Problem 4 (5 pts.). Plot the function f( heta) = 1.5...

Problem 4 (5 pts.). Plot the function f( heta) = 1.5 sin( heta) - 1 over the interval $-2pi le heta le 2pi$. Problem 5 (5 pts.). Plot the function f( heta) = csc( heta)$ over the interval $-2pi le heta le 2pi$. Problem 6 (6 pts.). Graph the functions T = cos( heta), T = cos(2 heta), and T = cos( heta/2) over the interval $-2pi le heta le 2pi$ on the same plot. Make sure you clearly label each function.

Transcript

00:01 All right, since there's three problems going on here, i'm going to jump right into this.

00:05 With the very first thing that's happening with number four is that we have this shift down one, and it's a sign curve with an amplitude of 1 .5.

00:17 So i put a little tick mark in here, here, and call this 1.

00:22 So what needs to happen, a little tick mark, negative 2, and then another small tick mark to get to negative 3.

00:31 So what needs to happen for this curve is it starts at negative 1 and goes up to, you need to go up to 0 .5, and then it comes back down.

00:45 So this ordered pair would be like pi over 2, 0 .5.

00:51 And then back down, and then over here at it would be 3 pi over 2, negative 2 .5.

01:02 And i didn't really write a few of these down, but like at 2 pi, you're back to negative 1.

01:08 So i'm just trying to clarify where i'm at, and then we also need to go backwards to, they also want you to get to negative 2 pi.

01:18 So hopefully that's enough for this graph for f of theta is equal to 1 .5 sine of theta minus 1.

01:32 I appreciate that they put the parentheses in there, but hopefully that makes sense.

01:36 Okay, let's look at number five then, where they just want you to graph the co -sequent curve.

01:42 Well, because co -secant is the reciprocal of sign, what i would typically do is draw the sign curve.

01:51 And so the normal sign curve, as you can see in the first one, starts at the origin, goes up to one, down to negative one, and then at two pi is back to zero.

02:04 Same thing on the other side, except, you know, negative 2 pi, comma, zero.

02:12 So what's the kosekin curve got to look like? well, if there's a zero for sign, because it's a reciprocal, it has to be a vertical asymptote.

02:24 So these are all vertical asymptotes.

02:26 And then if it's a max for sign, it's a minimum for koseken.

02:31 Because if you think about, well, what's the reciprocal of one? it's still one.

02:35 What's a reciprocal of one half, though? it's two.

02:39 So this is what the graph of cosecant should look like from negative 2 pi to 2 pi.

02:45 I think that's enough in that one...

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