Is a trigonometric function applied to a rational multiple of $\pi$ always algebraic? [duplicate]

Solution 1:

Let $z=\cos\frac{a\pi}b+i\sin\frac{a\pi}b=e^{\frac{a\pi i}{b}}$. Then $z^{2b}=1$ and hence $z$ is algebraic. Finally $\cos\frac{a\pi}b=\frac12(z+z^{-1})$ is also algebraic.

Solution 2:

To prove that $\cos\frac\pi7$ is algebraic:

Using the sum formulae a bunch of times, calculate $\cos7x$ as a polynomial in terms of $\cos x$. (It turns out that $64\cos^7x-112\cos^5x+56\cos^3x-7\cos x=\cos7x$.)

That means that, setting $x=\frac\pi7$, you'll have a polynomial in terms of $\cos\frac\pi7$ equal to $-1$. Now, just add $1$ to both sides, to get a polynomial in $\cos\frac\pi7$ equal to $0$, showing it's algebraic.

This works for any rational.