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?
 A: Not sure that this is what you want, but a neat way to do it is noticing that if $0 < \theta < \pi$:
$|1+e^{2i\theta}+...+e^{2i(n-1)\theta}|=\frac{|\sin (n\theta)|}{\sin (\theta)}$ and then use the triangle inequality on LHS
A: 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
$$|\sin(n\theta)| = |\sin(\underbrace{\theta + \theta + \ldots + \theta}_{\text{n times}})| \le \underbrace{|\sin(\theta)| + \ldots + |\sin(\theta)|}_{\text{n times}} = n|\sin(\theta)| = n \sin(\theta).$$
A: Let $z = \cos(θ) + i \sin(θ) $
$\displaystyle
{\sin(n\,θ)\over \sin(θ)} 
= {\Im(z^n) \over \Im(z)} 
= {z^n - 1/z^n \over z - 1/z} 
= z^{n-1} + z^{n-3}+\cdots+{1\over z^{n-3}} + {1\over z^{n-1}}$
RHS can be group in pairs (except possibly center 1, if n is odd) :
${\displaystyle z^k + {1 \over z^k} = 2 \cos(k\,x) }$
Sum of pair absolute value at most reached 2.
With RHS having n terms, $\displaystyle \left|{\sin(n\,θ)\over \sin(θ)}  \right| \le n$
