Let's say you plug-in $t=t_1,t_2,t_3,t_4$ then you'll have
$\begin{array}{cc} c_1 t_1 + c_2\sin(t_1)+c_3 \cos(2t_1)+c_4\sin(t_1)\cos(t_1) & =0 \\ c_1 t_2 + c_2\sin(t_2)+c_3 \cos(2t_2)+c_4\sin(t_2)\cos(t_2) & =0 \\ c_1 t_3 + c_2\sin(t_3)+c_3 \cos(2t_3)+c_4\sin(t_3)\cos(t_3) & =0 \\ c_1 t_4 + c_2\sin(t_4)+c_3 \cos(2t_4)+c_4\sin(t_4)\cos(t_4) & =0 \end{array}$
and so the corresponding matrix equation is
$ \begin{pmatrix} t_1 & \sin(t_1) & \cos(2t_1) &\sin(t_1)\cos(t_1) \\ t_2 & \sin(t_2) & \cos(2t_2) &\sin(t_2)\cos(t_2) \\ t_3 & \sin(t_3) & \cos(2t_3) &\sin(t_3)\cos(t_3) \\ t_4 & \sin(t_4) & \cos(2t_4) &\sin(t_4)\cos(t_4) \end{pmatrix} \begin{pmatrix} c_1\\ c_2\\ c_3\\ c_4 \end{pmatrix} = \begin{pmatrix} 0\\ 0\\ 0\\ 0 \end{pmatrix}$
By choosing the "right" values for $t_1,t_2,t_3,t_4$ you'll have a matrix which is invertible (so $c_1=c_2=c_3=c_4=0$ is the only solution). You can check this is case by making sure the determinant is non-zero. [I believe the values $t_1=0$, $t_2=1$, $t_3=2$, $t_4=3$ work, but it is not pretty.]
An alternate approach is to differentiate your equation 3 times and then plug-in a single value (like $t=0$). In this case, the coefficient matrix you'll end up with is called a Wronskian. If it's determinant is non-zero, then the matrix is invertible so the corresponding system only has the trivial solution and hence your functions are linearly independent.