How to show that $\{t, \sin t , \cos 2 t , \sin t \cos t \}$ is a linearly independent set of functions on $\mathbb{R}$? I have this homework question that I have no idea how to do:

Show that $\{t, \sin(t), \cos(2t), \sin(t)\cos(t) \}$ is a linearly independent set of functions defined on $\mathbb{R}$. Start by assuming that
  $$c_1 t + c_2 \sin(t) + c_3 \cos(2t) + c_4 \sin(t)\cos(t) = 0$$
  for all $t$. Choose specific values of $t$ ($t=0, 1, 2\dots$) until you get a system with enough equations to determine that all the $c_i$'s must be $0$. 

My only guess is to set up a matrix based on this polynomial somehow, then row reduce it to find the pivot columns. Subbing in different $t$'s would allow for some value to be attached to each $c$. Would that be the matrix? Shouldn't it be formed by the $c$ values?
 A: 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.
A: For instance, if you let $t=0$ then $t=0$, $\sin(0)=0$, and $\sin(0)\cos(0)=0$.  But, $\cos(2\cdot 0)=\cos(0)=1$ and your equation becomes $c_3=0$.  As The Chaz suggested, you might want to try values from the unit circle where you know the various function values of $\sin$ and $\cos$.
A: At $t = 0$  we have
$t = 0, \sin(t) = 0,\,\cos(2t) = 1$ and $\sin(t)\cos(t) = 0$. So if I choose $c_3 = 0$ and allow  $c_1, c_2, c_4$ to take any other value, $$c_1 t + c_2 \sin(t) + c_3 \cos(2t) + c_4 \sin(t)\cos(t) = 0$$ is satisfied at $t = 0$. I think that implies that the functions are linearly dependent at $t = 0$. This would hold true for $t = k\pi$.
