Hey everyone. In working with the unit circle, we found that multiple different angle measures could all be located at the exact same position, like, say, π/6 and 13π/6. These were referred to as coterminal angles, and we found them by adding or subtracting multiples of 2π. The same exact idea actually applies to points on our polar coordinate system. One single point can be represented by multiple different ordered pairs, and this is actually something that you'll be explicitly asked to find: multiple different ordered pairs that all map back to the same point. This might sound like it's going to be complicated, but all we're going to do here is continue using our knowledge of coterminal angles along with what we know about polar coordinates in order to find an infinite number of ordered pairs all located right at the same point. I'm going to show you exactly how to find these different ordered pairs. So let's go ahead and get started.

Taking a look at our graph here, I see that I have the point (4, π/6). Looking at this point, what if I wanted to represent it with a different ordered pair? How could I go about that? Well, from my knowledge of coterminal angles, I know that π/6 and 13π/6 are both located at the same position. What if I instead represented this point with the ordered pair (4, 13π/6)? If I locate this angle 13π/6, I know that I'm going to end up right along that line. Then if I count 4 units out, I do end up right back at the same point. These are two different ordered pairs, both located at the exact same point. I can continue to find even more ordered pairs by going more rotations around, just adding multiples of 2π the same way that we did with coterminal angles. For some point in polar coordinates, r, θ, it will be located at the exact same point as r, θ ± 2πn. This is not the only way that we can go about finding multiple ordered pairs for the same point because we can also change r by making it negative. If I make r negative, that means that I need to take my angle θ and add π to it in order to ensure that it's located at the same position as my original point. Then I can continue to add or subtract multiples of 2π as we did before. Seeing this all written out can look a little bit complicated, but it's actually rather simple. So let's think about this. If we're given a point r θ and we're asked to find multiple ordered pairs that are all located at this point, we can keep r the same. Keeping r the same, we can simply add or subtract multiples of 2π from our angle θ in order to get multiple ordered pairs. If we want to change θ by making it negative, all we have to do is add π to our angle. Then again, we can continue adding or subtracting multiples of 2π. With this in mind, let's go ahead and find even more coordinates for this point (4, π/6).

About this criterion to find these coordinates: r should be greater than or equal to 0. It should be positive, as it was in our original point. Our original point (4, π/6), I'm going to keep r the same positive 4. What should θ be? We're told that θ should be between -2π and 0. If we keep θ the same or if we keep r the same, we can take θ and add or subtract multiples of 2π. If I take my angle π/6 and I subtract 2π to give me -11π/6, this is my new angle θ within the specified interval. I have my new ordered pair (4, -11π/6). This should be located at the exact same point as my original point. Coming back up to my graph here, measuring to my angle -11π/6 clockwise from that polar axis, I end up right along this line. Then if I count 4 units out since my r value is still 4, I do end up right back at that same point. Let's find one final set of coordinates here.

Now here, our criterion tells us that r should be less than or equal to 0. It should be negative. I'm going to make my r value negative 4. Then I want θ to be between 0 and 2π. Looking at my original angle, π/6, I know that this is already within my specified interval. So what if I just kept it the same? That gives me my point (-4, π/6). We want this to be located at the same point, so let's verify that that's true. Coming to π/6 on my graph here, I know that I end up on this line. Since that r value is negative, I would actually be counting out in the opposite direction from the pole. I would count out in this direction and end up right about here in quadrant 3. This is definitely not at the same point as my original ordered pair (4, π/6). So what exactly happened? Well, remember that whenever we change r to be negative, we have to add π to our original angle θ. If I take my original angle π/6 and add π to it in order to give me 7π/6, that gives me the ordered pair (-4, 7π/6). Locating 7π/6 radians on my graph here, I end up along this line. But since this r value is negative, I'm going to count out in the opposite direction from the pole, 4 units. And now I do end right back up at that same point. We need to be really careful and remember to add our factor of π if we're changing the sign of r. Here, I have a bunch of different coordinates that are all located at the exact same point. Whenever you're asked to find ordered pairs, you may not always be given specific criteria to find these ordered pairs, and that's totally fine. You can continue to find them in this way or you can just take your angle θ and add or subtract as many multiples of 2π as you want to get more angles. Now that we know how to find different ordered pairs that all map back to the same point, let's continue practicing together. Thanks for watching, and I'll see you in the next one.