, because the a on this side cancels out, is equal to b sine of alpha over a. And if we divide both sides of this equation by b, we get sine of beta over b is equal to sine of alpha over a."}, {"video_title": "Proof Law of sines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So we know that a sine of beta is equal to x, which is also equal to b sine of alpha. If we divide both sides of this equation by a, what do we get? We get sine of beta, because the a on this side cancels out, is equal to b sine of alpha over a. And if we divide both sides of this equation by b, we get sine of beta over b is equal to sine of alpha over a. So this is the law of sines. The ratio between the sine of beta and its opposite side, and it's the side that it corresponds to, this b, is equal to the ratio of the sine of alpha and its opposite side. And then a lot of times in the books, let's say if this angle was theta and this was c, then they would also write that's also equal to the sine of theta over c. And the proof of adding this here is identical."}, {"video_title": "Proof Law of sines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And if we divide both sides of this equation by b, we get sine of beta over b is equal to sine of alpha over a. So this is the law of sines. The ratio between the sine of beta and its opposite side, and it's the side that it corresponds to, this b, is equal to the ratio of the sine of alpha and its opposite side. And then a lot of times in the books, let's say if this angle was theta and this was c, then they would also write that's also equal to the sine of theta over c. And the proof of adding this here is identical. I mean, we picked b arbitrarily. b is the side we could have done the exact same thing with theta and c, but instead of dropping the altitude here, we would have had to drop one of the other altitudes, and I think you could figure out that part. But the important thing is we have this ratio."}, {"video_title": "Proof Law of sines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And then a lot of times in the books, let's say if this angle was theta and this was c, then they would also write that's also equal to the sine of theta over c. And the proof of adding this here is identical. I mean, we picked b arbitrarily. b is the side we could have done the exact same thing with theta and c, but instead of dropping the altitude here, we would have had to drop one of the other altitudes, and I think you could figure out that part. But the important thing is we have this ratio. And of course, you could have written it, since it's a ratio, you could flip both sides of the ratio, and you could write it b over the sine of b is equal to a over the sine of alpha. And this is useful, because if you know one side and its corresponding angle, the angle opposite it, that kind of opens up into that side, and say you know the other side, then you could figure out the angle that opens up into it. Or, I mean, well, if you know three of these things, you can figure out the fourth."}, {"video_title": "Proof Law of sines Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "But the important thing is we have this ratio. And of course, you could have written it, since it's a ratio, you could flip both sides of the ratio, and you could write it b over the sine of b is equal to a over the sine of alpha. And this is useful, because if you know one side and its corresponding angle, the angle opposite it, that kind of opens up into that side, and say you know the other side, then you could figure out the angle that opens up into it. Or, I mean, well, if you know three of these things, you can figure out the fourth. And that's what's useful about the law of sines. So maybe now I will do a few law of sines word problems. I'll see you in the next video."}, {"video_title": "Showing relationship between cosine and sine of complements Trigonometry Khan Academy.mp3", "Sentence": "And so we have two other angles to deal with. And what I want to explore in this video is the relationship between the sine of one of these angles and the cosine of the other, the cosine of one of these angles and the sine of the other. So to do that, let's just say that this angle, I guess we could call it angle A, let's say it's equal to theta. If this is equal to theta, if its measure is equal to theta degrees, say, what is the measure of angle B going to be? Well, the thing that will jump out at you, and we've looked at this in other problems, is the sum of the angles of a triangle are going to be 180 degrees. And this one right over here, it's a right triangle. So this right angle takes up 90 of those 180 degrees."}, {"video_title": "Showing relationship between cosine and sine of complements Trigonometry Khan Academy.mp3", "Sentence": "If this is equal to theta, if its measure is equal to theta degrees, say, what is the measure of angle B going to be? Well, the thing that will jump out at you, and we've looked at this in other problems, is the sum of the angles of a triangle are going to be 180 degrees. And this one right over here, it's a right triangle. So this right angle takes up 90 of those 180 degrees. So you have 90 degrees left. So these two are going to have to add up to 90 degrees. This one and this one, angle A and angle B, are going to be complements of each other."}, {"video_title": "Showing relationship between cosine and sine of complements Trigonometry Khan Academy.mp3", "Sentence": "So this right angle takes up 90 of those 180 degrees. So you have 90 degrees left. So these two are going to have to add up to 90 degrees. This one and this one, angle A and angle B, are going to be complements of each other. They're going to be complementary. Or another way of thinking about it is B could be written as 90 minus theta. If you add theta to 90 minus theta, you're going to get, you're going to get, or theta to 90 degrees minus theta, you're going to get 90 degrees."}, {"video_title": "Showing relationship between cosine and sine of complements Trigonometry Khan Academy.mp3", "Sentence": "This one and this one, angle A and angle B, are going to be complements of each other. They're going to be complementary. Or another way of thinking about it is B could be written as 90 minus theta. If you add theta to 90 minus theta, you're going to get, you're going to get, or theta to 90 degrees minus theta, you're going to get 90 degrees. Now why is this interesting? Well, let's think about, let's think about what the sine, let's think about what the sine of theta is equal to. Sine is opposite over hypotenuse."}, {"video_title": "Showing relationship between cosine and sine of complements Trigonometry Khan Academy.mp3", "Sentence": "If you add theta to 90 minus theta, you're going to get, you're going to get, or theta to 90 degrees minus theta, you're going to get 90 degrees. Now why is this interesting? Well, let's think about, let's think about what the sine, let's think about what the sine of theta is equal to. Sine is opposite over hypotenuse. The opposite side is BC. So this is going to be the length of BC over the hypotenuse. The hypotenuse is side AB."}, {"video_title": "Showing relationship between cosine and sine of complements Trigonometry Khan Academy.mp3", "Sentence": "Sine is opposite over hypotenuse. The opposite side is BC. So this is going to be the length of BC over the hypotenuse. The hypotenuse is side AB. So the length of BC over the length of AB. Over the length of AB. Now, what is that ratio if we were to look at this angle right over here?"}, {"video_title": "Showing relationship between cosine and sine of complements Trigonometry Khan Academy.mp3", "Sentence": "The hypotenuse is side AB. So the length of BC over the length of AB. Over the length of AB. Now, what is that ratio if we were to look at this angle right over here? Well, for angle B, BC is the adjacent side, and AB is the hypotenuse. From angle B's perspective, this is the adjacent over the hypotenuse. Now what trig ratio is adjacent over hypotenuse?"}, {"video_title": "Showing relationship between cosine and sine of complements Trigonometry Khan Academy.mp3", "Sentence": "Now, what is that ratio if we were to look at this angle right over here? Well, for angle B, BC is the adjacent side, and AB is the hypotenuse. From angle B's perspective, this is the adjacent over the hypotenuse. Now what trig ratio is adjacent over hypotenuse? Well, that's cosine. So ka toa, let me write that down. Sine, doesn't hurt."}, {"video_title": "Showing relationship between cosine and sine of complements Trigonometry Khan Academy.mp3", "Sentence": "Now what trig ratio is adjacent over hypotenuse? Well, that's cosine. So ka toa, let me write that down. Sine, doesn't hurt. Sine is opposite over hypotenuse. We see that right over there. Cosine is adjacent over hypotenuse, ka, and toa."}, {"video_title": "Showing relationship between cosine and sine of complements Trigonometry Khan Academy.mp3", "Sentence": "Sine, doesn't hurt. Sine is opposite over hypotenuse. We see that right over there. Cosine is adjacent over hypotenuse, ka, and toa. Tangent is opposite over adjacent. So from this angle's perspective, taking the length of BC, which is BC is its adjacent side, and the hypotenuse is still AB. So from this angle's perspective, this is adjacent over hypotenuse."}, {"video_title": "Showing relationship between cosine and sine of complements Trigonometry Khan Academy.mp3", "Sentence": "Cosine is adjacent over hypotenuse, ka, and toa. Tangent is opposite over adjacent. So from this angle's perspective, taking the length of BC, which is BC is its adjacent side, and the hypotenuse is still AB. So from this angle's perspective, this is adjacent over hypotenuse. Or another way of thinking about it, it's the cosine of this angle. So that's going to be equal to the cosine of 90 degrees minus theta. That's a pretty neat relationship."}, {"video_title": "Showing relationship between cosine and sine of complements Trigonometry Khan Academy.mp3", "Sentence": "So from this angle's perspective, this is adjacent over hypotenuse. Or another way of thinking about it, it's the cosine of this angle. So that's going to be equal to the cosine of 90 degrees minus theta. That's a pretty neat relationship. The sine of an angle is equal to the cosine of its complement. So one way to think about it, the sine of, we could just pick any arbitrary angle. Let's say the sine of 60 degrees is going to be equal to the cosine of what?"}, {"video_title": "Showing relationship between cosine and sine of complements Trigonometry Khan Academy.mp3", "Sentence": "That's a pretty neat relationship. The sine of an angle is equal to the cosine of its complement. So one way to think about it, the sine of, we could just pick any arbitrary angle. Let's say the sine of 60 degrees is going to be equal to the cosine of what? And I encourage you to pause the video and think about it. Well, it's going to be the cosine of 90 minus 60. It's going to be the cosine of 30 degrees."}, {"video_title": "Showing relationship between cosine and sine of complements Trigonometry Khan Academy.mp3", "Sentence": "Let's say the sine of 60 degrees is going to be equal to the cosine of what? And I encourage you to pause the video and think about it. Well, it's going to be the cosine of 90 minus 60. It's going to be the cosine of 30 degrees. 30 plus 60 is 90. And of course, you could go the other way around. You could think about the cosine of theta is going to be equal to the adjacent side to theta, to angle A, I should say."}, {"video_title": "Showing relationship between cosine and sine of complements Trigonometry Khan Academy.mp3", "Sentence": "It's going to be the cosine of 30 degrees. 30 plus 60 is 90. And of course, you could go the other way around. You could think about the cosine of theta is going to be equal to the adjacent side to theta, to angle A, I should say. So the adjacent side is right over here. That's AC. So it's going to be AC over the hypotenuse, adjacent over hypotenuse."}, {"video_title": "Showing relationship between cosine and sine of complements Trigonometry Khan Academy.mp3", "Sentence": "You could think about the cosine of theta is going to be equal to the adjacent side to theta, to angle A, I should say. So the adjacent side is right over here. That's AC. So it's going to be AC over the hypotenuse, adjacent over hypotenuse. The hypotenuse is AB. But what is this ratio from angle B's point of view? Well, the sine of angle B is going to be its opposite side, AC over the hypotenuse, AB."}, {"video_title": "Showing relationship between cosine and sine of complements Trigonometry Khan Academy.mp3", "Sentence": "So it's going to be AC over the hypotenuse, adjacent over hypotenuse. The hypotenuse is AB. But what is this ratio from angle B's point of view? Well, the sine of angle B is going to be its opposite side, AC over the hypotenuse, AB. So this right over here, from angle B's perspective, this is angle B's sine. So this is equal to the sine of 90 degrees minus theta. So the cosine of an angle is equal to the sine of its complement."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "The temperature is typically halfway between the daily high and the daily low at both 10 a.m. and 10 p.m., and the highest temperatures are in the afternoon. Write a trigonometric function that models the temperature, capital T, in Johannesburg, lowercase t, hours after midnight. So let's see if we can start to think about what a graph might look like of all of this. So this, let's say this is our temperature axis in Celsius degrees, so that is temperature, temperature, and I'm actually gonna first, I'm gonna do two different functions. So that's my temperature axis, and then this right over here is my time in hours. So that's lowercase t, time in hours, and let's think about the range of temperatures. So the daily low temperature's around three degrees Celsius, so let's actually, and the high is 18, so let's make this right over here 18, this right over here is three, and we can also think about the midpoint between 18 and three that we hit at both 10 a.m. and 10 p.m. 18 plus three is 21, divided by two is 10.5."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "So this, let's say this is our temperature axis in Celsius degrees, so that is temperature, temperature, and I'm actually gonna first, I'm gonna do two different functions. So that's my temperature axis, and then this right over here is my time in hours. So that's lowercase t, time in hours, and let's think about the range of temperatures. So the daily low temperature's around three degrees Celsius, so let's actually, and the high is 18, so let's make this right over here 18, this right over here is three, and we can also think about the midpoint between 18 and three that we hit at both 10 a.m. and 10 p.m. 18 plus three is 21, divided by two is 10.5. So the midpoint, or we could say the midline of our trigonometric function is going to be 10.5 degrees Celsius, so let's, let me draw the midline. So we're gonna essentially oscillate around this right over here. We're gonna oscillate around this."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "So the daily low temperature's around three degrees Celsius, so let's actually, and the high is 18, so let's make this right over here 18, this right over here is three, and we can also think about the midpoint between 18 and three that we hit at both 10 a.m. and 10 p.m. 18 plus three is 21, divided by two is 10.5. So the midpoint, or we could say the midline of our trigonometric function is going to be 10.5 degrees Celsius, so let's, let me draw the midline. So we're gonna essentially oscillate around this right over here. We're gonna oscillate around this. The daily high is around 18 degrees Celsius. The daily high is around 18 degrees Celsius, and the daily low is around three degrees. The three degrees Celsius, just like that."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "We're gonna oscillate around this. The daily high is around 18 degrees Celsius. The daily high is around 18 degrees Celsius, and the daily low is around three degrees. The three degrees Celsius, just like that. So we're gonna oscillate around this midline. We're gonna hit the lows and the highs. Now to simplify things, because we hit this 10.5 degrees at 10 a.m. and 10 p.m., to simplify this, I'm not going to tackle their question that they want immediately, the hour in terms of t hours after midnight."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "The three degrees Celsius, just like that. So we're gonna oscillate around this midline. We're gonna hit the lows and the highs. Now to simplify things, because we hit this 10.5 degrees at 10 a.m. and 10 p.m., to simplify this, I'm not going to tackle their question that they want immediately, the hour in terms of t hours after midnight. I'm gonna define a new function, f of t, f of lowercase t, which is equal to the temperature, temperature in Johannesburg, where we're assuming everything is in Johannesburg, temperature t hours after, I'm gonna say t hours after 10 a.m. The reason why I'm picking 10 a.m. is because we know that the temperature is right at the midline at 10 a.m., t hours after 10 a.m. Because if I want to graph f of t at t equals zero, that means we're at 10 a.m., that means that we're halfway between, they tell us, we're halfway between the daily low and the daily high. Now what is the period of this trigonometric function going to be?"}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "Now to simplify things, because we hit this 10.5 degrees at 10 a.m. and 10 p.m., to simplify this, I'm not going to tackle their question that they want immediately, the hour in terms of t hours after midnight. I'm gonna define a new function, f of t, f of lowercase t, which is equal to the temperature, temperature in Johannesburg, where we're assuming everything is in Johannesburg, temperature t hours after, I'm gonna say t hours after 10 a.m. The reason why I'm picking 10 a.m. is because we know that the temperature is right at the midline at 10 a.m., t hours after 10 a.m. Because if I want to graph f of t at t equals zero, that means we're at 10 a.m., that means that we're halfway between, they tell us, we're halfway between the daily low and the daily high. Now what is the period of this trigonometric function going to be? Well, after 24 hours, we're back to, we're going to be back to 10 a.m. So our period is going to be 24 hours. So let me put 24 hours there, and then this is, halfway is 12 hours."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "Now what is the period of this trigonometric function going to be? Well, after 24 hours, we're back to, we're going to be back to 10 a.m. So our period is going to be 24 hours. So let me put 24 hours there, and then this is, halfway is 12 hours. So what happens after 12 hours? After 12 hours, we're back at 10 p.m., where we're back at the midway between our lows and our highs. And then, after 24 hours, we're back at 10 a.m. again."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "So let me put 24 hours there, and then this is, halfway is 12 hours. So what happens after 12 hours? After 12 hours, we're back at 10 p.m., where we're back at the midway between our lows and our highs. And then, after 24 hours, we're back at 10 a.m. again. So those are going to be points on f of t. And now let's think about what'll happen as we go beyond, as we start at 10 a.m. and go forward. So as we go, start at 10 a.m. and go forward, they tell us that the hottest part, the hottest part, the highest temperatures are in the afternoon. The afternoon is going to be around here."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "And then, after 24 hours, we're back at 10 a.m. again. So those are going to be points on f of t. And now let's think about what'll happen as we go beyond, as we start at 10 a.m. and go forward. So as we go, start at 10 a.m. and go forward, they tell us that the hottest part, the hottest part, the highest temperatures are in the afternoon. The afternoon is going to be around here. So we should be going up in temperature, and the highest point is actually going to be halfway between these two. So it's going to be six hours after 10 a.m., which is 4 p.m. So that's going to be the high at 4 p.m."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "The afternoon is going to be around here. So we should be going up in temperature, and the highest point is actually going to be halfway between these two. So it's going to be six hours after 10 a.m., which is 4 p.m. So that's going to be the high at 4 p.m. So let me draw a curve, draw our curve like this. So it'll look like this. And then our low, so now we're at 10 p.m. And then you go six hours after 10 p.m., you're now at 4 a.m., which is gonna be the low."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "So that's going to be the high at 4 p.m. So let me draw a curve, draw our curve like this. So it'll look like this. And then our low, so now we're at 10 p.m. And then you go six hours after 10 p.m., you're now at 4 a.m., which is gonna be the low. This is 18 hours after 10 a.m. After 10 a.m., you're gonna be at your low temperature, roughly right over there. And your curve will look something like this. So what would be, before we even try to model T of t, what would be an expression, and obviously we keep going, we keep going like that, and we could even go hours before 10 a.m."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "And then our low, so now we're at 10 p.m. And then you go six hours after 10 p.m., you're now at 4 a.m., which is gonna be the low. This is 18 hours after 10 a.m. After 10 a.m., you're gonna be at your low temperature, roughly right over there. And your curve will look something like this. So what would be, before we even try to model T of t, what would be an expression, and obviously we keep going, we keep going like that, and we could even go hours before 10 a.m. This is obviously, this keeps on cycling on and on and on forever. Now, what would be an expression for F of t? And I encourage you once again to pause the video and try to think about that."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "So what would be, before we even try to model T of t, what would be an expression, and obviously we keep going, we keep going like that, and we could even go hours before 10 a.m. This is obviously, this keeps on cycling on and on and on forever. Now, what would be an expression for F of t? And I encourage you once again to pause the video and try to think about that. Well, one thing that you could say, well, you say this could be a sine or a cosine function. Actually, you could model it with either of them, but it's always easiest to do the simplest one. Which function is essentially at its midline, at its midline, when the argument to the function is zero?"}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "And I encourage you once again to pause the video and try to think about that. Well, one thing that you could say, well, you say this could be a sine or a cosine function. Actually, you could model it with either of them, but it's always easiest to do the simplest one. Which function is essentially at its midline, at its midline, when the argument to the function is zero? Well, the sine of zero is zero, and if we didn't shift this function up or down, the midline of just a sine function, without it being shifted, is zero. So sine of zero is zero, and then sine begins to increase and oscillate like this. So it feels like sine is a good candidate to model it with."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "Which function is essentially at its midline, at its midline, when the argument to the function is zero? Well, the sine of zero is zero, and if we didn't shift this function up or down, the midline of just a sine function, without it being shifted, is zero. So sine of zero is zero, and then sine begins to increase and oscillate like this. So it feels like sine is a good candidate to model it with. Once again, you could model it with either, but I have a feeling this is gonna be a little bit simpler. Now, let's think about the amplitude. Well, how much do we vary?"}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "So it feels like sine is a good candidate to model it with. Once again, you could model it with either, but I have a feeling this is gonna be a little bit simpler. Now, let's think about the amplitude. Well, how much do we vary? What's our maximum variance from our midline? So here we are 7.5 above our midline. Here we are 7.5 below our midline."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "Well, how much do we vary? What's our maximum variance from our midline? So here we are 7.5 above our midline. Here we are 7.5 below our midline. So our amplitude is 7.5. And actually, let me just do that in a different color, just so you see where things are coming from. So this is 7.5, this is 7.5."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "Here we are 7.5 below our midline. So our amplitude is 7.5. And actually, let me just do that in a different color, just so you see where things are coming from. So this is 7.5, this is 7.5. So our amplitude looks like it's 7.5. And now, what is our period? Well, we've already talked about it."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "So this is 7.5, this is 7.5. So our amplitude looks like it's 7.5. And now, what is our period? Well, we've already talked about it. Our period is 24, 24 hours. This distance right over here is 24 hours, which makes complete sense. After 24 hours, you're at the same point in the day."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "Well, we've already talked about it. Our period is 24, 24 hours. This distance right over here is 24 hours, which makes complete sense. After 24 hours, you're at the same point in the day. So we would divide two pi by the period, divided by 24 times t. And if you forget, hey, divide two pi by the period here, you could just remind yourself that what t value will make us go from, so when t is equal to zero, the whole argument to the sine function is going to be zero. That's when we're over here. And then when t is equal to 24, the whole argument's going to be two pi."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "After 24 hours, you're at the same point in the day. So we would divide two pi by the period, divided by 24 times t. And if you forget, hey, divide two pi by the period here, you could just remind yourself that what t value will make us go from, so when t is equal to zero, the whole argument to the sine function is going to be zero. That's when we're over here. And then when t is equal to 24, the whole argument's going to be two pi. So we would have made one rotation around the unit circle if we think about the input into the sine function. Now, we're almost done. If I were to just graph this, this would have a midline around zero, but we see that we've shifted everything up by 10.5."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "And then when t is equal to 24, the whole argument's going to be two pi. So we would have made one rotation around the unit circle if we think about the input into the sine function. Now, we're almost done. If I were to just graph this, this would have a midline around zero, but we see that we've shifted everything up by 10.5. So we have to shift everything up by 10.5. Now, this is, we've just successfully modeled it, and we could simplify a little bit. We could write this as pi over 12 instead of two pi over 24."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "If I were to just graph this, this would have a midline around zero, but we see that we've shifted everything up by 10.5. So we have to shift everything up by 10.5. Now, this is, we've just successfully modeled it, and we could simplify a little bit. We could write this as pi over 12 instead of two pi over 24. But this right over here models the temperature in Johannesburg T hours after 10 a.m. After 10 a.m. That's not what they wanted. They want us to model, they want us to model the temperature T hours after midnight. So what would T of T be?"}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "We could write this as pi over 12 instead of two pi over 24. But this right over here models the temperature in Johannesburg T hours after 10 a.m. After 10 a.m. That's not what they wanted. They want us to model, they want us to model the temperature T hours after midnight. So what would T of T be? We're gonna have to shift this a little bit. So let's just think about it a little. Let me just write it out."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "So what would T of T be? We're gonna have to shift this a little bit. So let's just think about it a little. Let me just write it out. So T of T, so T of T, this is now we're modeling T hours after midnight. So we're gonna have the same amplitude. We're just gonna have the same variance from the midline."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "Let me just write it out. So T of T, so T of T, this is now we're modeling T hours after midnight. So we're gonna have the same amplitude. We're just gonna have the same variance from the midline. So it's gonna be 7.5 times sine of, actually, we do the same color so you see what I'm changing and not changing, times the sine of, instead of two pi over 24, I'll just write pi over 12. Instead of writing T, I'm gonna shift T either to the right or the left. And actually, you could shift in either direction because this is a periodic function."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "We're just gonna have the same variance from the midline. So it's gonna be 7.5 times sine of, actually, we do the same color so you see what I'm changing and not changing, times the sine of, instead of two pi over 24, I'll just write pi over 12. Instead of writing T, I'm gonna shift T either to the right or the left. And actually, you could shift in either direction because this is a periodic function. We're gonna have to think about how much we're shifting it. So T is gonna be plus or minus something right over here. I'm gonna shift it plus 10.5, plus 10.5."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "And actually, you could shift in either direction because this is a periodic function. We're gonna have to think about how much we're shifting it. So T is gonna be plus or minus something right over here. I'm gonna shift it plus 10.5, plus 10.5. Now this is always a little bit, at least in my brain, I have to think about this in a lot of different ways so that I make sure that I'm shifting it in the right direction. So here at 10 a.m., we were at this point. When T is equal to zero, this is zero hours after 10 a.m."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "I'm gonna shift it plus 10.5, plus 10.5. Now this is always a little bit, at least in my brain, I have to think about this in a lot of different ways so that I make sure that I'm shifting it in the right direction. So here at 10 a.m., we were at this point. When T is equal to zero, this is zero hours after 10 a.m. But in this function, when is 10 a.m.? Well, in this function, 10 a.m., let me write it this way. 10 a.m. is 10 hours after midnight."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "When T is equal to zero, this is zero hours after 10 a.m. But in this function, when is 10 a.m.? Well, in this function, 10 a.m., let me write it this way. 10 a.m. is 10 hours after midnight. So T, capital T of 10, this is 10 hours after midnight, should be equal to, should be equal to f of zero because here, the argument is hours after 10 a.m. So this is 10 a.m., this right over here represents temperature at 10 a.m. And this over here, because capital, this capital T function, this is hours after midnight. This is also temperature at 10 a.m."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "10 a.m. is 10 hours after midnight. So T, capital T of 10, this is 10 hours after midnight, should be equal to, should be equal to f of zero because here, the argument is hours after 10 a.m. So this is 10 a.m., this right over here represents temperature at 10 a.m. And this over here, because capital, this capital T function, this is hours after midnight. This is also temperature at 10 a.m. So we want T of 10 to be the same thing as f of zero. Or a same, another way of thinking about it, when f of zero, this whole argument is zero. So we want this whole argument to be zero when T is equal to 10."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "This is also temperature at 10 a.m. So we want T of 10 to be the same thing as f of zero. Or a same, another way of thinking about it, when f of zero, this whole argument is zero. So we want this whole argument to be zero when T is equal to 10. So how would we do that? Well, if this is T minus 10, notice, T of 10, you put a 10 here, this whole thing becomes zero. This whole thing becomes zero, and you're left with 10.5."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "So we want this whole argument to be zero when T is equal to 10. So how would we do that? Well, if this is T minus 10, notice, T of 10, you put a 10 here, this whole thing becomes zero. This whole thing becomes zero, and you're left with 10.5. And over here, f of zero, well, the same thing. This whole thing becomes zero, and all you're left with over here is 10.5. So T of 10 should be f of zero."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "This whole thing becomes zero, and you're left with 10.5. And over here, f of zero, well, the same thing. This whole thing becomes zero, and all you're left with over here is 10.5. So T of 10 should be f of zero. So if we wanted to graph it, we've already answered their question. If we put a 10 here, the argument to the sign becomes zero. These two things are going to be equivalent."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "So T of 10 should be f of zero. So if we wanted to graph it, we've already answered their question. If we put a 10 here, the argument to the sign becomes zero. These two things are going to be equivalent. But let's actually graph this. So T of 10, so if we're graphing capital T, T of 10, so this is six, 12, let's see. So this is, let me be, so 10 is going to be someplace around here."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "These two things are going to be equivalent. But let's actually graph this. So T of 10, so if we're graphing capital T, T of 10, so this is six, 12, let's see. So this is, let me be, so 10 is going to be someplace around here. So T of 10 is going to be the same thing as f of zero. So it's going to be like that. And then it's just going to, and then we've essentially just shifted everything to the right, everything to the right by 10."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "So this is, let me be, so 10 is going to be someplace around here. So T of 10 is going to be the same thing as f of zero. So it's going to be like that. And then it's just going to, and then we've essentially just shifted everything to the right, everything to the right by 10. And that makes sense. Because zero after, whatever hours you are after 10 a.m., it's going to be 10 more hours to get to that same point after midnight. So your curve is going to look, so this is going to be shifted by 10, this is going to be shifted by 10, and your curve's going to look something like, let me see, this is going to be shifted by 10, so you're going to get, this is going to be at 16 hours, so let's see, it's going to look something like that."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy (2).mp3", "Sentence": "And then it's just going to, and then we've essentially just shifted everything to the right, everything to the right by 10. And that makes sense. Because zero after, whatever hours you are after 10 a.m., it's going to be 10 more hours to get to that same point after midnight. So your curve is going to look, so this is going to be shifted by 10, this is going to be shifted by 10, and your curve's going to look something like, let me see, this is going to be shifted by 10, so you're going to get, this is going to be at 16 hours, so let's see, it's going to look something like that. And of course, it'll keep oscillating like that. So essentially, we just have to shift it to the right by 10. The argument, we have to replace T with T minus 10 to do it, and this was the logic why."}, {"video_title": "Applying inverse trig function with model Trigonometry Khan Academy.mp3", "Sentence": "But we weren't done. They want us to figure out how many days after January 7th is the first spring day when the temperature reaches 20 degrees Celsius. And I told you to be careful. Pay attention to this whole notion of the first spring day. And the reason is because there's actually two days where the temperature reaches 20 degrees Celsius. So let's say that this is 20 degrees Celsius right over here. Notice you have this day right over here, and then you have this day right over there."}, {"video_title": "Applying inverse trig function with model Trigonometry Khan Academy.mp3", "Sentence": "Pay attention to this whole notion of the first spring day. And the reason is because there's actually two days where the temperature reaches 20 degrees Celsius. So let's say that this is 20 degrees Celsius right over here. Notice you have this day right over here, and then you have this day right over there. And which one is in the spring or the first spring day? Well, if this is that we're in summer right over here, we're in the southern hemisphere. So our summer is going to be when it's the winter in the northern hemisphere."}, {"video_title": "Applying inverse trig function with model Trigonometry Khan Academy.mp3", "Sentence": "Notice you have this day right over here, and then you have this day right over there. And which one is in the spring or the first spring day? Well, if this is that we're in summer right over here, we're in the southern hemisphere. So our summer is going to be when it's the winter in the northern hemisphere. Summer, what season comes after summer? Well, this is going to be the fall. This is going to be the winter."}, {"video_title": "Applying inverse trig function with model Trigonometry Khan Academy.mp3", "Sentence": "So our summer is going to be when it's the winter in the northern hemisphere. Summer, what season comes after summer? Well, this is going to be the fall. This is going to be the winter. Now this is going to be the spring. And then, of course, you go back to the summer. So we want this value, not this value."}, {"video_title": "Applying inverse trig function with model Trigonometry Khan Academy.mp3", "Sentence": "This is going to be the winter. Now this is going to be the spring. And then, of course, you go back to the summer. So we want this value, not this value. This one will be the day in fall when the average high temperature is 20 degrees Celsius. This is the first day of spring where the average high temperature is 20 degrees Celsius. The first spring day."}, {"video_title": "Applying inverse trig function with model Trigonometry Khan Academy.mp3", "Sentence": "So we want this value, not this value. This one will be the day in fall when the average high temperature is 20 degrees Celsius. This is the first day of spring where the average high temperature is 20 degrees Celsius. The first spring day. Oh, I guess it's not necessarily the first day of spring. It's the first spring day when the temperature reaches 20 degrees Celsius. So this is happening in the spring right over here."}, {"video_title": "Applying inverse trig function with model Trigonometry Khan Academy.mp3", "Sentence": "The first spring day. Oh, I guess it's not necessarily the first day of spring. It's the first spring day when the temperature reaches 20 degrees Celsius. So this is happening in the spring right over here. So this is the value that we want. So let's just think about that as we try to manipulate this a little bit. So we want to get to 20 degrees Celsius."}, {"video_title": "Applying inverse trig function with model Trigonometry Khan Academy.mp3", "Sentence": "So this is happening in the spring right over here. So this is the value that we want. So let's just think about that as we try to manipulate this a little bit. So we want to get to 20 degrees Celsius. So we could write 20 is equal to 7.5 times cosine of 2 pi over 365 times the days plus 21.5. Now we could subtract 21.5 from both sides. And we get negative 1.5 is equal to, and I'll just copy and paste all of this, is going to be equal to that."}, {"video_title": "Applying inverse trig function with model Trigonometry Khan Academy.mp3", "Sentence": "So we want to get to 20 degrees Celsius. So we could write 20 is equal to 7.5 times cosine of 2 pi over 365 times the days plus 21.5. Now we could subtract 21.5 from both sides. And we get negative 1.5 is equal to, and I'll just copy and paste all of this, is going to be equal to that. So copy and paste. It's going to be equal to that right over there. Now I could divide both sides by 7.5."}, {"video_title": "Applying inverse trig function with model Trigonometry Khan Academy.mp3", "Sentence": "And we get negative 1.5 is equal to, and I'll just copy and paste all of this, is going to be equal to that. So copy and paste. It's going to be equal to that right over there. Now I could divide both sides by 7.5. Notice I'm trying to solve for cosine and eventually solve for d. But we're going to take a little pause once we have this in terms of cosine. We have to be careful here. So we're going to divide both sides by 7.5."}, {"video_title": "Applying inverse trig function with model Trigonometry Khan Academy.mp3", "Sentence": "Now I could divide both sides by 7.5. Notice I'm trying to solve for cosine and eventually solve for d. But we're going to take a little pause once we have this in terms of cosine. We have to be careful here. So we're going to divide both sides by 7.5. And we're going to get, let's see, actually, I don't even need a calculator for this. 1.5 divided by 7.5. This is 1 fifth."}, {"video_title": "Applying inverse trig function with model Trigonometry Khan Academy.mp3", "Sentence": "So we're going to divide both sides by 7.5. And we're going to get, let's see, actually, I don't even need a calculator for this. 1.5 divided by 7.5. This is 1 fifth. 5 times 15 is, or 5 times 1.5 is 7.5. So this is negative 1 fifth. Or I could write it as negative 1 fifth."}, {"video_title": "Applying inverse trig function with model Trigonometry Khan Academy.mp3", "Sentence": "This is 1 fifth. 5 times 15 is, or 5 times 1.5 is 7.5. So this is negative 1 fifth. Or I could write it as negative 1 fifth. Or I could write it as negative 0.2 is equal to cosine of 2 pi over 365 times days after January 7. Now this is where we have to be very, very careful. Instead of just blindly applying the inverse cosine function, we have to make sure which angle we are actually getting, that we're getting the right angle right over here."}, {"video_title": "Applying inverse trig function with model Trigonometry Khan Academy.mp3", "Sentence": "Or I could write it as negative 1 fifth. Or I could write it as negative 0.2 is equal to cosine of 2 pi over 365 times days after January 7. Now this is where we have to be very, very careful. Instead of just blindly applying the inverse cosine function, we have to make sure which angle we are actually getting, that we're getting the right angle right over here. Because remember, we want the argument to the cosine that doesn't give us this point, that gives us this point right here, or that corresponds to this point right over there. So let's draw a unit circle just to make sure we know what's going on. And I actually do this all the time, especially if I'm trying to apply the inverse trigonometric functions in kind of a applied context, where I just can't blindly plug it in into my calculator."}, {"video_title": "Applying inverse trig function with model Trigonometry Khan Academy.mp3", "Sentence": "Instead of just blindly applying the inverse cosine function, we have to make sure which angle we are actually getting, that we're getting the right angle right over here. Because remember, we want the argument to the cosine that doesn't give us this point, that gives us this point right here, or that corresponds to this point right over there. So let's draw a unit circle just to make sure we know what's going on. And I actually do this all the time, especially if I'm trying to apply the inverse trigonometric functions in kind of a applied context, where I just can't blindly plug it in into my calculator. So let me draw a unit circle right over here, x-axis, y-axis. So circle of radius 1 centered at the origin. So that looks, well, you get the picture."}, {"video_title": "Applying inverse trig function with model Trigonometry Khan Academy.mp3", "Sentence": "And I actually do this all the time, especially if I'm trying to apply the inverse trigonometric functions in kind of a applied context, where I just can't blindly plug it in into my calculator. So let me draw a unit circle right over here, x-axis, y-axis. So circle of radius 1 centered at the origin. So that looks, well, you get the picture. We've done this many times before. And January 7, that corresponds to this point right over here. That's that point right over here, that we are in the summer."}, {"video_title": "Applying inverse trig function with model Trigonometry Khan Academy.mp3", "Sentence": "So that looks, well, you get the picture. We've done this many times before. And January 7, that corresponds to this point right over here. That's that point right over here, that we are in the summer. We are in the summer. Then as the days go by, our argument to the cosine increases, the angle increases. And this right over here will be the fall."}, {"video_title": "Applying inverse trig function with model Trigonometry Khan Academy.mp3", "Sentence": "That's that point right over here, that we are in the summer. We are in the summer. Then as the days go by, our argument to the cosine increases, the angle increases. And this right over here will be the fall. So this point right over here. So we're at the fall right over here. And then we move on to the winter right over there."}, {"video_title": "Applying inverse trig function with model Trigonometry Khan Academy.mp3", "Sentence": "And this right over here will be the fall. So this point right over here. So we're at the fall right over here. And then we move on to the winter right over there. And then finally, we go to the spring. This is the spring right over there. And we want the angle that gets us negative 0.2."}, {"video_title": "Applying inverse trig function with model Trigonometry Khan Academy.mp3", "Sentence": "And then we move on to the winter right over there. And then finally, we go to the spring. This is the spring right over there. And we want the angle that gets us negative 0.2. So this is negative 1. Negative 0.2, it's a fifth of the way, so it's negative 0.2. And notice, there's two angles that get us there."}, {"video_title": "Applying inverse trig function with model Trigonometry Khan Academy.mp3", "Sentence": "And we want the angle that gets us negative 0.2. So this is negative 1. Negative 0.2, it's a fifth of the way, so it's negative 0.2. And notice, there's two angles that get us there. There's this angle. There's that angle right over there. And then there's also, let me draw a little dotted line here."}, {"video_title": "Applying inverse trig function with model Trigonometry Khan Academy.mp3", "Sentence": "And notice, there's two angles that get us there. There's this angle. There's that angle right over there. And then there's also, let me draw a little dotted line here. There's that angle. But then there's also this angle, which is going even further. Or another way you could think about it, you could go backwards to get to that angle."}, {"video_title": "Applying inverse trig function with model Trigonometry Khan Academy.mp3", "Sentence": "And then there's also, let me draw a little dotted line here. There's that angle. But then there's also this angle, which is going even further. Or another way you could think about it, you could go backwards to get to that angle. And then if you wanted to go all the way around to the next spring, you could add 2 pi to it. So which one do we want? Well, of course, we want the one in the spring."}, {"video_title": "Applying inverse trig function with model Trigonometry Khan Academy.mp3", "Sentence": "Or another way you could think about it, you could go backwards to get to that angle. And then if you wanted to go all the way around to the next spring, you could add 2 pi to it. So which one do we want? Well, of course, we want the one in the spring. But if we just blindly apply the inverse cosine of negative 0.2, that's going to give us this one. That's going to give us this one right over here. And we can verify that."}, {"video_title": "Applying inverse trig function with model Trigonometry Khan Academy.mp3", "Sentence": "Well, of course, we want the one in the spring. But if we just blindly apply the inverse cosine of negative 0.2, that's going to give us this one. That's going to give us this one right over here. And we can verify that. So let's see. Inverse cosine of negative 0.2 is 1.77. Remember, this is 0."}, {"video_title": "Applying inverse trig function with model Trigonometry Khan Academy.mp3", "Sentence": "And we can verify that. So let's see. Inverse cosine of negative 0.2 is 1.77. Remember, this is 0. This is pi right over here. So this is 3.14159. And this is 1.77."}, {"video_title": "Applying inverse trig function with model Trigonometry Khan Academy.mp3", "Sentence": "Remember, this is 0. This is pi right over here. So this is 3.14159. And this is 1.77. So notice, it's a little bit more than half of pi, which is exactly what it gave. It gave this angle right over here. So this is approximately 1.77 radians is this angle, is that angle right over there."}, {"video_title": "Applying inverse trig function with model Trigonometry Khan Academy.mp3", "Sentence": "And this is 1.77. So notice, it's a little bit more than half of pi, which is exactly what it gave. It gave this angle right over here. So this is approximately 1.77 radians is this angle, is that angle right over there. But we don't want that one. We want this one. So how do we figure that out?"}, {"video_title": "Applying inverse trig function with model Trigonometry Khan Academy.mp3", "Sentence": "So this is approximately 1.77 radians is this angle, is that angle right over there. But we don't want that one. We want this one. So how do we figure that out? Well, we could view it as we could go all the way around to 2 pi and then subtract 1.77. So we could say 2 pi minus 1.77, roughly, to get this angle. So let's do that."}, {"video_title": "Applying inverse trig function with model Trigonometry Khan Academy.mp3", "Sentence": "So how do we figure that out? Well, we could view it as we could go all the way around to 2 pi and then subtract 1.77. So we could say 2 pi minus 1.77, roughly, to get this angle. So let's do that. So let's take 2 times pi and then subtract this number. And I'm going to do second answer. So answer is just the previous answer that I got."}, {"video_title": "Applying inverse trig function with model Trigonometry Khan Academy.mp3", "Sentence": "So let's do that. So let's take 2 times pi and then subtract this number. And I'm going to do second answer. So answer is just the previous answer that I got. So that way, I have good precision here. So I get 4.511. And we can kind of make sure that that's the right thing, because that's going to be between pi and 2 pi."}, {"video_title": "Applying inverse trig function with model Trigonometry Khan Academy.mp3", "Sentence": "So answer is just the previous answer that I got. So that way, I have good precision here. So I get 4.511. And we can kind of make sure that that's the right thing, because that's going to be between pi and 2 pi. 2 pi is 2 times 3.14159. So it's going to be 6.28 something. This is 3.14159."}, {"video_title": "Applying inverse trig function with model Trigonometry Khan Academy.mp3", "Sentence": "And we can kind of make sure that that's the right thing, because that's going to be between pi and 2 pi. 2 pi is 2 times 3.14159. So it's going to be 6.28 something. This is 3.14159. This is the right angle. Now, we're not done yet. That's just the angle."}, {"video_title": "Applying inverse trig function with model Trigonometry Khan Academy.mp3", "Sentence": "This is 3.14159. This is the right angle. Now, we're not done yet. That's just the angle. That's the argument that we need to give in here to get to that point. But what's the days going to be? Well, 2 pi over 365 days is going to be equal to this number, 4.511."}, {"video_title": "Applying inverse trig function with model Trigonometry Khan Academy.mp3", "Sentence": "That's just the angle. That's the argument that we need to give in here to get to that point. But what's the days going to be? Well, 2 pi over 365 days is going to be equal to this number, 4.511. So let me write that down. So this right over here is going to be approximately equal to 4.511. So we scroll down a little bit."}, {"video_title": "Applying inverse trig function with model Trigonometry Khan Academy.mp3", "Sentence": "Well, 2 pi over 365 days is going to be equal to this number, 4.511. So let me write that down. So this right over here is going to be approximately equal to 4.511. So we scroll down a little bit. We can say 2 pi over 365 times the days after January 7 is approximately going to be equal to 4.511. To solve for days, we can just multiply both sides times the reciprocal of the coefficient. So we're going to multiply both sides times 365 over 2 pi."}, {"video_title": "Applying inverse trig function with model Trigonometry Khan Academy.mp3", "Sentence": "So we scroll down a little bit. We can say 2 pi over 365 times the days after January 7 is approximately going to be equal to 4.511. To solve for days, we can just multiply both sides times the reciprocal of the coefficient. So we're going to multiply both sides times 365 over 2 pi. That's going to cancel. That's going to cancel. And now we can use our calculator for this."}, {"video_title": "Applying inverse trig function with model Trigonometry Khan Academy.mp3", "Sentence": "So we're going to multiply both sides times 365 over 2 pi. That's going to cancel. That's going to cancel. And now we can use our calculator for this. So that's actually much better precision there. So let's take our previous answer times 365. We deserve a drum roll here."}, {"video_title": "Applying inverse trig function with model Trigonometry Khan Academy.mp3", "Sentence": "And now we can use our calculator for this. So that's actually much better precision there. So let's take our previous answer times 365. We deserve a drum roll here. Divided by 2 pi. And we get 200, and if we round to the nearest day, 262 days after January 7. So 262 days."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "And just as a review, let's just remind ourselves of the relationship. And I always do this before I have to convert between the two. If I do one revolution of a circle, how many radians is that going to be? Well, we know one revolution of a circle is 2 pi radians. And how many degrees is that? If I do one revolution around a circle, well, we know that that's 360. I can either write it with a little degree symbol right like that, or I could write it just like that."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "Well, we know one revolution of a circle is 2 pi radians. And how many degrees is that? If I do one revolution around a circle, well, we know that that's 360. I can either write it with a little degree symbol right like that, or I could write it just like that. And this is really enough information for us to think about how to convert between radians and degrees. If we want to simplify this a little bit, we can divide both sides by 2, and you could have pi radians are equal to 180 degrees, or another way to think about it, going halfway around a circle in radians is pi radians, or the arc that subtends that angle is pi radiuses. And that's also 180 degrees."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "I can either write it with a little degree symbol right like that, or I could write it just like that. And this is really enough information for us to think about how to convert between radians and degrees. If we want to simplify this a little bit, we can divide both sides by 2, and you could have pi radians are equal to 180 degrees, or another way to think about it, going halfway around a circle in radians is pi radians, or the arc that subtends that angle is pi radiuses. And that's also 180 degrees. And if you want to really think about, well, how many degrees are there per radian, you can divide both sides of this by pi. So if you divide both sides of this by pi, you get one radian. I have to go from plural to singular."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "And that's also 180 degrees. And if you want to really think about, well, how many degrees are there per radian, you can divide both sides of this by pi. So if you divide both sides of this by pi, you get one radian. I have to go from plural to singular. One radian is equal to 180 over pi degrees. So all I did is I divided both sides by pi. And if you wanted to figure out how many radians are there per degree, you could divide both sides by 180."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "I have to go from plural to singular. One radian is equal to 180 over pi degrees. So all I did is I divided both sides by pi. And if you wanted to figure out how many radians are there per degree, you could divide both sides by 180. So you'd get pi over 180 radians is equal to 1 degree. So now I think we are ready to start converting. So let's convert 30 degrees to radians."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "And if you wanted to figure out how many radians are there per degree, you could divide both sides by 180. So you'd get pi over 180 radians is equal to 1 degree. So now I think we are ready to start converting. So let's convert 30 degrees to radians. So let's think about it. So I'm going to write it out. And actually, this might remind you of kind of unit analysis that you might do when you first did unit conversion, but it also works out here."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "So let's convert 30 degrees to radians. So let's think about it. So I'm going to write it out. And actually, this might remind you of kind of unit analysis that you might do when you first did unit conversion, but it also works out here. So if I were to write 30 degrees, and this is how my brain likes to work with it, I like to write out the word degrees. And then I say, well, I want to convert to radians. So I really want to figure out how many radians are there per degree."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "And actually, this might remind you of kind of unit analysis that you might do when you first did unit conversion, but it also works out here. So if I were to write 30 degrees, and this is how my brain likes to work with it, I like to write out the word degrees. And then I say, well, I want to convert to radians. So I really want to figure out how many radians are there per degree. So let me write this down. I want to figure out how many radians do we have per degree. And I haven't filled out how many that is, but we see just the units will cancel out."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "So I really want to figure out how many radians are there per degree. So let me write this down. I want to figure out how many radians do we have per degree. And I haven't filled out how many that is, but we see just the units will cancel out. If we have degrees times radians per degree, the degrees will cancel out and I'll be just left with radians. If I multiply the number of degrees I have times the number of radians per degree, we're going to get radians. And hopefully that makes intuitive sense as well."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "And I haven't filled out how many that is, but we see just the units will cancel out. If we have degrees times radians per degree, the degrees will cancel out and I'll be just left with radians. If I multiply the number of degrees I have times the number of radians per degree, we're going to get radians. And hopefully that makes intuitive sense as well. And here we just have to think about, well, if I have pi radians, how many degrees is that? Well, that's 180 degrees. It comes straight out of this right over here."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "And hopefully that makes intuitive sense as well. And here we just have to think about, well, if I have pi radians, how many degrees is that? Well, that's 180 degrees. It comes straight out of this right over here. Pi radians for every 180 degrees or pi over 180 radians per degree. And this is going to get us to 30 times pi over 180, which we'll simplify to 30 over 180 is 1 over 6. So this is equal to pi over 6."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "It comes straight out of this right over here. Pi radians for every 180 degrees or pi over 180 radians per degree. And this is going to get us to 30 times pi over 180, which we'll simplify to 30 over 180 is 1 over 6. So this is equal to pi over 6. Actually, let me write the units out. This is 30 radians, which is equal to pi over 6 radians. Now let's go the other way."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "So this is equal to pi over 6. Actually, let me write the units out. This is 30 radians, which is equal to pi over 6 radians. Now let's go the other way. Let's think about if we have pi over 3 radians, and I want to convert that to degrees. So what am I going to get if I convert that to degrees? Well, here we're going to want to figure out how many degrees are there per radian."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "Now let's go the other way. Let's think about if we have pi over 3 radians, and I want to convert that to degrees. So what am I going to get if I convert that to degrees? Well, here we're going to want to figure out how many degrees are there per radian. And one way to think about it is, well, think about the pi and the 180. For every 180 degrees, you have pi radians. 180 degrees over pi radians, these are essentially the equivalent thing."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "Well, here we're going to want to figure out how many degrees are there per radian. And one way to think about it is, well, think about the pi and the 180. For every 180 degrees, you have pi radians. 180 degrees over pi radians, these are essentially the equivalent thing. Essentially, you're just multiplying this quantity by 1, but you're changing the units. The radians cancel out, and then the pi's cancel out, and you're left with 180 over 3 degrees. 180 over 3 is 60, and we could either write out the word degrees, or you can write degrees just like that."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "180 degrees over pi radians, these are essentially the equivalent thing. Essentially, you're just multiplying this quantity by 1, but you're changing the units. The radians cancel out, and then the pi's cancel out, and you're left with 180 over 3 degrees. 180 over 3 is 60, and we could either write out the word degrees, or you can write degrees just like that. Now let's think about 45 degrees. So what about 45 degrees? And I'll write it like that just so you can figure it out as they're."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "180 over 3 is 60, and we could either write out the word degrees, or you can write degrees just like that. Now let's think about 45 degrees. So what about 45 degrees? And I'll write it like that just so you can figure it out as they're. Figure it out with that notation as well. How many radians will this be equal to? Well, once again, we're going to want to think about how many radians do we have per degree."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "And I'll write it like that just so you can figure it out as they're. Figure it out with that notation as well. How many radians will this be equal to? Well, once again, we're going to want to think about how many radians do we have per degree. So we're going to multiply this times, well, we know we have pi radians for every 180 degrees, or we could even write it this way, pi radians for every 180 degrees. And here, this might be a little less intuitive, the degrees cancel out, and that's why I'd like to usually write out the word, and you're left with 45 pi over 180 radians. Actually, let me write this with the words written out."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "Well, once again, we're going to want to think about how many radians do we have per degree. So we're going to multiply this times, well, we know we have pi radians for every 180 degrees, or we could even write it this way, pi radians for every 180 degrees. And here, this might be a little less intuitive, the degrees cancel out, and that's why I'd like to usually write out the word, and you're left with 45 pi over 180 radians. Actually, let me write this with the words written out. Maybe that's more intuitive when I'm thinking about it in terms of using the notation. So 45 degrees times, we have pi radians for every 180 degrees. So we are left with, when you multiply, 45 times pi over 180, the degrees have canceled out, and you're just left with radians, which is equal to what?"}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "Actually, let me write this with the words written out. Maybe that's more intuitive when I'm thinking about it in terms of using the notation. So 45 degrees times, we have pi radians for every 180 degrees. So we are left with, when you multiply, 45 times pi over 180, the degrees have canceled out, and you're just left with radians, which is equal to what? 45 is half of 90, which is half of 180, so this is 1 4th. This is equal to pi over 4 radians. Let's do one more over here."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "So we are left with, when you multiply, 45 times pi over 180, the degrees have canceled out, and you're just left with radians, which is equal to what? 45 is half of 90, which is half of 180, so this is 1 4th. This is equal to pi over 4 radians. Let's do one more over here. So let's say that we had negative pi over 2 radians. What's that going to be in degrees? Well, once again, we have to figure out how many degrees are each of these radians."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "Let's do one more over here. So let's say that we had negative pi over 2 radians. What's that going to be in degrees? Well, once again, we have to figure out how many degrees are each of these radians. We know that there are 180 degrees for every pi radians, so we're going to get the radians cancel out, the pi's cancel out, and so you have negative 180 over 2. This is negative 90 degrees, or we could write it as negative 90 degrees. Anyway, hopefully you found that helpful, and I'll do a couple more example problems here, because the more example for this, the better, and hopefully it will become a little bit intuitive."}, {"video_title": "Interpreting trigonometric graphs in context Trigonometry Algebra Khan Academy (2).mp3", "Sentence": "Now what I wanna focus on this video is some features of this graph, and the features we're going to focus on, actually the first of them, is going to be the midline. So pause this video and see if you can figure out the midline of this graph, or the midline of this function, and then we're gonna think about what it actually represents. Well, Alexa starts off at five meters above the ground, and then she goes higher and higher and higher, gets as high as 25 meters, and then goes back as low as five meters above the ground, then as high as 25 meters, and what we can view the midline as is the midpoint between these extremes, or the average of these extremes. Well, the extremes are, she goes as low as five and as high as 25. So what's the average of five and 25? Well, that would be 15. So the midline would look something like this, and I'm actually gonna keep going off the graph, and the reason is is to help us think about what does that midline even represent?"}, {"video_title": "Interpreting trigonometric graphs in context Trigonometry Algebra Khan Academy (2).mp3", "Sentence": "Well, the extremes are, she goes as low as five and as high as 25. So what's the average of five and 25? Well, that would be 15. So the midline would look something like this, and I'm actually gonna keep going off the graph, and the reason is is to help us think about what does that midline even represent? And one way to think about it is it represents the center of our rotation in this situation, or how high above the ground is the center of our Ferris wheel? And to help us visualize that, let me draw a Ferris wheel. So I'm going to draw a circle with this as the center, and so the Ferris wheel would look something like, would look something like this, and it has some type of maybe support structure."}, {"video_title": "Interpreting trigonometric graphs in context Trigonometry Algebra Khan Academy (2).mp3", "Sentence": "So the midline would look something like this, and I'm actually gonna keep going off the graph, and the reason is is to help us think about what does that midline even represent? And one way to think about it is it represents the center of our rotation in this situation, or how high above the ground is the center of our Ferris wheel? And to help us visualize that, let me draw a Ferris wheel. So I'm going to draw a circle with this as the center, and so the Ferris wheel would look something like, would look something like this, and it has some type of maybe support structure. So the Ferris wheel might look something like that, and this height above the ground, that is 15 meters, that is what the midline is representing. Now, the next feature I want to explore is the amplitude. Pause this video and think about what is the amplitude of this oscillating function right over here, and then we'll think about what does that represent in the real world, or where does it come from in the real world?"}, {"video_title": "Interpreting trigonometric graphs in context Trigonometry Algebra Khan Academy (2).mp3", "Sentence": "So I'm going to draw a circle with this as the center, and so the Ferris wheel would look something like, would look something like this, and it has some type of maybe support structure. So the Ferris wheel might look something like that, and this height above the ground, that is 15 meters, that is what the midline is representing. Now, the next feature I want to explore is the amplitude. Pause this video and think about what is the amplitude of this oscillating function right over here, and then we'll think about what does that represent in the real world, or where does it come from in the real world? Well, the amplitude is the maximum difference or the maximum magnitude away from that midline, and you can see it right over here, actually right when Alexa starts, we have starting 10 meters below the midline, 10 meters below the center, and this is when Alexa is right over here. She is 10 meters below the midline, and then after a, looks like 10 seconds, she is right at the midline, so that means that she's right over here. Maybe the Ferris wheel is going this way, at least in my imagination, it's going clockwise, and then after another 10 seconds, she is at 25 meters, so she is right over there, and you can see that."}, {"video_title": "Interpreting trigonometric graphs in context Trigonometry Algebra Khan Academy (2).mp3", "Sentence": "Pause this video and think about what is the amplitude of this oscillating function right over here, and then we'll think about what does that represent in the real world, or where does it come from in the real world? Well, the amplitude is the maximum difference or the maximum magnitude away from that midline, and you can see it right over here, actually right when Alexa starts, we have starting 10 meters below the midline, 10 meters below the center, and this is when Alexa is right over here. She is 10 meters below the midline, and then after a, looks like 10 seconds, she is right at the midline, so that means that she's right over here. Maybe the Ferris wheel is going this way, at least in my imagination, it's going clockwise, and then after another 10 seconds, she is at 25 meters, so she is right over there, and you can see that. She is right over there. I drew that circle intentionally of that size, and so we see the amplitude in full effect, 10 meters below to begin the midline and 10 meters above, and so it's the maximum displacement or the maximum change from that midline, and so over here, it really represents the radius of our Ferris wheel, 10 meters, and then from this part, she starts going back down again, and then over here, she's back to where she started. Now, the last feature I want to explore is the notion of a period."}, {"video_title": "Interpreting trigonometric graphs in context Trigonometry Algebra Khan Academy (2).mp3", "Sentence": "Maybe the Ferris wheel is going this way, at least in my imagination, it's going clockwise, and then after another 10 seconds, she is at 25 meters, so she is right over there, and you can see that. She is right over there. I drew that circle intentionally of that size, and so we see the amplitude in full effect, 10 meters below to begin the midline and 10 meters above, and so it's the maximum displacement or the maximum change from that midline, and so over here, it really represents the radius of our Ferris wheel, 10 meters, and then from this part, she starts going back down again, and then over here, she's back to where she started. Now, the last feature I want to explore is the notion of a period. What is the period of this periodic function? Pause this video and think about that. Well, the period is how much time does it take to complete one cycle?"}, {"video_title": "Interpreting trigonometric graphs in context Trigonometry Algebra Khan Academy (2).mp3", "Sentence": "Now, the last feature I want to explore is the notion of a period. What is the period of this periodic function? Pause this video and think about that. Well, the period is how much time does it take to complete one cycle? So here, she's starting at the bottom, and let's see, after 10 seconds, not at the bottom yet, after 20 seconds, not at the bottom yet, after 30 seconds, not at the bottom yet, and then here she is, after 40 seconds, she's back at the bottom and about to head up again, and so this time right over here, that 40 seconds, that is the period, and if you think about what's going on over here, she starts over here, five meters above the ground, after 10 seconds, she is right over here, and that corresponds to this point right over here, after 10 more seconds, she's right over there, that corresponds to that point, after 10 more seconds, she's over here, that corresponds to that, and after 10 more seconds, or a total of 40 seconds, she is back to where she started. So the period in this example shows how long does it take to complete one full rotation? Now we have to be careful sometimes when we're trying to visually inspect the period, because sometimes it might be tempting to say, start right over here and say, okay, we're 15 meters above the ground, all right, let's see, we're going down, now we're going up again, and look, we're 15 meters above the ground, maybe this 20 seconds is a period, but when you look at it over here, it's clear that that is not the case."}, {"video_title": "Interpreting trigonometric graphs in context Trigonometry Algebra Khan Academy (2).mp3", "Sentence": "Well, the period is how much time does it take to complete one cycle? So here, she's starting at the bottom, and let's see, after 10 seconds, not at the bottom yet, after 20 seconds, not at the bottom yet, after 30 seconds, not at the bottom yet, and then here she is, after 40 seconds, she's back at the bottom and about to head up again, and so this time right over here, that 40 seconds, that is the period, and if you think about what's going on over here, she starts over here, five meters above the ground, after 10 seconds, she is right over here, and that corresponds to this point right over here, after 10 more seconds, she's right over there, that corresponds to that point, after 10 more seconds, she's over here, that corresponds to that, and after 10 more seconds, or a total of 40 seconds, she is back to where she started. So the period in this example shows how long does it take to complete one full rotation? Now we have to be careful sometimes when we're trying to visually inspect the period, because sometimes it might be tempting to say, start right over here and say, okay, we're 15 meters above the ground, all right, let's see, we're going down, now we're going up again, and look, we're 15 meters above the ground, maybe this 20 seconds is a period, but when you look at it over here, it's clear that that is not the case. This point represents this point at being 15 meters above the ground, going down, that's getting us to this point, and then after another 10 seconds, we get back over here. Notice, all this is measuring is half of a cycle, going halfway around. In order to go all the way around, not only do we have to get to the same exact height, but we have to be moving in the same direction."}, {"video_title": "Example Graphing y=-cos(\u03c0\u22c5x)+1.5 Trigonometry Algebra 2 Khan Academy.mp3", "Sentence": "So pause this video and think about how you would do that. And just to explain how this widget works, if you're trying to do it on Khan Academy, this dot right over here helps define the midline. You can move that up and down. And then this one right over here is a neighboring extreme point. So either a minimum or a maximum point. So there's a couple of ways that we could approach this. First of all, let's just think about what would cosine of pi x look like?"}, {"video_title": "Example Graphing y=-cos(\u03c0\u22c5x)+1.5 Trigonometry Algebra 2 Khan Academy.mp3", "Sentence": "And then this one right over here is a neighboring extreme point. So either a minimum or a maximum point. So there's a couple of ways that we could approach this. First of all, let's just think about what would cosine of pi x look like? And then we'll think about what the negative does and the plus 1.5. So cosine of pi x, when x is equal to zero, pi times zero is just going to be zero, cosine of zero is equal to one. And if we're just talking about cosine of pi x, that's going to be a maximum point when you hit one."}, {"video_title": "Example Graphing y=-cos(\u03c0\u22c5x)+1.5 Trigonometry Algebra 2 Khan Academy.mp3", "Sentence": "First of all, let's just think about what would cosine of pi x look like? And then we'll think about what the negative does and the plus 1.5. So cosine of pi x, when x is equal to zero, pi times zero is just going to be zero, cosine of zero is equal to one. And if we're just talking about cosine of pi x, that's going to be a maximum point when you hit one. Just cosine of pi x would oscillate between one and negative one. And then what would its period be if we're talking about cosine of pi x? Well, you might remember one way to think about the period is to take two pi and divide it by whatever the coefficient is on the x right over here."}, {"video_title": "Example Graphing y=-cos(\u03c0\u22c5x)+1.5 Trigonometry Algebra 2 Khan Academy.mp3", "Sentence": "And if we're just talking about cosine of pi x, that's going to be a maximum point when you hit one. Just cosine of pi x would oscillate between one and negative one. And then what would its period be if we're talking about cosine of pi x? Well, you might remember one way to think about the period is to take two pi and divide it by whatever the coefficient is on the x right over here. So two pi divided by pi would tell us that we have a period of two. And so how do we construct a period of two here? Well, that means that as we start here at x equals zero, we're at one, we want to get back to that maximum point by the time x is equal to two."}, {"video_title": "Example Graphing y=-cos(\u03c0\u22c5x)+1.5 Trigonometry Algebra 2 Khan Academy.mp3", "Sentence": "Well, you might remember one way to think about the period is to take two pi and divide it by whatever the coefficient is on the x right over here. So two pi divided by pi would tell us that we have a period of two. And so how do we construct a period of two here? Well, that means that as we start here at x equals zero, we're at one, we want to get back to that maximum point by the time x is equal to two. So let me see how I can do that. If I were to squeeze it a little bit, that looks pretty good. And the reason why I worked on this midline point is I liked having this maximum point at one when x is equal to zero, because we said cosine of pi times zero should be equal to one."}, {"video_title": "Example Graphing y=-cos(\u03c0\u22c5x)+1.5 Trigonometry Algebra 2 Khan Academy.mp3", "Sentence": "Well, that means that as we start here at x equals zero, we're at one, we want to get back to that maximum point by the time x is equal to two. So let me see how I can do that. If I were to squeeze it a little bit, that looks pretty good. And the reason why I worked on this midline point is I liked having this maximum point at one when x is equal to zero, because we said cosine of pi times zero should be equal to one. So that's why I'm just manipulating this other point in order to set the period right. But this looks right. We're going from this maximum point, we're going all the way down and then back to that maximum point, and it looks like our period is indeed two."}, {"video_title": "Example Graphing y=-cos(\u03c0\u22c5x)+1.5 Trigonometry Algebra 2 Khan Academy.mp3", "Sentence": "And the reason why I worked on this midline point is I liked having this maximum point at one when x is equal to zero, because we said cosine of pi times zero should be equal to one. So that's why I'm just manipulating this other point in order to set the period right. But this looks right. We're going from this maximum point, we're going all the way down and then back to that maximum point, and it looks like our period is indeed two. So this is what the graph of cosine of pi x would look like. Now, what about this negative sign? Well, the negative would essentially flip it around."}, {"video_title": "Example Graphing y=-cos(\u03c0\u22c5x)+1.5 Trigonometry Algebra 2 Khan Academy.mp3", "Sentence": "We're going from this maximum point, we're going all the way down and then back to that maximum point, and it looks like our period is indeed two. So this is what the graph of cosine of pi x would look like. Now, what about this negative sign? Well, the negative would essentially flip it around. So instead of whenever we're equaling one, we should be equal to negative one. And every time we're equal to negative one, we should be equal to one. So what I could do is I could just take that and then bring it down here."}, {"video_title": "Example Graphing y=-cos(\u03c0\u22c5x)+1.5 Trigonometry Algebra 2 Khan Academy.mp3", "Sentence": "Well, the negative would essentially flip it around. So instead of whenever we're equaling one, we should be equal to negative one. And every time we're equal to negative one, we should be equal to one. So what I could do is I could just take that and then bring it down here. And there you have it, I flipped it around. So this is the graph of y equals negative cosine of pi x. And then last but not least, we have this plus 1.5."}, {"video_title": "Example Graphing y=-cos(\u03c0\u22c5x)+1.5 Trigonometry Algebra 2 Khan Academy.mp3", "Sentence": "So what I could do is I could just take that and then bring it down here. And there you have it, I flipped it around. So this is the graph of y equals negative cosine of pi x. And then last but not least, we have this plus 1.5. So that's just going to shift everything up by 1.5. So I'm just gonna shift everything up by, shift it up by 1.5, and shift it up by 1.5. And there you have it."}, {"video_title": "Example Graphing y=-cos(\u03c0\u22c5x)+1.5 Trigonometry Algebra 2 Khan Academy.mp3", "Sentence": "And then last but not least, we have this plus 1.5. So that's just going to shift everything up by 1.5. So I'm just gonna shift everything up by, shift it up by 1.5, and shift it up by 1.5. And there you have it. That is the graph of negative cosine of pi x plus 1.5. And you can validate that that's our midline. We're still oscillating one above and one below."}, {"video_title": "Example Graphing y=-cos(\u03c0\u22c5x)+1.5 Trigonometry Algebra 2 Khan Academy.mp3", "Sentence": "And there you have it. That is the graph of negative cosine of pi x plus 1.5. And you can validate that that's our midline. We're still oscillating one above and one below. The negative sign, when cosine of pi times zero, that should be one, but then you take the negative of that, we get to negative one. You add 1.5 to that, you get to positive 0.5. And so this is all looking quite good."}, {"video_title": "Applying angle addition formula for sin Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So I'm assuming you've had a go at it, and in doing that, you might have realized that, okay, this line, that's one of the sides of this right triangle that I have right over here, and we're given this alpha and beta, but if we consider the combined angle, if we consider the combined angle alpha plus beta, then this side right over here, we can just take out our traditional trig functions, so our SOH CAH TOA definition of the basic trig functions, and we know that sine is opposite over hypotenuse. So if we're considering alpha plus beta, this angle right over here, opposite over hypotenuse, that's going to be this length over the hypotenuse, which is one. So sine of alpha plus beta is going to be this length right over here. So that seems interesting, so let me write that down. Sine of alpha plus beta, sine of alpha plus beta, plus beta, is essentially what we're looking for. Sine of alpha plus beta is this length right over here. Sine of alpha plus beta, it's equal to the opposite side, that, over the hypotenuse."}, {"video_title": "Applying angle addition formula for sin Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So that seems interesting, so let me write that down. Sine of alpha plus beta, sine of alpha plus beta, plus beta, is essentially what we're looking for. Sine of alpha plus beta is this length right over here. Sine of alpha plus beta, it's equal to the opposite side, that, over the hypotenuse. Well, the hypotenuse is just going to be equal to one, so it's equal to this side. So another way of phrasing the exact same problem that we first tried to tackle is how do we figure out the sine of alpha plus beta? And if you're familiar with your trig identities, something might be jumping out at you, that hey, we know a different way of expressing sine of alpha plus beta."}, {"video_title": "Applying angle addition formula for sin Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Sine of alpha plus beta, it's equal to the opposite side, that, over the hypotenuse. Well, the hypotenuse is just going to be equal to one, so it's equal to this side. So another way of phrasing the exact same problem that we first tried to tackle is how do we figure out the sine of alpha plus beta? And if you're familiar with your trig identities, something might be jumping out at you, that hey, we know a different way of expressing sine of alpha plus beta. We know that this thing is the same thing as, we know it's the same thing as the sine of alpha, plus, or sine of alpha times the cosine of beta, plus the other way around. The cosine of alpha, cosine of alpha, plus, or times the sine of beta, times the sine of beta. Let me draw a line here so we don't get confused."}, {"video_title": "Applying angle addition formula for sin Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And if you're familiar with your trig identities, something might be jumping out at you, that hey, we know a different way of expressing sine of alpha plus beta. We know that this thing is the same thing as, we know it's the same thing as the sine of alpha, plus, or sine of alpha times the cosine of beta, plus the other way around. The cosine of alpha, cosine of alpha, plus, or times the sine of beta, times the sine of beta. Let me draw a line here so we don't get confused. So if we're trying to figure this out, and we know that this can be re-expressed this way, it all boils down to can we figure out what sine of alpha is, cosine of beta, cosine of alpha, and sine of beta? And when you look at this, you see that you actually can figure those things out. So let's do that."}, {"video_title": "Applying angle addition formula for sin Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Let me draw a line here so we don't get confused. So if we're trying to figure this out, and we know that this can be re-expressed this way, it all boils down to can we figure out what sine of alpha is, cosine of beta, cosine of alpha, and sine of beta? And when you look at this, you see that you actually can figure those things out. So let's do that. Sine of alpha, I'll write it over here, sine of alpha is equal to, this is alpha, sine is opposite over hypotenuse, so it's 0.5 over one. So this is equal to 0.5. So that is 0.5."}, {"video_title": "Applying angle addition formula for sin Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So let's do that. Sine of alpha, I'll write it over here, sine of alpha is equal to, this is alpha, sine is opposite over hypotenuse, so it's 0.5 over one. So this is equal to 0.5. So that is 0.5. Cosine of beta, cosine of beta, this is beta, cosine is adjacent over hypotenuse. So this is beta, the adjacent side is 0.6 over the hypotenuse of one, so it's 0.6. 0.6, 0.6."}, {"video_title": "Applying angle addition formula for sin Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So that is 0.5. Cosine of beta, cosine of beta, this is beta, cosine is adjacent over hypotenuse. So this is beta, the adjacent side is 0.6 over the hypotenuse of one, so it's 0.6. 0.6, 0.6. Cosine of alpha, cosine of alpha, adjacent over hypotenuse, it's square root of three over two over one. So that's just square root of three over two. So this is just square root of three over two, and then finally, sine of beta, sine of beta, opposite over hypotenuse, is 0.8."}, {"video_title": "Applying angle addition formula for sin Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "0.6, 0.6. Cosine of alpha, cosine of alpha, adjacent over hypotenuse, it's square root of three over two over one. So that's just square root of three over two. So this is just square root of three over two, and then finally, sine of beta, sine of beta, opposite over hypotenuse, is 0.8. This is 0.8. And actually, let me write that as, I'm gonna write that as 4 5ths, just so that that's the same thing as 0.8, just because I think it's gonna make it a little bit easier for me to simplify right over here. So what is all of this equal to?"}, {"video_title": "Applying angle addition formula for sin Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So this is just square root of three over two, and then finally, sine of beta, sine of beta, opposite over hypotenuse, is 0.8. This is 0.8. And actually, let me write that as, I'm gonna write that as 4 5ths, just so that that's the same thing as 0.8, just because I think it's gonna make it a little bit easier for me to simplify right over here. So what is all of this equal to? Well, this is going to be equal to 0.5 times 0.6, this part right over here is 0.3, 0.3, and square root of three over two times 4 5ths, well, let's just multiply them. Well, four divided by two is two, so it's two square roots of three over five. So this is equal to, or so plus, two square roots of three over five."}, {"video_title": "Applying angle addition formula for sin Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So what is all of this equal to? Well, this is going to be equal to 0.5 times 0.6, this part right over here is 0.3, 0.3, and square root of three over two times 4 5ths, well, let's just multiply them. Well, four divided by two is two, so it's two square roots of three over five. So this is equal to, or so plus, two square roots of three over five. So this is essentially our answer. I feel a little uncomfortable having it in these two different formats, where I have a fraction here and I have a decimal here, so let me just write the whole thing as a rational expression. So 0.3 is obviously the same thing as 3 10ths, 3 10ths, so that's the same thing as three over 10 plus, now this, if I want to write it over 10, this is the same thing as four square roots of three over 10, and of course, if we add these two, we are going to get three plus four square roots of three, plus four square roots of three, all of that over, all of that over 10, and we are done."}, {"video_title": "Inverse tan domain and range Trigonometry Khan Academy.mp3", "Sentence": "So I've got my little scratch pad here to try to work that through. So let's figure out what g inverse of x is. This is g of x. So g inverse of x, I'm essentially, let me just read, this is g of x right over here. g of x is equal to tangent of x minus 3 pi over 2 plus 6. So g inverse of x, I essentially can swap the, I can replace the x with the g inverse of x and replace the g of x with an x, and then solve for g inverse of x. So I could write that x is equal to tangent of g inverse of x minus 3 pi over 2 plus 6."}, {"video_title": "Inverse tan domain and range Trigonometry Khan Academy.mp3", "Sentence": "So g inverse of x, I'm essentially, let me just read, this is g of x right over here. g of x is equal to tangent of x minus 3 pi over 2 plus 6. So g inverse of x, I essentially can swap the, I can replace the x with the g inverse of x and replace the g of x with an x, and then solve for g inverse of x. So I could write that x is equal to tangent of g inverse of x minus 3 pi over 2 plus 6. So let's just solve for g inverse of x. And I actually encourage you to pause this video and try to work through this out, or work it out on your own. So let's subtract 6 from both sides to at least get rid of this 6 here."}, {"video_title": "Inverse tan domain and range Trigonometry Khan Academy.mp3", "Sentence": "So I could write that x is equal to tangent of g inverse of x minus 3 pi over 2 plus 6. So let's just solve for g inverse of x. And I actually encourage you to pause this video and try to work through this out, or work it out on your own. So let's subtract 6 from both sides to at least get rid of this 6 here. And so I'm left with x minus 6 is equal to the tangent of g inverse of x minus 3 pi over 2. Now let's take the inverse tangent of both sides of this equation. So the inverse tangent of the left-hand side is the inverse tangent of x minus 6."}, {"video_title": "Inverse tan domain and range Trigonometry Khan Academy.mp3", "Sentence": "So let's subtract 6 from both sides to at least get rid of this 6 here. And so I'm left with x minus 6 is equal to the tangent of g inverse of x minus 3 pi over 2. Now let's take the inverse tangent of both sides of this equation. So the inverse tangent of the left-hand side is the inverse tangent of x minus 6. And on the right-hand side, the inverse tangent of tangent, if we restrict the domain in the proper way, and we'll talk about that in a little bit, is just going to be what the input into the tangent function is. So if you restrict the domain in the right way, the inverse tangent of the tangent of something, of say theta, is just going to be equal to theta. Because again, if we restrict the domain, if we restrict what the possible values of theta are in the right way."}, {"video_title": "Inverse tan domain and range Trigonometry Khan Academy.mp3", "Sentence": "So the inverse tangent of the left-hand side is the inverse tangent of x minus 6. And on the right-hand side, the inverse tangent of tangent, if we restrict the domain in the proper way, and we'll talk about that in a little bit, is just going to be what the input into the tangent function is. So if you restrict the domain in the right way, the inverse tangent of the tangent of something, of say theta, is just going to be equal to theta. Because again, if we restrict the domain, if we restrict what the possible values of theta are in the right way. So let's just assume that we're doing that. And so the inverse tangent of the tan of this is going to be just this stuff right over here. It's just going to be that."}, {"video_title": "Inverse tan domain and range Trigonometry Khan Academy.mp3", "Sentence": "Because again, if we restrict the domain, if we restrict what the possible values of theta are in the right way. So let's just assume that we're doing that. And so the inverse tangent of the tan of this is going to be just this stuff right over here. It's just going to be that. It's going to be g inverse of x minus 3 pi over 2. And now we're in the home stretch. To solve for g inverse of x, we could just add 3 pi over 2 to both sides."}, {"video_title": "Inverse tan domain and range Trigonometry Khan Academy.mp3", "Sentence": "It's just going to be that. It's going to be g inverse of x minus 3 pi over 2. And now we're in the home stretch. To solve for g inverse of x, we could just add 3 pi over 2 to both sides. So we get, and actually let me just swap both sides. We get g inverse of x is equal to the inverse tangent of x minus 6. And then we're adding 3 pi over 2 to both sides."}, {"video_title": "Inverse tan domain and range Trigonometry Khan Academy.mp3", "Sentence": "To solve for g inverse of x, we could just add 3 pi over 2 to both sides. So we get, and actually let me just swap both sides. We get g inverse of x is equal to the inverse tangent of x minus 6. And then we're adding 3 pi over 2 to both sides. So this side is now on this side, so plus 3 pi over 2. So let me actually type that in. Let me see if I can remember it, because I'm about to lose this on my screen."}, {"video_title": "Inverse tan domain and range Trigonometry Khan Academy.mp3", "Sentence": "And then we're adding 3 pi over 2 to both sides. So this side is now on this side, so plus 3 pi over 2. So let me actually type that in. Let me see if I can remember it, because I'm about to lose this on my screen. So inverse tangent of x minus 6 plus 3 pi over 2. So let me write that down. So let me type this."}, {"video_title": "Inverse tan domain and range Trigonometry Khan Academy.mp3", "Sentence": "Let me see if I can remember it, because I'm about to lose this on my screen. So inverse tangent of x minus 6 plus 3 pi over 2. So let me write that down. So let me type this. So g inverse of x is going to be the inverse tangent. So I could write it like this. The inverse tangent of x minus 6."}, {"video_title": "Inverse tan domain and range Trigonometry Khan Academy.mp3", "Sentence": "So let me type this. So g inverse of x is going to be the inverse tangent. So I could write it like this. The inverse tangent of x minus 6. And yes, it interpreted it correctly. Inverse tangent, you can view that as arc tangent of x minus 6 plus 3 pi over 2. And it did interpret it correctly."}, {"video_title": "Inverse tan domain and range Trigonometry Khan Academy.mp3", "Sentence": "The inverse tangent of x minus 6. And yes, it interpreted it correctly. Inverse tangent, you can view that as arc tangent of x minus 6 plus 3 pi over 2. And it did interpret it correctly. But then we have to think about what is the domain of g inverse? What is the domain of g inverse of x? Let's think about this a little bit more."}, {"video_title": "Inverse tan domain and range Trigonometry Khan Academy.mp3", "Sentence": "And it did interpret it correctly. But then we have to think about what is the domain of g inverse? What is the domain of g inverse of x? Let's think about this a little bit more. The domain of g inverse of x. So let's just think about what tangent is doing. So the tangent function, if we imagine a unit circle, so that's a unit circle right over there."}, {"video_title": "Inverse tan domain and range Trigonometry Khan Academy.mp3", "Sentence": "Let's think about this a little bit more. The domain of g inverse of x. So let's just think about what tangent is doing. So the tangent function, if we imagine a unit circle, so that's a unit circle right over there. I guess we can imagine it to be a unit circle. My pen tool is acting up a little bit. I'm putting these little gaps and things, but I think we can power through that."}, {"video_title": "Inverse tan domain and range Trigonometry Khan Academy.mp3", "Sentence": "So the tangent function, if we imagine a unit circle, so that's a unit circle right over there. I guess we can imagine it to be a unit circle. My pen tool is acting up a little bit. I'm putting these little gaps and things, but I think we can power through that. So let's just say, for the sake of argument, that that's a unit circle. That's the x-axis and that's the y-axis. If you form an angle theta, if you form some angle theta right over here, the tangent of theta is essentially the slope of this terminal ray of the angle."}, {"video_title": "Inverse tan domain and range Trigonometry Khan Academy.mp3", "Sentence": "I'm putting these little gaps and things, but I think we can power through that. So let's just say, for the sake of argument, that that's a unit circle. That's the x-axis and that's the y-axis. If you form an angle theta, if you form some angle theta right over here, the tangent of theta is essentially the slope of this terminal ray of the angle. Or I guess you could call it the terminal ray of the angle. The angle is formed by that ray and this ray along the positive x-axis. So the tangent of theta is the slope right over there."}, {"video_title": "Inverse tan domain and range Trigonometry Khan Academy.mp3", "Sentence": "If you form an angle theta, if you form some angle theta right over here, the tangent of theta is essentially the slope of this terminal ray of the angle. Or I guess you could call it the terminal ray of the angle. The angle is formed by that ray and this ray along the positive x-axis. So the tangent of theta is the slope right over there. And you can get a tangent of any theta except for a few. So you can find the tangent of that. You could find the slope there."}, {"video_title": "Inverse tan domain and range Trigonometry Khan Academy.mp3", "Sentence": "So the tangent of theta is the slope right over there. And you can get a tangent of any theta except for a few. So you can find the tangent of that. You could find the slope there. You could find the slope there. You could also find the slope there. You could find the slope there."}, {"video_title": "Inverse tan domain and range Trigonometry Khan Academy.mp3", "Sentence": "You could find the slope there. You could find the slope there. You could also find the slope there. You could find the slope there. But the place where you can't find the slope is when this ray goes straight up or this ray goes straight down. Those are the cases where you can't find the slope. There the slope you could say is approaching positive or negative infinity."}, {"video_title": "Inverse tan domain and range Trigonometry Khan Academy.mp3", "Sentence": "You could find the slope there. But the place where you can't find the slope is when this ray goes straight up or this ray goes straight down. Those are the cases where you can't find the slope. There the slope you could say is approaching positive or negative infinity. So the domain of tangent is essentially all real numbers, all reals except pi over 2 plus multiples of pi. Where k could be any integer. So you could also be subtracting pi because if you have pi over 2, if you add pi, you go straight down here."}, {"video_title": "Inverse tan domain and range Trigonometry Khan Academy.mp3", "Sentence": "There the slope you could say is approaching positive or negative infinity. So the domain of tangent is essentially all real numbers, all reals except pi over 2 plus multiples of pi. Where k could be any integer. So you could also be subtracting pi because if you have pi over 2, if you add pi, you go straight down here. You add another pi, you go up there. If you subtract pi, you go down here. Subtract another pi, you go over there."}, {"video_title": "Inverse tan domain and range Trigonometry Khan Academy.mp3", "Sentence": "So you could also be subtracting pi because if you have pi over 2, if you add pi, you go straight down here. You add another pi, you go up there. If you subtract pi, you go down here. Subtract another pi, you go over there. So this is the domain. But given this domain, you can get any real number. So the range here is all reals because you can get any slope here."}, {"video_title": "Inverse tan domain and range Trigonometry Khan Academy.mp3", "Sentence": "Subtract another pi, you go over there. So this is the domain. But given this domain, you can get any real number. So the range here is all reals because you can get any slope here. You can increase theta if you want a really high slope, decrease theta if you want a really negative slope right over there. So you can really get to anything. Now when you're talking about the inverse tangent, by convention you're going to make tangent invertible so that you don't have multiple elements of your domain all mapping to the same element of the range."}, {"video_title": "Inverse tan domain and range Trigonometry Khan Academy.mp3", "Sentence": "So the range here is all reals because you can get any slope here. You can increase theta if you want a really high slope, decrease theta if you want a really negative slope right over there. So you can really get to anything. Now when you're talking about the inverse tangent, by convention you're going to make tangent invertible so that you don't have multiple elements of your domain all mapping to the same element of the range. Because for example, this angle right over here has the exact same slope as this angle right over here. So if you have two thetas mapping to the same tangent, if you don't restrict your domain so that you only have one of them, it's not going to be invertible. So the convention is that to make tangent invertible, you restrict its domain."}, {"video_title": "Inverse tan domain and range Trigonometry Khan Academy.mp3", "Sentence": "Now when you're talking about the inverse tangent, by convention you're going to make tangent invertible so that you don't have multiple elements of your domain all mapping to the same element of the range. Because for example, this angle right over here has the exact same slope as this angle right over here. So if you have two thetas mapping to the same tangent, if you don't restrict your domain so that you only have one of them, it's not going to be invertible. So the convention is that to make tangent invertible, you restrict its domain. So you restrict the domain to the interval from negative pi over 2 to pi over 2 in order to construct the inverse tangent. So the inverse tangent, you can input any real number into it. So the inverse tangent's domain, this is just a convention, they could have restricted tangent's domains as long as for any theta, there's only one theta in that domain that maps to a specific element of the range."}, {"video_title": "Inverse tan domain and range Trigonometry Khan Academy.mp3", "Sentence": "So the convention is that to make tangent invertible, you restrict its domain. So you restrict the domain to the interval from negative pi over 2 to pi over 2 in order to construct the inverse tangent. So the inverse tangent, you can input any real number into it. So the inverse tangent's domain, this is just a convention, they could have restricted tangent's domains as long as for any theta, there's only one theta in that domain that maps to a specific element of the range. But the convention is to restrict tangent's domain between negative pi over 2 and pi over 2. So inverse tangent's domain is all reals, but its range is restricted. Its range, and this is by convention, it's going to be between negative pi over 2 and pi over 2 and not including them."}, {"video_title": "Inverse tan domain and range Trigonometry Khan Academy.mp3", "Sentence": "So the inverse tangent's domain, this is just a convention, they could have restricted tangent's domains as long as for any theta, there's only one theta in that domain that maps to a specific element of the range. But the convention is to restrict tangent's domain between negative pi over 2 and pi over 2. So inverse tangent's domain is all reals, but its range is restricted. Its range, and this is by convention, it's going to be between negative pi over 2 and pi over 2 and not including them. So let's go back to our original question right over here. What is the domain of G inverse? So let's look at the domain of G inverse."}, {"video_title": "Inverse tan domain and range Trigonometry Khan Academy.mp3", "Sentence": "Its range, and this is by convention, it's going to be between negative pi over 2 and pi over 2 and not including them. So let's go back to our original question right over here. What is the domain of G inverse? So let's look at the domain of G inverse. Well, G inverse, the domain of this, I could put any real number in here. Any real number here. Now, what this is going to pop out is going to be something between negative pi over 2 and pi over 2, but they're not asking us the range of G inverse."}, {"video_title": "Inverse tan domain and range Trigonometry Khan Academy.mp3", "Sentence": "So let's look at the domain of G inverse. Well, G inverse, the domain of this, I could put any real number in here. Any real number here. Now, what this is going to pop out is going to be something between negative pi over 2 and pi over 2, but they're not asking us the range of G inverse. Actually, it would have been a more interesting question. They're asking us what's the domain of G inverse, and I could put in any real number right here for x. So let's put that in here."}, {"video_title": "Inverse tan domain and range Trigonometry Khan Academy.mp3", "Sentence": "Now, what this is going to pop out is going to be something between negative pi over 2 and pi over 2, but they're not asking us the range of G inverse. Actually, it would have been a more interesting question. They're asking us what's the domain of G inverse, and I could put in any real number right here for x. So let's put that in here. So the domain of G inverse of x, it's negative infinity to infinity. But actually, just for fun, and let's just verify that we got the question right, and we did. But just for fun, actually, I am curious."}, {"video_title": "Inverse tan domain and range Trigonometry Khan Academy.mp3", "Sentence": "So let's put that in here. So the domain of G inverse of x, it's negative infinity to infinity. But actually, just for fun, and let's just verify that we got the question right, and we did. But just for fun, actually, I am curious. Let's just think about what the range of G inverse is. So the range of this thing right over here is going to be between negative pi over 2 to pi over 2. That's for this part right over here."}, {"video_title": "Inverse tan domain and range Trigonometry Khan Academy.mp3", "Sentence": "But just for fun, actually, I am curious. Let's just think about what the range of G inverse is. So the range of this thing right over here is going to be between negative pi over 2 to pi over 2. That's for this part right over here. And then we're going to add 3 pi over 2s to it. So the range for the entire function, so the range for this thing, the range is going to be what the low end, if we add 3 pi over 2 to this, this is going to give us 2 pi over 2, which is just going to be, so 3 pi over 2 minus pi over 2 is going to be 2 pi over 2, which is just pi, just pi, all the way to 3 pi over 2 plus another pi over 2 is going to be 4 pi over 2, or 2 pi. So the range of G inverse of x is pi to 2 pi, and it's an open interval, it doesn't include the boundaries, but its domain, you could put any value for x here, and it will be defined."}, {"video_title": "Inverse tan domain and range Trigonometry Khan Academy.mp3", "Sentence": "That's for this part right over here. And then we're going to add 3 pi over 2s to it. So the range for the entire function, so the range for this thing, the range is going to be what the low end, if we add 3 pi over 2 to this, this is going to give us 2 pi over 2, which is just going to be, so 3 pi over 2 minus pi over 2 is going to be 2 pi over 2, which is just pi, just pi, all the way to 3 pi over 2 plus another pi over 2 is going to be 4 pi over 2, or 2 pi. So the range of G inverse of x is pi to 2 pi, and it's an open interval, it doesn't include the boundaries, but its domain, you could put any value for x here, and it will be defined."}, {"video_title": "Inverse tan domain and range Trigonometry Khan Academy.mp3", "Sentence": "So the range of G inverse of x is pi to 2 pi, and it's an open interval, it doesn't include the boundaries, but its domain, you could put any value for x here, and it will be defined."}] \ No newline at end of file