Prove the following property of $f(x)$?

II believe that under the same assumptions the following result can be proved, that is, $|a_1+2a_2+\ldots+na_n|\leq 1$.

We will need the following result (which I believe can be proved using Calculus AB level): For every $k\in\mathbb{N}$, $\dfrac{\sin(kx)}{\sin(x)}$ can be extended to a continuous function. The proof goes like this:

Observe that $\dfrac{\sin(k(x+2\pi))}{\sin(x+2\pi)}=\dfrac{\sin(kx)}{\sin(x)}$. Observe that this function is continuous except possibly where $\sin(x)=0$. Because the above observation, we only need to worry about what happens at $x=0$. But in this case the following limit exists: $$ \lim_{x\to 0}\dfrac{\sin(kx)}{\sin(x)}=k $$ this limit follows using the well known result: $\lim_{x\to 0}\dfrac{\sin(x)}{x}=1$.

Since we have that $|a_1\sin(x)+a_2\sin(2x)+\ldots+a_n\sin(nx)|\leq |\sin(x)|$ for all $x$ we have that $$ |\sin(x)||a_1+a_2\dfrac{\sin(2x)}{\sin(x)}+\ldots+a_n\dfrac{\sin(nx)}{\sin(x)}|\leq|\sin(x)| $$ for all $x$. Subtracting $|\sin(x)|$ from both side we obtain the following inequality for all $x$: $$ |\sin(x)|(|a_1+a_2\dfrac{\sin(2x)}{\sin(x)}+\ldots+a_n\dfrac{\sin(nx)}{\sin(x)}|-1)\leq 0 $$ which implies that: $$ |a_1+a_2\dfrac{\sin(2x)}{\sin(x)}+\ldots+a_n\dfrac{\sin(nx)}{\sin(x)}|\leq 1 $$ for all $x\neq 2\pi m$. Taking $\lim_{x\to 0}$ in both sides we obtain: $$ |a_1+2a_2+\ldots+na_n|\leq 1 $$ Now if we take the inequality:

$$ |\sin(x)||a_1+a_2\dfrac{\sin(2x)}{\sin(x)}+\ldots+a_n\dfrac{\sin(nx)}{\sin(x)}|\leq|\sin(x)| $$ and divide by $x\neq 0$ both sides and take the $\lim_{x\to 0}$ we obtain: $$ |a_1+2a_2+\ldots+na_n|\leq 1 $$