Let $k\in\mathbb{Z}$ and $\theta>0$. Define $f(k,\theta)\triangleq \big [\cos k\theta~~~\sin k\theta \big]$. Then, it follows from direct calculation that \begin{align} f(k,\theta)-2\cos\theta f(k-1,\theta)+f(k-2,\theta)=0. \end{align} How can I prove or disprove by counter-example the following:
Let $a,b,c\in\mathbb{R}^2$. Then, \begin{align} f(k,\theta)a-2\cos\theta f(k-1,\theta)b+f(k-2,\theta)c=0,\qquad {\rm for~all~} k\in\mathbb{Z}, \end{align} if and only if $a=b=c$.