Linear independence of $sin(\pi*k_1*x)$ and $sin(\pi*k_2*x)$ $\{f_k: k \in \mathbb{N}\}$ $f_k(t) = sin(k*\pi*t)$ for $k \in \mathbb{N}$ 
How can I show that this set of functions is linear independent (on the interval $[-1, 1]$)?
I know how I can show it with two concrete functions but I have no clue how to prove it for the whole set.
What I tried so far is:
$\alpha * sin(\pi*t*1) + \beta * sin(\pi*t*2) = 0$ 
Now I choose $t = \frac{1}{2}$ to show that $\alpha$ has to be zero.
$\alpha * sin(\frac{\pi}{2}) + \beta * sin(\pi) = \alpha = 0$ 
Now I choose $t = \frac{1}{4}$ to show that $\beta$ has to be zero.
$\alpha * sin(\frac{\pi}{4}) + \beta * sin(\frac{\pi}{2}) = \frac{\sqrt{2}}{2}*\alpha + \beta= 0$ So $\beta$ needs to be -$\frac{\sqrt{2}}{2}*\alpha$ but $\alpha$ needs to be $0$. Therefore $\beta$ needs to be $0$ too.
How can I scale this argument to make it work for the whole set? I thought about using an inductive argument but I don't know how.
 A: An argument based on choosing points and showing the vectors of values there are linearly independent? You're best off choosing points uniformly in the interval, so you can use results about the Fourier matrix. Specifically, for $k=1,2,\dots,n-1$, use $2n$ points $\frac{j}{n}$ for $-n+1\le j\le n$. Then I claim that the $n-1$ vectors $v_k$ with entries $\sin\left(k\pi\frac{j}{n}\right)$ are linearly independent. Why? Because they're nonzero and orthogonal. Using the product-to-sum identity,
$$\sin\left(k_1\pi\frac{j}{n}\right)\sin\left(k_2\pi\frac{j}{n}\right)=\frac12\cos\left((k_1-k_2)\pi\frac{j}{n}\right)-\frac12\cos\left((k_1+k_2)\pi\frac{j}{n}\right)$$
$$\langle v_{k_1},v_{k_2}\rangle=\sum_{j=1-n}^n \sin\left(k_1\pi\frac{j}{n}\right)\sin\left(k_2\pi\frac{j}{n}\right) = \frac12\sum_j \cos\left((k_1-k_2)\pi\frac{j}{n}\right)-\cos\left((k_1+k_2)\pi\frac{j}{n}\right)$$
If $k_1-k_2$ is anything other than zero mod $2n$, the angles $(k_1+k_2)\pi\frac{j}{n}$ are distributed uniformly around the circle, and the sum of their cosines is zero. The same is true for the other term - those sum to zero unless $k_1+k_2\equiv 0\mod 2n$. With $k$ between $1$ and $n-1$, the only way to get either sum or poduct divisible by $2n$ is to have $k_1=k_2$, and $\langle v_{k_1},v_{k_2}\rangle=0$ unless $k_1=k_2$. Of course, $\langle v_k,v_k\rangle >0$.
Why do we use twice as many points as functions? Because we really should have been doing this for cosines at the same time, or for complex exponentials with both positive and negative $k$. Fewer points would suffice in theory, but they would be a lot harder to manipulate systematically.
There's also a continuous version of this - we can define an inner product $\langle f,g\rangle = \int_{-1}^1 fg$, and then $\langle \sin(k_1\pi x),\sin(k_2\pi x)\rangle = 0$ whenever $k_1\neq k_2$. Of course, a set of nonzero orthogonal vectors in any inner product space must be linearly independent.
