Linear Independence of an infinite set . Let $e_n=\sin nx$ ($x\in [-\pi,\pi])$and let $A=\{e_i|i\in \mathbb{N}\}$. Prove that A is a linearly independent set.
 A: One way to do it is to prove that $A$ forms an orthogonal set with respect to the inner product give by
$$\langle f,g\rangle=\int_{-\pi}^\pi f(x)g(x)dt.$$
If $A$ forms an orthogonal set, then $A$ is a linearly independent set. 
To prove that $A$ forms an orthogonal set, we need to prove that 
$$\langle e_n,e_m\rangle=\int_{-\pi}^\pi \sin(nx)\sin(mx)dt=0\mbox{ for }m\neq n.$$
Note that 
$$\int_{-\pi}^\pi \sin(nx)\sin(mx)dt=\frac{1}{2}\int_{-\pi}^\pi [\cos(m t-nt)-\cos(mt-nt)]dt=...$$
Can you prove that the last integral is zero when $m\neq n$?
Edit: If $A$ forms an orthogonal set, then we claim that $A$ is a linearly independent set. To see this, suppose we have 
$$\sum_{i=1}^na_ie_{n_i}(x)=0\mbox{ for all }x\in[-\pi,\pi]$$
for some $a_i$ where $1\leq i\leq n$. Taking the inner product with $e_{n_j}$ where $1\leq j\leq n$, we have
$$0=\langle \sum_{i=1}^na_ie_{n_i},e_{n_j}\rangle=\sum_{i=1}^na_i\langle e_{n_i}, e_{n_j}\rangle=a_j$$
for $1\leq j\leq n$. This implies that $A$ is a linearly independent set. 
A: This is my answer (which I would like you all to evaluate).
Let $z=\cos x+i\sin x$ for  $x\in [-\pi,\pi]$ and $A_k=\{e_i|i\in \mathbb{N},i\le k\}$
I will show that every $A_k$ is linearly independent.
If $\displaystyle \sum_{j=1}^{k}\lambda_je_j=0$ be a linear relation. Then I will show $\lambda_j=0,\forall j\le k$.
Now we know that $\sin nx=\frac{1}{2i}(z^n-\frac{1}{z^n})$.
So the relation becomes $\displaystyle \sum_{j=1}^{k}\frac{\lambda_j}{2i}(z^j-\frac{1}{z^j})=0\dots(1)$
Now $|z|=1$ which implies All the $z$ lies on the circle of unit radius.
So $z\ne 0\Rightarrow z^n\ne 0$
Multiplying eqn. (1) with $z^k $ we get,
$\displaystyle z^k\sum_{j=1}^{k}\frac{\lambda_j}{2i}(z^j-\frac{1}{z^j})=0$ 
$\Rightarrow \displaystyle \sum_{j=1}^{k}\frac{\lambda_j}{2i}(z^{j+k}-z^{k-j})=0$
SO we get a polynomial in $z$ valid for all $z$ such that $|z|=1$
As a polynomial of n degree cant have more that $n$ roots and as this polynomial has infinite roots so this must be the zero polynomial $\Rightarrow \lambda_j=0 ,\forall j\le k$ 
Now every finite subset will be a subset of some $A_k,k\in N$ .As all $A_k$ are linearly independent it implies that every finite subset will also be linearly independent.
