Showing that a set of trigonometric functions is linearly independent over $\mathbb{R}$ 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. 
 A: 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. 
A: For any fixed $k \in \mathbb R$, you can always write $\sin(t+k)$ as a linear combination of $\sin t$ and $\cos t$ (think angle addition formulas).  If I understand your question correctly, I don't think there's any hope of this set being linearly independent.
Note that everything in $A$ is a solution to the third-order linear equation $x''' + x' = 0$.  There isn't much room for those to be linearly independent.
