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 $a+b=2(\frac{\pi}{4})+\pi=\frac{3\pi}{2}=-\frac{\pi}{2}$
