Continued from Part 3, what happens when there is no "parallel", the rules for circles aren't what you thought they were, and so on.
So how does trigonometry work in this world?
See, I belatedly realized that spewing walls of equations like this is not actually going to be much use when you're stuck in a rowboat in the middle of the North Atlantic having to navigate by the stars with no cell phone and no GPS. Because, chances are, you also have No Internet, and then my blog entries with their handy tables go to waste.
It would be far better if I can teach you how to derive these relationships instead, i.e., in a way that you might actually be able to vaguely remember while sitting in a boat in the middle of the North Atlantic.
But first I'm going to introduce a bit of gratuitous extra notation. Write
— pronounce it "co-theta" if you want — to mean (90° − θ). I do this because:
- I can,
- it's less typing,
- it's way less degree vs. radian waffling, which I already do too much of,
- you get all of the following useful and amusing equivalences (no, really; read them aloud):
|sin ᶜᵒθ = cos θ|
|cos ᶜᵒθ = sin θ|
|tan ᶜᵒθ = cot θ||(= 1/tan θ, in case you've forgotten)|
|cot ᶜᵒθ = tan θ|
|sec ᶜᵒθ = csc θ||(= 1/sin θ, because nobody ever remembers that one)|
|csc ᶜᵒθ = sec θ|
You'd almost think they planned it this way.
Hopefully, it goes without saying that ᶜᵒ(ᶜᵒθ) = θ, except I had to go and say it, didn't I? (Damn.)
And now let's start with a right triangle, with vertices/angles and sides/lengths labeled a,b,c,A,B, the way you usually see it in trigonometry class, even though I'm going to tilt it weirdly so that I can do what follows in a single diagram.
If I take the vertex A as a pole, I can travel along side b, and then continue in the same direction by a distance by ᶜᵒb to arrive at the corresponding line/equator, which you can call "line A" if you want.
I can do the same thing with side c, ending up on line A at a distance A away from where I ended up coming down along side b.
I can then extend line A by a distance ᶜᵒA to arrive at a pole for line b, which I've actually labeled ᶜᵒb because…… we can also get there by traveling down the side a and continuing for a distance ᶜᵒa, and when we do that, the angle formed when we reach the pole corresponds to the distance between the points we started from on that equator (i.e., line b), namely ᶜᵒb.
We have thus successfully created a 2nd right triangle, whose measurements involve the same five numbers A, c, B, ᶜᵒa, ᶜᵒb, as the first triangle, but here we're using them in different places …
… the first lesson being that if you have a set of five numbers that work, you can rearrange them to get another set of five numbers that work.
Arguably this would be enough to derive things from, but rather than try to unscramble what's going on with this particular rearrangement, it turns out to be much easier if I just repeat the construction a second time.
Which gets us a 3rd right triangle, in which the same five numbers ᶜᵒb, A, c, B, ᶜᵒa, can now be easily seen to have been rolled by one position counter-clockwise from where/how they were used in the original triangle.
The bottom line here is that if I can discover any relationship at all amongst these five numbers, I can roll it four times and get four other relationships for free.
So we pick one that's easy to derive.
Well, okay, what you actually want for the middle-of-the-North-Altlantic scenario is to pick one that's easy to remember (i.e., if you're confident that Someone Else has already derived it correctly) in which case the (spherical) Pythagorean Theorem (below) wins by a mile.
But since we're not actually in the North Atlantic today (well okay, I'm not), and our job is to derive all of this, I'll instead start with one of the mysterious ones that, I promise you, will turn out to be true:
cos a = cos A/sin B
which just has the minor problem that we need to be referring to ᶜᵒa rather than a, but that's an easy fix:
sin B sinᶜᵒa = cos A
and now we have five relations for the price of one:
|rolled version is …||which rewrites to …||Name for this:|
|sin B||sinᶜᵒa||=||cos A||cos a = cos A/sin B||(later)|
|</td>||sinᶜᵒa||sinᶜᵒb||=||cos c||cos c = cos a cos b||The Pythagorean Theorem!|
|cos B||=||sinᶜᵒb||sin A||cos b = cos B/sin A||(later)|
|cosᶜᵒa||=||sin A||sin c||sin A = sin a/sin c||Sine Rule for A|
|cosᶜᵒb||=||sin c||sin B||sin B = sin b/sin c||Sine Rule for B|
But that's not all! Combining the first three rows gives us
sinᶜᵒa cos c sinᶜᵒb sin A = cos A sinᶜᵒa sinᶜᵒb cos B
which cancels down to
tan A cos c tan B = 1
which, unlike the relations above, which were all "two things vs. the thing opposite", is of the form "three things next to each other", so it's something new. We can now roll that four more times to get:
|rolled version is …||which rewrites to …||Name for this:|
|tan A||cos c||tan B||= 1||cos c = cot A cot B||(later)|
|tanᶜᵒb||cos A||tan c||= 1||cos A = tan b/tan c||Cosine Rule for A|
|tanᶜᵒa||cosᶜᵒb||tan A||= 1||tan A = tan a/sin b||Tangent Rule for A|
|cosᶜᵒa||tanᶜᵒb||tan B||= 1||tan B = tan b/sin a||Tangent Rule for B|
|tanᶜᵒa||tan c||cos B||= 1||cos B = tan a/tan c||Cosine Rule for B|
and we've now given you all 10 possible relationships of 3 things out of 5.
Meaning if you provide any two of the five numbers for a right triangle, you can derive the other three.
In particular, if we know both of the angles A and B, that in itself is enough to nail down the triangle, which is what the relations above marked "(later)" are about. They are entirely new and have no analogues in the Euclidean world. Though perhaps this is not surprising given what we said about the angle deficit determining the area, which roughly means if you know the angles, you know how big the triangle is, therefore you ought to know what the sides are.
Which means "similar triangles" are Not a Thing in this world.
Priming the Pump
Now that we know how to get lots of stuff via rolling, how do we get it all started?
One way is to zoom in on the B corner of the triangle and watch what happens when we increase the opposite side b by an infinitesimal amount db, while holding the other angles (A and the right angle) fixed.
Meaning we're sliding the side a ever-so-slightly to the right. This makes B move right and up and also increases that angle by dB (by an amount too small to actually see in the upcoming diagram). The c side increases as well, but we don't care.
The area increase is dB since A is fixed, and is essentially accounted for by an (insanely thin) attempted-rectangle thing on the right with base db, height a, area = sin a db — here we making use of the area and frontier formulas for Attempted Rectangles that we came up with last time. Since the height a might actually be a considerable distance, the frontier length shrinks from db to cos a db, but since db is so short, we'll ignore the curvature of the frontier and likewise for the area of the eentsy dark grey triangle at the upper right; these effects will be down in the O(db²) noise anyway.
There's enough here to derive a relation between B and a, at which point it's Differential Equation Party Time:
|tan B da||= cos a db = dB / tan a|
|cos B dB/sin B - sin a da/cos a||= 0|
|d(ln(sin B) + ln(cos a))||= 0|
|sin B cos a||= (some constant)|
And since this has to work when b ⟶ 0, a ⟶ 0, area ⟶ 0, B ⟶ ᶜᵒA, the constant thus has to be sin ᶜᵒA, and thence
cos A / sin B = cos a
and, oh hey, that looks useful …