# Proving inequality $|\sin(n\theta) |< n\sin\theta$

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?

• Did you mean $\leq$ ? The result is not true for $n=1$ – BaroqueFreak Sep 13 at 0:23
• Thanks for noticing. Fixed it – Abhi7731756 Sep 13 at 0:28
• Most straightforward way would be a Taylor expansion – rubikscube09 Sep 13 at 0:33
• How did you prove it via induction? What do you find unsatisfying in your proof? – Martin R Sep 13 at 7:29
• – Martin R Sep 13 at 7:36

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