Prove that $|\sin x|\le|x|$ without using the continuity property Without using any property about continuity of sine function, Prove that $|\sin x|\le|x|$.
First come to my mind is that using taylor series to approximate the sine function, where we know that $\sin x=\sum_{i=0}^{\infty}\frac{(-1)^kx^{2k+1}}{(2k+1)!}$, but i am not quite see how to prove the inequality.
 A: Without words... Well, nearly without words...

A: For $0\leq\phi\leq{\pi\over2}$ consider the two points $(\cos \phi,\pm\sin \phi)$ on the unit circle. Their distance is $2\sin \phi$, while the length of the circular arc between them is $2\phi$. Therefore we have
$$2\sin\phi\leq 2\phi\qquad(0\leq\phi\leq{\pi\over2})\ .$$
Putting $\phi:=|x|$ we obtain
$$|\sin x|\leq |x|\qquad(0\leq|x|\leq{\pi\over2})\ ,$$
and for $|x|\geq{\pi\over2}$ one obviously has $|x|\geq{\pi\over2}>1\geq|\sin x|$.
A: The sine function is an odd function, so it suffices to prove the inequality for nonnegative $x$.
For $0\le x\le 1$, rewrite $\sin x$ in two different ways:
\begin{align*}
\sin x &=\sum_{n=0}^\infty\left(\frac{x^{4n+1}}{(4n+1)!} - \frac{x^{4n+3}}{(4n+3)!}\right)\tag{1}\\
&=x - \sum_{n=0}^\infty\left(\frac{x^{4n+3}}{(4n+3)!} - \frac{x^{4n+5}}{(4n+5)!}\right).\tag{2}
\end{align*}
Since each bracket term in the two infinite series is nonnegative, $(1)$ implies that $\sin x\ge0$ and $(2)$ implies that $\sin x\le x$. Therefore $|\sin x|\le |x|$.
For $x>1$, note that $\sin x = |\operatorname{imag}(e^{ix})| \le |e^{ix}|$. If we can show that $|e^{ix}|\le1$, we are done. One way to prove this is to show that $\overline{e^{ix}}=e^{-ix}$ (easy) and $e^{ix}e^{-ix}=1$ (theorems about multiplication of absolutely convergent power series, i.e. convergence of Cauchy product, is needed here).
