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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. | Inverse trig functions arccos Trigonometry Khan Academy.mp3 |
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. | Solving triangle in unit circle Trigonometry Khan Academy.mp3 |
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. | Solving triangle in unit circle Trigonometry Khan Academy.mp3 |
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. | Solving triangle in unit circle Trigonometry Khan Academy.mp3 |
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. | Solving triangle in unit circle Trigonometry Khan Academy.mp3 |
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. | Solving triangle in unit circle Trigonometry Khan Academy.mp3 |
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. | Solving triangle in unit circle Trigonometry Khan Academy.mp3 |
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. | Solving triangle in unit circle Trigonometry Khan Academy.mp3 |
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? | Solving triangle in unit circle Trigonometry Khan Academy.mp3 |
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. | Solving triangle in unit circle Trigonometry Khan Academy.mp3 |
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. | Solving triangle in unit circle Trigonometry Khan Academy.mp3 |
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. | Solving triangle in unit circle Trigonometry Khan Academy.mp3 |
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. | Solving triangle in unit circle Trigonometry Khan Academy.mp3 |
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. | Solving triangle in unit circle Trigonometry Khan Academy.mp3 |
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. | Solving triangle in unit circle Trigonometry Khan Academy.mp3 |
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. | Solving triangle in unit circle Trigonometry Khan Academy.mp3 |
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. | Solving triangle in unit circle Trigonometry Khan Academy.mp3 |
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. | Solving triangle in unit circle Trigonometry Khan Academy.mp3 |
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. | Solving triangle in unit circle Trigonometry Khan Academy.mp3 |
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. | Solving triangle in unit circle Trigonometry Khan Academy.mp3 |
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? | Solving triangle in unit circle Trigonometry Khan Academy.mp3 |
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? | Solving triangle in unit circle Trigonometry Khan Academy.mp3 |
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. | Solving triangle in unit circle Trigonometry Khan Academy.mp3 |
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. | Solving triangle in unit circle Trigonometry Khan Academy.mp3 |
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. | Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3 |
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. | Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3 |
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. | Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3 |
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. | Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3 |
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. | Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3 |
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. | Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3 |
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. | Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3 |
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. | Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3 |
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. | Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3 |
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. | Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3 |
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. | Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3 |
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. | Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3 |
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. | Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3 |
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. | Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3 |
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. | Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3 |
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. | Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3 |
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. | Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3 |
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. | Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3 |
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. | Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3 |
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. | Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3 |
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. | Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3 |
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. | Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3 |
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. | Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3 |
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. | Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3 |
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. | Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3 |
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. | Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3 |
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. | Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3 |
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. | Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3 |
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. | Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3 |
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. | Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3 |
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. | Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3 |
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. | Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3 |
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. | Sine of non special angle Trig identities and examples Trigonometry Khan Academy.mp3 |
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. | Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy.mp3 |
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. | Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy.mp3 |
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. | Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy.mp3 |
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. | Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy.mp3 |
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. | Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy.mp3 |
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. | Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy.mp3 |
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. | Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy.mp3 |
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. | Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy.mp3 |
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. | Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy.mp3 |
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. | Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy.mp3 |
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. | Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy.mp3 |
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. | Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy.mp3 |
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. | Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy.mp3 |
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. | Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy.mp3 |
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. | Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy.mp3 |
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? | Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy.mp3 |
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. | Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy.mp3 |
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? | Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy.mp3 |
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. | Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy.mp3 |
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? | Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy.mp3 |
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. | Secant (sec), cosecant (csc) and cotangent (cot) example Trigonometry Khan Academy.mp3 |
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. | How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3 |
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? | How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3 |
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. | How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3 |
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. | How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3 |
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. | How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3 |
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. | How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3 |
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. | How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3 |
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. | How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3 |
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. | How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3 |
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. | How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3 |
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. | How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3 |
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. | How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3 |
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. | How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3 |
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. | How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3 |
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. | How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3 |
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. | How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3 |
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. | How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3 |
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. | How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3 |
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. | How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3 |
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. | How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3 |
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. | Sine and cosine of complements example Basic trigonometry Trigonometry Khan Academy.mp3 |
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. | Sine and cosine of complements example Basic trigonometry Trigonometry Khan Academy.mp3 |