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# Intro to arctangent

## Video transcript

In the last video, I showed you that if someone were to walk up to you and ask you what is the arcsine-- Whoops. --arcsine of x? And so this is going to be equal to who knows what. This is just the same thing as saying that the sine of some angle is equal to x. And we solved it in a couple of cases in the last example. So using the same pattern-- Let me show you this. I could have also rewritten this as the inverse sine of x is equal to what. These are equivalent statements. Two ways of writing the inverse sine function. This is more-- This is the inverse sine function. You're not taking this to the negative 1 power. You're just saying the sine of what-- So what question mark-- What angle is equal to x? And we did this in the last video. So by the same pattern, if I were to walk up to you on the street and I were to say the tangent of-- the inverse tangent of x is equal to what? You should immediately in your head say, oh he's just asking me-- He's just saying the tangent of some angle is equal to x. And I just need to figure out what that angle is. So let's do an example. So let's say I were walk up to you on the street. There's a lot of a walking up on a lot of streets. I would write -- And I were to say you what is the arctangent of minus 1? Or I could have equivalently asked you, what is the inverse tangent of minus 1? These are equivalent questions. And what you should do is you should, in your head-- If you don't have this memorized, you should draw the unit circle. Actually let me just do a refresher of what tangent is even asking us. The tangent of theta-- this is just the straight-up, vanilla, non-inverse function tangent --that's equal to the sine of theta over the cosine of theta. And the sine of theta is the y-value on the unit function-- on the unit circle. And the cosine of theta is the x-value. And so if you draw a line-- Let me draw a little unit circle here. So if I have a unit circle like that. And let's say I'm at some angle. Let's say that's my angle theta. And this is my y-- my coordinates x, y. We know already that the y-value, this is the sine of theta. Let me scroll over here. Sine of theta. And we already know that this x-value is the cosine of theta. So what's the tangent going to be? It's going to be this distance divided by this distance. Or from your algebra I, this might ring a bell, because we're starting at the origin from the point 0, 0. This is our change in y over our change in x. Or it's our rise over run. Or you can kind of view the tangent of theta, or it really is, as the slope of this line. The slope. So you could write slope is equal to the tangent of theta. So let's just bear that in mind when we go to our example. If I'm asking you-- and I'll rewrite it here --what is the inverse tangent of minus 1? And I'll keep rewriting it. Or the arctangent of minus 1? I'm saying what angle gives me a slope of minus 1 on the unit circle? So let's draw the unit circle. Let's draw the unit circle like that. Then I have my axes like that. And I want a slope of minus 1. A slope of minus 1 looks like this. If it was like that, it would be slope of plus 1. So what angle is this? So in order to have a slope of minus 1, this distance is the same as this distance. And you might already recognize that this is a right angle. So these angles have to be the same. So this has to be a 45 45 90 triangle. This is an isosceles triangle. These two have to add up to 90 and they have to be the same. So this is 45 45 90. And if you know your 45 45 90-- Actually, you don't even have to know the sides of it. In the previous video, we saw that this is going to be-- Right here. This distance is going to be square root of 2 over 2. So this coordinate in the y-direction is minus square root of 2 over 2. And then this coordinate right here on the x-direction is square root of 2 over 2 because this length right there is that. So the square root of 2 over 2 squared plus the square root of 2 over 2 squared is equal to 1 squared. But the important thing to realize is this is a 45 45 90 triangle. So this angle right here is-- Well if you're just looking at the triangle by itself, you would say that this is a 45 degree angle. But since we're going clockwise below the x-axis, we'll call this a minus 45 degree angle. So the tangent of minus 40-- Let me write that down. So if I'm in degrees. And that tends to be how I think. So I could write the tangent of minus 45 degrees it equals this negative value-- minus square root of 2 over 2 over square root of 2 over 2, which is equal to minus 1. Or I could write the arctangent of minus 1 is equal to minus 45 degrees. Now if we're dealing with radians, we just have to convert this to radians. So we multiply that times-- We get pi radians for every 180 degrees. The degrees cancel out. So you have a 45 over 180. This goes four times. So this is equal to-- you have the minus sign-- minus pi over 4 radians. So the arctangent of minus 1 is equal to minus pi over 4 or the inverse tangent of minus 1 is also equal to minus pi over 4. Now you could say, look. If I'm at minus pi over 4, that's there. That's fine. This gives me a value of minus 1 because the slope of this line is minus 1. But I can keep going around the unit circle. I could add 2 pi to this. Maybe I could add 2 pi to this and that would also give me-- If I took the tangent of that angle, it would also give me minus 1. Or I could add 2 pi again and it'll, again, give me minus 1. In fact I could go to this point right here. And the tangent would also give me minus 1 because the slope is right there. And like I said in the sine-- in the inverse sine video, you can't have a function that has a 1 to many mapping. You can't-- Tangent inverse of x can't map to a bunch of different values. It can't map to minus pi over 4. It can't map to 3-- what it would be? --3 pi over 4. I don't know. It would be-- I'll just say 2 pi minus pi over 4. Or 4 pi minus pi. It can't map to all of these different things. So I have to constrict the range on the inverse tan function. And we'll restrict it very similarly to the way we restricted the sine-- the inverse sine range. We're going to restrict it to the first and fourth quadrants. So the answer to your inverse tangent is always going to be something in these quadrants. But it can't be this point and that point. Because a tangent function becomes undefined at pi over 2 and at minus pi ever 2. Because your slope goes vertical. You start dividing-- Your change in x is 0. You're dividing-- Your cosine of theta goes to 0. So if you divide by that, it's undefined. So your range-- So if I-- Let me write this down. So if I have an inverse tangent of x, I'm going to-- Well, what are all the values that the tangent can take on? So if I have the tangent of theta is equal to x, what are all the different values that x could take on? These are all the possible values for the slope. And that slope can take on anything. So x could be anywhere between minus infinity and positive infinity. x could pretty much take on any value. But what about theta? Well I just said it. Theta, you can only go from minus pi over 2 all the way to pi over 2. And you can't even include pi over 2 or minus pi over 2 because then you'd be vertical. So then you say your-- So if I'm just dealing with vanilla tangent. Not the inverse. The domain-- Well the domain of tangent can go multiple times around, so let me not make that statement. But if I want to do inverse tangent so I don't have a 1 to many mapping. I want to cross out all of these. I'm going to restrict theta, or my range, to be greater than the minus pi over 2 and less than positive pi over 2. And so if I restrict my range to this right here and I exclude that point and that point. Then I can only get one answer. When I say tangent of what gives me a slope of minus 1? And that's the question I'm asking right there. There's only one answer. Because if I keep-- This one falls out of it. And obviously as I go around and around, those fall out of that valid range for theta that I was giving you. And then just to kind of make sure we did it right. Our answer was pi over 4. Let's see if we get that when we use our calculator. So the inverse tangent of minus 1 is equal to that. Let's see if that's the same thing as minus pi over 4. Minus pi over 4 is equal to that. So it is minus pi over 4. But it was good that we solved it without a calculator because it's hard to recognize this as minus pi over 4.