# How to prove $- |x| \leqslant \sin x \leqslant |x|\quad,\quad\forall x \in \mathbb{R}$?

How to formally prove $$- |x| \leqslant \sin x \leqslant |x|\quad,\quad\forall x \in \mathbb{R} \; ?$$ without using derivatives and graphs of real functions.

Thank you very much beforehand. Greetings.

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You can simplify your problem to proving that $-1 \leq \sin\,x \leq 1$... and due to periodicity and symmetry, you can in fact consider $x$ in the interval $[0,\pi/2]$... – J. M. Aug 4 '11 at 6:04
Tried to look at its series expansion ? en.wikipedia.org/wiki/… – M. Alaggan Aug 4 '11 at 6:05
I do not think is appropriate series expansion because it is a chapter belonging to the boundary – mathsalomon Aug 4 '11 at 6:46
And then, what else is Mr. J.M.? – mathsalomon Aug 4 '11 at 6:59

Draw a circle, radius 1, center $(0,0)$. Draw a radius making angle $\theta$ with the $x$-axis, and meeting the circle at $(x,y)$. Then $\sin\theta$ is $y$, while $\theta$ is the length of the arc of the circle from $(x,y)$ to $(1,0)$. Argue from the picture that $\sin\theta\lt\theta$ for $\theta\gt0$. That should get you started.