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.

Without words... Well, nearly without words...

