| Start | ok, so now we are going to learn some other stuff.
So I guess we did law of sines last time? Now we will do the law of cosines. isn't that fun?. Ok, I can show you where this comes from. But you know I already showed you where the law of sines comes from. One derivation is enough Remember if you want to find the sides of a right triangle |
| 0:35 | and you know an angle and side you can use SOH CAH TOA
and thats what the trig ratios are for.
If you have a non-right triangles you can use the law of sines. Law of sines if you remember. Basically you have a triangle that is not right. What we do is we label the three angle A, B, and C with capital letters. |
| 1:03 | and we use the lower case a, b, c to stand for sides that are opposite to those angles
Opposite, opposite, opposite.
Ok? The law of sines says--- For those of you missed class on Monday it was all review except for the law of sines part-- The law of sines said that sine of any angle divided by its opposite side |
| 1:38 | is equal to sine of the other angle divided by its opposite side and equal to the sine of the third angle divided by its opposite side.
ok? Where are we able to use this? We are able to use the law of sines for certain triangles but not for other triangles. And where some information are been given |
| 2:06 | For example, if you are given two of the sides and the included angle
you would not be able to use the law of sines.
So, when can I use the law of sines? You can use it if you have angle, side, angle (ASA) or you can use it if you have side, angle, angle (SAA). Do you know what I mean by that? Angle, side, angle. For example if I have angle A, side b, and angle C. |
| 2:34 | One thing that we know is that the sum of interior angles of a triangle is 180 degrees.
So if you know two of the angles you immediately know the third angle. In these two cases, given that you have two angles you can find the third angle. And since we have a pair angles and the sides you can find the other side. However, sometimes you don't have that information. Sometimes instead we have different sets of information. |
| 3:05 | Then you need a second formula. ok?
You will use this formula if you either have only sides, all three sides (SSS) and you have side, angle, side (SAS). Now, you use the law of cosines. I can derive the law of cosines, but you know. |
| 3:32 | Not that exciting! the law of cosines says kind like the pythagorean theorem.
It says c squared is equal to a squared plus b squared but since we don't have to have , since we don't have a right triangle you have to make an adjustment. You have to subtract 2 a b cos(C). Thats the law of cosines. Let me take a picture of that!. |
| 4:04 | I can put it up on America's favorite Instagram.
That is a capital C that goes there. Let me make it easy to see! minus -- not plus! -- 2ab cos(C). That is our two laws. So you use these when you want to solve a triangle. |
| 4:31 | In other words finding the missing information depending of what you have been given. So, |
| 5:10 | Suppose I want to find c.
The letters are arbitrary of course. You can call these any letters you want. If you want to find c, well we know that c squared is a^2 + b^2 - 2 ab cos(C). |
| 5:32 | You get that from the law of cosines.
So now you literarily just plug into the formula. c^2 is 10^2 -- Sorry 12 squared. Now 12^2 + 10^2 -2(10)(12)cos(60 degrees). This is more fun with the calculator of course. |
| 6:04 | c squared is 144 + 100 -240 (1/2). Which is 244 - 120.
Very important, these go together. ok? you don't do 244 -240. ok? |
| 6:34 | People tend to do that mistake a lot.
Remember this has to get multiplied by cosine first. ok? PEMDAS. This whole thing has to be figured out first. Alright, so c squared is 124. c is the square root of 124. Whatever the square root of 124 is. ok? Pretty straight forward right? |
| 7:15 | Ok, say we've got it that way.
see if you can figure out that angle C. Yes! [student] could it be a word problem? |
| 7:30 | it could be word problem or it could be just a written problem like this where it would just say find the missing side.
I will give you an example of word problem in just a minute. ok? So if you want to find angle C, what do we know? So, lets write the formula. c^2= a^2 +b^2 -2ab cos(C). So 11 squared is equal to 9 squared plus 8 squared minus 2 times 9 times 8 cosine C. |
| 8:07 | We are trying to figure out what cosine of C is.
121, 81, 64 minus 144 cos(C) . |
| 8:30 | 121 = 145 - 144 cos(C).
Now as I said, don't do 145 -144. Thats is wrong!. ok? Remember 144 goes with the cosine of C. They multiply together. So now if you want to figure out what cosine of C is, we can subtract 145. We get -24 = -144 cos(C). |
| 9:01 | Divide and you get 1/6
So cosine C is 1/6.
Cosine of what angle is 1/6? I have no idea! right? Do you know? Use your calculator. Or, you will say c is the inverse cosine of 1/6, and just leave it as this. ok? So if you were doing this on a test question, the answer will be the inverse cosine of 1/6. |
| 9:34 | I will get back to that in a couple of minutes.
We touched on the inverses back about a month ago, and now we'll spend a little more time on them. in about 5 minutes Do you understand what I did? It is just manipulating the formula. You make sure you are comfortable with that! Of course on web-assign you can use your calculator but you have to learn how to do without the calculator in the actual test.ok? |
| 10:00 | So what if we had word problems.
What kind of word problems would you get? Are we ready for this? Yes..? [student] How did I get -24? I took 145 and subtracted from both sides. I don't do like the teachers in high school where the put the number underneath and you add and subtract. It wasn't the way I was taught... so, I don't do that! Ok fine I am going to. |
| 10:37 | I know you all were taught that way.
It is a very bad way of learning this because then you became dependent on writing that way. but, whatever. It's not my job to fix 5th grade mathematics. What about word problems? What would a word problem look like? |
| 12:34 | ok, so you go to a tower.
You measure the distance and you say that tower is 1400 meters away. You use one of those cool laser things. If you look at a different tower, after having turned 30 degrees, and that tower is 1800 meters away. How far apart are the towers? This is actually how they measure how far away stars are. Ok. You measure the angle to where a star is |
| 13:01 | then you move, and now you have a new angle.
and measure angle to the star again. And then you can find out, since you know how far you mocws, you can find how far it is to where the star is-- that's called parallax, it is similar to this. Basically, you create a triangle. Ok, so lets think about what we did. You stood here. You said this tower is 1400 meters. this tower is 1800 meters this distance, this angle is 30 degrees. |
| 13:38 | How far apart were those two towers?
So take a minute and figure it out. Use the law of cosines. Alright, so if you want to figure out what x is, we use the law of cosines. The law of cosine says that x squared will equal to 1400^2 + 1800^2 -2 (1400)(1800)cos(30 degrees). ok? |
| 14:15 | So x^2 is 1960000+ 3240000 - 5040000 times (square root of 3 over 2).
|
| 14:41 | Ok? Minus, minus. So x^2 is 5200000 (not that this really matters) - 2520000 radical 3.
|
| 15:08 | And whatever. ok? Did I get that right?
Am I losing some zeros somewhere? Or did I gain a zero? One more zero, Here we go! ok? You can compute that, I don't care what the number is. Ok? |
| 15:31 | Thats how you do with the law of cosines.
Anyone wanna compute that for me? Make sure that you use degree, not radian mode on the calculator otherwise you will be embarrassed. could write a law of cosines program if it makes it easier. |
| 16:04 | 913.9 meters
In SigFigs, 920 meters right? 910 meters.
How many significant figures do I have? I've got 2. Yes, so really it is 910 meters. But this isn't a chemistry class. This isn't physics class. |
| 16:30 | So you can through those out the window.
It does not matter. Ok, none of that matters. Thats how you use word problems. Yes, [student] oh yeah, I mean this is good enough. ok? In fact, the truth is this is good enough with radical 3 over 2 substitute in. I don't really care if you square 1400. |
| 17:03 | Yes, [student] oh I took 5040000 divided by 2 and multiplied with radical 3. You know that is one of those math things, ok?
|
| 18:00 | ok? Practice one more!
Then we move on to inverse stuff. Alright that is long enough. For those who were paying attention when we did the law of coisnes formula. So one of the things we can do is looking at the information that you have and you try to use the law of sines or the law of cosines. Ok? How do we know when to use the law of sines? We could try to use the law cosines and doesn't work! We will be missing too many pieces of information. |
| 18:30 | Ok? Or if you noticed if you have side, side, angle, angle .
Side, angle, angle you use law of sines. So you will say (sin 60)/b is (sin 45)/10. ok? You cross multiply. you get 10 sin 60 = b sin 45. |
| 19:05 | So 10 time radical 3/2 = b times radical 2 over 2.
two cancels. (10radical 3) over radical 2 = b. You don't need to rationalize that. ok? |
| 19:32 | Of course we don't give you angles if we don't know the sines and cosines
You just need it in terms of sine and cosine. ok?
because this is really 10 sin 60 over sin 45 equal b. So, if we gave me that angles, angles that you don't automatically know, you just leave it in terms of that. OK? Enough with this. Lets move on to some other stuff. |
| 20:12 | ok, now we do inverse trig.
We did a little of inverse trig last time now we will do a little more. So, we have trig right? when we have a thing function and an angle. For example I tell you that this is sin pi/6 = 1/2. |
| 20:32 | Inverse trig is working backwards.
So like you know the square root of 4 is 2 because you know that 2 squared is equal to 4. Ok? so, the sin pi/6 =1/2 is the same as saying pi/6, inverse sine of 1/2. Often written as arcsin 1/2. ok? |
| 21:03 | So what is the inverse sine of 1/2 mean?
The sine of what angle equals 1/2. That is what the inverse sine means in general. So if you know that sin x= A, x is the inverse sine of A. When we use that we don't divide, we reverse. |
| 21:32 | We do the inverse of something.
You cannot divide by sine. That sine is not multiplied by x. That is sin "of" x. ok? So you need the inverse function so kind of what we had before where we had the cosine of an angle equal to 1/6. So, we want to know what angle has a cosine of 1/6. And that is the inverse cosine of 1/6. ok? So that is what inverse trig is used for. |
| 22:01 | So these is a complicated rule for inverse functions, inverse trig.
Here is the problem. You know that sin pi/6 is 1/2 right? The sine of 30 degrees is 1/2. ok? |
| 22:34 | So what is the inverse sine of 1/2?
It si pi/6 because after all the sine of pi/6 is 1/2. But we could also say that it is 5pi/6. After all isn't sine of 5pi/6 equal 1/2? Yea! Sine 150 is 1/2. How about 13pi/6? Sure! The sine of 13pi/6 is 1/2. |
| 23:01 | How about 17pi/6?
In other words there is infinite number of angles whose sine is equal to 1/2. ok? So, we want to use what we call the principle angle. otherwise we get an infinite number of answers for something that it is not a function. The only one is pi/6. So how do we know what to use? OK? How do we know whats going on? |
| 23:31 | Well, lets think about the problem.
The problem is, think about the sine graph. The sine graph looks like that. So if you say where is equal to 1/2, we will have all these places. Just keeps going for ever. And, when we do the inverse perhaps we will fail the vertical line test. |
| 24:03 | ok?Thats the problem that its going on.
So we only want to use part of the sine graphs. Here and in here. Which is equivalent of between here and here on the original graph. So we only take a piece of the sine graph between here and here. When you do the inverse we will only get a single answer. |
| 24:32 | Ok? so that is the theory of whats going on.
In other words, when you want to do the inverse function we want to make sure that it will pass the vertical line test. So, we only use piece of sine graph so when you flip it, and you do the inverse will pass the vertical line test. We will restrict what happens with inverse sine, inverse cosine, all of those. |
| 25:20 | Ok sin x, the domain is all real.
The range is -1 to 1. |
| 25:34 | Remember what the sine graph looks like.
It only goes up to 1 and down to -1. If I want to do the inverse of a function I just switch the domain and range. So now the domain, x will be between -1 and 1. But I don't want the range to come to be all reals. so i restrict it and I say I am only going to take values -pi/2 to pi/2. |
| 26:02 | In other words, only values in the first quadrant, quadrant I and quadrant IV.
So if I ask you, you have a calculator, what is the inverse sine radical 3/2, you only get the answer of 60 degrees or pi/3. Ok? If I ask for inverse sine of radical 3/2. |
| 26:30 | There is infinite number of angles that have sine radical 3 over 2.
So I restricted only those angles that come out in the IV quadrant and I quadrant. That means that when I do the inverse, you get a function. So you do something similar for the inverse cosine. Inverse cosine, again I only do for values between -1 and 1. Cosine only goes up to 1 and down to -1. And now the restricted values are between 0 and pi. |
| 27:06 | Why is the cosine between 0 and pi.
Well, The cosine graph looks like that. The inverse cosine graph, looks like that!. |
| 27:32 | This is 1 and this is -1. So we only take a piece between here and here.
Because then it would not fail the vertical line test. It only goes from 0 to pi. That is why on the previous page I restricted the cosine between 0 and pi. |
| 28:02 | alright! I am not going to do inverse tangent at the moment, but we will work with inverse sine and cosine for a couple of times.
We will do some fun stuff! Again, remember what the inverse trig functions are all about. If you think about a trig function, plug in the angle you get out a number. So, if you plug the sine of an angle, you get out some decimal. |
| 28:31 | So, if you plug the sine of a different angle, you get out a different decimal.
So the inverse trig, if you plug in the decimal you get out an angle. So the inverse trig function gives you an angle. If I ask you for the sine inverse of x, what we are getting is an angle. ok? So what if I asked you sin( cosine-inverse of 2/5)? |
| 29:00 | What does that mean?
So, inverse cosine will be positive in the first quadrant. And remember the inverse cosine is an angle. And what it is? Is the angle that has a cosine of 2/5. So lets call it x. Because it does not matter. So inverse cosine of 2/5 means the cosine of some angle is 2/5. |
| 29:32 | So, the inverse cosine of 2/5 is that x.
Then, I just need to find the sine of x. How can I find the sine of x? SOH CAH TOA. Opposite over hypothenuse. So I can find the missing side with pythagorean theorem. So the sine of x is square root of 21 over 5. |
| 30:03 | So, again this isn't hard, its just takes time to write down the concepts.
If I ask you the sine, the inverse cosine of 2/5. What you do is you say, I am going to draw a triangle and put in an x and you know that the cosine of x is 2/5. Now, I am looking for the sin x. Well, once I know two sides, I can find the third side using pythagorean theorem. And now that I have three sides, I can find the sine. |
| 30:32 | Ok? [students] What a great question!
What if I said, what is the sine cosine inverse of -2/5? You say well, I am going to go in which quadrant? |
| 31:00 | You go to the second quadrant.
Because the inverse cosine and its either going to be in the first quadrant or the second quadrant. You go here. That is an angle. Use cosine of -2/5. If you do pythagorean theorem you get square root of 21 again. It has to be positive, because its pointing up. So I still want the sine of x. In this case it will be square root of 21 over 5. It would be exactly the same because, well that how the problem is. Ok? |
| 31:41 | Lets do another one to make sure we get the concepts.
|
| 32:08 | Lets say I want to do the cosine sine inverse of 5/11.
Well, that means that I have a triangle. And the sine inverse is positive so I am in the first quadrant. And the sine of that angle is 5/11. |
| 32:33 | And now I just want to find cosine of x. ok?
How do I find cosine of x? Well, cosine is adjacent over hypothenuse. I can just use pythagorean theorem here. or 4 radical 6. |
| 33:05 | Yes, [student] Oh what did I choose that quadrant?
because that makes the inverse sine of positive angle ok? So remember what I do to the inverse sine, up 90 degrees in the first quadrant, that would be positive answers and down 90 degrees to the fourth quadrant for negative answers. I am just giving you a good way to remember that. |
| 33:34 | Now I am going to through in inverse tan now.
Having inverse tan makes it easier. Ok, Inverse tangent of x is an angle whose tangent is equal to x. |
| 34:10 | ok, The inverse sine, inverse cosine, inverse tangent.
And doing over a positive value, I will always going to use first quadrant for the angle. |
| 34:32 | In doing the inverse sine of a negative number, I will use the fourth quadrant.
Except we have two choices.Right? Where is sine positive? The sine is positive in the first and second quadrant. And it is negative in the third and fourth quadrant. So if I want to do the inverse sine of an negative angle, I need to use the third quadrant angle or a fourth quadrant angle. So we never use the third quadrant. |
| 35:01 | I am using the fourth quadrant angle for the inverse sine.
For the inverse cosine, I have two choices; I can use the second or the third quadrant. Thats where the cosine is negative. I never use the third. So I am going to use the second quadrant. The tangent, I actually have two choices. And I am going to use the fourth quadrant. That has to do more on what the tangent graph looks like. So, Lets put that back up!. |
| 35:34 | Ok, So in other ways I did inverse sine of a positive angle, a positive number.
My angle is located in the first quadrant. If I had done the inverse sine of a negative five over eleven, I will get the angle down here. |
| 36:03 | ok? That will still come to square root of 96.
So this will still be (cos x= radical 96/ 11) ok? Lets practice a couple. They are not that bad! Alright! What about if I ask you for, |
| 36:40 | what if I ask for the sine tan inverse of 2/7?
Well 2/7 is a positive number, we go to the first quadrant and draw a triangle. And you know that the opposite side is 2 and the adjacent side is 7. |
| 37:08 | then I can find the hypothenuse 2 squared plus 7 squared is 53.
So that is the square root of 53. So, I want the sine of x. That is 2 over the square root of 53. |
| 37:39 | You get it! [student] You want to know how I add these, I used pythagorean theorem.
So, side squared plus side squared is the hypothenuse squared. [student] yes, because we had the hypothenuse we were looking for a side. ok? So this is a squared plus b squared equals to c squared. |
| 38:07 | Why is in the first quadrant? Im a doing the inverse tan of a positive angle.
I am doing the inverse trig function of positive number. So, if you are doing the inverse trig function, of any positive number I am always use the first quadrant angle. So lets practice a couple. |
| 39:06 | Ok.there is three to play with!
Alright! Lets do each of these. Then I will do little more explaining. Cosine sine inverse of -5/8. Negative 5/8. You say alright, Im doing the sine inverse of a negative angle. negative number, we say negative angle. |
| 39:32 | So that means my angle is down here.
The sine of that angle is -5/8. You use pythagorean theorem, 8 squared minus negative five squared is 39. The square root of 39. Therefore, this is the square root of 39 over 8. |
| 40:10 | [students] this is 8 squared ok?
[student] don't confuse when you are solving for the hypothenuse and when you are solving for a leg. ok? Second one. |
| 40:36 | Tan cosine inverse of 3/7.
So we are in the first quadrant. With a positive angle, positive number. So the cosine is 3/7. That is square root of 40. ok? And that is x. Tangent of x is square root of 40 over 3. |
| 41:04 | We could ask you for the secant, cosecant, and cotangent.
They all the same rules. Yes! [student] Sine inverse tan of -5/12. Negative 5/12 it will be down here in the forth quadrant. Tangent is -5/12. If you do pythagorean theorem it come out to 13. |
| 41:32 | The sine is -5/13.
You can think about a radical 169 if you wanted. If you have a low radical 60 or radical 82, or whatever you can lose a couple of points. So make sure that you do pythagorean theorem. You know by heart right? One of my favorite theorems! Pretty much the only one we know right! |
| 42:01 | ok? why is inverse tangent in the fourth quadrant?
Well, we really need to go over what the tangent graph looks like. You know that the sine graph looks like. You know what the cosine graph look like. Tangent graph kind ok looks like this. |
| 42:30 | Then, more of these.
And then repeats again. The tangent is periodic. ok? The tangent graph looks like that. So, we have the same problem we had for sine and cosine. |
| 43:01 | So, if you pick some number.
There is an infinite number of possible answers. So we have to restrict which answers do we want to use. Since we have the asymptote in the middle, of the x that we used, we are going to restrict just using the values of tangent that are closest to the origin. So when you graph inverse tangent functions, |
| 43:45 | now we could, when we were flipping this,
we could pile up the y -axis and down to the y -axis.
So we only go from pi/2 to -pi/2. Thats why we only use values in the fourth quadrant and values in the first quadrant. |
| 44:02 | ok? So Thats a range.
domain is all reals. If you take the inverse tan of anything you want. Got the idea? |
| 44:38 | So far so good?
I've finished everything I need to do today. Do you mind if we end five minutes early? |