Thus both $\cos(k^2\theta) = \cos[k(k\theta)]$ and $\cos[(k^2-1)\theta] = \cos[(k-1)(k+1)\theta]$ are rational. By the Addition and subtraction formulas, we have
$$\cos[(k+1)\theta] = \cos k\theta \cos \theta - \sin k\theta \sin \theta \quad \text{and} \quad \cos(k^2\theta) = \cos[(k^2-1)\theta] \cos \theta - \sin[(k^2-1)\theta] \sin \theta.$$
Setting $r_1 = \cos k\theta$, $r_2 = \cos[(k+1)\theta]$, $r_3 = \cos[(k^2-1)\theta]$, $r_4 = \cos(k^2\theta)$, and $x = \cos \theta$ in the above equations yields
$$r_2 = r_1 x \pm \sqrt{(1 - r_1^2)(1 - x^2)} \quad \text{and} \quad r_4 = r_3 x \pm \sqrt{(1 - r_3^2)(1 - x^2)},$$
or
$$\pm\sqrt{(1 - r_1^2)(1 - x^2)} = r_2 - r_1 x \quad \text{and} \quad \pm\sqrt{(1 - r_3^2)(1 - x^2)} = r_4 - r_3 x.$$
Squaring these two equations and subtracting the resulting equations gives
$$2(r_1 r_2 - r_3 r_4)x = r_1^2 + r_2^2 - (r_3^2 + r_4^2).$$
Since $r_1, r_2, r_3, r_4$ are rational and $x$ is irrational, we must have $r_1r_2 - r_3r_4 = 0$ or
$$\cos k\theta \cos[(k+1)\theta] = \cos(k^2\theta) \cos[(k^2-1)\theta].$$
By the product-to-sum formulas, we derive
$$\frac{\cos[(2k + 1)\theta] - \cos \theta}{2} = \frac{\cos[(2k^2 - 1)\theta] - \cos \theta}{2}$$
or $\cos[(2k + 1)\theta] - \cos[(2k^2 - 1)\theta] = 0$. By the sum-to-product formulas, we obtain
$$2 \sin[(k - k^2 + 1)\theta] \sin[(k^2 + k)\theta] = 0,$$
implying that either $(k - k^2 + 1)\theta$ or $(k^2 + k)\theta$ is a integral multiple of $\pi$. Since $k$ is an integer, we conclude that $\theta = r\pi$ for some rational number $r$.
Considering Lemma 2 for $\alpha = k\theta$ and $\alpha = (k+1)\theta$, the possible values of $\cos k\theta$ and $\cos[(k+1)\theta]$ are $0, \pm 1, \pm \frac{1}{2}$. Consequently, both $k\theta$ and $(k+1)\theta$ is a integral multiple of $\frac{\pi}{6}$. Since $0 < \theta = k\theta - (k-1)\theta < \frac{\pi}{2}$, the only possible values of $\theta$ are $\frac{\pi}{3}$ and $\frac{\pi}{6}$. Since $\cos \theta$ is irrational, $\theta = \frac{\pi}{6}$.

(Based on the work by Kiran Kedlaya) We maintain the notations used in the first proof. Then $s = 2 \cos \theta$ is a root of $S_k(x) - 2r_1$ and $S_{k+1}(x) - 2r_2$ by the definition of $S_n$. Define
$$Q(x) = \gcd(S_k(x) - 2r_1, S_{k+1}(x) - 2r_2)$$
where the gcd is taken over the field of rational numbers. Then $Q(x)$ is a polynomial with rational coefficients, so the sum of its roots (with multiplicities) is rational. Since $s$ is assumed not to be rational, there must be at least one other distinct root $t$ of $Q(x)$.
Note that the $k$ distinct reals $2\cos(\theta + 2\pi a/k)$ for $a = 0, 1, \dots, k-1$ form $k$ roots of the degree $k$ polynomial $S_k(x) - 2r_1$, so they compose all of its roots. Similarly, all of the roots of $S_{k+1}(x) - 2r_2$ have the form $2\cos(\theta + 2\pi b/(k+1))$ for $b = 0, 1, \dots, k$. Note that $s$ and $t$ are roots of $Q(x)$. Therefore roots of both $S_k(x) - 2r_1$ and $S_{k+1}(x) - 2r_2$, and so they must have at least two distinct common roots. Each root $r$ of $Q(x)$ must thus satisfy
$$r = 2 \cos(\theta + 2\pi a/k) = 2 \cos(\theta + 2\pi b/(k+1))$$
for some $a$ and $b$. We either have $\theta + 2\pi a/k = \theta + 2\pi b/(k+1)$ and thus $r = 2 \cos \theta$ or $\theta + 2\pi a/k = -\theta - 2\pi b/(k+1)$ and thus
$$\theta = - \frac{\pi[(a+b)k+a]}{k(k+1)}.$$
In the first case, we obtain $s$, so $t$ must lead to the second value of $\theta$, as $s \neq t$.
Therefore, we can write $\theta = \frac{\pi c}{k(k+1)}$ for some integer $c$. By Lemma 2, $c/k$ and $c/(k+1)$ must both be multiples of $1/6$, since $\cos k\theta = \cos \frac{c\pi}{k+1}$ and $\cos(k+1)\theta = \cos \frac{c\pi}{k}$ are rational. Therefore, $\theta = \frac{c\pi}{k} - \frac{c\pi}{k+1}$ is a multiple of $\pi/6$. Since $t$ is not rational, $\theta$ can only be $\pi/6$.