Inverse Trigonometric Functions - Advanced Calculus 2 Video

Inverse Trigonometric Functions - Advanced Calculus 2DescriptionIn this section of Advanced Calculus 2, we continue to cover calculus topics, and learn about the use of inverse trigonometric functions.Transcript Hi! Welcome to this section of the Advanced Calculus Two Tutor.In this class, we are going to continue our study of Advanced Calculus Two and we are going to cover some topics in Calculus Two that are very important and things that typically trip up a lot of students. We are going to cover a lot of material and some of it is going to be a little bit disjunctive from one topic to the next because that is the way Calculus Two is there toward the end. You study one topic and then you will go off and you will study another topic and may not have much to do with the topic you studied right before, but we will do that because that is how the textbook usually go. But, everything in this class is very important to continuing in the http://math.etutorworld.com/ - http://math.etutorworld.com/ - Calculus Three and also into Differential Equations and all your topics in Advanced Engineering and Math and things like that.This section is going to start out with the Inversed Trigonometric Functions. Now, you all know when an inversed trig function is I think. Well, we will talk about it here in case you do not. But, in this section, we are going to focus on the inversed trig functions and how they kind of apply to Calculus. This is an introductory section explaining what the arcsin, and the inverse sine, and the inverse cosine and things like that are, but we are doing it from the point of view of what we will be using it for later which will be your Calculus when you are doing your intervals and your derivatives. You are going to need to know this stuff by the back of your hand. So, I am kind of putting this upfront, there is not that much Calculus in this section, but it is the essential baseline material that you will need to know.First, let us talk about our favorite trig function everybody knows and loves and that is called the sine, the sine usually. I am going to draw it up here on the board and we are going to review few things because that is what this section is about. So, you have your graph that we have all know and love, and here will be X and here will be sine of X. Here, X is just an angle, it is an angle one of radians, pi over two, two pi over two, three pi over two or whatever and when you put this angle into the sine function out spits the number and if you recall, the sine function always spits out a number between minus one and plus one. And by the way, a lot of the stuff that we are talking about in this section, I am covering it from a Calculus point of view which you will need to know. But, if you think you need a little more help on the basics of some other things that we will talk about, the unit circle and taking their inverse trig functions and such, go off and take a look at the Trig and Pre-Cal tutor. I know you guys are off in Cal Two now that is why you are looking at this course, so you probably know that. But, everybody needs to review every now and again and so, that is a really good class to review the unit circle and your radians, and how to take your sines and cosines and your cosecants and things like that.Let us continue on, what we have here--we are going to switch colors here. This is something you probably should already know. We have pi over two, pi, three pi over two and two pi. So, one full circle, because two pi radians is a full circle if you recall. And, on this side, we have negative pi over two, and we have negative pi, and then we have negative three pi over two, and I am going to go ahead and extend this over here and we will have negative two pi. What we have here is--and I am going to draw a little circle here because I am going to reference this circle as we talk about this, this is just the unit circle, nothing is special about this. So, just to refresh your memory because everybody needs a little refresher, this guy, this is what we call zero radians, as you trace under this circle, you will eventually hit pi over two radians and then two pi over two radians which just reduces the pi because two pi over two if pi. So, zero, pi over two, two pi over two, three pi over two, four pi over two, and four pi over two, you have completed one circle, four pi over two is just two pi. So, as we go here, we are just taking this graph and we are kind of unfolding it into kind of a linear fashion, so here, you started zero, you got your pi over two, your two pi over two reduces the pi, three pi over two, four pi over two reduces the two pi. So, what you have here is--and that is your angle stretch down here and then you have got your corresponding negative angles here.Switching colors again, because I am color happy in this section. Your sine function we call looks like this, it goes from positive one to negative one, that is the envelope and it starts out here and it makes--goes up there to the top, goes down here and it goes up here. And then, it continues negative and goes like this, on through the pi like this, so that is your favorite sine function you all know and love. You started out at zero, you reach a peak at pi over two because as you increase the angles up to pi over two, the sine which is by the way--do not forget, the sine function is the projection of this circle onto the Y-axis, which I have it labeled here, that is the Y-axis. So, as you take this circle and as you go up like this, the sine function is just looking at the projection of this thing on the Y-axis. So, when you are here at zero radians, the sine is zero because there is no projection on the Y-axis down here, so that is why it is zero. As you make your way over to pi over two radians, the projection all lies along the Y-axis, so it is equal to one. As you go around, you will see kind of make this shape, because as you turn around the circle, the projection along the Y-axis changes.That is our review, but I want it to put on the board because what you will see here in a minute is pretty important. So, here, you have your sine function that you have studied for a long, long time and used in Cal One and Trig and other classes. Now, what I want to call to your attention is sort of the obvious. We all know that if you take the sine of an angle, what you get back out of that operation is a number. Now, we all know that this number lies between positive one minus one. Because any angle that we put in, no matter if we go on to infinity or negative infinity, any angle we put in is going to go into this function, we call the sine function and it is going to spit out a number, that is always going to be between this. So, there is repetition here. You have one cycle of the sine function and if you keep going, you will get another cycle and another cycle, and another cycle, no matter what angle you put in, you always get something between positive and minus one.You will find that it is very useful and something you have to learn to define something called the inverse sine or the arcsin. So, you can define the opposite of the sine function as the arcsin of some number, meaning this number here, I am trying to draw an analogy here, is going to equal an angle, so it is in opposite you see. The arcsin, the inverse sine, and by the way, by analogy, the inverse cosine, the inverse tangent, the inverse cotangent, all of these things, all of these trig functions you know, they all have an inverse. They are basically an opposite and they all take a number which by the way must be between positive and negative one and they will map it and return back an angle to you, so, it exactly the opposite. Sine of an angle gives me a number between negative one and positive one and arcsin of some number between negative one and positive one gives me back some angel. If I try to put a number in here bigger than positive or minus one, it is invalid, it does not work. You put that in your calculator, arcsin of 34, you are going to get an error because these functions only define between negative and positive one. So, it does not make sense to put a number bigger than that in there, it just does not work.Making this a little bit more formal instead of having the word number around there, what you really have is the arcsin of some number X, we also call it, we also see it like sine with a negative one up here of X, these are called the inverse functions. And, this negative one by the way, this is not an exponent, this is not raise to a power, this is back in your Algebra when you talked about function inverses, this negative one up here and the X one that means an inverse. So, this arcsin which is the same thing as this, it is just written differently, different notation, it is an inverse, literally, the inverse of the sine function.There is something very, very important that I want to point out to you here, now that we have looked at this a little bit and might want to burn this in your head just to kind of keep it in your head. But, notice the following, if I take the arcsin of one, by the way, this is one, so remember, I am taking a number along this axis, I am popping it into this arcsin function and I am going again an angle back. One single angle is what I am going to get. If I take this and I put it in my calculator or you just look at the graph here, you can see the angle you are going to get back is pi over two. So, if I put out arcsin of positive one, I am going to get back an angle, the corresponding angle which is pi over two, that is what that is telling me. But also, notice that if I put in the arcsin of one, notice that an equally valid angle is negative three pi over two. Because look, if I go over here and map it over here, negative three pi over two is an equally valid angle. And likewise, if I do the arcsin of one, I can also get back five pi over two. And you can see that if you just go ahead and take this on to its logical conclusion, pi over two, two pi over two, three pi over two, four pi over two, five pi over two and ends up back up here wherever on this side, that is also corresponds to the value of one.So, you see what has happened here, we have a dilemma because a function or some relation that we are going to be used here and useful in Calculus, it is not very useful if you put the same number in and you get different values out. I mean, how would you program a computer to do that? Literally, look what we have here, we have taken a number one, we have put it in our function and we have shown via the graph that multiple values are valid as a result of that. So, how would you program a computer to do that, what value do you pick? The answer is, you cannot really program a computer to do that. So, if you put your calculator and you take the arcsin of one, you will not get three values back, you will only get one back, you will only get this one, pi over two. So, you can see here by looking at it that multiple values are valid when you take the arcsin. So, what we have to do in Calculus and what is done to kind of remove this ambiguity of what is the proper value to return, what we do in Calculus is we just look at a certain part of this graph.So, what we will do then is we will then look only at this part of the graph, let me just draw a little circle kind of around it. And then, the way you do it is, look, we know the sine function goes on forever and ever and we know that if we put a one--if we will look at the arcsin of anything up here, we know there is going to be multiple value spaced equally apart that is a valid angle that would satisfy that. But, we also know that that is not very useful to have infinite answers to a problem. So, what we are going to do when we define the arcsin function, is we are going to define it only to return values in a certain range and we are going to constrict that range to be right here. So, you see, if you put negative one into the arcsin function, you will get negative pi over two, you will never get anything over here because we are going to define the arcsin function only to return values from negative pi over two to positive pi over two. So, any number you put in along this axis and take the arcsin of it, you will get values, a point up on this curve corresponding to angles, but you will never get anything over here and you will never get anything over here and that is just simply a definition. It is nothing magical about it. It is just that the mathematicians had invented the arcsin function, realize this, and they said, we can have it returning. A ton of different answers, we are going to constrict it to look in a certain range and we are going to define it to only return values in this range.So, I do not think I need to write that down, but that is what we are doing, we are going to restrict it to do that. So, for instance, I am going to write the whole table over here to give you some help on all of the trig functions. So, for instance, arcsin of X is equal to Y, this is the definition of the arcsin function. When you take the arcsin, you are taking the arcsin of a number between negative one and positive one and you are getting back an angle. And, what you will see in the book is it returns values of Y that is negative pi over two to positive pi over two. Remember how to read this, Y is greater, reading it this way, greater than negative pi over two and Y is less than positive pi over two. So, what this just simply means is if you have negative pi over two here and if you have positive pi over two here, it is greater than negative pi over two, so it is over here and it is less than positive pi over two. So, basically, it only returns values in between here, which is just what we said, we are looking at that region of the graph between these points and that is all it is going to return.So, for the arcsin, we have shown with an example and talking through it that it is only going to be defined to return values in this range. And, it turns out that all of the trigonometric functions are really in this boat because all of them are cyclic and you are going to run into the same problems with multiplicity on all of them. So, next what I am going to do without a detailed proof of everyone of them, I am going to write down the table that you will see in your book of what all the trig functions returned in the ranges and now you will know at least why. You still have to memorize the ranges. There is no way around that, you have to memorize them, but, at least you will know why they exist instead of just being this mysterious thing.Here is a table of the inversed trig functions and the corresponding angles that they are going to return. So, we have already talked about the inverse sine function, it takes a number in and it spits out an angle and then returns an angle between negative pi over two and pi over two inclusive. And remember, these are just definitions, these are things that the brilliant Math guys before us defined to make these functions easy to use in Calculus when you are taking derivatives and intervals. So, these ranges here are just a range of angles like we talked about before with the sine, so that we can be guaranteed to get one single angle back given an input, instead of getting them a multiplicity of the angles back. You have to cut it off somewhere, so, this is the ranges that they have chosen and it makes the Math simpler as we go on by having these ranges.The inverse cosine, if you take the inverse cosine of a number and you get an angle between zero and pi inclusive, and it is a different range than you have with the sine function, but it is different because the cosine function is different. The cosine function is a shifted version of the sine function and so the ranges and angles that it is going to return, we could pick any range over a certain period of that cosine function, as long as we got a one to one angle back. But, this range is one that was chosen as the accepted standard and something that we are going to have to memorize and it makes the Math easier when you are talking about the inverse cosine function and the derivatives later on.Tangent, when you take the inverse tangent of an angle and you get a number and you get an angle back, it is going to return from negative pi over two to pi over two, but notice that that is not inclusive. It is not going to actually return negative pi over two in positive pi over two because we do not have these little equal signs underneath the inequalities here. And, the reason it does not return those is because if you look at the tangent function and you would graph the tangent function, you will see some singularities, some vertical isotopes there. And so, you obviously cannot define it to return angles that cross a vertical isotope like what happened at negative and positive pi over two, so you just define the function to not return those values and only return the values between them.Inverse cotangent, getting down to the lesser used functions. The inverse cotangent is going to return an angle between zero and pi not inclusive and if you are going to graph this cotangent function, you would find singularities and this range is chosen so as to never return an angle that crosses one of those singularity boundaries that is why it is done that way. And the same thing for secant and cosecant, the inverse secant function returns an angle between zero and pi over two and also between or, I should say or, pi and three pi over two and it was chosen that way. Because if you graph the secant function, you will see some singularities, so if you look at this range and this range compared to the secant graph, you will see that these ranges of angles cover a complete path along the secant function without crossing one of those isotopes. Because when you define it, you want to define it to return angles to you and you do not want it to define and to cross any boundary where you have any kind of thing that goes of to infinity. So, that is why the secant and the cosecant functions are split up like this with different ranges because you have some isotopes that fall in there and were choosing the ranges so that you never return an isotope.So, those are the inverse trig functions and the corresponding range of angles that they return. One thing I just want to point out to you here is you are going to have to memorize these things. I know it is a pain, but you are going to have to do that and that is why I put it here. You are going need it in your problems and you are going to need them on your tests. If I had to focus on some, I would focus on these because you will probably have fewer problems dealing with these, but certainly you will need to know all of them, but if I had to focus, I would focus on the first three here because those are more used and the other ones are not. One final thing I want to say, the secant and the cosecant, this range of angles here, remember, these are just arbitrarily chosen by the Math guys, so as to give us a function that returns one angle back for an input, that is what we talked about before.You can choose different ranges to do that. Most of the books, almost all of them are going to use this accepted range values as the definition of the inverse secant and the cosecant functions and not all of them do. It does not mean that your book is wrong. It just means that they chose a different standard. Now, these guys were chosen here by most textbook that you will find because later on, when we define the derivative of these functions, this definition here of the angles makes these derivatives simpler. So, when you are doing your complicated derivatives and your intervals is going to make life a little bit easier.Now, what we are going to do now that we have talked about the inverse trig functions and the angles that they produce when you pop a number in to the function, you get an angle back and we have talked about the different ranges of numbers that you will get back. And, the angles for the different trig functions, we are going to work a few problems. We are going to be ratifying a really short and sweet simple little problems to kind of illustrate what we are talking about. Let us just go right to it.The first one is going to be, what if you were given, find the inverse cosine or the arcosin of negative one. What is that going to equal? Well, the first thing you need to do, anytime you have the inverse of a trig function of a number, what I always do, every single time I do it is, this is going to return some angle, right, and what you need to do is you need to always rewrite it as what I always do. I always say, well, this is telling me, what this is asking me is the cosine of some angle which is what would have been here is going to equal negative one, that is what this is telling me. So, if you have the inverse cosine of a number returning an angle, that means that you are trying to find cosine of some angle though I do not know what this is, is going to give me a negative one. And, on top of that, we have already talked about the fact that these inverse cosines can only return angles on a certain range, so if you look back or rewind the tape over to the previous part, you will find that the angle there for cosine is going to always be between zero and pi. So, whatever angle you get has to be between zero and pi and that is going to be the angle that will give you the cosine of negative one.So, I think you can probably convince yourself because I chose a simple problem on purpose that the angle here that you are going to get is going to just be pi. And, the reason it is going to be pi is if you just go back and look at your unit circle, right, that we have all come to know and love, this is zero radians, this is pi over two radians, and this is two pi over two radians. Two pi over two reduces two pi, right. So, over here at pi radians, if I take the cosine of pi radians over here and cosine is the projection of whatever might align here that I am moving around onto the X-axis then it is going to be projected over here to minus one because this is plus one and this is minus one. So, as I move my unit circle radius over here and get over to pi radians, the projection to the X-axis is going to be minus one. So, that is why I said that the answer, the angle has to be pi. If I just chose it over here, then, that would give me a positive one, cosine of zero or cosine of two pi is positive one, well, that is not what we have. We have to make it equal to negative one. So, we had to be over on this axis and that is a pi radians.Now, the thing you always have to do when you get an answer is you have to say, okay, I found an angle, so that if I take the cosine of this angle it gives me negative one. But, is this angle in the range of the angles that the Math guys said that we defined, and the answer is yes, pi falls between zero and pi inclusive because that is the inclusive sign here. So, we are done. That is the answer to that question, we do not have to do anything more than that.Move on to another one, we will get some practice. What if we had the inverse tangent of square root of three and you wanted to find what angle this thing return. So, we are putting in a number into the inverse tangent function and we have to get an angle out and this angle is going to be the angle that we are going to want. Now, what this means is that the tangent of some angle is going to equal to square root of three. That is what this means, I am just rewriting it, tangent of some angle is going to give me that, that is what I am doing. So, I need to start walking around the unit circle to find an angle so that when I take the tangent of it, it is going to give me square root of three. But remember, any time you have a tangent, you really just need to start thinking in terms of sine over cosine, that is what I do every single time I do this, I always have to do this. But, I have to start at looking at angles, I start putting in angles here and taking the sine and the cosine and seeing which angles are going to give me or is there an angle that is going to give me square root of three.Now, one more thing, you just always need to keep in mind that for the tangent function, the angle that is going to be sent back to this has got to fall between negative pi over two and pi over two. I guess I should write this in terms of theta. It really does not matter, whatever label you put here, Y or theta or whatever it is, it is just saying that whatever number you get as a result of this operation, the inverse tangent on a number given you, theta has got to be between negative pi over two and positive pi over two. So, let us see if we can find that number, I will go and switch colors here. If theta is equal to pi over three, let us see what happens, because you are going to find it. When you do this, what you are going to end up doing is it takes practice like anything else. What you are going to do is you are going to start at some angles and start walking around until you found one that works and then eventually, you will get so good about it, you can just think of them, but let us start over at pi over three and let us see what happens.If I choose pi over three as my angle, then, what happens if I take the sine and the cosine? I have sine of pi over three and I will have cosine of pi over three and what is this going to equal? Well, the sine of pi over three is like the sine of 60 degrees so that is the square root of three over two and the cosine of the pi over three is just the same thing as the cosine of 60 degrees which is one half. And so, you have square root of three over two over one half and this two can just cancel with each other because they are both on the denominator, so, what you are going to have here is the square root of three, so we have got the right answer. So, never forget the force for the threes and do not forget that the answer we got which was pi over three in this case does fall between the range of negative pi over two and pi over two. So, that is correct, that is okay. We took a number, we plug them into the inverse tangent function and we got an angle back that fell between the range we needed to find as prescribed by the table there.The way we did it is just by working backwards. We just find an angle so that when you plug them into the tangent function, which the tangent function is the sine over the cosine function, you find an angle to give you the answer you want. I mean, you may have started off with something else like pi over four or pi over six and you would have gotten the wrong answer, you would have work your way around to the proper answer. That is what you have to do. So, if you look at some of these things and you are like man, it is not hitting me, just start with an angle and you will get good at your trig functions, you have to know this by the back of your hand.The next problem is going to be, find the inverse cosecant of square root of two. That is what the problem would say, find the inverse cosecant of square root of two and that is what you have to do. The first thing you need to do is rewrite it, that is what I do, and that means that the co--this is going to return an angle, the cosecant of some angle is going to equal square root of two and it is up to us to find what that angle is. So, when I am presented with a problem like this, when I have a cosecant or secant or cotangent or something like this, the first thing I do is I rewrite in terms of sine or cosine. And so, if you do not know the trick there, I would suggest you learn it, you just write the trig functions in order, sine, cosine, tangent, cotangent, secant, cosecant. Just memorize this order and then it is really easy because cotangent is one over tangent, secant is one over cosine, cosecant is one over sine. So, they are kind of form a rainbow like that. So, what I do is I say, one over the sine of an angle is equal to square root of two, that is what you are doing here. You are taking the cosecant and rewriting it in terms of sine because personally, I am more confident and comfortable and familiar with working with sine, cosine and tangent. So, I always rewrite my cosecants, secants, cotangents in terms of the other trig functions when I can do it. So, I have one over sine of theta and in order to get this in a form that I am more comfortable with, I will just flip both of these fractions over so I will say, sine of theta is equal to one over square root of two. So, I will just divide both sides, one over both sides and just basically divide both sides there to take the reciprocal and so you have this.Now, I have got the problem out in a form that is much easier, sine of an angle is one half, and do not forget here that now that we are talking about the sine function, the angle that can be returned from a sine function is negative pi over two to pi over two, inclusive. So, what I got to do now is work my way around the unit circle, find an angle that does this and that angle needs to fall between this range. So, what I am going to do is I am going to say, first, I am going to draw a unit circle because it is always a good reference to have. And, I am going to say, what if theta is equal to pi over four? What happens if theta is equal to pi over four? Well then, what will happen is I will have sine of pi over four, pi over four is here at 45 degrees and sine of pi over four you should know is square root of two over two. At first, you might look at that and you might say, well, that is not right because I was supposed to find an angle, so I took the sine of it, I get one over square root of two and this return the square root of two over two. But, look it here, I can actually take this, just to prove it to you, these are actually the same thing, I can take this fraction and multiply it by square root of two over square root of two, I am just multiplying by one, square root of two over square root of two is this one.So, if I do the simplification on the top, I am going to have square root of four, two times two is four, and on the bottom, we are going to have two times the square root of two. So, on the top, square root of four is two and on the bottom is two times the square root of two and you can see right here, the two is canceled so you have one over square root of two. So, it turns out that pi over four was the correct angle and pi over four does fall between the range http://privatetutoring.biz/ - http://privatetutoring.biz/ - of angles that inverse sine function is legally able to return. Obviously, I do not want to pick an angle outside of this range because that is illegal, it has to return an angle there. So, you start with the cosecant, return it into one over cosine and we just flip both sides over so that we were dealing with sine of an angle equal one over square root of two and then you just kind of work your way around the circle. You would find quickly but if you pick other angles, you are not going to get the right answer. Square root of pi over four here, we take the sine of it, we have got square root of two over two and we showed by this multiplication, an exercise here multiplying by root two over root two that it actually is the same thing as what the problem gave you.The next problem says cotangent inverse of negative square root of three, so I am taking the inverse cotangent or the arccotangent of negative square root of three. And, the way I always proceed next is I rewrite it, I say the cotangent of some angle, because remember, this is going to return some angle back. Cotangent of some angle is just equal to negative square root of three and that is what I want to do. And, the next thing I do is I like--when I am dealing with cotangents, secants or cosecants, I always like to rewrite them. Cotangent is just simply equal to cosine of theta over sine theta. That is what this is equal to, the reason it is equal to that is because the tangent function is sine over cosine and if you remember back to the rainbow, I already showed you that cotangent is equal to one over tangent, so you have here is one over sine over cosine, so you have cosine over sine. So really, to bring it down the theta here, just something easy for you to remember, the tangent function is sine over cosine, the cotangent function is cosine over sine, that is really the thing you need to remember.We wrote the cotangent function as cotangent over sine and the reason I am doing this is because personally I just like to work with these cotangents kind of split up because I know how to deal with cosines and I know how to deal with sines. It is much easier for me to figure out the answer if I am looking at these fundamental functions. So, I have got cosine over sine is equal to that and just to make life even a little bit easier, I am going to divide both sides, take the reciprocal of both sides, so I am going to say sine of an angle over cosine of an angle is equal to negative one over root three. It is just a personal preference, you do not have to do that, you could have started here and going around the unit circle and done it, but I prefer to do it this way. I prefer to have sine over cosine like this. And you can do that because I took the reciprocal of both sides, one divided by this, one divided by this and that is a legal thing to do as long as you are negative at both sides. So now, what you want to do is you want to look at your unit circle and you want to start moving around the unit circle starting at zero and kind of picking in the radiant values. And, you want to just basically play with it until you start getting an angle that is going to give you the right answer here and that is all you really have to do.What I want to do first is I want to say, what if the angle I pick is five pi over six? Remember, pi over six is like 30 degrees, so that will be around like here, this is pi over six, two pi over six. I know it pretty much has to be in this quadrant because I have a negative here and so I need to have some way to get a negative value. In here, my cosine is going to give me negative, my sine is going to give me positive so that is going to help me out. But, if that does make sense, I mean, really and truly, all you are doing is picking values until you find the right one. So, if you do not know and have an instincts, you just kind of start doing that. So, what happens if we do that? We take the sine of five pi over six over the cosine of five pi over six and what will that give me? The sine of five pi over six which is over here, pi over six, two pi over six, three pi over six, four pi over six, five pi over six and six pi over six would be a pi, so we know we are on track. So, this is five pi over six, the sine of it is like the sine of 30 degrees so it is like one half, so you are going one half on the top, positive one half because the projection on the Y-axis here. And, on the bottom is like the cosine of five pi over six which is like the cosine effectively of 30 degrees and it is negative because it is going to be over here. And so, what you are going to have here is negative square root of three over two because the cosine of some multiple of 30 degrees is always going to give you square root of three over two, it just depends on what quadrant you are in and if it is going to be negative or not. It is negative because we are over here, the projection is going to be negative over here.You can see right away the two we cancel with the two and so in there what you will have is one over square root of three and it is negative. So, the answer is five pi over six, just like we said. Now, you always have to ask yourself, did it fall in the right range. Notice that for the cotangent function, which was the original function, if you look back four for cotangent, the angles that you need to return was zero to pi and notice that this did fall between zero and pi because here is zero and here is pi, so this angle did fall where we were. Now, what if you screwed up here, so, we know this is the right answer because we got an angle, we plug in here and we have got the right answer that we got that we were asked to find up to the beginning here. But, the thing you always have to ask yourself is well, everybody might not have come to the five pi over six to begin with, what if he picked a different angle, what if instead of five pi over six, because you started at a different point, what if you chose negative pi over six? Well, your first instinct without thinking about it might have been to do that, if you take sine of pi over six. Well, let us just do it right here, sine of negative pi over six over cosine of negative pi over six, what would you have done if you have done that? Well, negative pi over six here, the sine of it is going to give you negative, so it is going to give you negative one half because you are projecting this guy over to the Y-axis negative here and it is going to be one half because it is basically 30 degrees, the sine of 30 is always one half. And for cosine, it is also taking the cosine of this, but the cosine is going to be positive because it is projecting over here on the X-axis, so that would be a positive square root of three over two because the cosine of 30 is square root of three over two and it is positive over in that quadrant. So, you would have gotten the same answer, negative one over root three.What I am trying to point out to you here is if you had started in a different place as I did and if you would have gone and say, well, I am going to try an angle of negative pi over six. And, I am going to take the sine of it and the cosine of it and I want to reduce it, and look at this, I got the right answer. Well, you see, that is not right, because you always have to look back at the beginning of the problem and realize that you were talking about a cotangent function and the cotangent function only can return angles between zero and pi by definition. So, this angle that you picked with that thinking to begin with was negative pi over six and it worked, but it is just not legal for what the inverse cotangent function can return because you have to get an angle between zero and pi. So, what we are writing into here with this little sub-example here is the multiplicity problem with all of these trig functions. There are multiple angles that you can find that you can legally put in and get the answer, but the inverse trig functions are only defined to return values in the specific range. So, when you are doing these inverse things and you are going around the unit circle and you are finding angles to satisfy them, you just have to make sure that you are picking angles in the range that you are supposed to. When you are dealing with a cotangent, you are going to start picking angles and start plugging them in, you want to make sure you pick angles in the right range which will be between zero and pi and they will start looking at negative. At different trig function, you might be perfectly okay to look at negative, you just have to look at that particular range. We started with the cotangent and we convert it into sine and cosine and I like to deal with sine over cosine so I flip things over and that is how we proceeded with that example.Now, the next question says, what if I am taking the inverse sine of negative one over square root of two? What this means just like everything else is it means that we are taking and this is returning some angle theta, is we are taking the sine of some angle and what we are getting back is negative one over square root of two. So, what we have to do is find angle, but do not forget that because we are talking about the inverse trig function and we are talking about some angle we are getting back, the angle is going to be between negative pi over two and pi over two, by definition. There is nothing magical, it is just the definition of it, so when you are picking your angles to see if it satisfies, you have to make sure you pick them between this range because we are talking about sine. So, you have your unit circle here, it is always helpful to draw that circle. It is kind of a crutch really more than anything else for me. So, if I pick an angle of negative pi over four, which will be down here at negative 45 degrees here, what would happen? Well, the sine of negative pi divided by four, what was going to happen here, well, at the sine at here, this is like a negative 45 degrees. It is going to be for the projection along the X-axis, which is negative down here, so it is going to be negative square root of two over two.Now, that looks different than what we started out with, but I think you remember from of the other problems, it really is exactly the same thing because I can take square root of two over two and I can multiply it by the fraction square root of two over square root of two. I am just proving to you that these are in fact the same thing. On the top, you will have square root of two times square root of two, square root of four, on the bottom, you will have two times the square root of two, on the top, you will have negative two, root four is two and on the bottom, you will have two square root of two and this will cancel. So, you will just have the negative one over the square root of two. I am just showing you that they are in fact the same thing. So, if you pick negative pi over four for your angle and you take the sine of it, you will get what you were supposed to get which is negative one over root two and you have to always make sure the angle that you chose is in the angle range here.The next problem is going to be the arctangent of negative square root of three over three, that is going to return some angle theta and we are trying to find out what that angle is. Now, the fact that this arctan is nothing fancy, it just means inverse tangent, same old thing, so what this mean is the tangent of some angle is going to be negative square root of three over three, that is what it means. And, when you look at the tangent function and what the range of angles that it is legally able to return, we said that it is going to return if you look back at the other thing, something between negative pi over two and pi over two, not inclusive of negative and positive pi over two. But, all the way up until that point and that is because the tangent function has singularities of those isotopes there. So, all we have to do is find an angle, we will just start walking around the unit circle until we find an angle in this range that satisfy this. That is all we really have to do when we do these problems.So, I like to draw a unit circle all of the time, just makes me feel better, and tangent do not forget is just the sine of theta over the cosine of theta, I always like to break up my tangents like this because it helps me find the answer, negative square root of three over three. So, what happens if I pick an angle, let us say if theta is equal to negative pi over six, let us see what happens. Well then, I would have sine of negative pi over six over the cosine of negative pi over six, and what will that do for me? Well, pi over six is like 30 degrees and it is negative, so it is like negative 30 degrees, so that is where your angle is at negative pi divided six roughly, that is roughly 30 degrees. So, the sine of this negative down here is also going to give you negative because that is projecting along the X-axis and the sine of 30 is one half, so it is going to be negative one half on the top. And, on the bottom, it is the cosine of the same angle, the cosine is going to give you a positive number because it is projecting along the X-axis that is positive and the cosine of 30 degrees is square root of three over two, so it is the square root of three over two. That is going to give you that and then this is going to cancel with this, the twos are going to cancel because they are both in the denominator. So, you are going to have negative one over square root of three.And, the first thing you should do is like I said, well, this give me what I was trying to get. It does not look like it does, but if you just multiply the top by square root of three and bottom by square root of three, you will find the following, you are going to have them because you multiplied it by one so have not changed anything, you are just manipulating it, you are simplifying it. One times square root of three is square root of three, and on the bottom, you are going to have square root of three times three is nine, and obviously, the square root of that is three, so you will negative square root of three over three which is the answer here and that matches exactly with this. So, that is actually is not the answer, the answer is the angle that we chose which is negative pi over six and you just want to make sure, just for sanity check, negative pi over six is between negative pi over two and positive pi over two. This is negative pi over two, this is positive pi over two, so it is in that range. So, what we did was we wrote it in terms of the function, use sine and cosine because that helps you walk around the circle and plug in values. We picked some angles, well, I did not pick more than one angle. I know what the answer is, but as you get comfortable with it, you will start to figure out and see patterns and the multiple of the angle that you will need. But, even if you have no idea, you could just start with random angles there and just kind of walk around until you get an answer that gives you what that inverse tangent is and you just have to make sure that that angle is in the range that you need.What if you were given a problem like the sine of the inverse sine of 0.7 what would that equal? Well, this is a little bit of a trick question because I am taking the inverse sine of a number between positive and negative one and I am going to get some angle from that and I do not even need to calculate that angle. But, what is going to happen is I am going to take the angle that I get here and I am going to immediately take the sine of it. So, you see, I am kind of canceling this sine with this inverse sine and I am going to get 0.7 as an answer. You can just of write that down because whatever this is, I am going to take the inverse sine of it. I am going to get an angle between positive and negative pi over two from our chart over there, from our definition. And, then I am going to take whatever angle that is and I am going to turn around and take the sine of it, so I am going to convert it right back into a number and I am going to get 0.7. That is kind of--where you can write those down pretty quick. But, what if I mix it up a little bit and I say what if the cosine of the inverse sine of square root of three over two? The cosine of the inverse sine, well, it is just like anything else and like you have to start small and go ahead and evaluate the inside and get the answer and then evaluate the outside.Let us work with the inside first, the inverse sine of square root of three over two, what that means is that the sine of some angle is equal to square root of three over two, and I just need to find out what that angle is. And, by the way, because we are talking about the sine, the angle by definition must be between negative pi over two and positive pi over two. I am drilling this into you every problem because it is really important that you keep that in mind. So, what angle is going to give me so that when I take the sine of it, it is going to give me the square root of three over two. I think you can convince yourself this is going to be pi over three because this is like 60 degrees and sine of 60 degrees is square root of three over two. This angle does fall in this range and that is what you will get if you put this in the calculator, inverse sine of square root of three over two, you would get that angle because that falls in the range there. So then, you just take that and you plug it in, cosine of pi over three because that is what the inside was and the cosine of pi over three. Again, if you just look at your unit circle, pi over three is like 60 degrees so that is up here because this is 30 degrees and this is 60 degrees, so this is pi over three up here. So, it is in the first quadrant, you are going to get a positive number because it projects down on the X-axis here and that is going to be one half. So, the answer to the problem is one half. So, if you were to take the inverse sine of square root of three over two and then take the answer and then just take the cosine of it, you would get one half because that is what you will get when you project down the X-axis there.And, if this stuff is not clear, I mean, we have done a lot of problems here, with inverse, taking the inverse sine, the inverse cosine and walking around the unit circle. And, if you have not done this stuff in a long time, it actually can be quite confusing even if you studied it before, you are willing to think about reviewing that, a good way to do that is to pick up the Trig and Pre-Cal DVD. I have got several hours of just refresher on that stuff if you need it.One final problem, what if you take the arcsin of the sine of five pi over four? Now, the first thing you must think is that the answer is going to be five pi over four because you are taking the arcsin of the sine. So, you might think, well, these are just going to cancel like before, but this is a little bit trickier and there is the reason why I did this problem is to show you the following. Let us work with the inside first. So, what this is saying, what is the sine of five pi over four? If you look at your unit circle, this is not an inverse sine, this is just a sine, this is the unit circle, this is pi over four, two pi over four, three pi over four, four pi over four, five pi over four. So, this is actually five pi over four right here, down there. So, what would the sine of that be? You know it is going to be square root of two over two because it is a multiple of pi over four, that is something you should know from your trig, that is going to be square root of two over two.But, because it is down here in this quadrant, it is projecting on the negative Y-axis there, so this is actually going to equal inside here, negative square root of two over two because it is projecting down here, negative square root of two over two. So, now, what we have to do is plug that in secant, what is the arcsin of negative square root of two over two? What would that be? So, that would be the answer. It is going to give you some angle back, and since we are doing all this arcsin stuff, we are just going to rearrange this and we are going to say the sine of some angle is going to give me negative square root of two over two. All we got to do is to find that angle, but do not forget because we are talking about the sine function, the angle, just to drill it in, is going to be between negative pi over two and pi over two and that is just from our table at the beginning of the lesson here. So, we have to find an angle in this range, that is all your calculator would ever return to you and all of the computer would ever return to you if you put this in there.So, what would it be? What angle would it be? What angle is going to give us that? Well, forgetting about this five pi over four now and just kind of get rid of this let us start with our unit circle. And by the way, we know it is going to be a multiple of pi over four because we are getting the square root of two over two out. What if you go up here two pi over four? Well, the sine of that is going to give you positive and what if you go to two pi over four, three pi over four here, well, that is going to give you a positive. Well, you see here, you are already outside of the balance. You can only pick angles from negative and positive pi over two by definition. So, let us go the other way, let us do negative pi over four. The answer is going to be negative pi over four. Because if you look down here, this is negative pi over four and if I take the sine of that angle, it is going to give me a negative, because I am projecting that here, negative square root of two over two, which is what I want and negative pi over four is in the range here.See, there was another angle maybe over here if you were to kept going over here, that would have satisfied it, but that is too far because you can only pick angles between negative and positive pi over two because you are dealing with the inverse sine. So, the reason I did this problem is because at first glance, you might think arcsin of a sine of an angle is just--they are going to cancel and you are going to get this back. But, because of the ranges on the angle that was chosen here, this--really to boil it down, the reason it work this way because this angle, five pi over four is down here which is outside of the range of what an arcsin can return which is going to be between negative and positive pie over two. So, when you take the sine of it, you map this thing back over here and then when you take the arcsin of that answer, it is going to basically map it over into one of these quadrants over here. So, because of that, you are not going to get the same angle back.The moral o the story is when you are taking arcsin over sine or sine over arcsin or something, you just work with the middle and do it one step at a time. In this case, we took the sine of the angle, we made sure we got the right answer there and then when we did the arcsin, we just made sure that the angle that was returned that we picked was inside of the range of angles that are legal for an arcsin.In this section, we cover quite a bit. Personally, I think this part of calculus is a little bit dry and confusing, but it is very important actually because you deal with sine and cosine, tangent and all of these things, all through calculus, all through engineering, all through physics and they are really amazingly prevalent all through nature. So, you are going to need to know them. When you start talking about the inverses, you just need to be aware of what range of angles the inverses can return because from the very beginning of the class, I showed you that because the sine and cosine functions go on and on forever and repeat themselves. For given input, they can return many, many outputs, so in order to prevent that ambiguity from coming up in your Math, you have to pick a range of angles that is going to return kind of over one path here. And, for each function, it is a little bit different and that is just by convention and that is just the way it is, you have to memorize them. And, when you do your problems, you make sure that you honor those ranges when you find the angles that those inverse trig functions are returning.
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