Proof that this is independent Prove that {$1, \sin(x), \sin(2x), \sin(3x),\ldots, \sin(nx)$} is an independent set.
I can think of the long way which is to differentiate this and put the differentiations into a matrix and row reduce it.
Please advice, thanks!
 A: Hint: Suppose that for some real numbers $k, \alpha_1, \alpha_2, \ldots \alpha_n$ we have
$$ k = \sum_{i=1}^n \alpha_i \sin (ix). $$
Then, multiply by $\sin(jx)$ and integrate from $0$ to $2 \pi$ to conclude that $\alpha_j = 0$. Hence $k=0$.
A: Consider the weighted sum
$$\begin{align}
1+\sum_{k=1}^{n}c_k\sin kx = c_0+c_1\sin x + c_2\sin2x+\cdots+c_n\sin nx = 0&&(1)
\end{align}$$
Then our goal is to prove all $c_0=c_1=\cdots=c_n=0$.
If we multiply $(1)$ with any $2\sin kx$ and integrate from $0$ to $2\pi$, where $1\le k\le2\pi$,
$$\begin{align}
2c_0\sin kx+2c_1\sin x\sin kx+\cdots + 2c_k \sin kx\sin kx + \cdots+2c_n\sin nx\sin kx &= 0\\
2c_0\sin kx+c_1[\cos (k-1)x-\cos (k+1)x] + \cdots + c_k [\cos (k-k)x-\cos (k+k)x] + \cdots+ c_n [\cos (k-n)x-\cos (k+n)x] &= 0\\
\int_0^{2\pi} \left\{2c_0\sin kx+c_1[\cos (k-1)x-\cos (k+1)x] + \cdots + c_k [\cos (k-k)x-\cos (k+k)x] + \cdots+ c_n [\cos (k-n)x-\cos (k+n)x]\right\}dx&=\int_0^{2\pi}0dx
\end{align}$$
Now, notice that $\int_0^{2\pi}\cos kx dx = 0$ if $k\ne0$, as we are integrating a sine over multiple times of its period. But if $k=0$, then $\int_0^{2\pi}\cos 0x dx = \int_0^{2\pi} dx = 2\pi$. Therefore, the above long equation becomes
$$\begin{align}
\int_0^{2\pi} c_k \cos (k-k)x dx = &\int_0^{2\pi}0dx\\
2\pi c_k = & 0\\
c_k =& 0
\end{align}$$
This shows every $c_k$ in $(1)$ is in fact zero. ($c_0 = 0$ can be proved similarly by integrating $(1)$ from $0$ to $2\pi$) Therefore, the set of functions are in fact independent.
