I want to understand this connection better.

Although it sounds the most involved, `arctan`

answers the simplest trigonometric question you could ask. “The full moon subtends seven degrees of my vision; so how big is the moon?” The `arctan`

(with radius = distance from Earth to moon) is the answer.

But what does that simple operation have to do with:

- a continuous sum of
- an inversion of
- one more than
- some continuous range of numbers
- times themselves?

That’s confusing.

Four years later…

This comment shows how to get the answer. (Pair with this on the derivative of logarithms.) A consequence of the chain rule (or perhaps another way to state it!) is that the derivative of ƒ⁻¹ at a point p is `flip( derivative( ƒ )) evaluated at ƒ⁻¹(p)`

, presupposing that all of the maps fit together right.

`arctan`

is defined, for simplicity, on a circle of unit radius. (This just means if you’re looking at the moon then use units of “one moon-distance”.) It takes as an argument a ratio of sides and returns an angle `θ`

. Since `derivative ( tan ) = derivative (sin⁄cos) = 1⁄cos² = sec × sec`

(reasoned with calc 101), by combining that `derivative ( tan )`

with JavaMan’s perspective on the derivatives of inverse-functions we can argue that

Java Man’s idea gets us to look at the triangle

which, since it’s constrained to a circle by the equivalence-classing of triangles to be just the ones with a certain angle (see fibration), limits us to just one free parameter `R`

(a ratio of opposite `O`

to adjacent `A`

side lengths of an equivalence-class of triangles). After following Java Man’s logic to see why `derivative( tan ) = sec²`

implies `derivative( inverse of tan ) = cos² [at the angle which is implied by tan R]`

, we’re left knowing that `A`

=the adjacent side implied by the `cos R = cos O⁄A`

of the original `O⁄A`

ratio we were given as the natural input space (ratios) which the `tan`

function accepts. This isn’t enough because the answer needs to be in `O⁄A`

terms to match the input. We have to do some Pythagorean jiu jitsu involving `A = A⁄1`

and `1=1²=A²+O²`

to get the answer into an `O⁄A`

form (since that was the information we were given). Using the `1=A²+O²`

is using the natural delimitation of the inscribing circle to make the two `A`

and `O`

move together the way they should on the circle, by the way. The algebraic jiu jitsu then yields `A = flip( 1+R² )`

, now using the proper input `R=O⁄A`

.

The point of all this mangling was merely to match up cosine’s output with tangent’s input. Sheesh with all the symbols!

But that’s merely deriving the correct answer algebraically, with a bit of “why” from the comment’s perspective on inverse functions generally. What about my original question? Why does this sequence of mappings, if iterated, subtend the moon?