# 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.

• $x$ is supposed to be a real number so induction is not applicable for the proof, unless you are restricting $x$ to say natural numbers. – Anurag A Oct 23 '19 at 20:18
• You can try prove it showing for $n\in \mathbb{N}$. After that, conclude for $r\in \mathbb{Z}$. Then, show that is true for $q\in \mathbb{Q}$. Use the fact that $\mathbb{Q}$ is dense in $\mathbb{R}$ and generalize to $\mathbb{R}$. – Matheus Nunes Oct 23 '19 at 20:23
• @AnuragA You can prove it by continuous induction, though. – conditionalMethod Oct 23 '19 at 20:36

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} 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.