let's take a detour. let's prove we can write cos(nx) as a polynomial in u = cos(x). for n = 1, this is trivial. now we assume we can do so for all 0 < k < n.
cos(2x) = 2cos^2(x) - 1 = 2u^2 - 1
cos(3x) = 4cos^3(x) - 3cos(x) = 4u^3 - 3u.
now for n ≥ 3, cos(nx) = cos(x)cos((n-1)x) - cos((n-2)x) = cos(x)[p(cos(x))] - q(cos(x)) = up(u) - q(u), by our induction assumption
(expand the RHS of the trig identity to prove the equality).
since up(u) - q(u) is clearly a polynomial in u, the assertion is shown.
in particular, this means we can write 1 = cos(2π) = f(cos(2π/n)), for some polynomial f.
hence if we set g(x) = f(x) - 1, then g(cos(2π/n)) = f(cos(2π/n)) - 1 = 1 - 1 = 0.
evidently cos(2π/n) is a root of g(x), so cos(2π/n) is algebraic.
(note: although this seems to only show a proof for natural numbers n, it actually shows it for any non-zero integer
since cos(2π/n) = cos(2π/-n), if n is nonzero).
I figured out a fairly straight forward solution. Seeing as the text in images doesn't show up as results when searching the forum; I'm going to restate the problem and give my proof:
Prove that for all non-zero integers , is an algebraic number.
Proof: From Euler's identity we know that and therefore . And from Euler's Formula we know that
So if we apply the Binomial theorem to the right, and Euler's identity to the left we have
Seperating this sum into two sums: one with terms having only even exponents, and another with terms having only odd exponents:
where . This sum consists of terms. If is odd, . Otherwise while . So after some algebra,
But the right side of this equation is a complex number of the form where and the lefthand side is of the form Therefore
and applying the trigonometric identity yields:
Replacing , with a variable gives
And after simplifying:
This is a polynomial with rational coefficients that evaluates to zero at for all non-zero integers .