$$\sin x<x\,(0<x<\frac{\pi}{2})$$
In most textbooks, to prove this inequality is based on geometry illustration (draw a circle, compare arc length and chord ), but I think that strict proof should be based on analysis reasoning without geometry illustration. Who can prove it? Thank you very much.

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**ps**:

1. By differentiation, monotonicity and Taylor formula, all are wrong, because $(\sin x)'=\cos x$ must use $\lim\_{x \to 0}\frac{\sin x}{x}=1$, and this formula must use $\sin x< x$. This is vicious circle.
2. If we use Taylor series of $\sin x$ to define $\sin x$, strictly prove $\sin x<x$ is very easy, but how can we obtain geometry meaning of $\sin x$?