Use the circle shown in the rectangular coordinate system to solv...

Question

Use the circle shown in the rectangular coordinate system to solve Exercises 81–86. Find two angles, in radians, between -2𝜋 and 2𝜋 such that each angle's terminal side passes through the origin and the given point.

F

Accepted Answer

Welcome back. I am so glad you're here. We are asked to determine two angles between negative two pi and two pi where the terminal side of each angle passes through the origin and point Z express your answer in radiance. Then we're given a graph. It's a coordinate plane with a vertical y axis and a horizontal X axis. They come together at the origin in the middle over the coordinate plane, there is a unit circle and one of the points marked on the unit circle is Z. We'll describe that point in a little more detail in a bit. Our answer choices are answer choice. A negative pi divided by four and 11 pi divided by six. Answer choice B negative pi divided by six and 11 pi divided by six. Answer choice C negative pi divided by four and seven pi divided by four and answer choice D negative pi divided by six and seven pi divided by four. All right, let's take a look at what's going on here. What information we have? All right, we've got an angle, two angles actually in standard position, meaning that the initial side has its vertex at the origin begins at the origin and then goes out along the positive part of the X axis. The terminal sign goes through the point Z. So point Z is on our unit circle. If we head clockwise from the initial side, it's the first point marked on our unit circle as we head down into quadrant four. So drawing our terminal side from the origin through point Z oh whoops, that wasn't through point C. Let me try that again. All right, there we go. So now what we need to do is determine two angles here and they're going to be co terminal angles. They need to be between negative two pi and two pi. What does that mean? Well, we recall from previous lessons that two pi is when we start on our initial side and we head counterclockwise and we go all the way around until we get back to where we started at the positive part of the X axis. That is two pie. Now, if we start in that same position, but we go clockwise and go all the way around heading back to where we started that is negative to pi so our two angles need to be between that and they need to be expressed in radiance. So we have a couple of ways to go about solving this one way is that this angle is on our unit circle. We can first determine what the positive value of this is. So heading in our counterclockwise direction, we can label the parts of the unit circle and then we'll know what angle Z is. So let's begin labeling our unit circle. So in the first quadrant, as we are heading up from our initial side, there are three hash marks marked on our unit circle and they represent going, it's 1/12 of the way around the whole thing which is 1/ of halfway or one third of a quarter of the way, one third of the quadrant. And so that is we know from previous lessons that half a straight line, half of the way around our unit circle is equal to pi that's the negative part of the X axis. So if we're going 1/6 of that, that's going to be pi divided by six and then the next hashmark on our unit circle here is one quarter of the way toward that straight line toward pi. So that will be labeled pi divided by four. The next mark on our unit circle in quadrant, one is two thirds of the way through our quadrant or one third of the way toward a straight line toward our pie. So this would be labeled pi divided by three and there are three hash marks in each quadrant, they follow this same pattern. And so I'm going to label the rest of this much more quickly if you need a quick review on unit circles, definitely check out previous lesson videos So the positive part of the y axis is pi divided by two. That almost looks like I wrote pi. So pi divided by two. Then in the second quadrant, we're continuing counterclockwise. The first hashmark after pi over two is two pi divided by three. The next is three pi divided by four. The next is five pi divided by six. Then we have pie now continuing into the third quadrant. The first mark is seven pi divided by six. The next is five pi divided by four. Then we have four pi divided by three. Now the negative part of the Y axis is the three pi divided by two. Now into the fourth quadrant, we're almost at our Z. So into the fourth quadrant, we have five pi divided by pi divided by four. And where our Z is that is 11 pi divided by six. And that is our positive angle here. And you can see that that is between negative two pi and two pie because we're not quite yet up to the positive two pie. Now to find the negative version of this, there's a couple ways to go about doing this. I'll show you both they'll be quick. So we could label our negative angle measurements in radiance. We know that starting from the positive part of the X axis heading down into the fourth quadrant, that would be 1/6 of the way toward the. Now negative pi which is the negative part of the x axis and 1/6 of that way would be negative pi divided by six. And we don't have to go far on this unit circle because that's the first thing. So we know that our other angle is negative pi divided by six, if it's less helpful for you to think of it in terms of the unit circle, and it's easier to think of it algebraically what you can also do so that you are between negative two pi and two pi is you can take the angle that we know the pi divided by six and you can subtract two pie and you know that that will still be within that range between negative two pi and two pi. And that will give us again, negative pi divided by six. So that is how we get those two angles that are co terminal, both going through point Z we look at our answer choices and this matches with answer choice B well done. We'll catch you on the next one.

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Use the circle shown in the rectangular coordinate system to solve Exercises 81–86. Find two angles, in radians, between -2𝜋 and 2𝜋 such that each angle's terminal side passes through the origin and the given point.
F