Are the functions $\sin(x)$ ; $\sin^2(x)$ ; ... ; $(\sin(x))^{2017}$ linearly independent?

Question

We have the functions $\sin(x)$ ; $\sin^2(x)$ ; ... ; $(\sin(x))^{2017}$ defined on $\Bbb R$. I tried to calculate the Wronskian, but it doesn't seem to do help in any way.

Answer 1

It seems you consider the $\mathbb R$- vector space $C(\mathbb R)$ of all continuous functions $f : \mathbb R \to \mathbb R$ and ask whether the functions $f_n(x) = \sin^n (x)$ are linearly independent. So assume that $$\sum_{n=1}^{2017}a_n f_n = 0$$ withe $a_n \in \mathbb R$. This means that $\sum_{n=1}^{2017}a_n (\sin (x))^n = 0$ for all $x \in \mathbb R$. Since each $y \in [-1,1]$ has the form $y = \sin(x)$, we see that $p(y) = \sum_{n=1}^{2017}a_n y^n = 0$ for all $y \in [-1,1]$. Howewer, if some $a_n \ne 0$, then $p(y)$ is a polynomial of degree $1 \le \deg(p) \le 2017$. This can have at most $\deg(p)$ zeros. In other words, in this case $p(y)$ cannot vanish on $[-1,1]$. Therefore all $a_n = 0$.

Answer 2

If we have that $$\sum_{i=1}^{2017}a_i\sin^i(x)=0$$ for all $x$, this is in particular true for all integer values of $x$. The numbers $\sin(n) $ as $n$ ranges over all integers are all different, and the polynomial $a_1x+\cdots +a_{2017}x^{2017}$ evaluates to $0$ on all of them. If this polynomial were nonzero it could have at most $2017$ zeros, hence it must be identically zero.