Graph two periods of the given cosecant or secant function.

Question

y = sec x/2

Accepted Answer

Hello there. Today we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Sketch the graph of the following cosicant function using two periods, Y equals 6 multiplied by the Cosicant of X divided by 3. Awesome. So it appears for this particular prong we're asked to create a graph for this particular cosicant function using two periods. So in order to do that, let us take a moment here to focus on its reciprocal function. So by using the reciprocal function, we'll be able to effectively and efficiently and quickly draw this cosicant function, or rather sketch this cosicant function more quickly. So, considering that, let us consider that the reciprocal function, as we should recall, states that Y is equal to 6, multiplied by the cosine of X divided by 3. So that is, that will be its reciprocal function. So we also now at this point need to recall and use that use from our information that we should know about sine and cosine graphs. We should be able to recall that Y equals A. Multiplied by some trig identity function, which is either gonna be a sign or cosine function, that's what T denotes. It's either a sign or cosine function, so some sort of trig function. Multiplied by B multiplied by X, so when you take B multiplied by X, so B multiplied by X, and then we need to add C, and then we need to add D outside of the parenthesis. So let's quickly recap here. So capital A denotes our amplitude. C denotes the phase shift and D denotes the vertical shift, and we should also recall that P the period is equal to 2 pi divided by B. Awesome. So when we consider our reciprocal function, so when Y equals 6 multiplied by the cosine of X divided by 3, you should be able to note that the amplitude is equal to 6, and our period P is going to be equal to 2 pi divided by 1/3, which is going to be equal to 6 pi. Fantastic. And now, our next step is we need to note and recall that the phase shift C is equal to 0, and the vertical shift D is also equal to 0. So now we need to recall and use the domain, which the domain for this particular problem as we should recall, will have to be 0 is less than or equal to X and X is less than or equal to. 12, and this has to be our domain because we need it to cover two periods. So in order to cover two periods, our domain has to be 0 is greater than, so 0 is less than or equal to X and X is less than or equal to 12 pi. OK. So now we need to evaluate Y at X values that are within the given interval. So considering once again that 0 is less than or equal to X and X is less than or equal to 12 pi. So when we plug in different values of X into our equation to solve for Y, we'll be able to put together the following table. So I've gone ahead and created a table. So when X equals 0, Y will equal 6. And etc. So when you plug in different values for X, you will get a different value for Y. So when you plug in a value for X, it will change what your value for Y will be. So, we need to note that we need to plot the points to make a sinusoidal wave that passes through these points, and it's optional because ultimately, you are going to be removing your reciprocal graphs, so you could graph your reciprocal graph, which you could graph Y equals 6 multiplied by cosine of X divided by 3, but I've gone ahead and just given you the graph for the cosicant function, because that's ultimately the only graph that we're concerned about. So, yeah, so right now we just need to focus on labeling our cosicant graph, and we need to note that for all X intercepts, we need to draw a vertical asymptote through them. And we need to make sure we draw smooth curves that pass through the vertices at the crests and troughs, where a crest is like the mountain and the trough is like the valley, and we need to make sure This is very, very important that our curves, which are represented in red in our graph, they're going to get very, very close to our vertical asymptotes, but they're never going to intersect them and they're never going to touch them. They're just going to get closer and closer and closer to the vertical asymptote for infinity, getting as close to it as possible, but they will never touch. And they, and it's also important to note that the vertical ascentoptes will extend vertically to infinity. And once again, just to reiterate, we're only interested in the graph of the Cosicant function, so we can remove the graph of our Y equals 6 multiplied by cosine of X divided by 3. So when we do that and we focus on just our cosicant function, we could see when we used our table to plot our different points on the graph, noting that we're using the following format where it's X Y. So when X equals 0, Y is 6, and when, when X is 6 pi, Y is 6, and when X is 12 pi, Y is 6, and when X is 3 pi, Y is -6, and when X is 9 pi, Y is -6. And as you can see, we have our vertical asyopes represented by dotted lines. That's standard. Usually when you do any sort of asymptotes, they are always going to be represented by a dotted line. And as you can see, we plotted our different points and then we drew our curves up to basically go towards our vertical acetote, like the stretch out towards them and follow them, but they're never going to cross intersect, or touch. Awesome. And as you can see, we have like a a parabola shape for our point that's plotted at 66. And etc. So this part, this type of problem is very easy to solve for as long as you have a strong understanding on how to plot trig identities or trig functions into a graph. So as long as you're able to plot trig functions graphically, this type of problem is very easy to solve. If anything that I said was confusing, I would suggest just watching more videos on this topic or to go back in your textbook and read up on it a little bit more, but It's fairly simple. All you need to do is be able to plug in your different values for X into your equation for Y and then be able to plot those points, and that's it. And we've officially solved for this problem. Hooray. So thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Graph two periods of the given cosecant or secant function.

y = sec x/2

Key Concepts

Cosecant and Secant Functions

Period of Trigonometric Functions

Graphing Techniques for Trigonometric Functions

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