Proving inequality $|\sin(n\theta) |\le n\sin\theta$

Solution 1:

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

Solution 2:

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 $$|\sin(n\theta)| = |\sin(\underbrace{\theta + \theta + \ldots + \theta}_{\text{n times}})| \le \underbrace{|\sin(\theta)| + \ldots + |\sin(\theta)|}_{\text{n times}} = n|\sin(\theta)| = n \sin(\theta).$$