I would like to determine under what conditions on $k$ the set $$ \begin{align}
A = &\{1,\cos(t),\sin(t), \\
&\quad \cos(t(1+k)),\sin(t(1+k)),\cos(t(1−k)),\sin(t(1−k)), \\
&\quad \cos(t(1+2k)),\sin(t(1+2k)),\cos(t(1−2k)),\sin(t(1−2k))\},
\end{align}$$
is linearly independent, where $k$ is some arbitrary real number.
As motivation, I know that the set defined by
$$
\{1, \cos wt, \sin wt\}, \quad w = 1, \dots, n
$$
is linearly independent on $\mathbb{R}$, which one generally proves by computing the Wronskian. I thought that I could extend this result to the set in question, but I haven't found a proper way to do so. My intuition tells me that $A$ will be linearly dependent when the arguments of the trig functions coincide, which will depend on the value of $k$.
Though, I'm at a loss for proving this is true. Computing the Wronskian for this set required an inordinate amount of time– I stopped running the calculation after a day. Is there perhaps a way to reduce the set in question so that the Wronskian becomes manageable?
I'm interested in any suggestions/alternative methods for proving linear independence that could help my situation. Note that I'd like to have a result that holds for any $m = 0, \dots, n,$ where $n \in \mathbb{Z}$ if possible.
Thanks for your time.
EDIT: The set originally defined in the first instance of this post was incorrectly cited. My sincere apologies.
Best Answer
The answer is $k = 0, \pm 1, \pm \frac{1}{2}$. This follows from the following result.
Claim: The functions $\{ 1, \sin rt, \cos rt \}$ for $r$ a positive real are linearly independent over $\mathbb{R}$.
Proof 1. Suppose that $\sum s_r \sin rt + \sum c_r \cos rt = 0$ is a nontrivial linear dependence. Consider the largest positive real $r_0$ such that $c_{r_0} \neq 0$. Take a large even number of derivatives until the coefficient of $\cos r_0 t$ is substantially larger than the remaining coefficients of the other cosine terms and then substitute $t = 0$; we obtain a number which cannot be equal to zero, which is a contradiction. So no cosines appear.
Similarly, consider the largest positive real $r_1$ such that $s_{r_1} \neq 0$. Take a large odd number of derivatives until the coefficient of $\cos r_1 t$ is substantially larger than the remaining coefficients of the other cosine terms (which come from differentiating sine terms) and then substitute $t = 0$; we obtain a number which cannot be equal to zero, which is a contradiction. So no sines appear.
So $1$ is the only function which can appear in a nontrivial linear dependence, and so there are no such linear dependences.
Proof 2. It suffices to prove that the functions are all linearly independent over $\mathbb{C}$. Using the fact that
$$\cos rt = \frac{e^{irt} + e^{-irt}}{2}, \sin rt = \frac{e^{irt} - e^{-irt}}{2i}$$
it suffices to prove that the functions $\{ e^{irt} \}$ for $r$ a real are linearly independent. This can be straightforwardly done by computing the Wronskian and in fact shows that in fact the functions $\{ e^{zt} \}$ for $z$ a complex number are linearly independent.
Proof 3. Begins the same as Proof 2, but we do not compute the Wronskian. Instead, let $\sum c_z e^{zt} = 0$ be a nontrivial linear dependence with a minimal number of terms and differentiate to obtain
$$\sum z c_z e^{zt} = 0.$$
If $z_0$ is any complex number such that $z_0 \neq 0$ and $c_{z_0} \neq 0$ (such a number must exist in a nontrivial linear dependence), then
$$\sum (z - z_0) c_z e^{zt} = 0$$
is a linear dependence with a fewer number of terms; contradiction. So there are no nontrivial linear dependences.