Proof by induction:$|\sin(x)| ≤|x|$, for any $x \in \mathbb N$ Firstly We verify for $1$ and holds, Secondly we know for $n$ and we prove it for $n+1.$ if I write $n+1$,  $|\sin(x+1)| ≤|x+1|$
$|\sin x\cos1+\cos x\sin1| \le|x+1|$ we this  $|\sin(x)| ≤|x|$ and how can I use this formula. Furthermore How we find this formula without induction.
 A: If $n\in \mathbb{N}$ then by the triangle inequality and by the inductive hypothesis
$$\begin{align}|\sin(n+1)|&=|\sin(n)\cos(1)+\cos(n)\sin(1)|\\
&\leq |\sin(n)|\underbrace{|\cos (1)|}_{\leq 1}+\underbrace{|\cos (n)|}_{\leq 1}|\sin (1)|\\
&\leq |\sin(n)|+|\sin (1)|\leq n+1.\end{align}$$
P.S. As pointed out by Peter Foreman, the inequality is trivial because for any real $x$ such that $|x|\geq 1$,
$$|\sin(x)|\leq 1\leq |x|.$$
However, by using the same approach we can prove this more general inequality (for $x=1$ we recover the previous one): for any $x\in\mathbb{R}$ and  any $n\in \mathbb{N}$,
$$|\sin(nx)|\leq n|\sin(x)|.$$
For the inductive step:
$$\begin{align}|\sin((n+1)x)|&=|\sin(nx)\cos(x)+\cos(nx)\sin(x)|\\
&\leq |\sin(nx)|\underbrace{|\cos (x)|}_{\leq 1}+\underbrace{|\cos (nx)|}_{\leq 1}|\sin (x)|\\
&\leq |\sin(nx)|+|\sin (x)|\\
&\leq n|\sin(x)|+|\sin (x)|=(n+1)|\sin(x)|.\end{align}$$
A: 
If the central angle is $x$ within a unit circle then the red segment is shorter than the arc which bounds it. That will give the desired inequality. 
