06 - Review of Essential Trigonometry (Sin, Cos, Tangent - Trig Identities & Functions) - By Math and Science
Transcript
00:0-1 | Hello . Welcome back to the Physics one course . | |
00:02 | This title of this lesson is called a review of | |
00:05 | essential trigonometry . So every single physics problem that you're | |
00:10 | gonna work is going to seem challenging to you in | |
00:12 | the sense that no longer will the problem tell you | |
00:15 | exactly what you need to do to find the answer | |
00:18 | . So a lot of times , almost all of | |
00:20 | the time when we get past the basic problems , | |
00:22 | there will be an angle of some kind in the | |
00:25 | problem . Maybe it's a plane like a wedge where | |
00:28 | a block is sliding down in the the plane has | |
00:30 | an angle to it . Maybe a baseball is thrown | |
00:33 | at an angle . Maybe you're pushing with a force | |
00:35 | at 35° or whatever . I can go on and | |
00:38 | on and on . There's almost always going to be | |
00:40 | an angle in every problem , but you're never gonna | |
00:43 | be told , hey , you need to do the | |
00:45 | sign of the coastline of the tangent of this angle | |
00:47 | . You're never gonna be told that you're going to | |
00:48 | need to draw a picture and figure out from your | |
00:51 | previous knowledge of trig in order what to do , | |
00:54 | you're gonna have to do that yourself and that's why | |
00:56 | these problems seem difficult . So in this lesson we're | |
00:58 | going to review basic essential trigonometry . It's important for | |
01:01 | you to know that I expect that you have seen | |
01:04 | this stuff before . If you haven't , then you | |
01:06 | need to stop and go to my trade class , | |
01:08 | my trigonometry course and review the basic idea of angles | |
01:11 | and stuff . And I am going to review as | |
01:14 | we go through as we solve the problems , I'm | |
01:16 | gonna baby step you as much as I can to | |
01:18 | review you , but still you you need to have | |
01:20 | been exposed to this material . So the first thing | |
01:23 | we're gonna do the most important things these these , | |
01:25 | this is not a review of everything in trigonometry . | |
01:27 | this is the review of the most important things to | |
01:30 | solve physics problems . Right ? So everything in this | |
01:33 | lesson is critical , fortunately none of it is really | |
01:36 | hard . So if you have a triangle , remember | |
01:38 | this guy , you talk about triangles all the time | |
01:40 | in trigonometry , that's what it's all about . So | |
01:43 | you have some kind of triangle and this is a | |
01:46 | right angle which means it's 90° over in the corner | |
01:49 | and there's some angle , like I told you , | |
01:51 | you might have a wedge a lot of times in | |
01:53 | physics problems , we have a wedge and we call | |
01:54 | this angle theta . So tha don't let it scare | |
01:57 | you . It's just another variable like X . Or | |
01:59 | why this could be 15 degree angle or this could | |
02:01 | be a 35 degree angle or a 70 degree angle | |
02:04 | or whatever , of course is the angle increases the | |
02:07 | the steepness of this wedge gets taller or whatever . | |
02:09 | But this is a general diagram here . Now , | |
02:12 | we label the different parts of this triangle . We | |
02:14 | say that relative to this angle , the opposite side | |
02:17 | of this guy , we're gonna label it opposite O | |
02:20 | . P . P . For opposite in relative to | |
02:22 | this angle . This side right here is adjacent . | |
02:25 | Uh A D . J . We call the adjacent | |
02:28 | . So adjacent means the side kind of close to | |
02:30 | the angle opposite means the side opposite the angle and | |
02:33 | then the very longest side of every right triangles called | |
02:36 | the high partners . And I know that you all | |
02:37 | know that from basic um from basic math . Now | |
02:43 | , also in basic math , you learn something called | |
02:46 | the Pythagorean theorem . Just , just don't forget Pythagorean | |
02:51 | theorem . Don't forget that with right triangles like this | |
02:55 | , you always have the pythagorean theorem at your disposal | |
02:58 | . The physics problem will never say use the Pythagorean | |
03:01 | theorem to solve it will never tell you that . | |
03:02 | You just have to know that you can use this | |
03:04 | for every right triangle . Pythagorean theorem . If the | |
03:08 | way you learn geometry is C squared is equal to | |
03:10 | a squared plus B squared . In this equation , | |
03:14 | C . Is always the hypotenuse , it's always the | |
03:17 | longest side of this triangle , the other two . | |
03:19 | Since they're added together , it really doesn't matter adjacent | |
03:22 | or opposite . It doesn't matter what you label them | |
03:24 | . So the way you read this , in terms | |
03:26 | of this triangle that we have right here , since | |
03:29 | C . Is the hypothesis will write it as hypotenuse | |
03:32 | squared is equal to And again , I told you | |
03:35 | it really didn't matter . So we'll just call it | |
03:36 | adjacent squared plus opposite . Sure , what does this | |
03:43 | equation mean ? That means for any right triangle , | |
03:45 | no matter what the angle here is Whether it's 15 | |
03:48 | degrees , 30 degrees seven degrees . Really slender triangle | |
03:52 | , whatever . Doesn't matter if you know the length | |
03:54 | of the hypotenuse when you square it and you know | |
03:56 | the length of this and this and you square them | |
03:58 | separately , and then you add these together , they | |
04:00 | equal each other always . So it allows you to | |
04:02 | find the third side of the triangle . Okay . | |
04:05 | The reason I'm bringing this up is because we're gonna | |
04:07 | use it A little bit later . But it's important | |
04:09 | for you to know that applies to all triangles , | |
04:11 | not all triangles . All right triangles with a 90° | |
04:14 | angle like this one here . Alright , The next | |
04:17 | thing we're gonna do , we're gonna talk about the | |
04:19 | fundamental trig functions . Now . In truth , there | |
04:26 | are more trig functions than the ones we're going to | |
04:28 | list here . But these are the trig functions that | |
04:30 | are most used in physics . We come across anything | |
04:33 | more complicated than this . I'll explain it as we | |
04:35 | go , the first one I know that you all | |
04:37 | heard of . It's the sign of some angle theta | |
04:41 | . Now , in a real problem , you'll put | |
04:42 | a number here like 45 degrees or something . But | |
04:44 | we're drawing everything generally . So we say the sign | |
04:47 | of whatever this angle here is is defined to be | |
04:51 | the opposite side over divided by the high partners . | |
04:53 | So you literally take however long the opposite side of | |
04:56 | the triangle is and you divide by the high partners | |
05:00 | , right ? That's defined as the sign of a | |
05:01 | single . Notice that the sine of the angle involves | |
05:03 | the opposite side . Remember that ? Because that's gonna | |
05:06 | be important . The co sign of the angle of | |
05:10 | this triangle is equal to the adjacent side . This | |
05:14 | is divided by the hypotenuse again , notice that the | |
05:18 | co sine of the angle involves the adjacent side of | |
05:21 | the triangle . That's gonna be important . So remember | |
05:23 | that . So just without going any further , the | |
05:25 | sign of this angle involves something to do with the | |
05:28 | opposite side of the triangle . The co sine of | |
05:30 | the angle has something to do with the adjacent side | |
05:32 | of the triangle . They both involve the hypotenuse , | |
05:34 | but that's kind of not so important for right now | |
05:38 | The 10 . Well , it's important but it's not | |
05:40 | gonna be um not trying , not trying to make | |
05:45 | you remember that right now , the tangent of the | |
05:48 | angle which actually we're not going to use quite as | |
05:50 | much in physics problems . Most of the time you'll | |
05:51 | probably use one of these , you'll see why as | |
05:53 | we solve vector problems is defined as the opposite side | |
05:56 | , divided by the adjacent side , so it's the | |
05:59 | opposite , divided by the adjacent side . So these | |
06:05 | are critically important if there's one thing that I would | |
06:08 | write down in this review of essential trig . First | |
06:12 | , it's called a review , I expect you have | |
06:14 | seen this before and it's essential trick . Which means | |
06:16 | it's not everything . It's the most important stuff for | |
06:17 | these kinds of problems is I would want you to | |
06:20 | know these three things . Sine cosine tangent , if | |
06:22 | that's all you remember . That sign is opposite over | |
06:25 | hypotenuse co signs adjacent over hypotenuse tangent is opposite over | |
06:28 | adjacent . If that's all you remember , then you | |
06:31 | can solve a great many problems because most everything else | |
06:33 | that I'm going to talk about in this lesson is | |
06:35 | just gonna be messing around with these functions to do | |
06:38 | useful things . So , if you only have to | |
06:40 | remember one thing , remember these and understand what it | |
06:43 | means . I know a lot of you looking at | |
06:45 | this . A sine theta , what does that really | |
06:47 | mean ? It's literally just a fraction sign of whatever | |
06:50 | this angle is . Don't worry about what the label | |
06:52 | is . It just means that something that's related to | |
06:55 | this angle is defined as a fraction defined on this | |
06:58 | triangle . And if you forget which is which opposite | |
07:02 | adjacent , whatever . Then just remember how to put | |
07:04 | this picture together and draw this picture on your test | |
07:08 | . For many years . I had to draw right | |
07:10 | triangles and label everything so that I would keep it | |
07:12 | straight now . I just remember it . But if | |
07:14 | you need to draw that picture please do draw it | |
07:16 | . Now we're gonna take this and we're gonna do | |
07:18 | something practical with it on the other board here . | |
07:21 | So let's solve a quick little problem . This is | |
07:23 | not exhaustive , but we're just gonna do a quick | |
07:25 | little problem just to illustrate what I'm talking about . | |
07:28 | Here is a triangle like this . Now inside this | |
07:32 | triangle , I'm going to again , I do not | |
07:35 | know what this angle is . Sometimes I do know | |
07:37 | what the angle is , but in this case I | |
07:38 | don't . But what I do know now is that | |
07:40 | this side is three m long , This guy is | |
07:43 | four m long and this guy is five m long | |
07:47 | and I'm saying this is a right triangle . First | |
07:50 | of all , you can always verify this is a | |
07:52 | right triangle . You should always make sure this is | |
07:54 | a right triangle . How ? Because we know that | |
07:56 | C squared is equal to a squared plus B squared | |
07:59 | . This is true for any triangle that has a | |
08:01 | 90 degree angle in it . C . C . | |
08:04 | Is the longest side . So that's five . Now | |
08:06 | we have to square it and we're gonna check and | |
08:07 | see if it's equal to this squared plus this , | |
08:11 | where notice that A . And B . Doesn't really | |
08:14 | matter because see I did four squared plus three squared | |
08:17 | . If I called a equal to three instead , | |
08:19 | then it would still be three squared plus four squared | |
08:21 | . So the right hand side would be the same | |
08:23 | no matter what . So the only one you really | |
08:25 | care about is to make sure the longest side goes | |
08:26 | into cease column here . So this is 25 this | |
08:30 | is 16 and this is nine , So 25 equals | |
08:34 | 25 . Check if you ever get in inequality , | |
08:38 | like if you have 25 and this was 22 , | |
08:40 | then something's wrong because the triangle has to be a | |
08:44 | right triangle in order to even use these sine cosine | |
08:47 | and tangent definitions . And if this thing doesn't work | |
08:50 | out , then this is not a right triangle , | |
08:51 | because the pythagorean theorem always is true for a right | |
08:54 | triangle . Okay , so this was just an aside | |
08:59 | . I wanted to show you the pythagorean theorem works | |
09:01 | , but what I mostly want to do is I | |
09:03 | want to ask you the question , what is the | |
09:08 | sine of theta ? Mhm . So literally when this | |
09:12 | happens , you go back to your definitions which we | |
09:14 | just wrote now . So I'm gonna write it down | |
09:16 | again . The sign of data is equal to you | |
09:18 | . Remember , I want you to remember if there's | |
09:20 | one thing in this lesson , remember that the sine | |
09:22 | of an angle involves the opposite side . The co | |
09:25 | sign of anyone involves the adjacent side . You're going | |
09:28 | to find out why that's so important as we solve | |
09:31 | problems . The sign involves the opposite side of this | |
09:34 | triangle . And of course , it's divided by the | |
09:36 | hot news . Okay , opposite side opposite to what | |
09:39 | its opposite of this angle . So that means it's | |
09:42 | three . The high partners is along the side , | |
09:45 | which is five . All right . So what this | |
09:49 | means is that the sign of this angle is equal | |
09:52 | to 3/5 of an exact fraction . If you put | |
09:55 | that in your calculator , you'll find that 0.6 . | |
09:57 | So if I asked you what is the sign of | |
09:59 | fada you would circle this . The sign of data | |
10:02 | is equal 2.6 . All right . Let's go work | |
10:05 | the next part of this . Little simple problem . | |
10:07 | And then I'll circle back and draw some conclusions at | |
10:10 | the end . So that was for sine of the | |
10:12 | angle . Let's calculate for part B . Let's calculate | |
10:18 | the co sign . You know , let's do it | |
10:20 | like this . Um Let's do like this . I | |
10:24 | want you to find the co signed of the angle | |
10:27 | . That's the question there . So what is the | |
10:29 | co sign of the angle And I told you before | |
10:33 | the co sign involves the adjacent side . So sign | |
10:36 | involved , opposite co sign involves adjacent . Try to | |
10:39 | burn that in your mind . So you're gonna write | |
10:40 | that down adjacent and it's gonna be divided by the | |
10:43 | same thing , the hypotenuse . So you go back | |
10:45 | to your triangle . The adjacent sides of this angle | |
10:48 | is four and your divided by the same high partners | |
10:51 | . five . So it's an exact fraction 4/5 . | |
10:54 | Yeah . Um so you would write down that the | |
10:56 | co sine of the angle theta is exactly 4/5 and | |
11:00 | if you want to put that on your calculator you | |
11:01 | can 0.8 , that's the answer . So notice , | |
11:05 | first of all , before we go any farther , | |
11:07 | the sine of the angle came out to be some | |
11:09 | kind of decimal less than one . And the co | |
11:12 | sine of the angle also turned out to be some | |
11:15 | decimal less than one . That's always going to be | |
11:18 | true . Right ? So the sign when you , | |
11:20 | if you go take your calculator out right now , | |
11:22 | make sure it's in degrees degree mode because they're different | |
11:25 | modes in the calculator , we're gonna be using degrees | |
11:27 | in physics almost all the time . If you take | |
11:30 | any degree angle you want , stick it in your | |
11:32 | calculator , hit the sign button , you're always going | |
11:34 | to get a number less than one . If you | |
11:36 | stick any degree number you want any number you want | |
11:39 | and put it in the calculator and hit the coastline | |
11:41 | button , you're always going to get a number less | |
11:43 | than one . That's gonna be important to understand when | |
11:46 | I wrap this all up in a nice bow at | |
11:48 | the end , but you should always get decimals for | |
11:50 | these guys . Why ? Because they're fractions because we | |
11:53 | know that the hypothesis is always the longest side of | |
11:55 | this triangle and the hypothesis is always going to be | |
11:57 | on the bottom , then whatever on the top is | |
11:59 | always going to be smaller than the bottom . And | |
12:01 | that means that the fraction will always be less than | |
12:03 | one side and coast . I never get bigger than | |
12:05 | one ever . That's the bottom line . Yeah . | |
12:08 | All right . What is let's look at part . | |
12:10 | See what is tangent of this angle ? Tangent is | |
12:14 | defined as a mixture of the two . Tangent is | |
12:17 | the one that involves the opposite side , but it | |
12:19 | also involves the adjacent side . But there's no hypothesis | |
12:21 | anywhere here . So we just write down the exact | |
12:24 | thing that we we we know it's the opposite . | |
12:26 | Alright . Almost type road down adjacent , it's the | |
12:29 | opposite side . Over the hypothesis , opposite side is | |
12:34 | three . Uh What's not over our partners , opposite | |
12:37 | over adjacent . Sorry about that . Opposite over adjacent | |
12:42 | , opposite is three adjacent is four . So 3/4 | |
12:45 | and you all know that that's 40.75 So you would | |
12:48 | write down that the tangent of some angle . Theta | |
12:51 | is 0.75 Or you could write it as 3/4 . | |
12:54 | Let's find too . When their exact decimals like this | |
12:57 | , I'm okay with decimals . So you can go | |
12:59 | put it there . All right . So now we | |
13:01 | know what the sine of the angle is , What | |
13:03 | the coastline of the angle is , What the tangent | |
13:05 | of the angle is also noticed . The tangent of | |
13:07 | the angle is also less than one . Okay , | |
13:09 | for this example the tangents are less than one . | |
13:13 | Now . It turns out that the tangent function is | |
13:16 | not always in between . Uh is not always like | |
13:20 | less than one or whatever . The tangent function can | |
13:22 | go off the rails when you plot it . I'll | |
13:23 | explain that some other time . But for sine and | |
13:26 | cosine they always stay between plus one and minus one | |
13:29 | . Okay . I didn't graphic for you but it's | |
13:31 | plus one and minus one . This guy can kind | |
13:33 | of go all over the place and um I don't | |
13:35 | want to get into the reason why right now . | |
13:36 | But that's just something you can begin observed when you | |
13:39 | plot it . So let's find out the most important | |
13:43 | thing here . What is the angle theta . In | |
13:47 | other words , is this angle um 13 degrees . | |
13:50 | Is this angle 17 degrees . Is this angle 35 | |
13:53 | degrees . What angle is it ? The first thing | |
13:55 | I want you to know before we actually calculate the | |
13:57 | angle is this angle is locked in place by the | |
14:01 | distances of this triangle . In other words , if | |
14:03 | you got some string out or a pencil and you | |
14:06 | measured five and you measured four and then you went | |
14:09 | up vertically and measured three . There would only be | |
14:12 | one way that you could put those together to make | |
14:14 | an actual triangle that connected that one triangle would have | |
14:18 | an angle here . That would be fixed . It's | |
14:19 | fixed because the size of the triangle are locked in | |
14:23 | place and there you are arranging them in one way | |
14:25 | because they have to make a right triangle . So | |
14:28 | for any given set of sides , there's only one | |
14:30 | angle . There's only one answer . I just don't | |
14:32 | know . Is that 35 degrees ? Or is it | |
14:34 | 42 degrees ? What is it ? And so the | |
14:36 | way I figure it out is there's lots of different | |
14:38 | ways , but the first one is Let's go back | |
14:41 | up here . We calculated what the sine of the | |
14:43 | angle the sine of the angle here was , and | |
14:46 | it was .6 . All right . So we know | |
14:49 | that the sign of this angle is 0.6 . We | |
14:53 | calculated that before . So remember from equations , from | |
14:58 | equations like algebra equations you do the opposite to solve | |
15:02 | for X right ? You might divide by something , | |
15:04 | multiplied by something . Ad by some to get X | |
15:06 | by itself . But the angle that we want to | |
15:08 | find is wrapped up in a sign . How do | |
15:11 | we do that ? Well , we say , well | |
15:13 | , the data is going to be equal to . | |
15:14 | We have to do the inverse function . In other | |
15:18 | words , we have to undo it with an opposite | |
15:20 | . So if sign is the function inverse side , | |
15:23 | that's what the negative one means . This is inverse | |
15:25 | sine or arc sine . You might see it un | |
15:27 | does it . So like the opposite of addition and | |
15:30 | subtraction . In the opposite of multiplication and division . | |
15:32 | When solving equations , the opposite of squaring something is | |
15:35 | a square root . The opposite of sine , inverse | |
15:38 | sine , right . So when we do inverse sine | |
15:40 | to both sides of this thing , in verse sign | |
15:42 | on the left and does it . So we have | |
15:44 | failed to inverse sine on the right . So what | |
15:46 | we're writing here , So if you stick 0.6 in | |
15:48 | your calculator and hit the inverse sine button , You're | |
15:51 | gonna get the angle back and the angle in this | |
15:53 | case is going to be 36.87 Degrees . So that's | |
16:00 | the angle . So if you built this triangle , | |
16:02 | this would be 36.87°. . And it can't be any | |
16:05 | other angle or else you can't make the triangle to | |
16:06 | begin with . Now we use we calculated data using | |
16:10 | the sign . Okay ? But we can actually calculate | |
16:13 | it Using the co sign . Because we also learned | |
16:16 | for this triangle that the co sine of the angle | |
16:19 | is 08 . So if we want to get fatal | |
16:22 | by itself , we have to inverse cosine both sides | |
16:24 | . That eliminates that kind of annihilates the left hand | |
16:27 | side leaving Fattah by itself . It will be the | |
16:29 | inverse coastline . 0.8 . So if you put 0.8 | |
16:32 | in your calculator and find the inverse coastline button and | |
16:35 | hit that one , what do you think you're gonna | |
16:37 | get ? You're gonna get 36.87 degrees . 36 27°. | |
16:44 | . Notice that matches exactly everything self consistent . Whether | |
16:48 | you use the sign to find the angle or the | |
16:50 | co sign to find the angle you get the same | |
16:51 | thing . Now what do you think is going to | |
16:53 | happen if we use the tangent ? So let's go | |
16:56 | here . Just kind of squeeze it in the bottom | |
16:58 | . If tangent of the angle is .75 , then | |
17:00 | the angle should also be , what do you think | |
17:03 | the inverse Tangent of 0.75 ? So if you put | |
17:08 | 0.75 and find the inverse tangent button , what do | |
17:10 | you think you're gonna get ? You're gonna get 36 | |
17:13 | 87 The purpose of this example was two things to | |
17:17 | show you how to calculate sine cosine tangent and to | |
17:19 | show you that you can take any one of these | |
17:21 | things and do its inverse to find the angle . | |
17:24 | Alright , Because you're gonna be finding the angle a | |
17:26 | lot , a lot of times , it'll be here's | |
17:28 | what the baseball is doing . What is the angle | |
17:30 | ? What angle did you throw it at ? Well | |
17:32 | , eventually you're going to construct a triangle and you're | |
17:34 | gonna end up having to figure out what this angle | |
17:36 | is , and you'll use one of these . So | |
17:38 | the second part of this is to tell you that | |
17:39 | it doesn't matter which one you use , they all | |
17:41 | give you the exact same answer . So there's not | |
17:44 | so much a right way to find the angle . | |
17:46 | It's just there's about 16 different ways . Not literally | |
17:48 | , but there's a bunch of different ways to find | |
17:50 | the angle . Okay . The only other caution I | |
17:53 | want to throw at you when you're dealing with triangle | |
17:55 | trig , like this is this angle is um When | |
18:01 | you put these numbers in the calculator like .8 , | |
18:03 | find the inverse co sign , you do this inverse | |
18:05 | tangent . So on when you do this stuff , | |
18:08 | the angle that your calculator is going to give you | |
18:10 | is always going to be the positive angle . It's | |
18:12 | always gonna be the angle as if the triangle was | |
18:13 | drawn here and it's gonna give you this positive and | |
18:16 | this is a positive angle because it's measured from the | |
18:18 | X . Axis like this , this is X axis | |
18:21 | Y axis . So this is a positive angle , | |
18:23 | right ? As we go and do more problems . | |
18:26 | I'm going to caution you when you're inverse tangent thing | |
18:29 | or inverse co signer inverse sine . You've got to | |
18:31 | be a little careful because you have to know that | |
18:33 | that calculators always gonna give you the positive angle because | |
18:36 | sometimes what if I'm throwing the ball down , right | |
18:40 | ? So the triangle is not oriented up like this | |
18:42 | , it's oriented down then that means that I'll actually | |
18:45 | have a negative value for the why and a positive | |
18:48 | value for the X . And so long story short | |
18:51 | , you gotta be careful to look at the quadrants | |
18:54 | of what you're actually doing when it returns that angle | |
18:57 | , you just have to know that it's always going | |
18:58 | to give you that positive angle . But your problem | |
19:01 | might actually be throwing the ball the other way and | |
19:03 | you might actually have to add 180° to that angle | |
19:06 | or something like that to get the correct angle you | |
19:08 | want . But when we get to that point , | |
19:10 | I will I will explain and caution you as we | |
19:13 | do more problems like that for now , just know | |
19:15 | when you inverse cosine inverse tangent , you're always gonna | |
19:17 | get that positive angle back . That's all I want | |
19:19 | you to remember at this point . All right . | |
19:23 | So , um this is very important coming up next | |
19:27 | if you have a triangle . Yes , as we | |
19:31 | have been talking about and it has some angle theta | |
19:36 | and it has an opposite side and it has an | |
19:38 | adjacent side and it has a hypotenuse as they all | |
19:41 | do . I'm kind of regurgitating over and over again | |
19:46 | that we had talked about the fact that the sine | |
19:49 | of the angle was equal to the opposite . Remember | |
19:51 | sign deals with opposite over hypotenuse and we talked about | |
19:55 | the fact that the co signing the angle had to | |
19:57 | do with the adjacent side . Remember co sign deals | |
20:00 | with adjacent over hypotenuse . Absolutely true . But there's | |
20:05 | actually a little bit , I don't want to call | |
20:07 | it easier , but it's a very , very useful | |
20:09 | way to write this down . If you were to | |
20:13 | take this and solve it for the opposite side , | |
20:16 | how would you do that ? Well , this is | |
20:17 | a fraction . You have to multiply left and right | |
20:19 | by the hypotenuse , right . So the if you | |
20:22 | solve this equation , you're gonna see that the opposite | |
20:25 | side is equal to the hypotenuse times signed data . | |
20:32 | Mhm . Okay , let me do this one and | |
20:35 | I'll explain what I'm talking about here . Also with | |
20:38 | this one , the adjacent side is equal to the | |
20:42 | hypotenuse times the co sign of data . It was | |
20:47 | just these equations are the same thing as these . | |
20:51 | There's no different . I'm not introducing anything new . | |
20:53 | I'm just telling you that if you take this relation | |
20:55 | and you solve it for the opposite side , you'll | |
20:57 | have to multiply both sides by the hypotheses . So | |
20:59 | it cancels on the right , giving the opposite side | |
21:02 | here . You multiply by the high partners . Same | |
21:04 | thing here here . You multiply by the high partners | |
21:06 | . All right . Um These are very useful . | |
21:10 | Okay . Uh They are the exact same thing . | |
21:16 | Mhm as these . Of course they are but they | |
21:20 | are very useful . And the reason they're useful is | |
21:22 | because in physics we deal with something called a vector | |
21:25 | . I'm actually gonna explain really briefly what a vector | |
21:27 | is to you now , but then we're gonna have | |
21:29 | an entire lesson actually , two or three lessons on | |
21:31 | vector . So don't stress out if you don't get | |
21:33 | everything right now . But the point is these particular | |
21:37 | relations are very very useful . Um Let me explain | |
21:40 | why they're useful because this is one of the very | |
21:42 | first things you'll be doing in physics . Let's say | |
21:45 | I throw a ball this direction , right ? How | |
21:50 | fast do I throw it 10 m per second ? | |
21:53 | Obviously I'm not throwing it horizontally , I'm throwing it | |
21:55 | up with some angle . So what I do is | |
21:57 | I define the angle and the way I define an | |
21:59 | angle as I draw a right triangle , And I'm | |
22:01 | always throwing the ball relative to the ground . This | |
22:04 | is the ground here . So let's say I throw | |
22:07 | that ball at 35°. . Let's put numbers in here | |
22:09 | instead of just data . Okay , you all know | |
22:13 | that if I throw a ball at an angle up | |
22:15 | , I have to throw it kind of in two | |
22:17 | directions . Right ? I'm when I do that I'm | |
22:19 | throwing it horizontally , that's the horizontal speed . And | |
22:23 | I'm also throwing it vertically straight up and down . | |
22:26 | The mixture of those two motions horizontal and vertical is | |
22:30 | what gives you the I know I know the path | |
22:32 | curves , forget about the curving , but right , | |
22:34 | when it leaves your hand , It's going at an | |
22:36 | angle which is a mixture of those two . And | |
22:39 | that mixture is reflected in the triangle . You can | |
22:41 | see that if I should throw it at 35°, , | |
22:44 | that the horizontal component of the velocity is bigger than | |
22:49 | the vertical component of the velocity . So I might | |
22:52 | have a vertical speed . Right ? I know I've | |
22:59 | been throwing around the word speed and velocity . We'll | |
23:01 | define these terms a little bit , but I think | |
23:04 | all of you know , uh that what speed speed | |
23:07 | , more or less is horizontal speed . Okay , | |
23:12 | So the reason I'm showing you and telling you that | |
23:14 | these relationships are very useful is because of the following | |
23:17 | thing . Usually I throw a baseball . This arrow | |
23:20 | represents how fast I'm throwing that ball in total , | |
23:23 | in the angle direction . 10 m per second . | |
23:26 | That's what that means . 10 m every second that | |
23:29 | ball goes . That's pretty fast . That's faster than | |
23:31 | I can throw . But anyway , it's a nice | |
23:34 | number . Okay , But let's say that I want | |
23:37 | to figure out what is the vertical part of the | |
23:38 | speed . In other words , how much of the | |
23:40 | speed exists only in the up and down direction , | |
23:43 | and how much of the speed exists only in the | |
23:45 | horizontal direction . Because the angled speed is a mixture | |
23:48 | of the two . All right , so here's what | |
23:52 | you do . The vertical speed . notice in this | |
23:59 | triangle the vertical speed is the opposite side of this | |
24:02 | triangle . To the angle , notice that this equation | |
24:05 | tells me the opposite side of this angle to opposite | |
24:08 | side of the triangle is the hypothesis times the side | |
24:11 | . Right . So I'm trying to find the opposite | |
24:14 | side of this triangle , which I know from this | |
24:17 | equation is the hypotenuse times the sine . Okay , | |
24:23 | Which means that the opposite side of this triangle is | |
24:27 | equal to what ? 10 times the sine of 35 | |
24:34 | degrees . Right ? So the vertical speed Is equal | |
24:41 | to if you take sign 35 and hit that on | |
24:43 | your calculator and multiply the answer by 10 you get | |
24:46 | 574m/s . It's the same units as whatever I threw | |
24:51 | it with . Let me kind of leave that alone | |
24:53 | . Let's calculate the horizontal speed and then I will | |
24:56 | wrap it all up together and and impress upon you | |
24:59 | something that you really must remember . Okay so the | |
25:02 | horizontal speed . Yes horizontal speed . What is that | |
25:08 | ? That's the according in this triangle , that's the | |
25:11 | adjacent side of this triangle . That's what I'm trying | |
25:13 | to find the adjacent side of this triangle . But | |
25:16 | we just said from this equation from the definition of | |
25:18 | co sign , I can find the adjacent sign of | |
25:20 | any triangle hypotheses times co sign . So it's the | |
25:23 | hypotenuse Times The co sign of . Now I know | |
25:27 | the angles 35° right . And so let me switch | |
25:32 | colors too . Kind of put the point home so | |
25:35 | the horizontal speed is equal to what it's one of | |
25:42 | the hypothesis here is 10 . So let me go | |
25:44 | and just write it down here . So it's 10 | |
25:46 | times the co sign of 35 degrees . So if | |
25:49 | you take 35 degrees , hit the coastline but multiplied | |
25:51 | by 10 , you'll get 8.19 meters per second . | |
25:57 | Yes . Now , first of all , let's see | |
25:59 | if this makes sense . We're saying that we're throwing | |
26:02 | a baseball at 10 m per second at some angle | |
26:04 | . And since the angle is pretty small , less | |
26:07 | than 45 degrees , were saying just from the triangle | |
26:09 | , the way it's drawn , the horizontal speed should | |
26:12 | be larger than the vertical part of the speed . | |
26:14 | The horizontal speed should be larger than the vertical part | |
26:17 | of the speed . Uh And that's true . So | |
26:21 | why did I bring this up ? Because the probably | |
26:23 | the most important thing in trigonometry that you will learn | |
26:26 | in physics and this is why I'm turning around and | |
26:29 | I'm looking right at you because I really really want | |
26:31 | you to remember this is we use triangle trigonometry to | |
26:35 | take what we call vectors and break them up into | |
26:38 | what we call components . This 10 m/s is called | |
26:42 | a vector . Please don't let the words scare you | |
26:44 | , vector is not a complicated thing . It means | |
26:46 | I throw a ball at a certain speed . I | |
26:48 | mean you know that that's not that hard to know | |
26:50 | . The length of this arrow is 10 . Okay | |
26:53 | , but maybe I don't want to just know what | |
26:55 | that angled length is . I want to know how | |
26:57 | much is in this direction , horizontal and how much | |
26:59 | is in this direction , vertical . So we take | |
27:02 | that vector and we break it into components meaning a | |
27:04 | horizontal direction in a vertical direction . And we use | |
27:08 | sine cosine tangent every single time to do that . | |
27:10 | Mostly sign and co sign to do that . Okay | |
27:13 | , So here's the equations that way I taught you | |
27:15 | and then we apply down here . The opposite side | |
27:18 | of a triangle is hypotenuse time sign . The adjacent | |
27:21 | side of the triangle here is hypotenuse times coastline , | |
27:24 | hypotenuse is the total speed in this example , 10m/s | |
27:28 | . And I'm throwing the baseball . Here's what I | |
27:30 | have written in my notes . I'm not gonna write | |
27:31 | it down , I'm just gonna say it about four | |
27:33 | times to make sure you understand when we use a | |
27:37 | sine function and multiply it by the total length of | |
27:41 | the baseball . Uh speed . In this case , | |
27:44 | what the sign is doing , listen to me here | |
27:47 | is it's taking the total speed 10 m per second | |
27:51 | and it's chopping it is what I have actually . | |
27:53 | This is the way I think about it . It's | |
27:55 | chopping it down . You throw the ball at 10 | |
27:57 | m per second , but I want to chop it | |
27:59 | and I want to know only how much is going | |
28:01 | in the vertical direction , that's the opposite side going | |
28:03 | up . So when you take a sign of an | |
28:06 | angle and multiply it by the hypotenuse what you're doing | |
28:09 | is you're taking the total speed and you're chopping it | |
28:12 | finding how much of it goes in the vertical direction | |
28:15 | because vertical is opposite . Remember I told you Sign | |
28:18 | always deals with opposites and co signs always deal with | |
28:21 | adjacent . And that's to help you remember that . | |
28:24 | When you take the sign of this angle , multiply | |
28:26 | it by this , it takes this number and it | |
28:28 | chops it only to give you the vertical part of | |
28:31 | the speed , right ? And when you take this | |
28:33 | number and take its co sign that deals with with | |
28:35 | adjacent . So when you take the coastline of this | |
28:38 | and multiplied by the same thing , you're chopping it | |
28:41 | also , but only giving the horizontal part , which | |
28:43 | is this . So here's the moral of the story | |
28:45 | , because in this case I've given you a speed | |
28:48 | at 10m/s at an angle . But this vector thing | |
28:51 | we haven't talked much about , it can represent lots | |
28:53 | of things . It can represent a force . Maybe | |
28:56 | I'm pushing on something at an angle and I'm pushing | |
28:59 | with £100 of force or 100 newtons of force . | |
29:01 | We'll talk about Newton's later . It's the unit of | |
29:03 | force . Okay . But I want to know not | |
29:06 | . How much am I pushing at an angle ? | |
29:07 | How much am I pushing horizontally ? And how much | |
29:10 | am I pushing vertically ? Well , I do that | |
29:12 | by using triangle trig . I take this when I | |
29:16 | multiply it by the co sign of this . It | |
29:18 | chops it and gives me only the horizontal part when | |
29:21 | I take this and I multiplied by the sine of | |
29:23 | this . It chops it and it only gives me | |
29:25 | the vertical . It's crucially important because all these physics | |
29:28 | problems , what we're gonna do , we're gonna break | |
29:30 | them all apart into X . Direction . And we're | |
29:33 | gonna solve those and then we're gonna break them separately | |
29:35 | into Y . Direction and we're gonna solve those . | |
29:37 | We're gonna break everything apart and solve the different directions | |
29:40 | separately . It's important to be able to take this | |
29:42 | and chop it this way and chop it this way | |
29:44 | to get the two different components . Mhm . So | |
29:47 | here is your mega mega mega summary of everything we | |
29:50 | learned in this section , summary summary . All right | |
29:57 | . Here's the summary when we have a triangle like | |
30:00 | that . The sine of an angle is the opposite | |
30:04 | . What's the opposite ? Over the hypotenuse ? The | |
30:09 | cosine of an angle is the adjacent . Over hypotenuse | |
30:14 | the tangent of an angle is the opposite over the | |
30:20 | adjacent . So it doesn't involve a hypotenuse there . | |
30:23 | Okay , now we can take these guys and we | |
30:26 | can we can find use these equations to find the | |
30:30 | angles and every one of these cases . How do | |
30:31 | we do it ? We can say that the angle | |
30:34 | is equal to the inverse sine of the exact same | |
30:38 | thing opposite over hypothesis . We can say that the | |
30:43 | angles also equal to the inverse cosine of the adjacent | |
30:48 | over the high partners . And we did an example | |
30:50 | showing exactly that . And we can say that this | |
30:53 | angle is also equal to the inverse tangent of opposite | |
30:58 | over adjacent . Yeah , literally you take the fraction | |
31:02 | , stick it in there if you want to find | |
31:03 | the angle , you just inverse it . That's all | |
31:04 | you do . Okay , which one of these is | |
31:07 | correct . They're all correct . You can find the | |
31:09 | angle using any one of these relations and we actually | |
31:11 | showed you that a minute ago . And then the | |
31:14 | final thing I want to point out is that we | |
31:16 | can take these equations . So that's a useful set | |
31:19 | of relations . We can take these set of equations | |
31:21 | and we can write even more useful set of relations | |
31:24 | . We can say that the opposite side of the | |
31:26 | triangle is just the hypotenuse times the sine of the | |
31:31 | angle that's going to give you the vertical . If | |
31:35 | you kind of think of your triangle , the way | |
31:37 | we've been drawing it and the adjacent side of that | |
31:40 | triangle is the hypotenuse times the co sign of the | |
31:44 | language . The sign chops this thing to give you | |
31:47 | the vertical part and the coastline chops this thing because | |
31:49 | you're multiplying by the high partners giving you the horizontal | |
31:53 | part . So when you look at a trig book | |
31:57 | or physics book , you'll see these equations , you'll | |
31:59 | see these equations and you'll see these equations . And | |
32:01 | it gets very confusing . Which ones do I remember | |
32:03 | ? Really all you really need to remember are these | |
32:06 | these are the core equations . These just come from | |
32:10 | these , these just come from these . So really | |
32:12 | just remember these . But what's gonna happen is you're | |
32:14 | gonna use these so much that you will remember them | |
32:18 | to find the vertical part of a vector . You're | |
32:20 | gonna take the hypothesis and chop it with a sign | |
32:23 | , chopping with a sign gives you that vertical part | |
32:25 | to find the horizontal part of some vector . Some | |
32:27 | arrows hypothesis of a triangle are gonna take the hypothesis | |
32:30 | and you're gonna chop it with the co sign that | |
32:32 | gives you the adjacent to the horizontal part . Those | |
32:35 | are the core things of trig . When you take | |
32:38 | a trade course , you'll do so much more than | |
32:39 | what we've done here . You will plot sine cosine | |
32:42 | and tangent as you'll see that sign and co sign | |
32:44 | go up and down like waves and you'll see that | |
32:47 | tangent looks even crazier and there are other trig functions | |
32:49 | besides sine cosine and tangent that you learn in a | |
32:52 | trade course in a trade course . You'll also learn | |
32:54 | or a pre calculus course , You'll learn about the | |
32:56 | unit circle and you'll learn about how to calculate sine | |
32:58 | cosine and tangent using the unit circle and you'll learn | |
33:00 | about radiance and you'll learn how to convert radiant measure | |
33:03 | two degrees because that's a unit of measure of , | |
33:05 | of angle also . But in physics which is what | |
33:08 | I'm focusing on . These are the most important things | |
33:11 | . These things are important . If you don't understand | |
33:14 | this , watch it a few times . If you | |
33:15 | don't understand anything about it then go review , trade | |
33:18 | using my trig and calculus class and then come back | |
33:21 | here and continue learning physics with me . We'll do | |
33:23 | everything step by step and give you plenty of practice | |
33:25 | to do really well and get a deep understanding of | |
33:28 | the topics . |
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