Proving inequality $|\sin(n\theta) |\le n\sin\theta$ I was trying to prove following inequality:
$$|\sin n\theta| \leq n\sin \theta \
\text{for all n=1,2,3... and } \ 
0<\theta<π $$
I succeeded in proving this via induction but I didn't get "feel" over the proof. Are there other proof for this inequality?

You can show that $|\sin(x)|$ is subadditive, i.e.
$$|\sin(x + y)| \le |\sin(x)| + |\sin(y)|.$$
To prove this, simply expand the left side:
\begin{align*}
|\sin(x + y)| &= |\sin(x)\cos(y) + \sin(y)\cos(x)| \\
&\le |\sin(x)| \cdot |\cos(y)| + |\sin(y)| \cdot | \cos(y)| \\
&\le |\sin(x)| + |\sin(y)|,
\end{align*}
as $|\cos(x)|$ and $|\cos(y)|$ are less than or equal to $1$.
How does this help? Note that, when $0 < \theta < \pi$, we have $\sin(x) \ge 0$, hence $|\sin(\theta)| = \sin(\theta)$. Using induction, we can use subadditivitiy to show that
$$|\sin(n\theta)| = |\sin(\underbrace{\theta + \theta + \ldots + \theta}_{\text{n times}})| \le \underbrace{|\sin(\theta)| + \ldots + |\sin(\theta)|}_{\text{n times}} = n|\sin(\theta)| = n \sin(\theta).$$
