Proof that $|\sin(nx)| \le n|\sin(x)|$ [duplicate]

Solution 1:

It is easy to see by induction. Keep the fact that, $|\cos x|\leq 1$. For $n=2$ $$|\sin(2x)| = |2\sin x \cos x|\leq 2|\sin x |$$

Assuming $|\sin(nx)| \leq n|\sin x|$ and using the addition formula we have

$$|\sin((n+1)x)| = |\sin nx ~\cos x +\sin x~ \cos nx|\\\overset{|\cos(\cdot)|\leq 1}{\leq}|\sin nx| + |\sin x| \overset{\text{by assumption}}{\leq}(n+1) |\sin x| .$$

For your worry

$$ |\sin(nx)| = |\int_0^{nx} \cos t\, dt\leq |nx|.$$

Solution 2:

You can use what you have already proved.

Apply it for "$x=nu$", then you get

$$\vert \sin(nu)\vert\leqslant \vert nu\vert=n\vert u\vert$$

since $n\geqslant 0$.

Edit.

For your new question, you should prove it by induction using the fact that

$$\vert a+b\vert \leqslant \vert a\vert+\vert b\vert.$$