Sentence
stringlengths 141
2.12k
| video_title
stringlengths 24
104
|
|---|---|
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. | Interpreting trigonometric graphs in context Trigonometry Algebra Khan Academy.mp3 |
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. | Interpreting trigonometric graphs in context Trigonometry Algebra Khan Academy.mp3 |
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. | Interpreting trigonometric graphs in context Trigonometry Algebra Khan Academy.mp3 |
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. | Interpreting trigonometric graphs in context Trigonometry Algebra Khan Academy.mp3 |
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. | Interpreting trigonometric graphs in context Trigonometry Algebra Khan Academy.mp3 |
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. | Interpreting trigonometric graphs in context Trigonometry Algebra Khan Academy.mp3 |
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? | Interpreting trigonometric graphs in context Trigonometry Algebra Khan Academy.mp3 |
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. | Interpreting trigonometric graphs in context Trigonometry Algebra Khan Academy.mp3 |
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. | Another example using angle addition formula with cosine Trigonometry Khan Academy.mp3 |
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. | Another example using angle addition formula with cosine Trigonometry Khan Academy.mp3 |
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. | Another example using angle addition formula with cosine Trigonometry Khan Academy.mp3 |
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. | Another example using angle addition formula with cosine Trigonometry Khan Academy.mp3 |
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. | Another example using angle addition formula with cosine Trigonometry Khan Academy.mp3 |
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. | Another example using angle addition formula with cosine Trigonometry Khan Academy.mp3 |
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. | Another example using angle addition formula with cosine Trigonometry Khan Academy.mp3 |
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. | Another example using angle addition formula with cosine Trigonometry Khan Academy.mp3 |
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. | Another example using angle addition formula with cosine Trigonometry Khan Academy.mp3 |
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. | Another example using angle addition formula with cosine Trigonometry Khan Academy.mp3 |
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. | Another example using angle addition formula with cosine Trigonometry Khan Academy.mp3 |
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. | Another example using angle addition formula with cosine Trigonometry Khan Academy.mp3 |
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. | Another example using angle addition formula with cosine Trigonometry Khan Academy.mp3 |
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. | Another example using angle addition formula with cosine Trigonometry Khan Academy.mp3 |
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. | Another example using angle addition formula with cosine Trigonometry Khan Academy.mp3 |
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. | Another example using angle addition formula with cosine Trigonometry Khan Academy.mp3 |
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? | Another example using angle addition formula with cosine Trigonometry Khan Academy.mp3 |
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. | Another example using angle addition formula with cosine Trigonometry Khan Academy.mp3 |
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. | Another example using angle addition formula with cosine Trigonometry Khan Academy.mp3 |
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. | Another example using angle addition formula with cosine Trigonometry Khan Academy.mp3 |
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. | Another example using angle addition formula with cosine Trigonometry Khan Academy.mp3 |
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. | Another example using angle addition formula with cosine Trigonometry Khan Academy.mp3 |
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. | Another example using angle addition formula with cosine Trigonometry Khan Academy.mp3 |
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. | Another example using angle addition formula with cosine Trigonometry Khan Academy.mp3 |
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. | Another example using angle addition formula with cosine Trigonometry Khan Academy.mp3 |
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. | Another example using angle addition formula with cosine Trigonometry Khan Academy.mp3 |
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. | Another example using angle addition formula with cosine Trigonometry Khan Academy.mp3 |
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. | Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3 |
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. | Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3 |
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. | Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3 |
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. | Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3 |
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. | Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3 |
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. | Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3 |
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? | Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3 |
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. | Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3 |
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. | Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3 |
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? | Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3 |
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. | Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3 |
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? | Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3 |
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. | Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3 |
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. | Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3 |
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. | Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3 |
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. | Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3 |
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. | Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3 |
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. | Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3 |
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. | Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3 |
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. | Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3 |
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. | Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3 |
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. | Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3 |
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. | Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3 |
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. | Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3 |
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. | Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3 |
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. | Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3 |
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. | Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3 |
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. | Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3 |
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. | Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3 |
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. | Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3 |
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. | Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3 |
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. | Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3 |
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? | Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3 |
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? | Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3 |
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. | Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3 |
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. | Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3 |
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. | Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3 |
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? | Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3 |
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. | Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3 |
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? | Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3 |
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? | Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3 |
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. | Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3 |
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. | Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3 |
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. | Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3 |
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. | Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3 |
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. | Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3 |
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. | Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3 |
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. | Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3 |
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. | Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3 |
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? | Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3 |
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. | Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3 |
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. | Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3 |
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. | Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3 |
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. | Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3 |
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. | Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3 |
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. | Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3 |
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? | Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3 |
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. | Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3 |
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. | Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3 |
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. | Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3 |
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. | Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3 |
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. | Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3 |
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. | Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3 |
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. | Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3 |
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. | Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3 |