Determining algebraicity of $\cos1^\circ$ and $\sin1^\circ$

Question

Let $a = \cos1^{\circ} $ and $b = \sin 1^{\circ}$

We say that a real number is algebraic if it is a root of a polynomial with integer coefficients.

Then-

A. $a$ is algebraic but $b$ is not algebraic

B. $b$ is algebraic but $a$ is not algebraic

C. both $a$ and $b$ are algebraic

D. neither $a$ nor $b$ is algebraic.

I am stuck whether the ratio of some arbitrary sums of integral powers of $\sin1^{\circ}$ with integer coefficients with another integral power can be represented in the form of a rational number. This question is meant to be solved with high school math techniques.

Answer 1

Since $\bigl(\cos(1^\circ)+i\sin(1^\circ)\bigr)^{360}=1$, the number $\cos(1^\circ)+i\sin(1^\circ)$ is algebraic. And the real part and the imaginary part of an algebraic number are always algebraic numbers.

Answer 2

$\cos nx = T_n(\cos x)$, where $T_n(t)$ is Chebyshev polynomial of the first kind. It means that $\cos 1^{\circ}$ is the root of $T_{360}(t) = \cos 360^{\circ} = 1$.

For even $n$, the multiple-angle formula for $\sin (nx)$ is

$$\sin nx = (-1)^{n/2-1}\cos x ~ U_{n-1}(\sin x)$$ where $U_{n-1}(t)$ is the Chebyshev polynomial of the second kind. By squaring both sides, we get $$\sin^2 nx = (1-\sin^2x)U_{n-1}^2(\sin x)$$ so $\sin 1^{\circ}$ is the root of $$(1-t^2)U_{359}^2(t) = \sin^2360^{\circ} = 0$$

Both numbers are algebraic