Irrationality with Cos functions

Definition: An rational number can be written as the ratio between two whole numbers .Statement: $\cos \frac {\pi}{2^n}$ is irrational for all n greater than or equal to 2. Proof: The proof is by induction: consider the basis case: $\cos \frac {\pi}{4}$ = $\frac{\sqrt{2}}{2}$ so the basis case holds. Assume that $\cos \frac {\pi}{2^k}$ is irational. $\cos \frac{2\pi}{2^{k+1}}$ = $\cos \frac {\pi}{2^k}$ . $\cos \frac {\pi}{2^k}$ = $2\cos ^2 (\frac{\pi}{2^{k+1}})-1$ . Suppose $\cos \frac {\pi}{2^{k+1}}$ = $\frac{a}{b},a,b\in \mathbb{Z}$ . Then $\cos \frac {\pi}{2^k}$ = $\frac{2p^2}{q^2}-1$ = $\frac{2p^2-q^2}{q^2}$ and we have a contradiction because $\cos \frac {\pi}{2^k}$ was assumed to be irrational. So $\cos \frac {pi}{2^n}$ is irrational for all n greater than or equation to 2 by the principle of mathematical induction