$\cos k\theta$ and $\cos(k+1)\theta$ are both rational only when $\theta=\pi/6$ Let $\theta$ be an angle in the interval $(0,\pi/2)$. Given that $\cos \theta$ is irrational and $\cos k\theta$ and $\cos (k+1)\theta$ are both rational for some positive integer $k$, show that $\theta=\pi/6$.
 A: It is natural in this context to think of Niven's theorem, and my solution
below uses this theorem. The link in Gyumin Roh's comment yields a solution that does not use it.
Here is a method for $k$ odd, $k=2t+1$ (the case $k$ even is similar). Let $c_j=\cos(j\theta)$ for $j\in{\mathbb Z}$. We have the fundamental identity :
$$
c_{i-j}+c_{i+j}=2c_ic_j \ (i,j\in{\mathbb Z}) \tag{1}
$$
Because of Chebyshev polynomials of the first kind (or by induction from (1) if you
prefer), if $c_t$ is rational then so is $c_{t'}$ for any integer multiple
$t'$ of $t$.
From the hypotheses, we see then that $c_{j}$ is
rational whenever $j$ is a multiple of $k$ or $k+1$. Next, we deduce by using
(1)  twice that
$$
\begin{array}{lcl}
2c_{1}c_{(2j+1)k} &=& c_{2jk+k+1}+c_{2jk+k-1} \\
2c_{k+1}c_{2jk} &=& c_{2jk+k+1}+c_{2jk-k-1} \\
2\bigg(c_{1}c_{(2j+1)k}-c_{k+1}c_{2jk}\bigg) &=& d_{j+1}-d_j, \ \text{where}\ d_j=c_{2jk-k-1} \\
\end{array}
$$  
Summing the last identity from $1$ to $t$, we obtain
$$
\sum_{j=1}^{t}c_{1}c_{(2j-1)k}-c_{k+1}c_{2jk}=
d_{t+1}-d_1=c_{(k-1)(k+1)}-c_{k+1}
$$
So if the sum $L=\sum_{j=1}^{t} c_{(2j-1)k}$ were nonzero, we could write $c_1=\frac{c_{(k-1)(k+1)}-c_{k+1}+\sum_{j=1}^{t}c_{k+1}c_{2jk}}{L}$ contradicting
the hypothesis that $c_1$ is irrational. This forces $L=0$, whence $\bar{L}=0,L+i\bar{L}=0$ which means that $\sum_{j=1}^{t} \gamma^j=0$
where $\gamma=e^{i(2k\theta)}$. We deduce $\gamma^t=1$, and now we can apply
Niven's theorem.
