I was trying to prove that for $n \geq 1$ and $n = 0$, $$\lvert \sin(nx)\rvert \leq n \cdot \lvert \sin(x) \rvert$$
Turns out it's actually not true...but I don't know why. My proof seems reasonable to me, please help me find the problem(s).
EDIT: This is for real numbers $n$
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Proof:
Since we're interested in the absolute value of $\sin(nx)$ we only have to consider $nx$ over the interval $[0, \frac{\pi}{2}]$, because any angle $kz \gt \frac{\pi}{2}$ can be expressed as $nx$ such that $\lvert \sin(nx) \rvert = \lvert \sin(kz) \rvert$.
We know that the rate of change of $\sin(x)$ with respect to $x$ decreases continuously as $x$ increases, for $x$ over $[0, \frac{\pi}{2}]$.
We also know that any angle $nx$ can be broken into the sum of $x$ and $(nx - x)$.
Therefore we know that $$\frac{\sin(x)}{x} \geq \frac{\sin(nx) - \sin(x)}{nx - x}$$
In other words, since $\frac{\mathrm{d}}{\mathrm{d}x} \sin(x)$ is continuously decreasing over the interval $[0, \frac{\pi}{2}]$, the slope over the interval $[0,x]$ is always greater than or equal to the slope over $[x, nx]$.
From this, we get
$$nx \cdot \sin(x) - x \cdot \sin(x) \geq x \cdot \sin(nx) - x \cdot \sin(x)$$
$$nx \cdot \sin(x) \geq x \cdot \sin(nx)$$
$$n \cdot \sin(x) \geq \sin(nx)$$
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I'm pretty sure what's wrong is my first assumption that: Since we're interested in the absolute value of $\sin(nx)$ we only have to consider $nx$ over the interval $[0, \frac{\pi}{2}]$, because any angle $kz \gt (\frac{\pi}{2})$ can be expressed as $nx$ such that $\lvert \sin(nx) \rvert = \lvert \sin(kz) \rvert$.
But I'm not sure exactly why. I'd really appreciate an explanation of why this part or others of my proof are wrong.
EDIT: A second question I have is whether my proof would actually work if we did indeed constain $nx$ to $[0, \frac{\pi}{2}]$. What about constraining just $x$ to $[0, \frac{\pi}{2}]$