In Exercises 29–44, graph two periods of the given cosecant or...

Question

In Exercises 29–44, graph two periods of the given cosecant or secant function. y = csc(x − π)

Accepted Answer

Welcome back, everyone. In this problem, we want to sketch the graph of Y equals the cos of negative pi plus two X using two periods. Now, what do we already know? Well, recall that the co is the reciprocal of the sine function. And what that means is that if I have my sine function, which would look something like this, then my second would look like it's reciprocal or look like it's inverse, it's multiplicative inverse in a sense. Now you might be looking at the graph and saying, well, that doesn't really look that like that exactly. But that's because the co function only works when the sign of the value is not equal to zero. And that makes sense because if the sign of the value equals zero, then the cos would be one divided by zero, which would give us an undefined value. So at the points on which the graph is undefined, then our Cosic tends towards infinity, but it never comes to that point. And if we're going to plot the graph of the cos, it means we can graph sine X figure out where those vertical lines would be for ASYM tos and then plot our graph of the co and remove our sine graph. So let's go ahead with that plan. So let's let Y be equal to the sign of two X minus pi I just rewrote negative pi minus two X in an easier way in an easier form to understand. OK. Now, since we have a trick function, what do we know about the grass of trick functions? Recall that for a trick function Y equals a multiplied by the trick function. In this case, the sign of BX minus C plus D where from our graph notice that A which is the coefficient of sine will be equal to one B is the coefficient of X which equals two C is the opposite value to our constant in the parentheses for or sign which in this case, it was C would have been pi and D would be equal to zero because there's no constant outside of the function. Then our amplitude for our graph would be equal to A which in this case is one or period equals two pi divided by B which in this case is two pi divided by two, which equals pi our phase shift. That is how much our graph shifts to the left or right would be equal to C divided by B which should be pi divided by two. And there would be no vertical shift because D equals zero. So all of this information tells us that our graphs, amplitude is one, meaning it won't get any taller or any smaller. Its period is pi that's how far apart the peaks will be. And its phase shift is a half of pi, meaning when we graph our ass sign function, we're going to move it to the right by a half of pi unit. So I'm going to go ahead and graph the sine function with a period of pi and an amplitude of one on the graph. Now, if you're not sure how we got this graph of the sign of two X, then I would recommend that you revisit one of those lessons on how we graph trigonometric functions, but this is what it would essentially look like. So we've done it with our amplitude of one and our period of pi. No, we just need to show our face shift of half of pi, which means that I'm just going to move my function or move my graph half five units to the right. So when I do that, oops, that's a little disconnected. Let me put it together properly. As I move it to a half of a pie units, it would end up right here. Now, there are two things we need to specify. First, let's show the points at which the function is undefined. That is the points at which the sign of X equals zero. And that would just be where it cuts the X axis. So notice that it would be at a half of pi, OK? It would be at pie. It would also be at three halves of pie, five, sorry two pie. We already passed that one and also at five halves of pie. OK. So those would be our vertical lines for which our asymptotes would tend towards. Also, let's make note of the peaks and the valleys. So that means our second will pass through these points when now that we have that all we need to do is to draw our C and remove our sine graph. So I'm going to put the C the cos in blue, my apologies, the Cosic in blue. Now let's draw it and it's going to come here. So that would be the first one at the top, the next one at the bottom between pie and three halves of pie. Then the next one at the top between three halves of pie and two pie and then the next one at the bottom between two pie and five halves of pie. So those would be the points that our C can graph would run through. Our would run through. Finally let me remove or a OK. And no, we just need to specify what points the graph runs through for our co. So notice that it runs through R point 3/4 of pi and ne and one, sorry, positive one, it runs through five fourths soft pi A negative one, 74 and one and nine fourths of pi and negative one and those would be the points four or co second of negative pi plus two X. Thanks a lot for watching everyone. I hope this video helped.

In Exercises 29–44, graph two periods of the given cosecant or secant function. y = csc(x − π)

Key Concepts

Understanding the Cosecant Function

Phase Shift in Trigonometric Functions

Period of the Cosecant Function

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