now that we've explored the definition and properties of inverse trig functions let's take a look at the kinds of expressions that frequently appear in analyses it's quite common to end up with an inverse sine of a frequently seen triangle let's see how that works out here if you're asked to evaluate exactly an expression like this you can use your calculator but that'll give you some potentially long decimal expression that isn't easy to interpret it's easier to be able to sketch it first just remembering what the definition of the arc sine is the arc sine gives us an angle however you're more familiar with sine of an angle equals a ratio so take advantage of that and recalling that the sign is opposite over hypotenuse you can actually draw a triangle with root three on the opposite side and two on the hypotenuse and the angle here being angle you're looking for now there's no way this is always going to work out exactly but this triangle here should hopefully bring some bells because this is the half of an equilateral triangle which is 2 1 1 root 3 here and that's the angle we're looking for and so that angle is exactly 60 degrees or in calculus pi over 3 radians the key here is recognizing that an arc sine with a ratio is something that you can sketch out now what if we do arc cos and we include a negative sign well you can think of the same kind of construction we're calling that we're looking for the angle which satisfies cosine of theta is minus root 3 over 2 and this is the adjacent over hypotenuse well if we think of our angle here and think of the adjacent being a negative quantity all we do is move over to this side and say negative root 3 here with a length of 2 and because this is a triangle we just imagine the lengths but in this side then we're talking about this angle here as one possible option that's our theta and if we take a look at the geometry here this is the 60 degree angle that's the 30. so this part here has to be 150 degrees or that's pi over 6 we start with pi so 5 pi over 6 radians so arc cos of negative root 3 over 2 is equal to 5 pi over 6. let's get a little more exotic here we have the arc secant of 2. now the immediate reaction for most people is that secants are less familiar so so let's write it out as an angle first to get started secant of some angle equals 2 but then quickly as quickly as possible turn it into a sine and cosine remembering that secant is one over cosine and that means that our cosine of theta is equal to one-half this gives us a scenario similar to the earlier examples where we have a triangle that we built and the adjacent is one and the opposite is the hypotenuse is two something like that and in fact that's our one two root three triangle so in here we would have 60 degrees or pi over 3 radians and so arc secant of 2 is pi over 3. in this example we have a sequence of trig and then inverse trig and on the surface this looks like an ideal sequence we have an angle to start with in radians when we take the sine of that we're going to get a ratio and then when we do the inverse sine we're going to get an angle back and so the temptation is to say because these are inverses we're going to get 8 pi over 3 back of course we wouldn't be asking a question if it was that simple recall that the range of arc sine the possible values that arc sine can give back because of its restrictions is only negative pi over 2 to pi over 2. and unfortunately 8 pi over 3 is outside of that so how do we deal with this there are two common approaches one is using the unit circle let's do that first and on the unit circle we imagine going around and around and around 2 8 pi over three counting counter clockwise that's by convention and just looking at the fractions eight pi over three is the same as six pi over three and two pi over three well that's just two pi that's a full circle we can ignore that so we can focus on the 2 pi over 3 element and 2 pi over 3 gets us over to here if we take a look back at the range of arc sine how does that play a role here well the thing to keep in mind is that the coordinates here are x and y's and the x and y's represent cosine of the angle and sine of the angle so if we look at the y coordinate where we ended up and find a matching point with its own y coordinate then we're going to be able to identify something in the range of minus pi over 2 to pi over 2 which is essentially these two quadrants and say this angle which is going to give us the same sine value is what we're looking for now here it just takes a little bit of practice looking at the sketches if we take a look at our arc and we cut it down the middle here we have 2 pi over three i'm just going to write two thirds and one third well that means that this is one-sixth and one-sixth because the symmetry here that leaves the angle that we actually want in this diagram here as the full one half pi over two minus one sixth half minus six is pi over three arc sine of sine of eight pi over 3 is actually equal to pi over 3 because of the limitation on the inverse sine if i said there were two ways to do this the other way to do this is with a graph if we do a graph of sine we just have to make sure we include all the way out to eight pi so breaking this down here's our two pi and remember eight pi over three is two pi plus two thirds of a pi that's a little bit more than another pi over 2 so there's 8 pi over 3 and then we take the point there and we back it off so it could be this value this is the same y value this gives us the same y values and so does that then we use a couple of different symmetries similar to what we did here we know that 2 pi over 3 is going to give us exactly the same sine value as 8 pi over 3 because it's 2 pi apart one full cycle that's great but unfortunately we can only use points that are between pi over 2 and the origin so we know we're going to need to find our way back to this point we can do the same kind of logic we did here this is pi over 2 this is 2 pi over 3 the difference is pi over 6. we know our point that we care about is going to be symmetric so it's going to be pi over 6 on this side and then what's left over is 1 pi over 3. to get from here to here is pi over 2 we went back a step pi over 6 that leaves pi over 3 at the end and the same answer here it makes no difference which way you approach these problems whichever way is more comfortable for you is fine now with that in mind we can actually make more generic replacements this is something that you'll frequently encounter if you imagine moving arms where you've got some ratio that's going to be conserved you find the angle but then you find some related property to the angle fortunately it turns out these are actually fairly simple in practice to work with because we can go right back to a triangle in the general case here and what we're going to do is build an arbitrary triangle with an angle theta recognizing that the cos of something is going to be an angle so another way to write this is that sine of that angle equals r and that equals our opposite over hypotenuse this doesn't look like a ratio but the lovely thing about the number one is you can always make a ratio by dividing by one so if we take this and look for a triangle that opposite the theta has an r length and a hypotenuse of one this triangle will have an arc sine that gives us r but then we find the cos of that angle and cos is just adjacent over hypotenuse we don't know what that is right away but we have pythagoras because this is a right angle triangle so this r and one we have let's call this length here a a squared plus r squared equals one so a squared equals one minus r squared a equals the square root of one minus r squared well then cos of theta is what we want is adjacent over hypotenuse one minus r squared all over one we usually just don't write that one anymore so we can put our final answer that cos of arc sine of r in general is going to be an interesting not even trig looking relationship but square root of 1 minus r squared and we can change this up let's take x as our input as the ratio find the angle and then take the cos of that angle again we're going to find that as theta draw an arbitrary triangle here and our tan is equal to x which is equal to opposite over adjacent now this time we can still do our one trick in the denominator so we define our angle we find the opposite as x the adjacent as one and then we ask what cosine theta equals well that's adjacent over hypotenuse we don't have the hypotenuse yet but we can find it through pythagoras we take our one squared plus let's give it a name let's call it b this time one squared plus x squared equals b squared and so that's it we want b b is going to equal the square root of one plus x squared okay we have our hypotenuse the adjacent is just one so we're gonna have one over one plus x squared all square rooted but that's exactly what we started with so the cos of arctan of x we can eliminate the trig and inverse trait functions completely out of that and get this relationship of 1 over 1 plus x squared all square rooted