Three tips for evaluating inverse trig functions

what's up guys in this video I want to go over three tips that you can use when you need to evaluate the composition of inverse functions the first tip is to know your restrictions now what I mean by knowing your restrictions is you need to know the restrictions on the domain when you're doing the inverse trigonometric functions so just a quick little recap you're going to want to make sure you have these so if you don't already have them or memorize make sure you go ahead and write them down okay so basically what that's saying is if I'm taking the sin inverse or the tangent inverse I'm going to get a value of y however that Y is going to be restricted between the values of netive Pies which is also like 90° and pies which would be a positive 90° for cosine when we're taking the inverse of cosine of a value we're going to get a y which is again these are going to represent a angle these y's represent angles that angle is going to be between 0 and Pi I always like to visualize this from the unit circle so for s or sorry for cosine it's going to be between 0o and pi and for S and tangent it's going to be between negative pies and pies so I always like to look at it like this this way okay so that is going to be the restrictions on that Angle now why is this so important why is this such a great tip because this is exactly what you need to make sure you remember when you're working on evaluating inverse tangents and you can also do things rather rather quickly because of that so there's two ways that the inverse or knowing your restrictions can come into play the first one is just going to be a composition when we're dealing with the unit circle so if we had something like this we had s of sin inverse of1 well again if we're trying to go ahead and evaluate well what is the sin inverse of ne1 the cool thing we need to know about trigonometric functions and their inverse functions as they're inverses of each other right they're going to undo each other so whenever you have your inverse is going to be inside of your other trigonometric function you always want to work inside out right so always going to work this way first so this problem is going to be fairly simple because the answer is just going to be NE 1 because the sign and the sign in are going to undo each other but if we work on this step by step you'll see why the Restriction is not going to come into play here if I just take the sin inverse of ne1 basically that's asking me what is going to be the angle on this restriction where Y is going to equal a Nega 1 right because remember on the unit circle the sign of value is represented of the y-coordinate on the unit circle so you can see over here we have 0 comma 1 well what is that angle within this restriction well it's going to be this angle down here which is going to be a negative pi Hales then if I take the S of netive Pi Hales that's just going to represent the y coordinate which again is equal to a -1 but here is where knowing your restrictions is going to come into play and be so important okay so now we have a composition where it's actually switch around now we have the S inverse of s of 3 pi/ 2 so again always work from the inside out so what is s of 3i 2 well if you were to graph that that'd be all the way over here now you might be thinking and saying no Mr mclan that's the exact same point that's equal to a 1 yeah but a lot of students what they'll do is they'll say oh these two cancel each other out I got1 so these two cancel each other out so therefore it's equal to a 3 pi/ 2 no we have to know our restrictions we have the sign of 3 pi/ 2 right that's equal to a NE 1 so s inverse of1 well we know that answer what is going to be the sin inverse of NE 1 that's equal to a pi halves so the answer here is a NE pies so make sure you understand your restrictions when you're doing that composition the other play where knowing your restrictions or the other reason why knowing your restrictions is so important or such a helpful hit tip is when you have to go ahead and graph a triangle so let's say I had something like this okay so in this case I have cosine of sin inverse of -3 over 5 now in this case we're going to draw a triangle but there's a problem we don't know what is this triangle going to look like it's a -3 over 5 right because remember it could be opposite over adjacent or I'm sorry opposite over hypotenuse or the triangle could look like this -3 and 5 but since I know my restrictions has to be either in the first and or the fourth quadrant this is not going to be my triangle this is going to be the triangle I'm going to want to work with now let's move on to tip number two all right now tip number two is actually one of those that can make your life so much easier and that is to look for no Solutions so there's something really really important to remember about these trigonometric functions if I have a right triangle right it's going to look something like this and here's going to my angle right and we have our hypotenuse we have the opposite side and we have our adjacent and I guess I should just label the hypotenuse well remember ladies and gentlemen the hypotenuse is always going to be larger than our opposite and our adjacent side so if I go ahead and rewrite the sign of let's say an angle and I get something that has a Rao of this 3 over two well let's go and draw what that would look like on a triangle and immediately you can see there's a problem right because now my opposite side because it's always opposite over hypotenuse my opposite side is larger than my hypotenuse therefore you can see that this is going to have no solution so anytime that we're dealing with a sign and a cosine of a value that's going to be larger than one we automatically know ladies and gentlemen there is going to be no solution however just notice that for tangent remember tangent is I don't know what this is going to be let's just let's do the triangle like this I know over here tangent is opposite over adjacent they don't make any reference to what the hypotenuse is so tangent can have a value of that is larger than one so if you need to find the inverse the reason why this is so important or the reason why I'm bringing this up is sometimes we'll get a problem that looks like this and you might say all right well let's go and evaluate well I go into this point right here and I recognize well what is going to be the cosine inverse of four well remember ladies and gentlemen I can never take the cosine inverse of a value that's going to be larger than one the sign and the cosine of any angle is always going to be less than one it's going to be between zero and one so therefore you can see or you can see a negative as well you can see that in this case since the cosine inverse of four that's not going to work within that restriction so therefore because technically this is saying the cosine of what angle equals four how can you have the adjacent over the hypotenuse where your adjacent side is larger than your hypotenuse it doesn't work so that's a quick little easy way to recognize oh I can't take the cosine inverse or the sin inverse of a number that's larger than one so therefore I immediately have no solution however be wary with Tangent you can go ahead and do that now on to tip number three so tip number three is to when you have values that are not on the unit circle look to create a triangle so when we're first teaching the composition of inverse trig functions you know a lot of times we're just dealing with values that we see on the in circle they might look something like this or they might look something like this either way you're always going to have some problems that you recognize that oh this is the value on the unit circle this is going to be an angle that's on the unit circle and so therefore I can use my unit circle and use my restrictions but remember I talked about in the first tip of knowing your restrictions restrictions are important when you're dealing with values that are on the unit circle and angles that are on the unit circle as well as when you're dealing with triangles so let's take a look at a problem where we would need to create a triangle let's actually use exactly this but let's just change up some of the problems okay so what if I had something on this now I know of 2 over2 that is a point that's on the unit circle I know what that cosine value is I can immediately go ahead and evaluate that however when I look over here I say the S cosine inverse of 3 over 4 well I don't know 34s as a value that's on the unit circle however what I can do is go ahead and create and actually let's make this a negative because let's have a little fun with those restrictions right so let's have a -3 over 4 now we know that the hypotenuse is never going to be uh negative so I have two triangles that's going to work I have a triangle in the second quadrant as well as a triangle in the third quadrant however we know our restrictions of cosine right are between zero and Pi so the only angle that would work would be dealing with this triangle right in here so I can immediately just go ahead and forget about this one right there I'm sorry that is a -3 and that's a four okay so now what we need to do is we need to find the sign of this well if I need to find the sign for this angle right here I need to know what this opposite side is so I have the adjacent side I have the hypotenuse what is the one thing that kind of comes to mind when we have the adjacent and the hypotenuse and we need to find the opposite that would come with the thing it starts with a P Pythagorean theorem all right so remember a 2 + b^2 = c^2 or what I like to summarize it as leg squ plus leg s equal hypotenuse squ so we have one leg we do not have the other leg and we have the hypotenuse so 3^ s is going to be a 9 plus a b^2 = a 16us a 9us 9 b^2 is going to equal that's going to be A7 now when I introduce the square root make sure you are including the plus or minus because again it does matter on the restrictions if you had a positive would be up here a negative would be down here now obviously we know the Restriction is going to be in these first and second quadrant so therefore my B is going to be my opposite side is going to be a positive value but sometimes you have to use a negative value if you have triangles down here so B is going to equal a positive sare < TK of 7 now we're not going to use the approximation 2. 645 7513 we'll just keep it as the square OT of 7 so this will be the square OT of 7 and then I want to say well what is the sign of this cosine inverse for this triangle that we just created well that's going to be theun of 7 over 4 right which is opposite over hypotenuse so always work from the inside side out so the sign of this angle right here is going to be a 7 over 4 hope that helps