Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Inverse sine functions if F of X equals sine X. We know that this is a function because it passes the vertical line test. In order for it to have an inverse function, it has to be 1 to one. Hence it has to pass the horizontal line test. Currently, it does not pass the horizontal line test. So what we're going to do is we're going to look at a restriction on the domain so that we can make this one to one. We always want to use the angle 0 in our restriction and we want, if possible, to make it continuous covering every single Y value from -1 to 1. So in this case, if we want to have zero and we want every single Y value from -1 to one, we're going to look at from negative π halves to Pi halves. So for our inverse function, we write it with an F to the -1 being sine inverse of X. And what that really does is it switches our X and our Y. So the range of the original is the domain of the inverse, and the domain of the original would be the range of the inverse. But in this case it wasn't 1:00 to 1:00, so we restrict it to make it 1 to one from negative π halves to Pi halves. When we say F inverse of X equals sine inverse of X, we could think of that as Y equaling sine inverse of X. So this is really saying that X equal sine of Y assuming that we have the right restrictions for the domain and the range. If we look at our cosine function, very similar F of X is cosine X. So the domain for the original function is negative Infinity to Infinity and the range is -1 to one. For the inverse we would switch our X and our Y. So this really means that our X equals cosine of Y. Our domain is going to be the range of the original function. Our range it currently is not one to one. We have to restrict the original domain to get it to be 1 to one. We always want to use the angle 0 and we want to try to cover all positives and negatives of Y and do it continuously. So in this case, we're going to go from zero to π for our range. So let's look at some examples sine inverse of root 3 / 2. What that's really asking is, within our restriction of negative π halves less than or equal to X less than or equal to π halves, what value for sine gives us out root 3 / 2? Well, we know that root 3 / 2 occurs at π thirds, so sine inverse of root 3 / 2 equals the angle π thirds. What about sine inverse of negative 1/2? Well, negative 1/2 occurs in lots of places that negative 1/2 occurs at 7 Pi 611 Pi 6. But seven Pi 6:00 and 11:00 Pi 6 aren't within our restriction. So we actually are going to use negative PI6. If we thought about our sine graph, negative 1/2 is going to be about here, and we want the angle that's in between negative π halves to Pi halves. So that angle would be the negative PI6. How about sine inverse of 0? Well, that's just going to be 0 sine inverse of -1. Once again, because of our restriction, we're going to use negative π halves sine inverse of negative root 2 / 2. Well, this time it's about here, and we know that π positive root 2 / 2 occurs at π force. This is an odd function, It's got symmetry. So if positive root 2 / 2 occurs at π force, negative root 2 / 2 is going to occur at negative π force. Let's look at some with cosine. Cosine's restriction was 0 less than or equal to X less than or equal to π. So when is cosine? What angle gives cosine of root 3 / 2? That's PI6. How about cosine inverse of -1 half? Well, if we're in between 0 and π, negative 1/2 is going to occur at 2π thirds. Cosine inverse of 0. Knowing our unit circle that's going to occur at Pi halves, we can also draw our graph of our function cosine inverse of -1 that's going to occur at π cosine inverse of negative root 2 / 2. That one is 3 Pi force. So we're really thinking about what angle gives us out whatever was in the parentheses, with that angle being restricted in between 0 and π. For cosine, we have identities that say if we have an inverse of the original function, it has to equal X if the X is in between negative π halves and Pi halves, or it has to equal X if we're taking the sine of the sine inverse where -1 less than or equal to X less than or equal to 1. Same thing applies with cosine. The cosine restriction was zero to π if we're inputting cosine X or -1 to one if we're starting with cosine inverse of X. So when we look at sine, sine inverse of negative Pi 12th negative π twelfths was within the original restriction. So our answer here is just negative π twelfths, sine inverse of negative root 3 / 2 and then the sine of that. That's also within our restriction. So it's just negative root 3 / 2, cosine of negative π twelfths. This one's tricky because cosine of negative π twelfths is not within the restriction. So what we need to do is we need to think about whatever that cosine value here at negative π twelfths, what would be its equivalent in between 0 and π. This is an even function. So we know that if we have negative π twelfths here, Pi twelfths is going to have the same exact Y value. So cosine inverse of cosine negative π twelfths, cosine of negative π twelfths is going to have the same value for the Y as cosine of positive Pi twelfths. Now that Pi twelfths is within this restriction, so we're going to get an answer of π twelfths. The last one cosine cosine inverse of 1.51 point 5 is not in between -1 and one. And there's no way I can force negative or 1.5 to be in between those two numbers. Like I forced the negative π twelfths a moment ago. So this one is not defined or not possible. We're going to also look at finding the actual inverse functions. To find an inverse function, we're going to switch our X and we're going to switch our Y. So we're going to add 1, divide by 4, and based on what we were just seeing, we need to get rid of this tangent. We're going to take the tangent inverse of X + 1 / 4, and that's going to equal the tangent inverse of tangent Y. If I do it to one side, we do it to the other, but the tangent inverse and the tangent are going to cancel out. Instead of Y, I'm going to use F inverse of X. My correct notation tangent inverse X + 1 / 4. We want to find the domain and range of this new function. Well, we know that the range is just the domain of the original, so the range is going to be negative π halves less than X less than π halves. The domain is the range of the original. So the range of a tangent function is all real values or negative Infinity to Infinity. So the domain of the inverse has got to be the same as the range of the original or negative Infinity to Infinity. Looking at this next one, switch our X and our Y and we want to solve for Y. So we're going to have X / 3 equals sine of two y + 1. The way we get rid of the sine is we take the sine inverse of each side and the sine inverse of the sign are going to cancel, leaving us just two y + 1. Then we're going to subtract the 1 and divide by 2. So sine inverse of X / 3 -, 1 and we could think of that as all being multiplied by 1/2. So our inverse function here is going to be 1/2 sine inverse X / 3 - 1. We want the domain and the range again. Well, the domain is the range of the original. If we look at the amplitude here, the original was going from 3 to -3. So the domain is going to be -3 to three, and the range is just what was given as the domain of the original, remembering that range is our Y values. So back here we actually should take that X off and make it AY. Finally, we're going to be able to solve. And if we solve, we want to get the trig function by itself. So sine inverse of X equal negative π thirds take the sign of each side. So sine of sine inverse of X equals sine of negative π thirds. So X equals the sine of negative π thirds. Sine and negative π thirds is going to be negative root 3 / 2. If we look at cosine 5, cosine inverse X + 3, Pi equal 2, cosine inverse X + 2π, we're going to get all the cosine inverses on one side. By subtracting that two cosine inverse X, we're going to get all the pi's on the other side. So 2π - 3 Pi is negative π. We're going to divide. Then by that 3 we're going to take the cosine of each side because the cosine of the cosine inverse will cancel. So we're going to get X equal the cosine of negative π thirds. The cosine of negative π thirds is 1/2 inverse tangent function. If we look at our graph F of X equal tangent X, we know that it's a function because it passes the vertical line test. For a function to have an inverse, it has to be 1 to 1, so it has to pass both the vertical line test and the horizontal line test. Looking at this currently, we realize it doesn't pass the horizontal line test. So what we're going to do is we're going to restrict its domain. We always want 0° as one of our points in our in our restriction. Then we need to cover all of the Y values from the negative Infinity all the way up to positive Infinity. And we want to do this with a smooth continuous curve if possible. So our inverse function tangent inverse X really says that Y equal tangent inverse of X if X equal tangent of Y. Based on the restrictions, the domain of the inverse is always the range of the original. So in this case, negative Infinity to Infinity, the range is the original domain, but we have to have those restrictions to make it 1 to 1. So negative π halves to Pi halves. So let's look at some examples. We want to think about what angle gives us out a tangent of one that's within our restriction. Remember, our restriction is negative π halves to Pi halves. If we thought about looking at our graph, negative π halves to Pi halves, so tangent inverse of 1 occurs at π force. How about tangent inverse of -1? If we have to be within this restriction, we know that -1 is going to occur at negative π force because this is an odd function. So if +1 occurred at positive Pi force -1 has to occur at negative π force. What about tangent inverse of negative root three? Well, we want to think about when is our Y value negative root three? Well, our Y value is going to be negative root 3 at negative π thirds, tangent inverse of 0 is just going to be 0 and tangent inverse of root 3 / 3. If we thought about root 3 / 3, we want to think about what's the angle there that would give us Pi 6th. We have the inverse composition of functions. So if the inverse of the original function, it will equal X if the original X was in between our restrictions and negative π halves to Pi halves. If we have the function of the inverse function, it's going to equal X if the original X was in between negative Infinity, Infinity. So when we look at this tangent in tangent of tangent inverse of 4/4 was in between negative Infinity and Infinity. So that's just going to be 4 tangent inverse of tangent. 4 Pi FIS. 4 Pi FIS is not within our restriction. 4 Pi FIS is past π halves. So we need to think about four Pi FIS. 4 Pi FIS is going to be somewhere in here. So tangent of four Pi FIS is going to be a negative value. So I need the same value that would give me AY of negative of four Pi FIS whatever that Y value was out. But it had to be in between negative π halves and Pi halves. So this is going to occur at negative π FIS. We could think about subtracting a full. In this case. So now we have tangent inverse of tangent of negative π FIS because the tangent value at 4 Pi FIS and negative π FIS will be the same tangent value. Now the negative π FIS is within the given restriction, so this is just going to equal negative π negative π FIS. This next one, tangent inverse of tangent and -3 Pi sevenths -3 Pi sevenths is within negative π halves to Pi halves. So this one's just going to be -3 Pi sevens, we can find inverse functions. So if we want to find an inverse function, we're going to switch the X and the Y and solve for Y. So we're going to start by subtracting 4 and then we're going to divide by 4. Then we need to get rid of that tangent. Well, the way we're going to get rid of the tangent is we're going to take the tangent inverse of each side because the tangent inverse of the tangent is going to cancel, and that's going to just leave us y + 1 on the right side. To get Y by itself, we're going to subtract 1. So the inverse function using the correct notation is going to be tangent inverse of X - 4 / 4 - 1. We also want to find the domain and the range. Well, the domain of the inverse was the range of the original, and the range of any tangent function is negative Infinity to Infinity. The range of the inverse is the domain of the original, and that was given so -1 - π halves less than X less than π halves -1 if we wanted to actually find values that make this true. For the next example, we're going to start by dividing each side by -6. Oh. Let's write this so that it makes some sense. We're going to start by dividing each side by -6 so we're going to get negative Pi 6. How do I get rid of a tangent inverse? I'm going to take the tangent of each side. If I take the tangent of one side, I need to take the tangent of the other side. The tangent of negative Pi 6 is going to be negative root 3 / 3. Thank you and have a wonderful day.