When $\cos x$ is transcendental?

Question

About the transcendence of trigonometric functions I know that:

1) if $x$ is an algebraic number $\ne 0$ than $\cos x$ is transcendental.

2) if $p=\dfrac{m}{2^n}$ with $m,n \in \mathbb{Z}$ than $\cos (p\pi)$ is a constructible number (using bisection formula).

After the comments and the answer it's also proved that:

3) For any rational number $q$, $\cos (q\pi)$ is algebraic ? (since $e^{iq \pi}$ is a root of unity, hence algebraic)

So the open questions are:

4) if $a$ is an algebraic number than $\cos (a \pi)$ is algebraic or transcendental?

5) there exists a transcendental number $\alpha$ such that $\alpha \ne r \pi , \forall r \in \mathbb{Q}$ and if $p=\dfrac{m}{2^n}$ with $m,n > \in \mathbb{Z}$ than $\cos (p\alpha)$ is a constructible number?

Someone know if there is some answer to this related questions?

Answer 1

(1) to show that $\cos(\frac {m}{n} \pi)$ is algebraic it is enough to show that $\cos(\frac{1}{n} 2\pi)$ is algebraic. So it comes down to showing that the real part of an $n$-th root of unity is algebraic.

(3) there is a transcendental $a$ not equal to any rational multiple of $\pi$. Just take square root of $(\pi -1)$, lest $\pi$ would be algebraic.

Answer 2

(4) $\cos(a\pi)=\frac{e^{ia\pi}+e^{-ia\pi}}{2}$ so $\cos(a\pi)$ is algebraic iff $e^{ia\pi}$ is algebraic.

Since $e^{ia\pi} = (e^{i\pi})^a$ and $e^{i\pi}=-1$ is algebraic, the Gelfond-Schneider theorem implies that $\cos(a\pi)$ is transcendental if $a$ is algebraic but not rational.