Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Graphs of cosecant and secant. We're going to start by graphing sine X and or cosine X and then we're going to look at the inverse of that. So if we're going to graph our sine X, we know that sine is zero at 0 and at π and at 2π negative π negative 2π, it's going to have a maximum of one at π halves and a minimum at three Pi halves. And we're going to just draw in a smooth continuous curve. We're going to know that it repeats. So we're going to also be able to put in our negative π halves all the way back to -2π. Now an inverse. We can never have 1 / 0, that's undefined. So we're going to put in asymptotes every time the sine graph crosses the X axis, because 1 / 0 is undefined, so we know that we have asymptotes there. If we look at the highest and the lowest points, 1 / 1 is really just going to be one. So we're going to share this highest point sign and cosecant both. How about 1 / -1? Well, that's -1 also. So we're going to share these lowest points. If we thought about something like our Pi 6 value that was 1/2, well, 1 / 1/2 is 2 and we know we've got to be close to our asymptotes. So 1 / a small number is going to be a bigger #1 / a small negative number is going to be a big negative number. So we're going to share the highest and the lowest points with our sign. And the graph of our cosecant is going to be that black graph that you see. We're going to do the cosine the same way. So cosine, we start at 01 at Pi, halves at zero. At 3 Pi halves at zero, negative π halves zero -3 Pi halves zero at π it's at its lowest, which is -1 two Pi, it's back up to its highest. So π negative π is going to be at a -1 and -2π is going to be at its highest. A smooth continuous curve. Now the same concept as what we did with the Co secant. Every time we cross the X axis, we're going to put in an asymptote because we can't divide by zero. And then we're going to share the highest and the lowest points. So we're going to get close to the asymptote. We're going to share the highest and lowest, close to the asymptote, share the highest and lowest. Let's look at the domain here. So the domain for the Co secret, it's not all real numbers because we can't have -2π or negative π or 0 or π or 2π. So the way we write this is we say it's all reals X not equal to K Pi where K is an integer. How about the range? Well, if we're looking at our Y values, we're going from negative Infinity to -1 we're including -1, and then we're unioning that one to Infinity. If we want to do the same thing for our secant, our secant now can't be -3 Pi halves, negative π halves, Pi halves, and three Pi halves. So it's going to be all real numbers X not equal to π halves plus K Pi where K is some integer. The range here is going to be negative Infinity to -1 including that -1 union one to Infinity. How about the period? How often does it repeat? Well, the period for the cosecant it repeats every 2π. The period for the secant it also repeats every 2π. Is cosecant an even or an odd function? If you picked it up and rotated at 180° you would realize that this is a odd function. How about cosecant or secant? Secant. If I fold it along the Y axis, it would fall right on top of each other. Hence secant would be an even function. Our cosecant and secant graphs are going to have stretches and shifts just like our sine and cosine. The A in front is going to talk about a vertical stretch or shrink. The range is going to be from negative Infinity to the opposite of the absolute value of a union. And that's going to be a bracket because we're actually going to include that point union, the absolute value of a comma Infinity. Our phase shift is going to be setting that parenthesis equal to 0 and seeing what happens. So our phase shift, if we subtract the C and divide by B is going to be negative C / b. Our vertical translation is just the D at the end up or down, and our new is going to be the old period which was 2π divided by the B. Now, if we want to graph something like Y equal 3 cosecant of 1/2 X, what we want to do is we want to come up with all the information first. So that three, three times the highest and lowest within a shift that that cosecant is going to be as one and -1 we're going out to Infinity and negative Infinity. So 3 * 1 would give me three, 3 * -1 would give me -3. That's how we share the highest and lowest point. Our period is going to be 2π / 1/2, or in this case, it's going to be 4 Pi. We're not going to have any phase shift, so a whole period's going to be 4 Pi, and cosecant is one over sine, so that's going to be 00. We're going to stretch it out between 2π and -2π. So at negative π we're going to be at -3, and at positive Pi we're going to be at three. This would be what the graph of Y equals sine of three. Sorry, Y equal 3 sine 1/2 X would look like to do our cosecant. Every time we cross the axis, we're going to put in an asymptote, and then we're going to remember that sine and cosecant share the highest and the lowest points, getting closer and closer and closer to those asymptotes. Y equal -2 secant 3X plus one. Remember, secant is just the inverse of cosine, so our new. Is going to be 2π / 3 or 2π thirds. Our amplitude for a cosine graph would be two to -2. The negative in front tells us it's a reflection through the X axis. So if we wanted to do our graph, we would have at 2π thirds, we would have completed an entire graph or an entire period. Now with the reflection, and we're stretching it -2 or two 2π thirds, so π thirds be about there doing some approximating here. So if we were graphing the cosine graph, it would look something like that. Now we want the secant, so we're going to take everywhere we see a crossing of the axis, we've got to put in an asymptote because that's where it's undefined. We know it's going to share the highest and the lowest points. And then the last piece of this is that we actually have to shift it up one. So our new graph is going to be that graph shifted up one and it's periodic. So we can make lots and lots of periods if we so desired. We're going to look at Y equal to cosecant, the quantity four X + π. So we need to realize that our period, the regular. For cosecant is 2π. Now we're dividing it by 4. So we're going to get π halves this problem. We're actually going to have a phase shift, and our phase shift, if we think about setting that parenthesis equal to 0, our phase shift is going to be negative π force. Sometimes students like to think about our regular. For sine would be in between 0 and 2π. So if we thought about subtracting π from each piece and then dividing by four, we know that a full. Is going to occur between negative π force and π force, and that distance is really a period of π halves. With our sine graph, we would start at the origin 00, but now we've shifted to negative π force by π force. I should be. They're totally through with one complete. So if I know that halfway between I need to have an X intercept. Also my highest and lowest points are going to be two and -2 for the sine graph. Now, if we want to graph cosecant, and we know this is periodic, so we could graph a bunch of them if we so desired, we know that every time we cross the X axis, we're going to have an asymptote when we put in the inverse. We also know we're going to share the highest and lowest points. So we're just going to sketch in our graph, sharing the highest, sharing the lowest, highest and lowest. Thank you and have a wonderful day.