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# Intro to arctangent

Sal introduces arctangent, which is the inverse function of tangent, and discusses its principal range. Created by Sal Khan.

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.