Prove the set $\{\sin{x},\sin{2x}\}$ in $F(\mathbb{R},\mathbb{R})$ is linearly independant Prove the set $\{\sin(x),\sin(2x)\}$ in $F(\mathbb{R},\mathbb{R})$ is linearly independent
I know this has been asked before, but I'm looking for a solution without integration that makes sense.
Essentially we must show that if 
$$r_1\sin(x)+r_2\sin(2x)=0,$$ then the only solution is $r_1=r_2 = 0$, where $r_1,r_2\in\mathbb{R}$.
This should hold for all $x$.
Can I get an answer that doesn't deal with integrating and such?
If we consider $x=0$, then the value of $r_1,r_2$ do not matter, as the equation still evaluates to $0$, but in this case $r_i\neq 0$, so does this mean its not linearly independent?

Take $x=\pi/4$ and $x=\pi/2$ and solve the equations.
$x=\pi/2$ you obtain $r_1\sin(\pi/2)+r_2\sin(\pi)=r_1=0$,
$x=\pi/4$, $r_1 \sin(\pi/4)+r_2\sin(\pi/2)=r_2=0$.
