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#### GendoIkari

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« Reply #975 on: January 17, 2019, 11:12:36 am »
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What do you all think of this problem?  The result is so unexpected when you first hear it.  I managed to solve it myself, but I still think that the result is fascinating.

Ha, I was just coming here to post this video.

This blew my mind. Like, Pi is an unending irrational number. There's no reason in the world that the first some number digits of Pi, without the entire thing, should ever mean anything.
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#### sudgy

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« Reply #976 on: January 17, 2019, 11:15:10 am »
+1

What do you all think of this problem?  The result is so unexpected when you first hear it.  I managed to solve it myself, but I still think that the result is fascinating.

Ha, I was just coming here to post this video.

This blew my mind. Like, Pi is an unending irrational number. There's no reason in the world that the first some number digits of Pi, without the entire thing, should ever mean anything.

Well, there's some reason.
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Quote from: sudgy on June 31, 2011, 11:47:46 pm

#### GendoIkari

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« Reply #977 on: January 17, 2019, 11:17:55 am »
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What do you all think of this problem?  The result is so unexpected when you first hear it.  I managed to solve it myself, but I still think that the result is fascinating.

Ha, I was just coming here to post this video.

This blew my mind. Like, Pi is an unending irrational number. There's no reason in the world that the first some number digits of Pi, without the entire thing, should ever mean anything.

Well, there's some reason.

I haven't put too much thought into solving it; just waiting for the part 2 video. But I assume it must be something where as one block approaches infinite size; some physics equation about collisions approaches something that involves a circle. But even then, you aren't ending up with Pi, you are ending up with Pi * 10^x.
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#### sudgy

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« Reply #978 on: January 17, 2019, 11:21:28 am »
+2

What do you all think of this problem?  The result is so unexpected when you first hear it.  I managed to solve it myself, but I still think that the result is fascinating.

Ha, I was just coming here to post this video.

This blew my mind. Like, Pi is an unending irrational number. There's no reason in the world that the first some number digits of Pi, without the entire thing, should ever mean anything.

Well, there's some reason.

I haven't put too much thought into solving it; just waiting for the part 2 video. But I assume it must be something where as one block approaches infinite size; some physics equation about collisions approaches something that involves a circle. But even then, you aren't ending up with Pi, you are ending up with Pi * 10^x.

I know that he says pi always involves a circle, but my solution doesn't actually use any geometry.  I just kept dealing with equations and eventually complex numbers showed up and then the solution was pretty simple.  I'm looking forward to his solution because I know it will be different than mine.

The way I see pi in the original problem is that pi always involves oscillations, and the block on the left is constantly oscillating back and forth.  This oscillation happens to be perfect enough that it make pi jump out.
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Quote from: sudgy on June 31, 2011, 11:47:46 pm

#### GendoIkari

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« Reply #979 on: January 17, 2019, 11:23:31 am »
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What do you all think of this problem?  The result is so unexpected when you first hear it.  I managed to solve it myself, but I still think that the result is fascinating.

Ha, I was just coming here to post this video.

This blew my mind. Like, Pi is an unending irrational number. There's no reason in the world that the first some number digits of Pi, without the entire thing, should ever mean anything.

Well, there's some reason.

I haven't put too much thought into solving it; just waiting for the part 2 video. But I assume it must be something where as one block approaches infinite size; some physics equation about collisions approaches something that involves a circle. But even then, you aren't ending up with Pi, you are ending up with Pi * 10^x.

I know that he says pi always involves a circle, but my solution doesn't actually use any geometry.  I just kept dealing with equations and eventually complex numbers showed up and then the solution was pretty simple.  I'm looking forward to his solution because I know it will be different than mine.

The way I see pi in the original problem is that pi always involves oscillations, and the block on the left is constantly oscillating back and forth.  This oscillation happens to be perfect enough that it make pi jump out.

I don't know the physics off-hand to know how the velocity of each object changes after collision.
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#### faust

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« Reply #980 on: January 17, 2019, 11:28:01 am »
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What do you all think of this problem?  The result is so unexpected when you first hear it.  I managed to solve it myself, but I still think that the result is fascinating.

Ha, I was just coming here to post this video.

This blew my mind. Like, Pi is an unending irrational number. There's no reason in the world that the first some number digits of Pi, without the entire thing, should ever mean anything.
It's kind of surprising, but I can see the reason. The number of collisions is an integer, so to calculate we probably need to do some sort of rounding, and since one object is 10^d times the mass of the other, it's 10^d * pi rounded down, which turns out to be the first d digits. The fact that the first digits turn out to be important here is simply because they are the digits to base 10 and we have a factor of 10^d involved.

(This is guesswork without actually going through the calculations.)
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Since the number of points is within a constant factor of the number of city quarters, in the long run we can get (4 - ε) ↑↑ n points in n turns for any ε > 0.

#### pacovf

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« Reply #981 on: January 17, 2019, 05:37:39 pm »
+2

@GendoIkari:

If you want to try it, you need to know conservation of momentum:

M1*V1 +M2*V2=M1*V1’+M2*V2’

And, for elastic collisions in one dimension:

(V1-V2) = -(V1’-V2’)

M denotes mass, V velocity, 1-2 denote the body, and the ’ denotes the value post-collision.
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#### qmech

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« Reply #982 on: January 18, 2019, 09:46:35 pm »
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(Unrelated to sudgy). A few years ago, I made a post here complaining about the way coordinate vectors are introduced in linear algebra. In particular, no-one ever explained me whether, if you write (2,3,4), what you mean is "the element (2,3,4) \in \R^3" or "the vector 2*b_1 + 3*b_2 + 4*b_3 for some basis {b_1, b_2, b_3}".

I saw your first post while catching up this thread and I'm glad you sorted it eventually.  This is taught badly all over the place and, as you've said, doing it right isn't hard!

One thing that might help as you go on is to keep in mind the distinction between an n-dimensional vector space over F, of which there are many, and the particular vector space F^n, which is just one very special example.  For a general vector space, choosing a basis to write coordinates with respect to is the same as fixing an isomorphism with F^n.  (If the basis is v_1, ..., v_n then the isomorphism is just (x_1, ..., x_n) |-> x_1v_1 + ... + x_nv_n.)

(And eventually you'll want to avoid choosing a basis/using coordinates if at all possible!)
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#### sudgy

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« Reply #983 on: January 18, 2019, 10:56:03 pm »
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Barely related, but I feel like matrix multiplication isn't "multiplication".  It's more like composition.
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Quote from: sudgy on June 31, 2011, 11:47:46 pm

#### Cuzz

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« Reply #984 on: January 18, 2019, 11:29:27 pm »
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Barely related, but I feel like matrix multiplication isn't "multiplication".  It's more like composition.

Eh, what is "multiplication" then? It's a binary operation and it distributes over addition, might as well call it that.
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#### sudgy

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« Reply #985 on: January 18, 2019, 11:42:37 pm »
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Barely related, but I feel like matrix multiplication isn't "multiplication".  It's more like composition.

Eh, what is "multiplication" then? It's a binary operation and it distributes over addition, might as well call it that.

Do you call the composition of two functions a multiplication?
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Quote from: sudgy on June 31, 2011, 11:47:46 pm

#### sudgy

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« Reply #986 on: January 19, 2019, 12:20:00 am »
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Barely related, but I feel like matrix multiplication isn't "multiplication".  It's more like composition.

Eh, what is "multiplication" then? It's a binary operation and it distributes over addition, might as well call it that.

Do you call the composition of two functions a multiplication?

Well, actually, function composition doesn't distribute, so my point is not as good.
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Quote from: sudgy on June 31, 2011, 11:47:46 pm

#### Cuzz

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« Reply #987 on: January 19, 2019, 12:48:09 am »
+1

Barely related, but I feel like matrix multiplication isn't "multiplication".  It's more like composition.

Eh, what is "multiplication" then? It's a binary operation and it distributes over addition, might as well call it that.

Do you call the composition of two functions a multiplication?

Well, I'm a dynamicist so kinda. By which I just mean that for me the notation f^2(x) generally means f(f(x)) and not [f(x)*f(x)] (exceptions being things like trig functions where the notation is standard but annoyingly inconsistent). But this is maybe confusing the issue...

To get at the heart of it, there are two things at play here:
(i) Terms in math mean whatever they are defined to mean and nothing else
(ii) Terms should ideally be chosen to be useful and consistent

Importantly, "multiplication" does not have a precise mathematical definition as a stand-alone term, so we can't appeal to (i). But there are certain binary operations on certain sets with certain properties that perhaps warrant using the same name for all of them, to highlight exactly those common properties.

On sets that already have a commutative binary operation which we've agreed to call "addition" and have agreed to denote with the symbol "+",  we often have another binary operation (which we maybe denote by *) satisfying a*(b+c) = a*b+a*c and (a+b)*c = a*c+b*c. Seems useful and consistent to refer to such operations as "multiplication."

For general functions, composition definitely does not satisfy this property (sin(a+b) =/= sin a +sin b), so it seems less useful to call it "multiplication." Could we? Sure, but then what property is shared by all "multiplications?"

You can debate these things of course. I've avoided using the word "ring" but maybe you want to restrict "multiplication" to only refer to a ring operation. But multiplication in rings must be associative. Is that important to you? Maybe, but then you lose the ability to call the cross product of vectors in R^3 "multiplication." Do you care?

Now of course in the realm of matrices, we have such a binary operation which distributes over addition, so we do tend to call it "multiplication". It also happens to relate to linear transformations of vector spaces, and in a sense coincides with the notion of "composition" in that context (where you should now think about what a useful and consistent definition of "composition" would be).

At the end of the day I'd argue that it's better to talk about multiplying matrices as opposed to composing them. You'd not be insane to take the operation we call matrix multiplication and call it something else, as long as it is defined precisely, but you should think about whether that term is useful and consistent. In this case, you'd be clashing with established nomenclature, though, which is hard to overcome even when it's not useful or consistent (like sin^2(x) v. sin^(-1)(x)).

As a final quiz, how should we define multiplication of ordered pairs of positive integers? For a,b,c,d positive integers, which is more natural (consistent and useful)?
(i) (a,b)*(c,d) = (ac,bd)

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#### faust

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« Reply #988 on: January 19, 2019, 02:56:27 am »
+2

Barely related, but I feel like matrix multiplication isn't "multiplication".  It's more like composition.
On the other hand, multiplication is also composition if we identify any element f of our field F with the map "⋅ f : F -> F ".
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#### sudgy

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« Reply #989 on: January 19, 2019, 04:27:52 pm »
+1

The main reason I don't like the term is that it's confusing to beginners.  It's usually the first time that something that is called multiplication doesn't behave like any other type of multiplication that they are used to.  I know I at first was wondering why in the world it was defined in the way that it is.  "Why not make it like matrix addition and use the component-wise product?  That has all of the standard properties of multiplication."  Of course we don't use the component-wise product because it's rarely useful, but back then it was radically different than any other kind of multiplication I had seen, and it seemed so arbitrary.

This kind of reminds me of tau vs. pi, or whether we should really call imaginary numbers "imaginary" numbers.  It's probably too late to change things now, but it would make more sense, especially for beginners, if the terms were different.
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#### Polk5440

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« Reply #990 on: January 19, 2019, 05:50:39 pm »
+1

Forward to today, I've been reading "Linear Algebra Done Right" and it immediately did what both lectures I visited didn't manage to. In the book, vectors are elements in F^n, not relative to anything. Matrices are something separate. There is a function M taking any linear map and the basises of two vector spaces and returning a matrix. Matrix multiplication is defined in the usual way, but there is no multiplication of a matrix with a vector. Instead, there is another function which takes a vector and a basis and returns a n*1 matrix representing the vector relative to that basis. That matrix can then be multiplied with another matrix following matrix multiplication as usual.

I find that super awesome, so I wanted to share it.

That's the textbook we used in Linear Algebra in college! I loved the book but had completely forgotten that bit! The book sets up a great theoretical environment for Linear Algebra.
« Last Edit: January 19, 2019, 05:59:17 pm by Polk5440 »
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#### shraeye

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« Reply #991 on: January 20, 2019, 01:50:34 am »
+2

Hmm; I'm not sure if I've ever posted in this thread.  Anyway, Linear Algebra is hard to teach.
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#### Cuzz

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« Reply #992 on: January 20, 2019, 12:28:40 pm »
+2

The main reason I don't like the term is that it's confusing to beginners.  It's usually the first time that something that is called multiplication doesn't behave like any other type of multiplication that they are used to.  I know I at first was wondering why in the world it was defined in the way that it is.  "Why not make it like matrix addition and use the component-wise product?  That has all of the standard properties of multiplication."  Of course we don't use the component-wise product because it's rarely useful, but back then it was radically different than any other kind of multiplication I had seen, and it seemed so arbitrary.

This kind of reminds me of tau vs. pi, or whether we should really call imaginary numbers "imaginary" numbers.  It's probably too late to change things now, but it would make more sense, especially for beginners, if the terms were different.

This is all totally valid. Two things I would say:

1. Some terminology, while being confusing to beginners, is very intuitive and natural in a higher level context, or sometimes it's just so well established that you just have to deal with the fact that it's what everyone uses. In either case, though, we preumably want some of these beginners to become non-beginners some day, and we should think about the trade-offs between using counter-intuitive terminology in the introduction of a topic vs. making people re-learn proper terminology later on so that they can communicate with the rest of the community. (As an aside, this reminds me a little bit of learning how to ski. We teach young kids to make wedge-turns and then have to have them un-learn this and make proper turns later on).

2. I like your point about how "we don't use the component-wise product because it's rarely useful, but back then it was radically different than any other kind of multiplication I had seen, and it seemed so arbitrary." I think in a well-taught class there should be some serious time spent on getting students to understand why a component-wise product would be rarely useful, and why the proper definition is in fact not so arbitrary. The problem is that so many classes just introduce matrices and show how to multiply them with no context as to why we might want to do such a thing.

You can define anything you want, but the things worth studying in mathematics have a reason for the definition being what it is and not something else. I can define a binary operation on the set of all functions from the reals to the reals by

(f#g)(x) = f(x)g(x-2)+f(-2x)[g(x)]^2

Perfectly good binary operation. I could even prove theorems about it. But what's it good for? Not much as far as I know. Whereas the binary operation defined by

(f*g)(x) = int_{-infty}^infty f(y)g(x-y)dy (when convergent)

is good for lots of things.

Hmm; I'm not sure if I've ever posted in this thread.  Anyway, Linear Algebra is hard to teach.

This is very true. It's also somewhat unique among introductory math courses in that there are two pretty radically different ways to approach the subject, which I generally think of as the "Begin with systems of linear equations"  approach and the "Begin with the definition of an abstract vector space" approach. Courses titled "Linear Algebra" might be either one and they have a very different feel to them.
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#### sudgy

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« Reply #993 on: May 03, 2019, 08:43:11 pm »
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I was working on a problem, and had to calculate cos(atan(45*sqrt(3)/28)/3), which wolfram alpha helpfully tells me is equal to 4/sqrt(19).  Does anybody know how it figured that out?
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#### faust

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« Reply #994 on: May 04, 2019, 02:37:23 am »
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I was working on a problem, and had to calculate cos(atan(45*sqrt(3)/28)/3), which wolfram alpha helpfully tells me is equal to 4/sqrt(19).  Does anybody know how it figured that out?
tan x = a/b in some rectangular triangle, and cos x = b/c. So atan(a/b) = acos(b/c). Also since everything is rectangular, c = sqrt(a² + b²).

thus atan(a/b) = acos(b/sqrt(a² + b²)) = acos(sqrt(b²/(a² + b²)) = acos(sqrt(1/(a²/b² + 1))). This can be written as atan(x) = acos(sqrt(1/(x² + 1))).

I think the rest should follow easily.
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Since the number of points is within a constant factor of the number of city quarters, in the long run we can get (4 - ε) ↑↑ n points in n turns for any ε > 0.

#### sudgy

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« Reply #995 on: May 04, 2019, 03:44:59 am »
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I was working on a problem, and had to calculate cos(atan(45*sqrt(3)/28)/3), which wolfram alpha helpfully tells me is equal to 4/sqrt(19).  Does anybody know how it figured that out?
tan x = a/b in some rectangular triangle, and cos x = b/c. So atan(a/b) = acos(b/c). Also since everything is rectangular, c = sqrt(a² + b²).

thus atan(a/b) = acos(b/sqrt(a² + b²)) = acos(sqrt(b²/(a² + b²)) = acos(sqrt(1/(a²/b² + 1))). This can be written as atan(x) = acos(sqrt(1/(x² + 1))).

I think the rest should follow easily.

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#### faust

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« Reply #996 on: May 04, 2019, 04:08:08 am »
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I was working on a problem, and had to calculate cos(atan(45*sqrt(3)/28)/3), which wolfram alpha helpfully tells me is equal to 4/sqrt(19).  Does anybody know how it figured that out?
tan x = a/b in some rectangular triangle, and cos x = b/c. So atan(a/b) = acos(b/c). Also since everything is rectangular, c = sqrt(a² + b²).

thus atan(a/b) = acos(b/sqrt(a² + b²)) = acos(sqrt(b²/(a² + b²)) = acos(sqrt(1/(a²/b² + 1))). This can be written as atan(x) = acos(sqrt(1/(x² + 1))).

I think the rest should follow easily.

Oh I didn't see the 3 was outside. Well you can use the trig rules for cos(x+y) to get that cos(3x) = 4cos³(x) - 3cos(x), and then solve for cos(x).
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Since the number of points is within a constant factor of the number of city quarters, in the long run we can get (4 - ε) ↑↑ n points in n turns for any ε > 0.

#### sudgy

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« Reply #997 on: May 04, 2019, 09:29:54 am »
+1

I was working on a problem, and had to calculate cos(atan(45*sqrt(3)/28)/3), which wolfram alpha helpfully tells me is equal to 4/sqrt(19).  Does anybody know how it figured that out?
tan x = a/b in some rectangular triangle, and cos x = b/c. So atan(a/b) = acos(b/c). Also since everything is rectangular, c = sqrt(a² + b²).

thus atan(a/b) = acos(b/sqrt(a² + b²)) = acos(sqrt(b²/(a² + b²)) = acos(sqrt(1/(a²/b² + 1))). This can be written as atan(x) = acos(sqrt(1/(x² + 1))).

I think the rest should follow easily.

Oh I didn't see the 3 was outside. Well you can use the trig rules for cos(x+y) to get that cos(3x) = 4cos³(x) - 3cos(x), and then solve for cos(x).

And solve a cubic, which is what I was trying to do in the first place?
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#### Cuzz

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« Reply #998 on: May 04, 2019, 02:57:56 pm »
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I was working on a problem, and had to calculate cos(atan(45*sqrt(3)/28)/3), which wolfram alpha helpfully tells me is equal to 4/sqrt(19).  Does anybody know how it figured that out?
tan x = a/b in some rectangular triangle, and cos x = b/c. So atan(a/b) = acos(b/c). Also since everything is rectangular, c = sqrt(a² + b²).

thus atan(a/b) = acos(b/sqrt(a² + b²)) = acos(sqrt(b²/(a² + b²)) = acos(sqrt(1/(a²/b² + 1))). This can be written as atan(x) = acos(sqrt(1/(x² + 1))).

I think the rest should follow easily.

Oh I didn't see the 3 was outside. Well you can use the trig rules for cos(x+y) to get that cos(3x) = 4cos³(x) - 3cos(x), and then solve for cos(x).

And solve a cubic, which is what I was trying to do in the first place?

It's a depressed cubic, and so at this point you can use e.g. Cardano's method to solve it explicitly.
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#### Dingan

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