Log in L. A. Wilson 11 years agoPosted 11 years ago. Direct link to L. A. Wilson's post “Could someone further ill...” Could someone further illustrate why arctan is restricted to QI and QIV, whereas arccos is restricted to QI and QII? I understand why arcsin is restricted to QI and QIV, and I understand why arccos is restricted to QI and QII. Why is arctan not restricted to QI, QIV and QII? For that matter, why not use all four quadrants for arctan (acknowledging that neither arcsin nor arccos use QIII)? • (31 votes) Jim Riley 10 years agoPosted 10 years ago. Direct link to Jim Riley's post “If arctan's range consist...” If arctan's range consisted of angles in QI, QII, and QIV, then what's arctan(-1)? There's an angle in QII, namely 135 degrees, whose tangent is -1, and there's an angle in QIV, namely 315 degrees (or -45 degrees, if you prefer) whose tangent is -1. In order for arctan to be a function, arctan(-1) must have just one value, and the same has to be true for arctan(x), no matter what real number x stands for. (44 votes) Albert 12 years agoPosted 12 years ago. Direct link to Albert's post “I have been trying to pra...” I have been trying to practice this concept but I find it very hard to solve without the table of angules, especially with arctan exercises, since is not always so easy find a slope of sqrt(3)/3 should we be solving them without any help? • (21 votes) Lauren 12 years agoPosted 12 years ago. Direct link to Lauren's post “For 45-90-45 and 30-90-60...” For 45-90-45 and 30-90-60 triangles, I try to memorize the SIDES of the triangles, not neccesarily the values of the trig ratios. That way I only have to memorize six SIDES, not over 20 ratios. With angles outside of these set triangles, just use a calculator. (30 votes) Jin Hee Kim 13 years agoPosted 13 years ago. Direct link to Jin Hee Kim's post “do we always have to conv...” do we always have to convert the answar to radians? • (17 votes) Amber Lane Dolezal 13 years agoPosted 13 years ago. Direct link to Amber Lane Dolezal's post “You will not always be re...” You will not always be required to answer in radians. Whether you answer in radians or degrees depends on what is asked of you. Neither one is better than the other. If you are doing a trig problem, and it was not specified whether to use radians or degrees, use whichever one you are more comfortable with. I personally work the problem out in both degrees and radians to check my answer. (25 votes) Venish 8 years agoPosted 8 years ago. Direct link to Venish's post “what does he mean by vani...” what does he mean by vanilla tangent? • (6 votes) John 8 years agoPosted 8 years ago. Direct link to John's post “The figure of speech "van...” The figure of speech "vanilla" refers to a popular flavor of ice cream, the idea being "something is as simple as you can get." So "plain vanilla" implies you're talking about the simplest or most common version of something. Other kinds (and combinations) of ice cream get more complex, sometimes much more complex: https://en.wikipedia.org/wiki/Ice_cream So comparing something to vanilla ice cream is a simplicity metaphor. (16 votes) Mike 12 years agoPosted 12 years ago. Direct link to Mike's post “I'm having a lot of troub...” I'm having a lot of trouble with this subject. Could somebody walk me through a detailed explanation of this problem; What is the principal value of sin^-1 (-1/2)? • (7 votes) gopika0908 12 years agoPosted 12 years ago. Direct link to gopika0908's post “here:look at the questio...” here: therefore ur answer is -30 degree (11 votes) Sneha Verma 11 years agoPosted 11 years ago. Direct link to Sneha Verma's post “At 4:29, why isn't the tr...” At • (7 votes) Sid 11 years agoPosted 11 years ago. Direct link to Sid's post “That's just a convention ...” That's just a convention - the principal values of arcsin and arctan are in Q1 & Q4 while the principal values of arccos are in Q1 & Q2. (6 votes) kittypuppy02 11 years agoPosted 11 years ago. Direct link to kittypuppy02's post “How did you know how to d...” How did you know how to draw a -1 slope at • (9 votes) Eman 10 years agoPosted 10 years ago. Direct link to Eman's post “First he needs to discove...” First he needs to discover the angle of the triangle, in this case it's 45º, and then he traces the line according to the angle. (2 votes) Skullcrusherj 4 years agoPosted 4 years ago. Direct link to Skullcrusherj's post “How would we find somethi...” How would we find something like "arctan(5)"? • (4 votes) obiwan kenobi 4 years agoPosted 4 years ago. Direct link to obiwan kenobi's post “This kind of thing you ha...” This kind of thing you have to put into the calculator. Maybe there are ways to do this by hand but they are long tedious calculations. It is much simpler to just use your calculator and it is much more efficient as well. Hope this helps! (7 votes) Leaven Bread 3 years agoPosted 3 years ago. Direct link to Leaven Bread's post “At 4:03 how does Sal know...” At • (5 votes) loumast17 3 years agoPosted 3 years ago. Direct link to loumast17's post “So Sal said tan = sin/cos...” So Sal said tan = sin/cos, but I am going to use tan = opposite / adjacent. So now we have a right triangle, and since we are trying to salve arctan(-1) this means we want some angle x where tan(x) = -1. Well, we know tan(x) = opposite/adjacent so -1 = opp/adj. To make a fraction equal 1, what do we need? we need the numerator and denominator to be the same. of course it's -1 so one will be negative. that means our right triangle has the two legs (non hypotenuse sides) as something like like 3 and -3. It can be any number, I just chose 3 at random. Now, if you do not know this it is good to know, if the length of each leg of a right triangle is the same then it has matching angles, aside fromt he right angle. Does that make sense? (4 votes) Tara Stogner 7 years agoPosted 7 years ago. Direct link to Tara Stogner's post “At 3:54, how does Sal kno...” At • (5 votes) jwinder47 7 years agoPosted 7 years ago. Direct link to jwinder47's post “Sal is following a conven...” Sal is following a convention that he hasn't introduced yet (it will come in a subsequent video). That's the only reason (at this point) for choosing the 4th quadrant, and for measuring the angle as -45 rather than 315. So you are right: 135 (second quadrant) is an equivalent answer, at this point. The problem is that "the inverse functions of sine, cosine, and tangent,...are not really invertible." This means we must restrict their domains to make them invertible, and this is why we are obliged (by convention, or definition) to designate -45 degrees as the (principle) value of arctan(-1). This will (hopefully) make more sense as you continue to study these inverse functions. (Thanks for the time-stamp: they really help with a question like this :-) (2 votes)Want to join the conversation?
Restricting the range of arctan to quadrants 1 and 4 isn't the only possible way to define it, but it work well because (a) it forms an interval, provided you think of it as -90 to 90 degrees (or -pi/2 to pi/2 radians), and leave out the endpoints -90 and 90; (b) no two angles in that interval have the same tangent (so we'll never get two answers for a question like "what's arctan(-1)?"); (c) no matter what real number x you choose, some angle in that interval has x as its tangent (so a question like "what's arctan(-1)?" will always have an answer); and (d) there aren't any angles in that interval whose tangent is undefined. Of those four properties, only (b) and (c) are absolutely necessary, but (a) and (d) make for simplicity.
The interval from 90 to 270 degrees (again not including 90 and 270) would also satisfy all four properties listed above, and in theory would work just as well as -90 to 90. However, if we defined arctan(x) to always be an angle in that interval, then we'd have to say arctan(0) = 180, which wouldn't be as satisfying as arctan(0) = 0!
if yes, why? if no, what it better?
(clarifying edits, typos - I swear I read these things before posting :-) )
look at the question in this way
sin^-1 (-1/2)=what angle of sin?(let that be theta)
now break down the question
sin theta =-1/2....now what angle will be equal to -1/2
which is - 30 degree
4:29 3:48 4:03 3:54
Video transcript
In the last video, I showed youthat if someone were to walk up to you and ask you whatis the arcsine-- Whoops. --arcsine of x? And so this is going to beequal to who knows what. This is just the same thingas saying that the sine of some angle is equal to x. And we solved it in a coupleof cases in the last example. So using the same pattern--Let me show you this. I could have also rewrittenthis as the inverse sine of x is equal to what. These are equivalentstatements. Two ways of writing theinverse sine function. This is more-- This is theinverse sine function. You're not taking this tothe negative 1 power. You're just saying the sine ofwhat-- So what question mark-- What angle is equal to x? And we did this inthe last video. So by the same pattern, if Iwere to walk up to you on the street and I were to say thetangent of-- the inverse tangent of x is equal to what? You should immediately in yourhead say, oh he's just asking me-- He's just saying thetangent of some angle is equal to x. And I just need to figureout what that angle is. So let's do an example. So let's say I were walkup to you on the street. There's a lot of a walkingup on a lot of streets. I would write -- And I wereto say you what is the arctangent of minus 1? Or I could have equivalentlyasked you, what is the inverse tangent of minus 1? These are equivalent questions. And what you should do is youshould, in your head-- If you don't have this memorized, youshould draw the unit circle. Actually let me just do arefresher of what tangent is even asking us. The tangent of theta-- this isjust the straight-up, vanilla, non-inverse function tangent--that's equal to the sine of theta over the cosine of theta. And the sine of theta is they-value on the unit function-- on the unit circle. And the cosine oftheta is the x-value. And so if you draw a line--Let me draw a little unit circle here. So if I have a unitcircle like that. And let's say I'mat some angle. Let's say that'smy angle theta. And this is my y-- mycoordinates x, y. We know already thatthe y-value, this is the sine of theta. Let me scroll over here. Sine of theta. And we already know that thisx-value is the cosine of theta. So what's the tangentgoing to be? It's going to be this distancedivided by this distance. Or from your algebra I, thismight ring a bell, because we're starting at the originfrom the point 0, 0. This is our change in yover our change in x. Or it's our rise over run. Or you can kind of view thetangent of theta, or it really is, as the slope of this line. The slope. So you could write slope isequal to the tangent of theta. So let's just bear that in mindwhen we go to our example. If I'm asking you-- and I'llrewrite it here --what is the inverse tangent of minus 1? And I'll keep rewriting it. Or the arctangent of minus 1? I'm saying what angle givesme a slope of minus 1 on the unit circle? So let's draw the unit circle. Let's draw the unitcircle like that. Then I have my axes like that. And I want a slope of minus 1. A slope of minus 1looks like this. If it was like that, itwould be slope of plus 1. So what angle is this? So in order to have a slopeof minus 1, this distance is the same as this distance. And you might already recognizethat this is a right angle. So these angles haveto be the same. So this has to be a45 45 90 triangle. This is an isosceles triangle. These two have to add up to 90and they have to be the same. So this is 45 45 90. And if you know your 45 45 90--Actually, you don't even have to know the sides of it. In the previous video, wesaw that this is going to be-- Right here. This distance is going to besquare root of 2 over 2. So this coordinate in they-direction is minus square root of 2 over 2. And then this coordinate righthere on the x-direction is square root of 2 over 2 becausethis length right there is that. So the square root of 2 over 2squared plus the square root of 2 over 2 squared isequal to 1 squared. But the important thingto realize is this is a 45 45 90 triangle. So this angle right here is--Well if you're just looking at the triangle by itself, youwould say that this is a 45 degree angle. But since we're going clockwisebelow the x-axis, we'll call this a minus 45 degree angle. So the tangent of minus 40--Let me write that down. So if I'm in degrees. And that tends tobe how I think. So I could write the tangent ofminus 45 degrees it equals this negative value-- minus squareroot of 2 over 2 over square root of 2 over 2, whichis equal to minus 1. Or I could write the arctangentof minus 1 is equal to minus 45 degrees. Now if we're dealing withradians, we just have to convert this to radians. So we multiply that times--We get pi radians for every 180 degrees. The degrees cancel out. So you have a 45 over 180. This goes four times. So this is equal to-- youhave the minus sign-- minus pi over 4 radians. So the arctangent of minus 1 isequal to minus pi over 4 or the inverse tangent of minus 1 isalso equal to minus pi over 4. Now you could say, look. If I'm at minus piover 4, that's there. That's fine. This gives me a value ofminus 1 because the slope of this line is minus 1. But I can keep goingaround the unit circle. I could add 2 pi to this. Maybe I could add 2 pi to thisand that would also give me-- If I took the tangent of thatangle, it would also give me minus 1. Or I could add 2 pi again andit'll, again, give me minus 1. In fact I could go tothis point right here. And the tangent would alsogive me minus 1 because the slope is right there. And like I said in the sine--in the inverse sine video, you can't have a function thathas a 1 to many mapping. You can't-- Tangent inverseof x can't map to a bunch of different values. It can't map tominus pi over 4. It can't map to 3-- what itwould be? --3 pi over 4. I don't know. It would be-- I'll justsay 2 pi minus pi over 4. Or 4 pi minus pi. It can't map to all ofthese different things. So I have to constrictthe range on the inverse tan function. And we'll restrict it verysimilarly to the way we restricted the sine--the inverse sine range. We're going to restrict it tothe first and fourth quadrants. So the answer to your inversetangent is always going to be something in these quadrants. But it can't be thispoint and that point. Because a tangent functionbecomes undefined at pi over 2 and at minus pi ever 2. Because your slopegoes vertical. You start dividing--Your change in x is 0. You're dividing-- Yourcosine of theta goes to 0. So if you divide bythat, it's undefined. So your range-- So if I--Let me write this down. So if I have an inverse tangentof x, I'm going to-- Well, what are all the values thatthe tangent can take on? So if I have the tangent oftheta is equal to x, what are all the different valuesthat x could take on? These are all the possiblevalues for the slope. And that slope cantake on anything. So x could be anywherebetween minus infinity and positive infinity. x could pretty muchtake on any value. But what about theta? Well I just said it. Theta, you can only gofrom minus pi over 2 all the way to pi over 2. And you can't even include piover 2 or minus pi over 2 because then you'd be vertical. So then you say your--So if I'm just dealing with vanilla tangent. Not the inverse. The domain-- Well the domain oftangent can go multiple times around, so let me notmake that statement. But if I want to do inversetangent so I don't have a 1 to many mapping. I want to crossout all of these. I'm going to restrict theta, ormy range, to be greater than the minus pi over 2 and lessthan positive pi over 2. And so if I restrict my rangeto this right here and I exclude that pointand that point. Then I can only get one answer. When I say tangent of whatgives me a slope of minus 1? And that's the questionI'm asking right there. There's only one answer. Because if I keep-- Thisone falls out of it. And obviously as I go aroundand around, those fall out of that valid range for thetathat I was giving you. And then just to kind ofmake sure we did it right. Our answer was pi over 4. Let's see if we get thatwhen we use our calculator. So the inverse tangent ofminus 1 is equal to that. Let's see if that's the samething as minus pi over 4. Minus pi over 4 isequal to that. So it is minus pi over 4. But it was good that we solvedit without a calculator because it's hard to recognizethis as minus pi over 4.
