Expressing cos(2π/n) without using trigonometric functions

[Yayness]

I've read a bit about constructibe polygons, and that a regular n-gon can be constructed with compass and straightedge if and only if trigonomatric functions of 2π/n can be expressed with square roots and basic arithmatic alone.
That is possible if and only if n is the product of distinct Fermat primes and/or a power of 2.
That means if n = 2^aF₀^b₀F₁^b₁F₂^b₂F₃^b₃... where a∈N₀, F_i are the Fermat primes, and b_i∈{0,1}.

If you place an n-gon in a coordinate system and let it be centered at the origin and place one of the vertices at (1,0), the coordinates to every vertex will be on the form (cos(2kπ/n),sin(2kπ/n)) where k is an integer.
If the coordinates to the vertex (cos(2π/n),sin(2π/n)) can be expressed with basic arithmetic and square roots alone, the coordinates to any vertex can.

But if you also allow roots of other degrees, like cubic roots and 5th roots, will the possible values of n be extended? If so, to what?

 

[micromass]

Yes, if you allow more elements, then the values of n will increase.

For example, the 9-gon cannot be constructed with straightedge and compass. This is because cos(2*pi/9) is the root of 8x^3-6x+1. But if you allow cube roots, then this becomes constructible.

A good question is the following: if you allow all radicals, are all n-gons constructible? I don't know the answer, but I don't think that even allowing all radicals should do it...

 

[micromass]

OK, I delved into the proof for you and I obtained this:

If you allow cube roots, then an n-gon is constructible if n=2^kp_1...p_s or n=2^k3p_1...p_s where p_1,...,p_s are distinct Fermat primes (and where it is possible that one of the p_i is 3).

For example, if n=9, then n=3.3. Thus it is of the form of n=3p, where p is a Fermat prime.
Another example, if n=15, then n is also constructible.

 

[Yayness]

Are you sure? When solving a cubic function, you need to find the cubic root of an imaginary number if the function has 3 real roots.
The cubic function you mentioned, has 3 real roots. These are cos(2π/9), cos(8π/9) and cos(14π/9). (Expressing those in other ways would require much space, so I won't do that here.)
Is there a way to remove the imaginary parts when expressing these roots with basic arithmetic, square roots and cube roots, or do we have a casus irreducibilis? (A case where imaginary numbers are used when expressing real numbers, and these imaginary numbers can't be removed.)

If you also allow casus irreducibilis, I know it would work for for n=7.
If there is a way to express cos(2π/9) with basic arithmetic, square roots and cube roots alone, without imaginary numbers in the expression, how?

 

[micromass]

I don't see why the imaginary numbers bother you so much... A constructible number is a number which can be constructed with straightedge and compass. And it can be shown that i is a constructible number. So working with imaginary numbers for solving the cubic is allowed

Also, in the case where you consider solvability by radicals, you add all solutions of x^n+\alpha=0. So, considering n=2 and \alpha=1 gives you i and -i. Thus, when considering solvability by radicals, it is allowed to work with (certain) imaginary numbers.

All I'm saying is that 8x^3-6x+1=0 can be solved by radicals. That is, the solutions can be expressed by addition, substraction,multiplication, division, exponentiation and nth-roots. But note that in this case, also the nth-roots of -1 are allowed. So imaginary numbers give no complication...

 

[Yayness]

Ok, we'll allow imaginary numbers.
Then, if n is a number on the form you mentioned in the quote below, the n-gon is "constructible".

If you allow cube roots, then an n-gon is constructible if n=2^kp_1...p_s or n=2^k3p_1...p_s where p_1,...,p_s are distinct Fermat primes (and where it is possible that one of the p_i is 3).

But that's not the only numbers it would work for. The solutions to 64x⁶-112x⁴+56x²-7=0 (or if u=x², then 64u³-112u²+56u-7=0) will be the real parts of the primitive 28th roots of unity, or said in another way:
The polynomial 64u³-112u²+56u-7 has these roots: cos(π/14), cos(3π/14), cos(5π/14), cos(9π/14), cos(11π/14) and cos(13π/14). That means an n-gon is also "constructible" if n=2^k7p_1...p_s or n=2^k21p_1...p_s (in addition to what you said).

But would it work for other numbers as well?

 

[micromass]

Ok, I think I made a mistake.

Anyway, if you allow cube roots, then it works for n if and only if \phi(n)=2^k3, where \phi is the Euler totient function. I'll let you figure out for yourself which values of n that are, because I apparently suck at it

 

[Yayness]

Hehe ok, thanks!
Hmm interesting, that means it'll work for n=13 as well.

Anyway, I believe it would work for any n-gon if you allow any a'th root (and if the following method is allowed). One could just take the the n'th root of i. This number to the power of 4, would then be cos(2π/n)+i sin(2π/n). (That is if you pick the n'th root with the largest real part to be "the n'th root".)
To remove the imaginary part from the number cos(2π/n)+i sin(2π/n), one could just add the number to its conjugate and divide it by 2.
Like this: \cos\frac{2\pi}{n}=\frac{\sqrt[n]{i}^4+\sqrt[n]{-i}^4}{2} but then one would have to specify which n'th root one's talking about, as there are n of them.

But I think that would be less interesting than if you limited the opportunities by not allowing any a'th roots other than square roots and cube roots (for instance).
Or you could ask: What's the largest a'th root required to express cos(2π/n) with radicals?

Anyway, thanks for the answer. It helped a lot.