This is my 50th daily post! Thanks for reading, everyone.
Today I watched the second and third episodes of the Lockdown Math series by 3blue1brown. They are the first two parts of a three-part subseries dedicated to the connections between trigonometry and complex numbers.
This evening I am finally ready to continue my exposition of the parabola problem. Below are two inverse trigonometric identities needed for the proof of the two remaining cases. I found these in an online table of identities, but I proved them myself.
Lemma: When $xy<1$, $$\arctan x + \arctan y = \arctan \frac{x+y}{1-xy}\text{,}$$ and when $xy>1$, $$\arctan x + \arctan y = \pi + \arctan \frac{x+y}{1-xy}\text{.}$$
Proof: Let us take as given the sum formula for tangent that is found in every trigonometry textbook: $$\tan (a+b)=\frac{\tan a + \tan b}{1-\tan a\tan b}\text{ when }\tan a \tan b\neq 1\text{.}$$
Using $\arctan x$ and $\arctan y$ as the two angles being added, it follows that $$\tan(\arctan x + \arctan y)=\frac{\tan(\arctan x)+\tan(\arctan y)}{1-\tan(\arctan x)\tan(\arctan y)}=\frac{x+y}{1-xy}\text{ when }xy\neq 1\text{.}$$
According to the definitions of the tangent, sine, and cosine functions, this implies that, when $xy\neq 1$, $\sin(\arctan x + \arctan y)=\frac{x+y}{h}$ and $\cos(\arctan x + \arctan y)=\frac{1-xy}{h}$ for some $h>0$.
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Now consider the range of the arctangent function, shown in red above. It dictates that $\frac{-\pi}{2}<\arctan x<\frac{\pi}{2}$ and $\frac{-\pi}{2}<\arctan y<\frac{\pi}{2}$ and therefore that $-\pi<\arctan x+\arctan y<\pi$.
Given any non-zero tangent $t$, there are two angles on the interval $(-\pi,\pi)$ with tangent $t$, $\theta$ and $\theta +\pi$, as shown below. Only one of these, however, is within the range of the arctangent function, and it is that one that will be found when the arctangent of $t$ is taken.
Since $\cos(\arctan x + \arctan y)=\frac{1-xy}{h}$ for some $h>0$, $\cos(\arctan x+\arctan y)$ is positive when $xy<1$ and $\cos(\arctan x+\arctan y)$ is negative when $xy>1$. Notice that a positive cosine corresponds to being within the range of the arctangent function, while a negative cosine corresponds to being outside it.
It follows that, when $xy<1$, $\arctan x+\arctan y$ will be the angle found by taking the arctangent of $\frac{x+y}{1-xy}$, while when $xy>1$, the angle found will be $(\arctan x+\arctan y)-\pi$.
Thus, when $xy<1$, $$\arctan x + \arctan y = \arctan \frac{x+y}{1-xy}\text{,}$$ and when $xy>1$, $$\arctan x + \arctan y = \pi + \arctan \frac{x+y}{1-xy}\text{.}$$
Addendum: Apropos of the identity at the beginning of this proof, note that when $\tan a \tan b=1$ is when $a+b=\pm\frac{\pi}{2}$, the points at which tangent is undefined.
When $\tan a \tan b=1$, $\tan a$ and $\tan b$ are reciprocals. Since tangent can be interpreted as the slope of the terminal side of an angle, this means that the terminal sides of $a$ and $b$ are either the reflections of one another across the line $x=1$ or else a terminal side and the “other end” of its reflection, which is the reflection rotated by $\pi$. The diagram below shows the two cases.
That they sum to $\pm\frac{\pi}{2}$ can be shown algebraically:
$a-\frac{\pi}{4}=\frac{\pi}{4}-b$ or $a-\frac{\pi}{4}=(\frac{\pi}{4}-b)+\pi$
$a+b=2(\frac{\pi}{4})=\frac{\pi}{2}$ or $a+b=2(\frac{\pi}{4})+\pi=\frac{3\pi}{2}=-\frac{\pi}{2}$
Hey, congrats on making it to 50 days! That’s a big milestone, and it’s so heartening to see you re-connect to math.
Thank you. I’m amazed I’ve been able to muster this much consistency.
Lovely, as always. I haven’t thought carefully about this; obviously the fact that $\arctan(\tan x)$ doesn’t need to be $x$ might get in the way, but what would happen if you took your equation
$$
\tan(\arctan x+\arctan y) = \frac{x+y}{1-xy}
$$
and took the arctan of both sides? Would that simplify things at all? Or am I just being careless?
Well, as you say, it is not necessarily the case that $\arctan(\tan\theta)=\theta$. Instead, $\arctan(\tan\theta)=\theta+n\pi$ for some integer $n$. So we would still need to prove that $n$ is either 0 or $\pm1$ and show under what conditions each occurs, which is the point of the section with the diagrams above. I don’t think taking the arctangent first would really make the proof any shorter.
It’s so fun having this kind of conversation with you again! And congratulations on 50 days.
Since you’re talking about arctangents and $pi$, I thought maybe I’d mention for you and for your readers something called Gregory’s series. You may know all about this, but I still think it’s fun.
You need to know the fact from calculus that
$$
\arctan x = \int\frac1{1+x^2}\,dx.
$$
You also need to know the formula for a geometric series,
$$
\frac1{1-y}=1+y+y^2+y^3+\dots.
$$
Replacing $y$ with $-x^2$ gives you
$$
\frac1{1+x^2}=1-x^2+x^4-x^6+\dots.
$$
As we already observed, you get the arctangent by integrating that:
$$
\arctan x=x-\frac{x^3}3+\frac{x^5}5-\frac{x^7}7+\dots
$$
From what you’ve told us about slopes, we know that $\arctan1=\pi/4$, so we can plug in $x=1$ and get
$$
\frac\pi4=\arctan1=1-\frac13+\frac15-\frac17+\dots
$$
a formula for $\pi$!
The trouble with this is that it converges very slowly, and you have to add really a lot of terms in order to get $\pi$ with any accuracy. I still think it’s cool, and you can use your arctangent identity to speed up the convergence by writing $\arctan1$ in terms of the arctangents of smaller numbers.
Okay, I left my laptop at school and I’m doing this on my phone. I’m going to hit submit and hope it’s at least halfway readable.
And of course there’s a $+c$ in that integral. I’ll let the experts work out whether I picked the right value of $c$ by ignoring it completely. đŸ™‚
Thank you for this (and for your heroic efforts typesetting it on a phone—gracious). I should have directed people to this comment in my Pi Day post.