# Prove $\sin(1/n)<1/n$ for all $n$

I need to prove $\sin(1/n)<1/n$ for all $n \in \Bbb N$ using mathematical induction.

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Starting is the easy part! Base case: $n=1$ $$\sin \left(1\over 1\right)<1 \, \color{green}\checkmark$$ – Git Gud Apr 21 '13 at 20:33
Why did you take it so literally? – mopdf Apr 21 '13 at 20:44
Because he is a mathematician. – Fixed Point Apr 21 '13 at 21:02
@mopdf Boys will be boys. – Git Gud Apr 21 '13 at 21:05
(You can easily show $\sin(x)<x$ for all positive $x$ using the Taylor series for $\sin$.) – Potato Apr 21 '13 at 21:07

Because of $\sin(x)=x-\frac{x^3}{3!}+ \frac{x^5}{5!}-\frac{x^7}{7!} +\dots$ As $x^n$ is monotone decreasing on $[0,1]$ in respect to $n$. Hence $\sin(x)\leq x$ for $x$ in $[0,1]$. As $\sin(x)\leq 1$ the inequality is even true for all $x\in [0,\infty)$.
We can see that when $x-\sin(x)> 0$ holds for $x\in (0,1]$ this implies that $\sin\big(\frac{1}{n}\big) < \frac{1}{n}$ holds for all $n \in \mathbb{N}$. At first we note that for $x=0$ $\sin(x)-x=0$. The derivative of $x - \sin(x)$ is $1-\cos(x)$. With the fundamental theoroem of calculus we know that $x-\sin(x) = (1-\cos(\xi)) \cdot x$ where $\xi \in (0,x)$. As $\cos (x) \leq 1$ and $x>0$ we have that $x-\sin(x)$ is positive and hence our inequality holds.