Let's prove this by combining the proof by contradiction and mathematical induction\footnote{Both methods are discussed in the book \textit{Art of Thinking} written by the same author.}. Assuming $\cos 1^\circ$ is rational. Then $\cos 2^\circ$ will be rational too because $\cos 2^\circ = 2\cos^2 1^\circ -1$. Meanwhile, if both $\cos n^\circ$ and $\cos (n-1)^\circ$ are rational where $n$ is a positive integer greater than $1$, so will be $\cos (n^\circ+1^\circ)$ because $$\cos(n^\circ + 1^\circ) + \cos(n^\circ - 1^\circ)=2\cos n^\circ\cos 1^\circ$$ Therefore, by the principle of mathematical induction, for any positive integer $n$, $\cos n^\circ$ is rational. However, this conclusion is clearly false because $\cos 30^\circ$ is irrational. This follows that $\cos 1^\circ$ cannot be rational, i.e., it must be irrational.