# Inductive proof of the inequality $|\sin (nx)| \leq n\sin(x)$

I am trying to prove this inequality by induction:

For all $x$ in the interval $x\in [0, \pi]$, prove that: $$|\sin (nx)| \leq n\sin(x) \textit{, n a nonnegative integer}$$

The base case is pretty trivial so after I assumed that this inequality holds for $k$ I want to show that it holds for $P(k+1)$ then

$$|\sin[(k+1)x]| \leq (k+1)\sin x = k\sin x + \sin x$$

I am not really sure how to proceed from here, or if my method is not very good. Anyways any help would be really great, thanks!

• $\sin((n+1)x)=\sin(nx+x)=\sin(nx)\cos x + \cos(nx)\sin x$ also $|\cos nx|\le 1$ and $|\cos x|\le 1$, use induction hypothesis $|\sin(nx)| \le nx$, you will get the result :) – r9m Mar 12 '14 at 20:41
• So what I have then is $|sin(k+1)x| = |sin(kx+x)| = |sin(kx)cos(x)+cos(kx)sin(x)| \leq |sin(kx)cos(x)| + |cos(kx)sin(x)| = |sin(kx)||cos(x)|+|cos(kx)||sin(x)|$ we know $|cos(nK)| \leq 1$ and $|cos(x)\leq 1$, then we have $|sin(kx)||cos(x)|+|cos(kx)||sin(x)| \leq |sin(kx)| + |sin(x)| \leq ksin(x) + sin(x)$ (by induction hypothesis) so that $(k+1)sin(x)$ which is what I wanted to prove! – InsigMath Mar 12 '14 at 20:49
• absolutely that :) – r9m Mar 12 '14 at 20:50
• Thank you! That helped a lot! – InsigMath Mar 12 '14 at 20:51

Hint: We have $\sin(a+b)=\sin a\cos b+\cos a\sin b$. Let $a=kx$ and $b=x$.