# 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.

EDIT: The set originally defined in the first instance of this post was incorrectly cited. My sincere apologies.

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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.

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These are very nice. Just to be sure, the $k$ values you cited would be the values for which the set is linearly dependent, right? Also, I'm not sure that I follow the argument in the third proof. The contradiction is deriving a new dependency relation with fewer terms, which contradicts the assumption that the set had a dependency relation, right? It seems like this argument could always be constructed in a case where differentiating yields the same functions in the set. – unbounded_despair Aug 5 '12 at 11:58
@unbounded: yes. This is a special property of functions like the exponential, sine, and cosine. It is very hard for them to be linearly dependent because they are all essentially eigenfunctions of differentiation. A more general statement is true for eigenvectors of any linear operator (the ones with distinct eigenvalues are linearly independent). – Qiaochu Yuan Aug 5 '12 at 14:04
Makes much sense. Thanks kindly, Qiaochu. – unbounded_despair Aug 5 '12 at 14:20

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.

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Erick, is there then an explicit way to determine the dependency relation? – unbounded_despair Aug 5 '12 at 1:42
Erick, I incorrectly defined the original set (sorry about this). The question has been corrected, if you're still interested in it. Cheers – unbounded_despair Aug 5 '12 at 3:53