linear algebra - Showing: family is linearly independent

The Following text is not a mathematical proof, yet, it's just a collection of ideas I have.
I need your help to make it a math. proof :)

$(f_n)_{n \in N}$ is given by $f_n : R \to R, x \to sin(2^{-n}x).$

Show that the family $(f_n)_{n \in N}$ is linearly independent.

Well.. it seems as if every f_n is zero for x=0.
However, each $f_n$ does have a zero-point that $f_{n-1}, f_{n-2}, ..., f_0$ do not have in common.

I now took a look at:

$C_0*f_0 + C_1*f_1 + ... + C_n*f_n = 0$

For this proof I have to show: this equation can only be solved for $C_0 = C_1 = ... = C_n = 0$.

Well.. With whatever Constant C i multiplicate my function f_n, the zer-point remains the same.

So: let's call the point, where $f_0, f_1, ..., f_{n-1}$ have a common zero-point (!but $f_n$ doesnt!!) $N_1$.

If $(f_n)$ would be linearly dependent, then there must be a way to solve
$C_0+f_0 + C_1+f_1 + ... + C_n*f_n = 0$ at $N_1$.
So there are actually many ways for $C_0, ..., C_{n-1} \not = 0$, since all of those functions are 0 at $N_1$, the choice of C doesnt matter.
However: I cannot find a $C_n \not = 0$ so that $C_n*f_n = 0$, since $f_n \not = 0$ at $N_1$

Answer

You have the right idea, but you really want a statement of the opposite type: for each $n\ge0$, the point $x_n=2^{n-1}\pi$ has the property that $f_0(x_n) = \cdots = f_{n-1}(x_n)=0$ but $f_n(x_n)=1$.

Now let $c_0f_0(x) + \cdots + c_nf_n(x) = 0$ be an identically vanishing linear combination of $\{f_0,\dots,f_n$. Plugging in $x_n$ to both sides, we see that $c_n=0$. Next plugging in $x_{n-1}$ to both sides, we see that $c_{n-1}=0$. Continuing in this way, we eventually show that all the $c_j$ equal $0$. Therefore the set $\{f_0,f_1,\dots\}$ is linearly independent.