trigonometric inequality

$\forall n\in \mathbb{N}-\left\{ 0,\left. 1 \right\} \right.$ , $x\in \mathbb{R}-\left\{ \pi \mathbb{Z} \right\}$ , $\left| \sin \left( nx \right) \right|<n\left| \sin x \right|$

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Jesse, what have you tried? – JavaMan Mar 10 '11 at 23:23
One other hint: $\sin(a + b) = \sin(a)\cos(b) + \cos(a) \sin(b)$. – JavaMan Mar 10 '11 at 23:29
I have tried...to show it step by step... give n=1,2...,n... soit il a vraie jusqu'à n+1.. – Jesse Mar 11 '11 at 11:52

Hint: Have you heard of induction?

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I know how to establish now~ – Jesse Mar 11 '11 at 11:53

What is the range of $\sin(z)$ What is the range of $\sin(n z)$?

All the restrictions on $n$ and $x$ just remove annoying counterexamples.

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I spent some time on solving this. Here is my solution based on your suggestions. I would like to verify it's correctness.

We'll prove by induction that $\forall n \in \Bbb N : |sin(nx)| \le n|sin(x)|$.

For n=1 this is obviously true. We'll assume that this statemnet is true for n and prove that it's also true for n+1. So

(using the aforementioned trig. identity)

$|\sin((n+1)x)|=|\sin(nx+x)|=|\sin(nx) \cos(x)+\cos(nx)\sin(x)|$

(using the triangle inequality and the bounds of sin and cos)

$\le |\sin(nx)\cos(x)|+|\cos(nx)\sin(x)| \le 1 \cdot |\sin(nx)|+1 \cdot |\sin(x)|$

$\le n|\sin(x)|+|sin(x)| = (n+1)|\sin(x)|$

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