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?
 A: Take $x=\pi/4$ and $x=\pi/2$ and solve the equations.
$x=\pi/2$ you obtain $r_1\sin(\pi/2)+r_2\sin(\pi)=r_1=0$,
$x=\pi/4$, $r_1 \sin(\pi/4)+r_2\sin(\pi/2)=r_2=0$.
A: If $\sin x$ and $\sin 2x$ were linearly dependent, then there would exists real numbers $r_1$ and $r_2$ such that the equation 
$$r_1 \sin(x) + r_2 \sin(2x) = 0$$
would be true for all $x$
\begin{align}
   r_1 \sin(x) + r_2 \sin(2x) &= 0 \\
   r_1 \sin(x) + 2 r_2 \cos(x) \sin(x) &= 0 \\
   \sin(x) (r_1 + 2 r_2 \cos(x)) &= 0 \\
\hline\
   \sin(x) &= 0 \\
   \cos(x) &= -\dfrac{r_1}{2r_2}
\end{align}
The above equations imply that, for all $x$, either $\sin(x)=0$ or
$\cos(x) = -\dfrac{r_1}{2r_2}$. Since that is clearly false, then $\sin x$ and $\sin 2x$ are linearly independent.
A: $$\quad{r_1\sin(x)+r_2\sin(2x)=0\\
r_1(\frac{e^{ix}-e^{-ix}}{2i})+r_2(\frac{e^{2ix}-e^{-2ix}}{2i})=0\\
r_1(e^{ix}-e^{-ix})+r_2(e^{2ix}-e^{-2ix})=0\\\times e^{2ix}\implies\\
r_1e^{3ix}-r_1e^{ix}+r_2e^{4ix}-r_2=0 \\\forall x \in \mathbb{R} :r_1e^{3ix}-r_1e^{ix}+r_2e^{4ix}-r_2=0 \\\text{ for example } x\to -\infty\\\underbrace{r_1e^{3ix}-r_1e^{ix}+r_2e^{4ix}}_{\to 0 }-r_2=0 \implies r_2=0}$$ then apply $r_2=0 \implies r_1=0$
