ArcTan(1) + ArcTan(2) + ArcTan(3) = pi
Math
Learning how to solve hard math problems
A couple of years ago, my brother in law showed me this cool proof that the arctangent of one plus arctangent of two plus arctangent of three is pi. I don't know if this is exactly how he did it, but it's pretty close. I'm going to try to go pretty fast with this video. It doesn't take forever. Just let me remind you if you forgot what arctangent is. That means the angle that has a tangent of one. So it's 45°. So arctangent of one is 45°. And as a picture what that means, tangent is basically slope. So that's 45°. Okay, and then the arctangent. That means an angle that has basically a slope of two. That's a little bit steeper, so that's 63°. It looks like I'm going to store that one. And then the angle that has a slope of three, the angle that has a tangent of three is 71°. So what this is saying is that the arctangent of three plus the arctangent of two, which I stored in B, plus the arctangent one which was 45 is pi, well, before in degrees a 180°.
So if I switch the mode over to radians, it's just a little bit more cool in radians. Now the arctangent of one plus the arc, the angle it has a tangent of two, plus the angle that has a tangent of three, is a 180°, but if we're in radians, now it's pi. Pretty awesome. Okay, now this tune on the calculator certainly does not constitute a proof. So let's see if we can actually prove that. Okay, first I'm going to start out with a square. It could be any size square, but it's definitely a square, so I've got a right angle there, and these two legs are the same. We'll call that angle a. Again, I'm going to add another square to that. Same size exact same square. And add this angle right there in red. I'll call that one angle B and add a third one right there, and we'll call that one in green angle C so what I would need to show is that angle a black one plus angle B, the red one plus angle C the green one. Would have to actually add up to a 180° or add up to pi.
Okay, so I'm going to do that with there's the top same squares only now I have 6 of them. So there is angle C now just to make sure you agree that that's angle C notice the green has a rise of one and a run of three. So here it is up one over three. So there's angle C and there's angle B I've got a rise of one and a run of two. I'll speak of absolute values. It could be. So it's down one over two, so it's really negative one negative one half. But it doesn't matter. I just trying to get the same angle right there for a for B so see the red one up one over two. Vertically one horizontally two. So there's angle B not really all I have to show is that this one in here is also angle a and then they add up to a straight line so that that would mean that they add up to a 180° or pi radians. Okay, there's my original box so I'm going to try to show that this has got to be angle a so I'm going to reproduce angle a all by itself. So we started out with the box with angle a making this a right angle and splitting that a half, so that's got to be 45°. So the arctangent of this is the 45°. The issue is, is that the same one that's right there. Okay, it's a square so you can make the legs whatever you want.
I'll just make it a unit square. So how about one and one? And then what I want to do is put the same thing over here. I'm going to put this leg right there. So I'm trying to show that this leg right here, let me change it to black. I'm trying to show that this angle right here is that angle right there. If I can show those two things, then yes, that's angle a okay. Let me give that a try. Whoop. All right, consider this red. I'm going for angle a now so consider this red one. See how in the red one that's one and two. So if we do the Pythagorean theorem, two squared is four, plus one squared is one, so that must mean this hypotenuse is square root of 5. And then take a look at this one over here. That's one, and that's two. So we got the square root of 5 again. So that's the square root of 5, and that's the square root of 5. So it's going to be similar. Maybe before I make the claim that it's going to be similar, let me get the length of the green one. That's going to be one and that leg is going to be three.
So if I do the Pythagorean theorem there, I got three squared plus one squared so that green one is the squared at ten of do me a little Pythagorean theorem here. The square root of 5 plus the square root of 5 is that equal to square root of 5 squared, a squared plus B squared is that the square root of ten squared is of course you can see it is. 5 plus 5 equals ten so what that means is this is a right angle. Okay, so I got these are now similar triangles. I've got one one and the square root of two here, square root of 5 square root of 5 squared of ten there. There's a right angle. That's this red angle right there. So that angle a now is that angle a right there by similar triangles and corresponding parts and side angle sides, hiding aside, who got all kinds of ways now, we can show that that's that angle a is that angle a so that gives us angle C plus angle a plus angle B is equal to 180° or pi so now from the picture we can see we can see from the picture that the tangent of angle C so right there C is opposite over adjacent, so that's three over one.
We can see that the tangent of angle a right here is angle a if I do opposite over adjacent, that's square root of 5 over square root of 5. Or one. And the tangents of angle B from my picture, right? There's B so it would be opposite over adjacent is two over one. So the tangent of C is three, the tangent of a is one. And the tangent of B is two. So now we can write it the other way around. If the tangent of C is three, then the arctangent of three. Plus the arctangent of one, that one right there, plus the arctangent of two, that's angle C plus angle a plus angle B, must equal pi. From that picture right there. Now we could do this, we could do this. For the pre cow part, we could do it all in arc sign also. The inverse sine of this, let's go, let me go a, B, and C this time instead of C a and B, it doesn't matter what order we go in. Okay, so angle a right here is angle if I do sine, sine is opposite over hypotenuse. So that would be the square root of 5 over the square root of ten. Or we could just do it in this form too, but I'll do it from this form right here. Opposite over hypotenuse.
How about one over the square root of two? That would be the same thing as the square root of 5 over the square root of ten. Plus B arcsine, that would be two over the square root of 5. And then C would be opposite three over the square root of ten. That's also equal to pi, those also add up, make that thing. So if you did it the other way around, that would be the arcs are if we rationalize this, that would be the arc sine of square root of two over two. Plus the arcsine of two square roots of 5 over 5, that would be that one simplified, or rationalized, and the arc sign of three square roots of ten over ten equals pi. That's also true. That's the same statement as this one. Only it doesn't look as cool. This one is tangent one plus tangent two plus tangent three. This has got a bunch of weird numbers. And just to extend this into a pre Cal lesson, you can show this. The arc sign of the square root of two over two plus the arcsine of two square roots of 5. Over 5 plus the arcsine of three square roots of ten. Over ten should be pi. Ta-da. Just like the arctangent of one plus the arctangent of two plus the arctangent of three was pine. We could certainly do it with cosine also, just use the other legs and do arc cosine. So very fun. Arctangent of one plus arc tangent two plus arctangent of three happens to be pi.