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 collection $\{v_1,\dotsc,v_n\}$ in a vector space $V$ is linearly independent if the equation
$$
\lambda_1v_1+\dotsb+\lambda_nv_n=0
$$
is only solved by $\lambda_1=\dotsb=\lambda_n=0$.
In your case, you have the collection $\{f,g,h\}$, which can be viewed as a collection in the vector space of functions $\Bbb R\to\Bbb R$.
Now, note that
$$
1\cdot f+0\cdot g+0\cdot h=f=0
$$
Do you see why this proves that your collection is linearly dependent?
Best Answer
Yes, you have proved that your functions are pairwise orthogonal, hence independent (as @vadim123 has also pointed out).