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?