Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Some examples of inverse trig functions. Remember that an inverse trig function is really, really an angle. And so whenever I see sine inverse or cosine inverse or tangent inverse or secant inverse, we're talking about what angle they have to be 1 to one to be an inverse trig function. So our sine is going to be restricted between negative π halves and Pi halves. So when I look at this, I think about what sine value, what angle gives me out of half that's in between these restrictions. And our answer is going to be Pi 6. This comes from knowing our unit circle well, the next one, sine inverse of negative root 3 / 2. Once again our restrictions between negative π halves and Pi halves. So our answer is going to be negative π thirds. If we thought about our graph of Y equals sine X. The reason it's restricted between the negative π halves and Pi halves is I needed to cover all the Y values exactly one time. Cosine inverse is going to be restricted between 0 and π. It's the same reasoning if we thought about our Y equal cosine X graph, we want to hit every Y value exactly once to make it 1 to 1. So what value in between 0 and π? Give us out cosine or give us out negative root 3 / 2 as our angle value and our answer is going to be 5 Pi 6 tangent is going to get restricted in between negative π halves and Pi halves because we want to cover every single Y value and if we can, we want it to be continuous. So when we look at this one, when do we get out negative root 3? And that's going to be at negative π thirds sine inverse of 0. When is sine 0 at 0 sine inverse of -1? What angle gives us out -1 for our sine value? And that'd be negative π halves because we have that restriction of negative π halves to Pi halves. Cosine inverse of 1 is going to be 0 because the angle 0 cosine of 0 is 1. Cosine inverse of 0 is going to give us Pi halves. Once again, these are from knowing our unit circle very well. Sine, sine inverse of .7. We need to remember that the sign is restricted between negative π halves and Pi halves. So the first question is, is .7 in between negative π halves and Pi halves? And the answer is yes. So a function and its inverse will cancel each other out as long as the domain restriction is OK. So we're going to get .7 as our final answer. This next one, what is cosine of π thirds? Well, cosine of π thirds is 1/2. And what's cosine inverse of 1/2? It is π thirds. So this next one, cosine of four Pi thirds, cosine of four Pi thirds is going to be negative 1/2. Cosine inverse of negative 1/2 is going to be 2π thirds because remember our restriction for cosine has to be 0 to π. So I can't just say the cosine inverse and the cosine cancel here because that four Pi thirds wasn't within the cosine inverse. So we have to think about what is cosine of four Pi thirds. It was negative 1/2. Then take the cosine inverse of that to get 2π thirds sine of sine inverse of π. Sine had to be in between negative π halves and Pi halves for its inverse. So is sine inverse of π in between here? No. So this one's not possible. Well, this next one it's tricky because it is possible. What is sine of π? Well, sine of π is really just 0. So now what is sine inverse of 0 in between negative π halves to Pi halves for our sine inverse restriction. So sine inverse of 0 is 0. We're going to move into ones that we don't know immediately off of our unit circle. So if we have sine of tangent inverse of 720 fours, this is really really really just an angle. So when we think about that being an angle, we could think about the tangent is the opposite side over the adjacent side. If I know two sides of a right triangle, can I find the third side? Yes, by Pythagorean theorem, right now we know sqrt 24 ^2 + 7 ^2 is going to equal that last side. Well, 24 ^2 + 7 ^2 is going to give us the square root of 625, which is going to be 525 South. Now when I want the sign of that, we're just going to take the opposite side over the hypotenuse to get 720 fifths, cosine, sine inverse of -4 fifths, the negative 4/5. Remember it had to be in between negative π halves and Pi halves for sine inverse. So we're going to be down here. This is all really, really just some angle. So sine opposite over hypotenuse. If we know two sides of our right triangle, we can find our third. So 5 ^2 -, 4 ^2 which is going to be 3 when we compute it. So looking at this triangle, I want cosine, so I want the adjacent side over the hypotenuse. This next one cosine inverse of negative 1/4. Cosine inverse is restricted between 0 and π cosine is adjacent over hypotenuse. So we're going to find our third side by 4 ^2 - 1 ^2 or sqrt 15. Now we want the tangent of this triangle. So the tangent of this triangle is going to be the opposite side over adjacent and tangent in the second quadrant negative. So we get negative square roots of 15. We can do the same thing if we just had variables tangent, cosine, inverse of X. Now here we're going to make some assumptions that our X is positive, so that cosine inverse is really just an angle. Cosine is adjacent over hypotenuse. If I know two sides of the triangle, I can find the third. So 1 ^2 -, X ^2 or sqrt 1 -, X and now I just want the tangent of this. So when I look the tangent's the opposite over the adjacent this next one sign. So we know the opposite side and we know the hypotenuse. We can find our third side by the hypotenuse squared and that square and that square root are going to cancel minus our leg squared. So when we compute this out, we get the square root and the square to cancel. So X ^2 + 4 -, X ^2 or sqrt 4 or two. When we want secant, secant is the hypotenuse over the adjacent, so sqrt X ^2 + 4 / 2. If we wanted to do one more, let's say we wanted the cosine of tangent inverse of X / sqrt 3, we would draw our triangle remembering that the tangent inverse is really just some angle opposite over adjacent. So our third side would be X ^2 + sqrt 3 ^2 or sqrt X ^2 + 3, and when I want the cosine, it's adjacent over hypotenuse. We would rationalize this so sqrt X ^2 + 3 / sqrt X ^2 + 3. So we'd get as a final answer, square root of three X ^2 + 9 / X ^2 + 3. Thank you and have a wonderful day.