11 - Learn ArcSin, ArcCos & ArcTan (Inverse Sin, Cos & Tan) - Part 1 - By Math and Science
Transcript
00:00 | Well , welcome back . The title of this lesson | |
00:02 | is called Arc Sine . Our co sign Arc tangent | |
00:05 | . This is part one . In other words , | |
00:07 | title of this lesson is inverse . Sine inverse Cosine | |
00:11 | inverse tangent . So we have kind of a a | |
00:14 | cousin to each of the trig functions and that would | |
00:17 | be the inverse , the inverse of the sign , | |
00:19 | the inverse of the co sign , the inverse of | |
00:20 | the tangent . Now , I'm really excited to teach | |
00:22 | this lesson because Arc Sine Arcos in our candidate gets | |
00:25 | very , very confusing with almost every book I've ever | |
00:28 | seen . It's really not the fault of the book | |
00:31 | . It's just that it's actually really hard to draw | |
00:33 | on paper what you know , what's going on here | |
00:36 | and get it across without actually talking to somebody . | |
00:39 | So what I'm gonna do , I think in the | |
00:40 | beginning is I've drawn some things myself and I want | |
00:43 | to show you the punch line of what's actually happening | |
00:46 | without any real explanation . Then I'm gonna walk you | |
00:49 | through a detailed , uh I don't want to call | |
00:52 | it a derivation but a detailed sequence of of of | |
00:54 | problems that shows you exactly why the ark assigned the | |
00:58 | arc sine . The arc tangent behave the way that | |
01:00 | they do because it is not obvious the first time | |
01:02 | you look at it . But I can tell you | |
01:04 | for those of you who think well , I'll just | |
01:06 | blow through this . No problem . It's just the | |
01:08 | opposite of the co signer , opposite of the sign | |
01:10 | . I really encourage you to watch the entire lesson | |
01:12 | and practice the problems because there are several gouaches in | |
01:15 | here that will come back to haunt you in the | |
01:17 | future . I can say that you will use this | |
01:19 | stuff all throughout algebra , all throughout trig and pre | |
01:23 | calculus , all throughout calculus and definitely all throughout Physics | |
01:26 | and engineering . So take a moment here and listen | |
01:29 | to what we're talking about . So you understand exactly | |
01:31 | how these things behave . Okay , up until now | |
01:35 | we have been taking the sign , the co signed | |
01:37 | the tangent whatever of an angle . So for instance | |
01:40 | , sign of 30 is one half , which is | |
01:42 | the same thing as sine of pi over six radiance | |
01:45 | is one half because five or six radiance is 30 | |
01:48 | degrees . Okay , so in your mind think Sine | |
01:50 | of pi over six is one half . Sine of | |
01:52 | 30 is one half . That's the direction we've been | |
01:54 | going . So there is an opposite operation to all | |
01:57 | of these functions called the inverse . Remember the inverse | |
02:00 | of a function just does the opposite . It kind | |
02:02 | of undoes the original function . Right ? So basically | |
02:05 | the arc sine does the opposite operation of the sign | |
02:08 | . So for instance , if you know , and | |
02:10 | I'm gonna write all this on the board , so | |
02:11 | don't stress out here in the beginning , but verbally | |
02:13 | just follow me here . If you know that the | |
02:15 | sine of pi over six is one half , then | |
02:18 | you know that the arc sine of one half is | |
02:22 | giving you back an angle of pi over six . | |
02:25 | Let me say that one more time when you do | |
02:26 | the arc sine operation of the number , you're asking | |
02:29 | for the angle back . So instead of taking the | |
02:32 | sine of an angle to get a number , we're | |
02:34 | gonna take the arc sine of some number and get | |
02:37 | the angle back . So it's the way that we | |
02:39 | find the angles , how we reverse , solve for | |
02:42 | the angles in a triangle or an equation or whatever | |
02:44 | is going to be basically the arc sine , the | |
02:46 | arcos in the arc tangent , it goes the opposite | |
02:48 | direction . So we're going to be taking the ark | |
02:51 | assigned the arc tangent , the dark side of some | |
02:53 | number and getting angle back from it . However , | |
02:56 | there's a huge gotcha here and that is the way | |
02:58 | the unit circle works . There's always multiple angles around | |
03:01 | the unit circle that can give you the same sign | |
03:05 | . So there's always gonna be multiple choices . And | |
03:07 | so we have uh in your books , you'll see | |
03:09 | exactly how we lock it down and and to know | |
03:12 | what angles the calculator the computer is going to give | |
03:15 | back to you . So we have to talk through | |
03:17 | that . So again , let me give you the | |
03:19 | punchline . Then we're gonna go through a quite a | |
03:21 | detailed set of problems so that you understand exactly where | |
03:24 | it's coming from here is the punchline for those of | |
03:26 | you who want to know the punch line . I | |
03:28 | am not gonna go through all of this right now | |
03:30 | because I'm going to go through it in detail in | |
03:32 | the lesson . But we have this function called the | |
03:35 | inverse sine . Also called the ark sign . It | |
03:37 | is the opposite operation of a sign . You feed | |
03:40 | numbers into the arc sine operation and you get angles | |
03:44 | back . The numbers that you feed into the arc | |
03:47 | sine can only be from plus one to minus one | |
03:49 | . I'll explain why later you put those numbers into | |
03:52 | the arc sine . You press the button on your | |
03:54 | calculator , outcomes and angle . But the calculator will | |
03:57 | only give you angles from negative pi over 22 pi | |
03:59 | over to these angles over here . If you hit | |
04:02 | any inverse sine of anything on the calculator when it | |
04:05 | gives you an answer back , it will never be | |
04:06 | outside of this range , negative 90 to 90 degrees | |
04:09 | or negative pi over 22 pi over two radiance . | |
04:12 | I will explain exactly why . As we go through | |
04:14 | the lecture here , then we have an inverse cosine | |
04:17 | operation called also called arc Co sign . Because the | |
04:21 | co sign can only be between negative one and one | |
04:24 | . I'll explain why in a minute . Then the | |
04:26 | input to the art co sign function can only be | |
04:29 | between negative one and positive one . And because of | |
04:32 | that , the angles that the calculator will give back | |
04:34 | as the baseline angles can only fall between zero and | |
04:38 | pi . You see negative one and one fall on | |
04:40 | the X axis here . So the angles that you | |
04:42 | will get back will only be up here in the | |
04:43 | top half of the unit circle zero to pi or | |
04:46 | 0 to 180 . That's what this means . But | |
04:49 | greater than zero . Less than pi . Greater than | |
04:51 | zero . Less than 1 80 . If you stick | |
04:54 | a number in your calculator and our co sign , | |
04:56 | inverse co sign . You will never get an angle | |
04:58 | outside of that range . Okay , arc tangent is | |
05:01 | the one that's really bizarre and weird and hard to | |
05:03 | understand it . First there's an inverse tangent called arc | |
05:06 | tangent operation . The inputs to the arc tangent can | |
05:09 | be way outside of the range of negative one , | |
05:11 | positive one . They can be negative infinity . Up | |
05:13 | to positive infinity . I will explain as we go | |
05:15 | through the lesson why that's the case . The numbers | |
05:18 | that you can feed into this operator . This is | |
05:20 | like a function essentially here that you can put into | |
05:23 | there can be any number at all , from negative | |
05:25 | infinity to positive infinity . But the angles that you | |
05:27 | get out of the arc tangent operation will only be | |
05:30 | between negative pi over two and pi over two negative | |
05:34 | pi over 22 pi over two or negative 90 deposited | |
05:36 | 90 . So this is the punch line . I | |
05:38 | don't expect you to understand anything other than the fact | |
05:42 | that when you do an arc sine operation or you | |
05:44 | do an art co sign operation or you do an | |
05:47 | arc tangent operation . What you're getting out as an | |
05:49 | answer is an angle and the angle that you get | |
05:52 | back out of each one of those operations has limits | |
05:55 | to it . Your calculator or computer will never give | |
05:58 | you an angle outside of those limits . For an | |
06:00 | arc sine , you will always get an angle between | |
06:03 | negative pi over two or pi over two back which | |
06:05 | is negative 90 deposited 90 For anarcho sign , you | |
06:08 | will always get an angle between zero and pi , | |
06:10 | which means zero and 100 and 80 for an arc | |
06:12 | tangent . You will always get an angle between negative | |
06:15 | pi over 22 pi over two Which is negative 92 | |
06:18 | positive 90°. . That's the punchline . Now , the | |
06:21 | question is why is that ? We have to dive | |
06:24 | into it . I don't want you to just memorize | |
06:26 | things . I want you to understand things . So | |
06:28 | , let's cover this up . We will revisit this | |
06:30 | in great detail . I promise you . And let's | |
06:32 | go down and talk exactly through the logic of why | |
06:35 | all this works . It is more then a one | |
06:38 | or a two minute thing . I can't compress it | |
06:40 | into three minutes . So please watch the whole thing | |
06:42 | and you will understand exactly the reasons why . All | |
06:45 | right . So , we know we know for instance | |
06:47 | that the sign of pi over six , which is | |
06:51 | 30 degrees right ? Remember five or six or 30 | |
06:53 | degrees is what you should all remember . Sign of | |
06:56 | 30 is the famous one half . I've been telling | |
06:58 | you sign of 30 is one half . Sine of | |
06:59 | 30 is one half . Now that we know radiant | |
07:02 | sine of pi over six is one half . So | |
07:04 | how do we go backwards ? If we know the | |
07:06 | sine of pi over six is one half , then | |
07:09 | we should be able to say the following thing . | |
07:12 | Arc sine which is inverse sine or the opposite operation | |
07:16 | of the sign of whatever the kind of the answer | |
07:20 | is one half . Let me do it this way | |
07:22 | , let me do it this way arc sine of | |
07:25 | um yeah , let me do it this way . | |
07:27 | Arc sine of one half is going to equal what | |
07:31 | angle like this is what this is basically asking you | |
07:33 | to do Arc sine of one half is what angle | |
07:36 | . Another way to write this , You can kind | |
07:40 | of translate this problem . I mean we've written it | |
07:41 | down here but it's the same thing as saying the | |
07:43 | sign of some angle is equal to one half . | |
07:47 | When you say arc sine of one half . What | |
07:49 | you're really asking is a sign of what unknown angle | |
07:52 | is equal to one half . And you know in | |
07:54 | your mind that sign of 30 is one half . | |
07:56 | So it makes sense . Right sign of 30 is | |
07:58 | 1/2 30°. . So we can then say that this | |
08:01 | angle is 30°, , but we don't like to talk | |
08:03 | about degrees too much now . So we say it's | |
08:05 | actually five or six radiance . Right ? So one | |
08:08 | way to write it is called arc sine . That's | |
08:11 | the cleanest way to write it . That means inverse | |
08:14 | of the sign operation going backwards . Give me the | |
08:17 | number , I give you the angle , that's what | |
08:19 | it is , right . But you can also write | |
08:21 | this in the following way . You can write it | |
08:24 | in a different way and sometimes you'll see it written | |
08:26 | like this in different books . You might see it | |
08:28 | written as the following sign with a little negative one | |
08:31 | up here of one half Is equal to Pi over | |
08:36 | six . This representation is exactly the same as this | |
08:40 | representation . So when you see sign to the negative | |
08:43 | one power , it does not mean that it's raised | |
08:46 | to the negative one power . It doesn't okay , | |
08:48 | it does not mean one over sine . Okay . | |
08:52 | It doesn't mean that the negative one is not an | |
08:55 | exponent . The negative one is if you remember back | |
08:58 | we talked about inverse functions a long time ago we | |
09:01 | said that negative one up there can mean inverse function | |
09:04 | . So this literally means inverse function of the sine | |
09:07 | function . So this is the inverse sine of some | |
09:10 | number , gives me some angle again . Going backwards | |
09:13 | . This is very confusing to write down because the | |
09:15 | negative one looks like an exponent . So it confuses | |
09:18 | a lot of people so it's much more clean to | |
09:20 | say arc sine , but you can write it either | |
09:23 | way and you'll see it either way in lots of | |
09:25 | books , you'll see in calculus or whatever . You'll | |
09:27 | see it written both ways , depending on the author | |
09:30 | . Okay , so it returns an angle back . | |
09:33 | Let's give another quick example and then I'm gonna show | |
09:36 | you the huge gotcha in here about why it's so | |
09:39 | confusing in just a second . So let's go and | |
09:40 | take a look at another one that we know very | |
09:42 | , very easily are very , very rapidly . What | |
09:45 | is the ark co sign of square root of 3/2 | |
09:52 | ? So we want to figure out we have the | |
09:54 | inverse of the cosine function . That means we want | |
09:57 | to take as an input , not an angle , | |
09:59 | we want to take , kind of what the unit | |
10:01 | circle gives you around the outside and red the number | |
10:04 | coming off the unit circle . And I want to | |
10:05 | get the angle back . That's what the arc co | |
10:08 | sign the arc sine does for . So what you | |
10:10 | can do in order to kind of translate this in | |
10:13 | your mind is you can then say that this is | |
10:15 | the same thing as saying that the co sign of | |
10:18 | some unknown angle is the square of 3/2 and the | |
10:21 | unknown angle is what you're kind of getting here . | |
10:23 | So the co sign of what angle here is going | |
10:26 | to do that . The coastline of pi over six | |
10:29 | degrees is a square to 3/2 , which means that | |
10:33 | kind of coming up in translating appear the arco sign | |
10:35 | of square to 3/2 is the angle pi over six | |
10:38 | or 30 degrees . How do we know that ? | |
10:40 | Because we know the sign of 30 degrees is one | |
10:42 | half the co sign of 30 degrees is the other | |
10:45 | number square root of 3/2 . So the co sign | |
10:47 | of 30 degrees square to 3/2 , which means the | |
10:49 | art co sign of the answer is the angle coming | |
10:52 | back . Yeah . Okay . And of course , | |
10:55 | another way to write this that you might see in | |
10:58 | other books down the road . Another way to write | |
11:00 | this is the co sign inverse . This means inverse | |
11:04 | . Cosine does not mean one over co sign or | |
11:06 | co signs of the negative . One power of the | |
11:09 | number square root of 3/2 is the angle coming back | |
11:14 | , pi over six . Okay , pi over six | |
11:20 | . So that's another example . Let's do another quick | |
11:24 | example here that we can talk about . What about | |
11:27 | arc tangent of square root of three ? Arc tangent | |
11:33 | of square root of three is equal to what another | |
11:35 | way to do this is to translate it in your | |
11:37 | mind when you see arc tangent of a number of | |
11:39 | the way you do it in your mind . Or | |
11:40 | on your paper Actually as you say , the tangent | |
11:44 | of some unknown angle that I don't know is going | |
11:47 | to be equal to the square root of three . | |
11:49 | Okay , now the angle that actually works here is | |
11:52 | pi over pi over three . So the answer to | |
11:57 | the arc tangent of this going backwards is the angle | |
11:59 | pi over three . Right ? How do I know | |
12:02 | it's pi over three . Well , I I know | |
12:04 | the answer but the reason is because the tangent is | |
12:06 | the same as the sign . So put sine of | |
12:09 | pi over three over co sign of Pi over three | |
12:16 | . What is the sine of pi over three ? | |
12:18 | That's 60 degrees right . Sine of pi over three | |
12:20 | is 60 degrees . So you no sign of 60 | |
12:22 | degrees . Is the square root of 3/2 . And | |
12:25 | then the co sign of 60 degrees , you know | |
12:28 | , is one half . If you flip and multiply | |
12:30 | the twos will cancel and what you get back out | |
12:32 | of it is the square root of three . So | |
12:34 | , you have to bust this tangent up into a | |
12:36 | sign and a co sign to figure it out . | |
12:37 | Unless you have it memorized . I don't actually have | |
12:39 | them memorized . So I have to go figure it | |
12:41 | out . That the square root of three is the | |
12:43 | is the result when you take the tangent of pi | |
12:45 | over three . So , because of that , the | |
12:47 | arc tangent of the pi over three squared of three | |
12:50 | . Uh Number is the angle pi over three . | |
12:54 | All right . And you can also write this in | |
12:57 | another way you can write this as or uh tangent | |
13:02 | Inverse of Square root of three is high over three | |
13:08 | , however , three . All right . So arc | |
13:11 | tangent , square root of three is the same thing | |
13:13 | as inverse tangent written with the negative one of square | |
13:16 | root of three . Do not write this or think | |
13:19 | that the -1 is an exponent . Do not try | |
13:21 | to add exponents . Do not try to subtract exponents | |
13:24 | don't raise the exponent to a square root and cancel | |
13:27 | them . Don't do any of that stuff because it's | |
13:29 | not a real exponent . It's just a notation that | |
13:31 | means arc tangent or arc sine or cosine . That's | |
13:34 | what it means . Now , the last thing I | |
13:36 | want to leave you with before we go into the | |
13:38 | details of why this is kind of so confusing is | |
13:42 | the following thing . You know . A long time | |
13:44 | ago I introduced the concept of a function as a | |
13:47 | box . A function is a box . You stick | |
13:50 | numbers coming into this box . Inside the box is | |
13:53 | a computation could be any kind of function X squared | |
13:57 | could be the function three , X plus four could | |
13:58 | be the function could be a linear function , could | |
14:00 | be a quadratic function , could be a square root | |
14:03 | as a function . All kinds of things can be | |
14:04 | in the box . Now we're learning a new kind | |
14:06 | of function basically , and it's called the inverse sine | |
14:09 | or the inverse co sign . So basically what you | |
14:12 | have is you feed numbers , not angles , just | |
14:15 | numbers . You feed numbers into this box . Right | |
14:21 | . And what is inside of this box ? This | |
14:24 | box I'm going to represent right now is arc sine | |
14:27 | or arc co sign or arc tangent . So it's | |
14:33 | , I'm putting three different functions here as examples . | |
14:36 | But the punch line is what comes out of the | |
14:39 | box , angles come out of the box . It's | |
14:43 | actually really easy to get confused . So I'm writing | |
14:46 | it down here and of course you can get degrees | |
14:48 | if you're working , injuries or radiance . So when | |
14:52 | you're taking an arc sine or an art co sign | |
14:53 | , you need to expect degrees coming out or you | |
14:56 | need to expect pi over 63 pi over four radiance | |
14:58 | coming out . Keep it in your mind because even | |
15:00 | though I know that you know this it gets confusing | |
15:03 | when you start doing art sign our coastline , all | |
15:05 | this stuff that you expect angles to come out . | |
15:07 | Sometimes we forget what we're doing . So numbers come | |
15:10 | in . We're gonna talk a lot about the numbers | |
15:13 | that are allowed to come into these boxes . The | |
15:15 | boxes contain arc sine or cosine or tangent and the | |
15:18 | output of these boxes are basically angles are getting degrees | |
15:21 | or radiance . All right , so we're done with | |
15:23 | the first part here . Now , the tricky part | |
15:26 | is what's coming up next . The punchline , the | |
15:28 | gotcha that I really want everybody to um to comprehend | |
15:32 | . Let's go back to our example , sign of | |
15:34 | some angle is one half . We already did it | |
15:37 | . We said a sign of some angles . One | |
15:39 | half means arc sine of one half is 30 degrees | |
15:42 | or pi over six . Okay , so let's go | |
15:45 | back to sign Of some angle is equal to 1/2 | |
15:52 | . Now remember back from basic algebra , you know | |
15:56 | you have inverse functions , inverse operations to solve equations | |
16:01 | . Right ? So if you have an equation like | |
16:03 | three x equal six . You're multiplying on the left | |
16:07 | hand side by three . So to get rid of | |
16:09 | it , you do kind of the inverse operator . | |
16:10 | And you divide opposite of multiplication is division . If | |
16:13 | you have a square root you might undo the square | |
16:16 | root by squaring both sides . So it un does | |
16:18 | the square root , I can go on and on | |
16:20 | . If you have a long algorithm on one side | |
16:22 | then to undo the log rhythm , you might raise | |
16:24 | both sides of the equation to a power of the | |
16:27 | base of the law algorithm . To undo the logarithms | |
16:29 | . So here's kind of a basic equation . I | |
16:31 | know it looks so simple because we know that sign | |
16:33 | of 30 is one half . So we can solve | |
16:35 | this equation really easily . But really the real way | |
16:38 | that you solve this equation is you undo the sine | |
16:41 | function and try to solve for data and get the | |
16:44 | function or the variable data by itself and then you | |
16:47 | should have some kind of angle . As a result | |
16:50 | we already know the answer . The answer is 30 | |
16:51 | degrees or pi over six . But really the way | |
16:54 | that you sign it , the way that you solve | |
16:56 | it is you have to undo the sign operation . | |
16:59 | How do you undo and assign function ? You do | |
17:01 | the inverse the arc sine to both sides of the | |
17:04 | equation . So to solve this simple equation , what | |
17:06 | we would do is apply the arc sine to both | |
17:10 | sides of the equation , we apply it to the | |
17:12 | left , which means we're applying the dark side to | |
17:15 | the side guy . And on the right we have | |
17:17 | to do the same thing to both sides . Just | |
17:19 | like we have to do the same thing to both | |
17:20 | sides of an equation all the time . So to | |
17:22 | solve this equation , we already know the answer . | |
17:24 | But basically we can apply the opposite functions of the | |
17:26 | left and then apply the opposite function to the right | |
17:30 | . All right now , what ends up happening is | |
17:32 | the arc sine is the exact opposite of the sign | |
17:35 | , so it under does the sign . And so | |
17:38 | all you have left on the left is data and | |
17:41 | on the right you have arc sine of one half | |
17:45 | . And we already talked about the fact that our | |
17:46 | sign of one half means that it's a 30 degree | |
17:49 | angle . So then data 30 degrees or pi over | |
17:53 | six , which we've already talked about . So I'm | |
17:56 | introducing arc sine our coastline but I want you to | |
17:58 | understand the big picture . We obviously want to go | |
18:01 | backwards so that we can solve lots of times when | |
18:04 | we want to figure out what the angle is . | |
18:06 | But also because later we're gonna have trig equations which | |
18:09 | are really big equations with trig and metric functions . | |
18:12 | The goal of them is always going to be to | |
18:14 | figure out what data is . So you have to | |
18:16 | have a way to rip open that box and get | |
18:18 | the data that's inside . And the way you do | |
18:20 | it is to is to basically annihilate the sign with | |
18:23 | its inverse , which is the dark side to annihilate | |
18:26 | a co sign , you use an arc cosine function | |
18:29 | to annihilate a tangent . You use an arc tangent | |
18:31 | and we're applying it to both sides . Like we | |
18:33 | solve all of these equation here . So we already | |
18:37 | said the angle here is Pi over six . So | |
18:40 | let's draw this and talk about it a little bit | |
18:43 | and I'll use the unit circle that we have on | |
18:45 | the other board as well . I think I want | |
18:46 | to do it probably right over here . Let's draw | |
18:49 | this function right here or this unit circle right here | |
18:52 | , we'll do a little sketch . And I also | |
18:53 | use the real unit circle that we have over there | |
18:55 | . What were essentially saying is that if we go | |
18:58 | to pi over six here here is about roughly speaking | |
19:01 | a 30 degree angle , it's not perfect , but | |
19:03 | this is roughly a 30 degree angle . So I'll | |
19:06 | put feta this is pie Over six , which is | |
19:09 | 30°. . What we're saying is that the sign of | |
19:13 | this angle of pi over six is one half and | |
19:16 | that's why it is the solution of the equation . | |
19:18 | Let's go check in on the inner circle . Here's | |
19:21 | pi over six , which is 30 degrees . The | |
19:24 | co sign is the first number of the sign is | |
19:25 | the second number . So here we're saying the sign | |
19:28 | of 30 is one half to sign up . Five | |
19:29 | or six is one half , which means the projection | |
19:31 | here is right because we're talking about the sign , | |
19:35 | the projections on the y axis , which is exactly | |
19:37 | in the middle , which means it's at one half | |
19:39 | exactly the case . But let me ask you this | |
19:42 | , are there any other angles around the unit circle | |
19:45 | that also have a sign that give you one half | |
19:49 | ? I mean think about what you're asking yourself , | |
19:51 | you're over here saying our sign of some angle is | |
19:55 | one half . Give me give me the angles at | |
19:56 | work and we figured out through doing all this arc | |
19:59 | sine arc sine of one half , we're getting the | |
20:01 | angles right and we got an angle five or six | |
20:03 | we said , hey , it works , it's the | |
20:04 | angle , that's the answer . But it turns out | |
20:07 | there's tons of other answers , There's tons of other | |
20:09 | angles that work because in order to solve this , | |
20:12 | all you need to do is figure out the angles | |
20:14 | such that the sine of the angle is one half | |
20:16 | here is one of those angles but walk with me | |
20:19 | over here and look at this . If this is | |
20:21 | pi over six , then right here would be to | |
20:24 | five or six , then here would be 35 or | |
20:26 | six , then here would be 45 or six , | |
20:29 | then here would be five pi over six . What | |
20:33 | is the sign of this angle 55 or six ? | |
20:36 | It projects onto the Y axis is exactly the same | |
20:38 | location here , it's going to be positive one half | |
20:41 | . So what we have figured out is that there's | |
20:44 | actually multiple answers to this . Pi over six is | |
20:47 | not the only answer , but let's just go through | |
20:49 | it . It's a sign of pi over six is | |
20:53 | indeed one half , so it does satisfy this equation | |
20:55 | . However , the sign of five , pi over | |
21:00 | six is actually also equal to one half . So | |
21:03 | it looks like there's another answer that lives over here | |
21:06 | . If you go back to the unit circle , | |
21:07 | the sign of this is one half . But look | |
21:10 | at this , the sign of this angle is also | |
21:12 | one half , because both of these angles project onto | |
21:14 | the exact same location . But wait , there's more | |
21:19 | , there's possible to have more angles . In fact | |
21:21 | , there's an infinity number of angles that will give | |
21:24 | you that same sign of one half . What happens | |
21:26 | if I go all the way around the unit circle | |
21:29 | and continue counting until I get here . What's gonna | |
21:32 | happen there ? Let's just take a look . You | |
21:34 | have five or 6 to 5 or 63 than uh | |
21:37 | Four than 5 , 5 , 6 and six and | |
21:39 | 7 and eight . Then nine . Then 10 10 | |
21:42 | , 11 , 12 , 13 , 5 or six | |
21:44 | is right here , 13 5 or six . So | |
21:47 | we can say that sign of 13 pi over six | |
21:52 | is also equal to one half . So this is | |
21:55 | 13 5 or six years , 14 , 15 , | |
21:57 | 16 . Here's 17 pi over six . It gives | |
22:00 | you the same projection here . So we also know | |
22:02 | that sign of 17 pi over six is equal to | |
22:07 | one half . So you see , I can play | |
22:08 | this game again and again . Any time I land | |
22:10 | here , any time I land here , I'm going | |
22:13 | to get the sign of these angles . Both of | |
22:15 | them are going to give me one half . Now | |
22:17 | , if I land on angles down here , all | |
22:19 | of these angles are gonna project to negative one half | |
22:23 | . So the sign of those angles down there don't | |
22:25 | work . Only the angles in the upper half plane | |
22:28 | give me the sign the same exact sign with the | |
22:30 | same exact sign . Once the positive one half . | |
22:33 | Also notice you can go in the negative direction . | |
22:37 | There's negative pi over six , negative 256 negative 356 | |
22:40 | negative 456 negative 556 negative 656 negative 75 or six | |
22:45 | is right here negative 75 or six is the same | |
22:48 | location . So the sign of negative seven pi over | |
22:52 | six is also equal to one half . I encourage | |
22:54 | you get a calculator or a computer and put all | |
22:57 | these angles in there and then hit the sign button | |
23:00 | . It's going to give you the same thing one | |
23:01 | half and that's because it's a unit circle and I | |
23:04 | can keep counting over and over again . Not only | |
23:07 | can I keep going in circles to land on the | |
23:09 | same place , but there's actually two different kind of | |
23:11 | fundamental angles that also give me the same sign . | |
23:14 | So the question is whenever I take the arc sine | |
23:17 | of one half , which is what I'm doing to | |
23:19 | solve this equation , what angle should I get back | |
23:22 | ? It seems like they're infinity number of angles . | |
23:25 | They're an infinity number of angles . What angle should | |
23:27 | the calculator return ? And this is where I come | |
23:30 | back to what we talked about in the beginning that | |
23:34 | certain , there's only a certain range of angles that | |
23:36 | the calculator will return . So I need to walk | |
23:38 | through . Why ? But in the back of your | |
23:40 | mind , before we jump into all this uh stuff | |
23:42 | that we have to talk about , just remember when | |
23:44 | I take the ark sign of anything , I can | |
23:47 | get an infinite number of angles because all I have | |
23:49 | to do is figure out the fundamental ones . And | |
23:51 | then I can keep spinning around the unit circle , | |
23:53 | finding an infinite number of additional angles for any arc | |
23:56 | sine or any art co sign or any arc tangent | |
23:58 | of any number , I can find tons of angles | |
24:01 | that all work . So the calculator is not going | |
24:03 | to give you an infinite number of answers . So | |
24:05 | how does it know what to give you ? This | |
24:07 | is what we have decided by convention in math . | |
24:10 | The way it's gonna work , let's first talk about | |
24:12 | only the inverse sine . It's called the inverse sine | |
24:15 | . It's important for you to know that the sine | |
24:17 | of an angle always lies between negative one and one | |
24:21 | always . How do you know ? Because it's a | |
24:23 | unit circle . The circle only has a radius of | |
24:27 | one . So if I take the sign of any | |
24:29 | number around here , it can only give me a | |
24:33 | maximum up here . The sign up here would be | |
24:35 | one positive one . Any angle over here will just | |
24:39 | be a fraction of that projecting onto this axis , | |
24:41 | it'll be a fraction of that . The sign over | |
24:43 | here is zero . The sign over here is also | |
24:46 | zero . The sign over here is positive one and | |
24:49 | the sign down here is negative one . Any angle | |
24:52 | down here is going to be projected here . It | |
24:54 | will be less than well , larger than negative one | |
24:57 | , basically anywhere between plus one and minus one in | |
25:01 | that range , larger than negative one and smaller than | |
25:04 | positive one . It's impossible to get a sign go | |
25:07 | type in any angle you want . Your calculator , | |
25:09 | hit the sign button , you will never get a | |
25:10 | number larger than one or smaller than negative one . | |
25:14 | Like you'll never get negative two or negative three or | |
25:16 | negative four because it doesn't work like that the way | |
25:18 | the projections work . Okay , So it's important for | |
25:22 | , you know , that the input , that the | |
25:24 | sine function can only spit out numbers between plus or | |
25:28 | -1 . That means that the ark sign , which | |
25:32 | is the opposite uh inverse . The opposite function of | |
25:35 | the sign can only take as inputs to the bit | |
25:39 | numbers between negative one positive one . Why ? Because | |
25:42 | the arc sine is going backwards , I'm feeding numbers | |
25:45 | in from the outside of the unit circle and I'm | |
25:47 | getting angles back . So because the sign can only | |
25:51 | give numbers between plus and minus one , then feeding | |
25:54 | numbers into the arc sine . Remember this picture I | |
25:57 | drew for you , feeding numbers into the arc sine | |
25:59 | can only be between plus or minus one because those | |
26:02 | are the only values that the sine function can give | |
26:05 | out . Those are the only numbers on the outside | |
26:07 | of the unit circle . So because of that , | |
26:09 | the only numbers that can go into this arc sine | |
26:12 | are actually between plus and minus one . If you | |
26:14 | go press let's say five , that's way outside that | |
26:17 | range and hit arc sine . You'll get an error | |
26:20 | . It won't do it because there is no angle | |
26:23 | . It has a sign that gives you a five | |
26:26 | or something because sign can only go between plus and | |
26:28 | minus one . So the point is the inputs to | |
26:30 | the arc sine can only be between plus or minus | |
26:32 | one . But what I'm trying to show here is | |
26:34 | the sign is as a projection on the y axis | |
26:37 | . So the numbers between negative one on the y | |
26:40 | axis down here and positive one on the y axis | |
26:43 | here . Any number that I give into this function | |
26:46 | is only going to be on this part of the | |
26:47 | y axis between here and here . So the fundamental | |
26:51 | angle , the basic angle that is returned from a | |
26:53 | calculator is going to be the smallest angles that have | |
26:58 | that as a sign . And so the smallest angles | |
27:00 | would be from zero up to pi over to and | |
27:03 | from zero down to negative pi over two . That | |
27:06 | is by mathematical definition , those are the only angles | |
27:09 | that are going to come back and that's going to | |
27:10 | cover all possibilities . Well , there's still other angles | |
27:14 | around the unit circle that will give you that sign | |
27:16 | . But what I'm saying is the calculator is only | |
27:18 | going to give you the fundamental angle back . We | |
27:21 | all know that we can spin around the unit circle | |
27:23 | and get additional angles that have the same sign . | |
27:25 | But as far as what is that fundamental angle that | |
27:28 | the calculator will give you back at the computer or | |
27:30 | by definition of what the function is able to give | |
27:32 | you back is always going to be in this range | |
27:35 | . Because if you think about it , the sign | |
27:38 | of , let's take some angles here . The sine | |
27:40 | of pi over two is going to give you one | |
27:42 | and the sign of negative pi over two is going | |
27:44 | to give you negative one and the sign of zero | |
27:46 | is going to give you zero . So any angle | |
27:49 | that I pick in this shaded blue region is going | |
27:51 | to give me a sign , appear at one or | |
27:53 | a sign at negative one And the sign only goes | |
27:55 | from -1 positive one . So by convention , when | |
27:59 | I put negative one positive one in the calculator , | |
28:01 | it's not going to give me an infinity of angles | |
28:03 | back , it's gonna give me the smallest angle possible | |
28:06 | that satisfies the thing . So when we go back | |
28:08 | over here and we said , hey , we're trying | |
28:10 | to find what angle gives me . Some of the | |
28:13 | sign is one half . It is true . There's | |
28:14 | an infinite number of angles , but there's only one | |
28:17 | fundamental angle and that's going to be an angle . | |
28:19 | This first one it's going to be between negative pi | |
28:22 | over two and pi over two . So when you | |
28:24 | hit the inverse sine that's the angle you're going to | |
28:26 | get back . You're never gonna get five pi over | |
28:28 | six because five pi over six is over here . | |
28:31 | 123455 or six . It's outside the range over here | |
28:35 | . You're only going to get that fundamental back angle | |
28:38 | back . So that's the sine function . Let's go | |
28:40 | have a similar discussion for the cosine function . The | |
28:43 | coastline functions inverse . Cosine called our coastline . It | |
28:47 | also can only give you values . The co sign | |
28:50 | function can only give you values between negative one and | |
28:52 | one . For the same exact reason any angle I | |
28:55 | pick is going to project onto the X . Axis | |
28:57 | . Which can only go between negative one and positive | |
29:00 | one . So because of that the arc co sign | |
29:04 | , the opposite function can only take his input values | |
29:07 | between negative one and positive one inputs to the arc | |
29:10 | cosine function but those negative one positive one . Those | |
29:13 | are projections on the X . Axis before it was | |
29:16 | projections on the Y axis . But for coziness projections | |
29:19 | on the X axis . So those go from negative | |
29:21 | one all the way to positive one along the X | |
29:23 | axis . So these are the inputs to the function | |
29:27 | . The output angles then are going to be the | |
29:29 | smallest angles that give me that projection and the smallest | |
29:34 | angles there is going to go from zero to pi | |
29:37 | write these down here . These angles aren't gonna ever | |
29:39 | give you a return value because um well you see | |
29:44 | down here in any angle down here would project to | |
29:46 | the same axis as this angle projecting down here . | |
29:49 | So you have to have unique angles . I mean | |
29:51 | think about it . If you sweep angles through here | |
29:54 | , what's the the co sign of zero ? It's | |
29:56 | going to be a one . What's a co sign | |
29:59 | of uh pie ? It's gonna give you negative one | |
30:03 | . Right ? What's the coastline of pi ? Over | |
30:04 | 20 So you see I've already said that the coastline | |
30:08 | can only go between negative one and one . So | |
30:09 | all of the angles required to do that up in | |
30:12 | the top here because the co sign of any of | |
30:15 | these angles here are going to go between negative one | |
30:17 | and positive one . So the pot the punch line | |
30:19 | is you can only feed negative one positive one in | |
30:22 | . But the angles that come out are going to | |
30:24 | be in the upper half playing up here between zero | |
30:26 | and pi . It's important for you to realize when | |
30:28 | you take an art co sign , hit the button | |
30:30 | on your calculator . You will always get an angle | |
30:32 | up here . You'll never ever get an angle down | |
30:34 | here , even though those angles down there can still | |
30:38 | give you the same co sign . We kind of | |
30:40 | lock the function down into like a base angle return | |
30:44 | . It's like the base angle that comes back , | |
30:46 | the most fundamental . The smallest angle that satisfies it | |
30:49 | will be up here . So , so far arc | |
30:52 | . Sine always gonna give you angles in the right | |
30:55 | half plane like this between negative pi over two and | |
30:57 | pi over to our co sign is always going to | |
30:59 | give you angles back between zero and pi . Okay | |
31:02 | , now the tangent function is the one that's hardest | |
31:05 | to understand and I really want you very much to | |
31:08 | understand it , so I'm going to have to do | |
31:10 | a little bit more talking now . What's going on | |
31:12 | here with the tangent ? Is that the tangent can | |
31:15 | actually go between negative and positive infinity . Uh in | |
31:19 | other words , the sign and the coastline . When | |
31:21 | you take signing coastline of angles , you always get | |
31:23 | between plus or -1 . However with tangent you can | |
31:26 | get numbers way outside of that . Why ? Because | |
31:30 | the tangent is the sign divided by the coastline . | |
31:33 | It's the sign divided by the coastline . So let's | |
31:34 | go through a couple of quick examples to remind you | |
31:37 | of this . What is the tangent of negative pi | |
31:42 | over two ? Negative pi over two . It is | |
31:45 | the sign of negative pi over two Divided by the | |
31:50 | co sign of Negative Pi over two . Right , | |
31:54 | so negative pi over two is down here , right | |
31:57 | down here on the unit circle . So what is | |
31:59 | the sign of that ? Well , the projection onto | |
32:01 | the y axis is negative one . What is the | |
32:04 | co sign down here ? Way down here , the | |
32:05 | coastline is zero . Okay so what do you have | |
32:08 | , you have negative 1/0 . And so what happens | |
32:11 | is the tangent of this angle is actually negative infinity | |
32:14 | . And if as you get away as you get | |
32:17 | away from this negative pi over two like angles really | |
32:19 | close , what happens is it gets really really , | |
32:21 | really big because the denominator is getting really really close | |
32:24 | to zero . So go ahead and do it , | |
32:26 | go ahead and put an angle very close to negative | |
32:29 | pi over two in but not quite there . And | |
32:31 | you'll see the tangents like you know , 100 million | |
32:33 | or something . It's because the top part is approaching | |
32:36 | negative one and the bottom part is getting closer and | |
32:39 | closer to zero . So we say at negative pi | |
32:41 | over two , you actually get negative infinity . Uh | |
32:44 | There for the tangent . Now let's take another example | |
32:47 | . We took an angle down here . Let's take | |
32:49 | the Angle zero . The tangent of zero radiance or | |
32:53 | zero degrees . Whatever is the sign of zero divided | |
32:58 | by the co sign of zero . What is the | |
33:00 | sign of zero ? At zero here , there's no | |
33:02 | projection on why ? So the sign of zero is | |
33:05 | in fact zero . Okay . And what is the | |
33:07 | co sign of zero ? The projections on the X | |
33:10 | axis . The coastline is one . So because it's | |
33:12 | 0/1 , actually the tangent of zero is zero . | |
33:15 | Certainly it's not infinity or anything . It's just a | |
33:17 | number , it's it's zero . Right now let's go | |
33:20 | back to the upper part . So we said this | |
33:23 | is negative pi over two . This is zero , | |
33:25 | this is positive pi over two . Let's take a | |
33:27 | look at what the tangent Of positive pi over two | |
33:31 | is . It's the sign of Pi over two divided | |
33:36 | by the co sign Of Pi over two . What | |
33:40 | is the sine of pi over two will appear . | |
33:42 | The sign is positive one positive one . What is | |
33:46 | the co sign of pi Over two will appear to | |
33:48 | co sign is zero . No projection onto the X | |
33:51 | axis . So you get a zero positive over zero | |
33:54 | , give you a positive infinity . So what I'm | |
33:56 | trying to prove to you through an example is the | |
33:58 | first time . It seems weird , students say , | |
34:00 | well , how can tangent go to positive and negative | |
34:02 | infinity ? But Sine and Cosine can't . Sine and | |
34:05 | Cosine are pure projections on the X and Y . | |
34:07 | Axis of the unit circle , So they can never | |
34:10 | be bigger or smaller than plus or -1 . But | |
34:12 | tangent is not a projection . It well , it | |
34:15 | is kind of but it's the division of two projections | |
34:17 | . So because it's sign over coastlines , sometimes the | |
34:20 | co sign on the bottom can go to zero . | |
34:22 | And because of that it can drive the tangent sometimes | |
34:25 | to negative infinity and sometimes all the way up to | |
34:28 | positive infinity . So because of that , I've proven | |
34:30 | to you the tangent function doesn't go between zero negative | |
34:34 | one and positive one . It goes between negative infinity | |
34:36 | , positive infinity . Start typing in angles into your | |
34:39 | calculator . Hit tangent over and over two different angles | |
34:41 | . You'll see that they go way outside the plus | |
34:44 | and minus one range . So because of this , | |
34:46 | because tangent can actually give values between negative infinity and | |
34:50 | positive affinity . What it means is the arc tangent | |
34:53 | . The opposite operation can accept as an input value | |
34:57 | values that are between negative infinity and positive infinity . | |
35:00 | And those values are going into the function . And | |
35:03 | then what I'm telling you here is that the angles | |
35:05 | that come out of the function are also limited to | |
35:08 | between negative pi over two and pi over two or | |
35:10 | negative 90 and positive 90 . Why are those the | |
35:14 | special angles ? Because remember the calculator or the computer | |
35:17 | or whatever is going to give you the smallest range | |
35:19 | of angles possible to satisfy the input . Kind of | |
35:23 | like the input values possible . So if I put | |
35:25 | negative infinity in or positive infinity in what are the | |
35:29 | range of angles at work ? I actually just proved | |
35:31 | it to you . If we put a negative pi | |
35:33 | over two in for the for the tangent , we | |
35:35 | actually get negative infinity . And if we put a | |
35:37 | zero and we get something in the middle which is | |
35:39 | zero . And if we put a positive pi over | |
35:41 | two and we actually get a positive infinity . So | |
35:43 | these angles here from negative pi over two up to | |
35:46 | positive pi over to cover all the possibilities of where | |
35:50 | tangent can go from negative infinity . Up to positive | |
35:52 | infinity . Think about it . Anything from here . | |
35:54 | On on up to hear the tangent here is negative | |
35:57 | infinity . Make sure I'm right here . Yeah , | |
35:59 | the tangent here is negative infinity . The tangent here | |
36:02 | is zero and the tangent here is positive infinity . | |
36:05 | So these angles cover all the possibilities any number you | |
36:08 | can think of sticking into a calculator , hitting inverse | |
36:11 | tangent . These angles will cover it and they're the | |
36:14 | smallest angles that will cover that range . So we | |
36:17 | say the inverse tangent can take as an input any | |
36:20 | number you want to put in and they can spit | |
36:22 | out as an output anything between negative pi over two | |
36:24 | and pi over two . Now , usually you see | |
36:26 | up here I have the equal signs and I have | |
36:28 | the equal signs . But for the tangent we don't | |
36:31 | usually put the equal signs because we don't like to | |
36:33 | deal with infinities in real life . So we say | |
36:35 | the inverse tangent . I mean your calculator really never | |
36:38 | gonna give you infinity back . I mean infinity is | |
36:40 | an infinite number , it never stops . So we | |
36:42 | really say the inverse tangent goes between negative pi over | |
36:45 | two and pi over two as its output functions as | |
36:48 | its output angles . Um But not exactly including like | |
36:52 | it gets incredibly close to negative pi over two . | |
36:54 | It drives it up super high basically to infinity all | |
36:57 | the way up to positive pi over two . So | |
37:00 | let me read my notes and make sure I didn't | |
37:01 | forget anything . Anytime you take the inverse trig function | |
37:04 | of a number , the calculator returns a base angle | |
37:07 | which is like the smallest angle possible to satisfy the | |
37:12 | smallest angle possible to satisfy what you gave it as | |
37:14 | an input . Okay , other angles exist around the | |
37:17 | unit circle that have the same sign or co sign | |
37:19 | or whatever . But the calculator doesn't return all those | |
37:22 | angles because there's an infinite number of them . It | |
37:24 | returns the smallest angle possible , which are these ranges | |
37:27 | I've written on the board here . So sometimes when | |
37:29 | you're solving equations , what you'll do is you'll get | |
37:32 | the base angle back The fundamental angle and sometimes you | |
37:35 | might have to add or subtract 180° or 360° to | |
37:39 | go into another quadrant depending on what the problem is | |
37:42 | telling you to do and we will solve problems like | |
37:44 | that later . But the fundamental base angle of what | |
37:46 | your calculator will give you is going to be given | |
37:49 | by these terms here . So what I want to | |
37:52 | do , I think this was a really long lesson | |
37:54 | , but I want to go through it all real | |
37:56 | quick at lightning speed to make sure we're all on | |
37:58 | the same page . Okay , so we know that | |
38:00 | the sign of 30 degrees five or six is one | |
38:02 | half . So we define an inverse going backwards where | |
38:05 | the arc sine of the right hand side gives us | |
38:07 | an angle back And so it gives us an angle | |
38:10 | back . And we can also write it as the | |
38:12 | inverse sine with the -1 . But that is not | |
38:14 | an exponent . Don't treat it like an exponent because | |
38:16 | it's not but you might see it in books like | |
38:18 | this . The art co sign of a number gives | |
38:20 | us an angle back , can be written in a | |
38:22 | similar fashion . So numbers come in , angles come | |
38:24 | out . Same thing with the arc tangent numbers come | |
38:26 | in , angles come out and we can write it | |
38:28 | with the -1 up there just like before . So | |
38:32 | graphically numbers come in arc inverse sine Cosine tangent . | |
38:36 | Arc sine cosine tangent are in here and angles come | |
38:39 | out degrees or radiance . So then we start asking | |
38:42 | ourselves the question , going back to our fundamental example | |
38:45 | , sine of an angle is one half what angle | |
38:48 | works ? So we apply the arc sine to both | |
38:50 | sides , solve and we basically want to figure out | |
38:52 | what the arc sine of one half is . That | |
38:53 | means what angle exists So that I take the sign | |
38:57 | of that angle and I get one half but you | |
38:58 | know it's 30 degrees , you know it's pi over | |
39:00 | six . So we put five or six there which | |
39:02 | is right here in the unit circle . But then | |
39:04 | we realized , wait a minute , there's other angles | |
39:06 | because this angle also has the sign equal to positive | |
39:09 | one half . And then we go even further and | |
39:12 | say well wait a minute . Not only is the | |
39:15 | sine of pi over six equal to one half and | |
39:17 | the sign of five pi over six also equal to | |
39:19 | one half . But if we spin around the unit | |
39:21 | circle to here then we see the sign of 13 | |
39:24 | 5 or six is one half . And if we | |
39:25 | spin around the unit circle to hear another revolution , | |
39:28 | we see that 17 5 or six , the sign | |
39:30 | of that's one half , even going negative to negative | |
39:34 | seven pi over six , the sign of it , | |
39:36 | the projection is always in the same place one half | |
39:38 | . So what angle does the calculator give you ? | |
39:41 | That's what the rest of the lesson was about . | |
39:43 | And we said that for the sine function it only | |
39:49 | gives values back between negative one and positive ones . | |
39:51 | So the inputs to the arcs and can only be | |
39:53 | from negative one positive one which are projections on the | |
39:56 | y axis between negative one and positive one . That | |
39:59 | means the angles the smallest ones that will work for | |
40:02 | this projection are always going to be between negative pi | |
40:04 | over two and positive pi over two . Just like | |
40:06 | this for the ark assign , it also gives numbers | |
40:09 | between negative one and positive one . So the inputs | |
40:12 | to the arc co sign is on the X axis | |
40:15 | between negative one and positive one . These are the | |
40:17 | projections on the X axis and the angles the smallest | |
40:19 | set of angles that work are going to be between | |
40:21 | zero and pi which is zero and 100 and 80 | |
40:24 | degrees . Then we said the tangents , the weird | |
40:26 | one . It actually can take values as inputs between | |
40:29 | negative infinity to positive infinity . Or I should say | |
40:31 | the tangent gives values back between negative infinity and positive | |
40:35 | infinity . So the input to the ark tan can | |
40:37 | be numbers from negative infinity to positive infinity . And | |
40:40 | we've already described why here it's because it's the ratio | |
40:43 | of two trade function . So sometimes that denominator goes | |
40:46 | to zero which drives the tangent to infinity . So | |
40:49 | if the tangent can go between negative infinity positive affinity | |
40:53 | , it turns out that the set of angles that | |
40:55 | allow that to happen are also between negative pi over | |
40:58 | two . Up to positive pi over two . Not | |
41:00 | inclusive because we really don't want to get to infinity | |
41:03 | . But anyway , those are the boundaries there . | |
41:05 | Why is that ? Because the tangent of negative pi | |
41:08 | over two is the sign over the coastline which is | |
41:10 | negative infinity . And the tangent of positive pi over | |
41:13 | two is the sign over the coastline which is positive | |
41:15 | infinity . So this set of angles from here to | |
41:17 | here is the smallest set of angles such that the | |
41:21 | tangent goes from negative infinity to positive infinity . Now | |
41:24 | that is a ton of talking , it's a lot | |
41:26 | of talking and that's why every book I've ever opened | |
41:29 | . It makes it incredibly hard to understand what they're | |
41:31 | doing because it's hard to describe in words but fundamentally | |
41:35 | that's what we're gonna do . So as we solve | |
41:36 | problems with the inverse sine or the inverse co sign | |
41:39 | or the inverse tangent , what we're gonna do is | |
41:41 | try to figure out the angles that make the equation | |
41:44 | work . We want to solve the angle because the | |
41:47 | angle is what comes out . But the angles that | |
41:49 | come out of an arc sine are only gonna be | |
41:51 | in this range and the angles that come out of | |
41:53 | anarcho sign are only going to be in this range | |
41:55 | , and the angles that come out of an arc | |
41:56 | tangent are only gonna be between this range . And | |
41:59 | so we're gonna have to keep track of that as | |
42:01 | we solve our problems and we will be doing that | |
42:03 | in the future lessons . So make sure you understand | |
42:05 | this as well as you can watch it a couple | |
42:07 | times . If you need to follow me on to | |
42:08 | the next lesson , we're gonna crank through a ton | |
42:10 | of problems involving the inverse trig and metric functions . |
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