In this video, we restrict the domains of the sine and cosine functions so that the graphs satisfy the horizontal line tests and therefore become graphs of invertible functions and use reflection in the line y equals x to visualize the inverse sine function. We also sketch the graph of the tan function and repeat this process to describe the inverse tan function. Here are the graphs of the two circular functions y equals sine x and y equals cos x, both producing sinusoidal curves and wave patterns that replicate infinitely often as one moves back and forth along the real number line. There sine x and cos x are defined for all real numbers x interpreted as angles that wind around the unit circle. The domain in both cases is R, the whole real line. Because we moved from points on the unit circle across the vertical axis of sine x and down or up to the horizontal axis of cos x, the values are trapped between plus and minus 1. So the range is the interval from negative 1 to 1 including both endpoints. Graphs of all functions automatically satisfy the vertical line test. What about the horizontal line test in these cases? In fact, any horizontal line that touches either graph passes through infinitely many points. So, the horizontal line test fails spectacularly for both graphs. Remember, for a function to be invertible, we need the graph to pass the horizontal line test. So, neither of these look like they have any hope of being invertible. Well, there is hope. We can severely restrict their domains to that we get a fragment of each graph that passes the horizontal line test. For the sine curve, we box in this tiny part of the curve ignoring everything else. It's standard to restrict the domain to the interval from negative pi on 2 to pi on 2 including both end points, and then we get this fragment of the original sine curve. Notice that the range remains the same the interval from minus 1 to 1. This fragment has a very pleasant rotational symmetry about the origin. The curve coincides with itself if we rotate it a 180 degrees and instance of being an odd function, a concept that's important in future videos. For the cosine curve, it's not so clear what to choose but everyone agrees to box in this part of the curve, restricting the domain to the interval from zero to pi producing this fragment and again the range is unchanged, the interval from minus 1 to 1. Here are the two fragments, and you can see that horizontal lines pass through those curves at most once. They've been deliberately chosen to both pass the horizontal line test. Let's see what happens to this fragment of the sine curve when we reflect in the line y equals x. To simulate the reflection, we can take a transparency with this fragment and flip it over. Here are the fragment, the line y equals x and the reflected image all on the same diagram, and here's the reflected image on its own. The roles of input and output have been interchanged so the labels plus and minus pi on 2 now become points on the vertical axis and the labels plus or minus 1 now become points on the horizontal axis, and we have the graph of y equals inverse sine of x. Remember, inversion interchanges domain and range. So, the domain of the inverse sine is the range of sine which is the interval from minus 1 to 1 , and the range of inverse sine is the domain of our fragment to the sine function which is the interval from minus pi on 2 to pi on 2. Just think of the sine function as taking angles to numbers and inverse sine function is bringing numbers back to angles. Now, of course, angles are numbers if measured in radians but it can help to have some kind of mental image of the geometric angle that is the output of inverse sine. For example, sinusoidal degrees or sine upon 6 radians is a half. So, inverse sine of half gets you back to 30 degrees or pi on 6 radians. Maybe you conjure up in your mind an image of a 30-degree right angled triangle with side length one-half, sine of 60 degrees or pi on 3 radians is root 3 on 2. So, inverse sine of root 3 on 2 get you back to 60 degrees, or pi on 3 radians. Sine of 90 degrees or pi on 2 radians is 1, so inverse sine of one gets you back to 90 degrees or pi on 2 radians. Sine of minus pi on 2 is minus 1. Summing to sine of minus 1 gets you back to minus pi on 2. Here's an application. Suppose the kite is flying 40 meters directly above the ground but fastened to the ground by a piece of string which is 50 meters long. Assuming the string is approximately straight, we want to estimate the angle of elevation which we call theta. In the diagram, you can see a right angle triangle with angle theta opposite side then 40 meters and hypotenuse represented by the string of length 50 meters. The adjacent side is the ground but we don't need to know what its length is to solve the problem. To solve this, observe that sine theta is the opposite of the hypotenuse which is 40 out of 50 equal to 0.8. To get back to the angle theta, we take the inverse sine of 0.8. If your calculator is in radian mode, then you should get 0.93 radians to two decimal place. If it's in degree mode, you should say 53 degrees to the nearest degree. This checks out because 53 multiplied by the conversion factor power on a 180 is approximately 0.93 radians. The analysis for inverting the fragment to the cosine curve is similar and explained in the notes. I'd like to spend the remaining time discussing the inverse tan function which plays a central role later in integral calculus. You might recall from the last video that tan of x heads off towards infinity as x moves towards pi on 2 or 90 degrees in the first quadrant of the unit circle. We can graph the relationship between x and tan x for these values of x and it looks like this. We have similar behavior in the negative direction if we move in the fourth quadrant from zero towards negative pi on 2. The lines x equals plus or minus pi on 2 become vertical asymptotes to the curve. Because tan x is sine x on cos x, the periodic behaviors of sine and cosine mean that this picture reproduces itself at infinite on both to the right and to the left with infinitely many vertical asymptotes, which in fact occur at odd multiples of plus or minus pi on 2. Again, the horizontal line test fail spectacularly, so you have to drastically restrict the domain in order to produce the graph of an invertible function. The convention is to restrict the domain to the interval from minus pi on 2 to pi on 2 but not including the endpoints with tan is undefined. So, we focused on this very nice space of the tan curve which satisfies the horizontal line test. We can reflect in the line y equals x to invert the function, producing the graph of y equals inverse tan of x. A beautiful shape called a sigmoid curve sandwiched in between two horizontal asymptotes. Sigmoid curves are important in computer science and the mathematics of neurons used to model behavior in the brain. Remember, inversion interchanges domain and range, so the domain of inverse tan is the range of tan which is the entire real number line, and the range as in this tan is the domain of our fragment of tan which is the interval from negative pi on 2 to pi on 2 but not including the end points. Again, just think of tan as taking angles to numbers and inverse tan is bringing numbers back to angles. For example, tan of 45 degrees or pi of 4 radians is 1. So, inverse tan takes 1 back to 45 degrees or pi on 4 radians. Maybe you conjure up in your mind an image or a 45-degree right angled triangle with equal opposite and adjacent side things. Tan of negative pi on 4 is negative 1 so inverse tan takes minus 1 back to minus pi on 4. Here's an application. The statue of liberty is 46 meters tall standing on a pedestal at the same height. Find the angle theta subtended by the statue if you're viewing it from a distance of 250 meters. Though we haven't been asked to find it, let us denote the angle subtended by the pedestal by another angle called phi. So, there is a right angle triangle with angle phi with opposite side length of 46 meters and adjacent side length 250 meters. So, tan of phi is the fraction 46 over 250. So, I'm doing this because as phi equal to inverse tan of that fraction. There's another right angle triangle, with angle phi plus theta with opposite side length 46 plus 46 equals 92 meters and adjacent side length again 250 meters. So, tan of phi plus theta is the fraction 92 over 250. I'm doing this gives us phi plus theta equals inverse tan of that fraction. Taking phi away gives us theta equals this very nice explicit expression involving fractions and inverse tan. We can evaluate this using a calculator to get approximately 0.17 radians, which is about 9.8 degrees. This solves our original problem. The angle subtended by the statue from one-quarter of a kilometer away is almost 10 degrees. This is in fact the last video for module two. Over the course of this module, we have introduced and discussed so many different and contrasting ideas all glued together by the unifying concept of a function and associated ideas of domain, range, and graph. We've talked about linear functions whose graphs of lines and quadratic functions whose graphs of parabolas, and discussed techniques involving completing the square and the quadratic formula. We talked about polynomial functions built out of non-negative integer powers of x which can be evaluated using simple arithmetic. On the way, I mentioned fractional powers which are a neat way of interpreting square roots and other types of roots. We showed how to create new functions by means of composition and inversion both geometrically by reflecting in the line y equals x and by algebraic manipulation. By changing our points of view with regard to the roles of constants and variables, we moved from power functions to exponential functions and discuss the most important base Euler's number E. We described logarithmic functions and discussed exponential and logarithmic laws and applied them to exponential growth and decay. Recently, we introduce trigonometry and the sine, cosine, and tangent functions. Just now we discussed the inverse sine and inverse tangent functions and briefly illustrated how that can be used to interpret numerical information in terms of angles. Please read and digest the notes, and when you're ready, please attend the exercises. Thank you very much for watching, and I look forward to seeing you again soon.