Prove the following property of $f(x)$? Let $$f(x)=|a_1\sin(x)+a_2\sin(2x)+a_3\sin(3x)+...+a_n\sin(nx)|.$$ 
Given that $f(x)$ is less than or equal to $|\sin(x)|$ for all $x$, prove that $|a_1+a_2+a_3+....|$ is less than or equal to 1.Please keep this at a calculus AB level because that's where I got the problem from. Thanks! 
 A: Suppose that the $a_i$ are all nonnegative.  Under this restriction:
Let $x$ be small and positive.  Using the angle-sum formula for sine, we get $\sin(mx)=(\sin x)(\cos(m-1)x) + (\sin(m-1)x)(\cos x) \ge (\sin x)(\cos(m-1)x)$.  Hence we have $|\sin(x)|\ge f(x)\ge |\sin(x)| |a_1+a_2\cos(x)+a_3\cos(3x)+\cdots+a_n\cos((n-1)x)|$.  Dividing we get $1\ge |a_1+a_2\cos(x)+a_3\cos(3x)+\cdots+a_n\cos((n-1)x)|$ for all such $x$, which implies the desired result by continuity.
A: 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:
$$
|a_1+2a_2+\ldots+na_n|\leq 1
$$
