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

Specifically, just to talk about cosine, is it true that $\cos(\frac{a\pi}{b})$ is algebraic for integers $a$ and $b$? Looking at this post and the link to trigonometric constants in the comments, it seems likely that this is true. But most of the values calculated there are the result of sum/difference of angle formulas for existing algebraic values of sine and cosine.

This came up when looking at $\cos(\frac{\pi}{7})$. If this is algebraic, how can we calculate it? If it is not, for which arguments will sine and cosine be algebraic and for which arguments will they be transcendental?

## marked as duplicate by quid♦, Gerry Myerson, Jonas Meyer, projectilemotion, Community♦Feb 9 '17 at 0:19

• "This question has been asked before" XD. – Xam Feb 9 '17 at 14:57
• @Xam, yeah right? :P But the newer one did get more attention and some really good answers, so ... – Mike Pierce Feb 9 '17 at 21:28

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.
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.