For which rational degree is its tangent rational?

For which rational number `r in QQ` is its tangent `tan r^o` rational? What about `cos r^o` and `sin r^o`?

We know the obvious ones, `tan 45^o = 1` is rational, is there another rational angle with its tangent rational?

Rationality of `tan 1^o`.

 Is `tan 1^o` a rational number?

Constructibility of trigonometric values.

If we can construct the lengths `cos alpha` or `sin alpha`, can we then construct `tan alpha`? How about the converse, if `tan alpha` is constructible, is `cos alpha` or `sin alpha` constructible? 

Probability of three segments forming a triangle.

Consider the unit interval `I = [0,1]` and pick two points `x,y` uniformly from `I` independently at random. This creates three segments. What is the probability that these three segments can form the sides of a triangle?

Would you guess this is "likely", `> 50%`, or "less likely", `< 50%`?

Probability of a self-intersecting brokenline.

On the plane starting at some point `O`, pick a random angle uniformly and draw a unit segment `OA`. Again pick a random angle uniformly and draw a unit segment `AB`. Repeat two more times to get `BC` and `CD`. What is the probability of this broken line `OABC` self-intersects?

Test post: jxgbox

Testing jxgbox to draw with javascript library jxgbox

Self-intersections: 0

Non-constructible angle of `20^o`.

 In the popular lore of straightedge and compass construction, we have

One cannot trisect an arbitrary angle.

It is often used as an example that `60^o` cannot be trisected into `20^o`. And this is demonstrated by the fact that the length `cos 20^o` is not a constructible length. By the way, the quoted lore above about "arbitrary angles cannot be trisected" itself has nuances, but I will explain it in another post.

As for `cos 20^o` is not constructible, we shall use the lore that 

Given the length `1` on the plane, then the length `r` is constructible by straightedge and compass, if and only if `r` lies in a finite tower of quadratic extensions from `QQ`.

As a corollary,

If `r` is a constructible length, then `r` is a root of some irreducible polynomial over `QQ` of degree `2^m`, for some `m`. 

(Note the converse is not necessarily true.)

So it suffices to show that `r = cos 20^o` is a root of some irreducible polynomial of an inappropriate degree. Indeed, recall the triple angle formula, we have 
$$
\cos 3\theta = 4\cos^3\theta - 3\cos \theta, 
$$ (cf. Chebyshev polynomials) we see that `r` is a root of the polynomial equation `1/2 = 4 r^3 - 3r` or `8r^3 -6r-1 =0`. Since this is a cubic polynomial, if it is reducible over `QQ` then it would have a rational root. But by rational root test one sees that `8r^3-6r-1` has no rational root. Hence it is irreducible of degree 3. Since `3 ne 2^m` for any non-negative integer `m`, we conclude that `cos 20^o` is not constructible!

Construction of a regular pentagon.

It is well-known that a regular pentagon is constructible. Here is the simplest way that I know of.

Note that in a regular pentagon inscribed in a unit circle shown on the right, the angle `angle BAC = 36^o`. This makes the length `AB = (1+sqrt(5))/2`, as `triangle ABC` is a right triangle.

So all we need is to be able to construct the length `(1+sqrt(5))/2`. To do so, observe the diagram on the right, where two smaller circle of radius `1/2` is inscribed in a unit circle. If we extend the segment of `AP` to `AQ`, then we will find that `AQ = (1+sqrt(5))/2`, precisely what we need!

Constructible angles of integral degrees.

For which positive integer `n` is `n^o` constructible (classically with straightedge and compass)? There is a satisfactory answer:

Let `n` be a positive integer. Then the angle `n^o` is constructible if and only if `n` is a multiple of `3`.

The proof is straightforward as well, provided we believe some other facts. First note the classic fact that

A regular pentagon is constructible, hence `72^o` is constructible.

And that

If `A` and `B` are any two constructible angles, then so is any integer combination `xA + yB`, where `x, y in ZZ`.

So, since we can also construct an equilateral triangle, we can construct `60^o`. Hence we can construct `72^o - 60^o = 12^o`. Now, using the fact that

If `A` is a constructible angle, then so is its bisection `A/2`.

This shows we can construct `6^o` and `3^o`. Now using the fact of integer combinations of constructible angles, we have the first direction:

For any integer `x`, the angle `3x^o` is constructible.

Now we show if integer `n` is not divisible by `3`, then `n^o` is not constructible.

 

We show this by contradiction. Suppose to the contrary that some integer `n` is not divisible by `3` but is constructible. Then by repeated subtraction of `3^o`, we can construct the angles `1^o` or `2^o`, as `n mod 3` is `1` or `2`. And since we can bisect, constructing `2^o` means we can construct `1^o`. So far, this says if such integer `n` that is not a multiple of `3` but `n^o` is constructible, then we can construct `1^o`. But wait, this is crazy! Because of the venerated fact that

The angle `20^o` is not constructible.

We should not be able to construct `1^o`, otherwise by integer combination of constructible angles, we could construct any integral angles!