TITLE: how to strictly prove $\sin x<x$ for $0<x<\frac{\pi}{2}$
QUESTION [23 upvotes]: $$\sin x<x\,(0<x<\frac{\pi}{2})$$
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.

ps:

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. 
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 $\sin x$?

REPLY [3 votes]: "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."

The geometrical figure accompanying such a proof is an illustration to help you understand the argument. The proof is not dependent on the geometrical illustration but is based on certain basic facts about arc-length.
The concept of arc length itself is based on analytic notions of supremum and the fact that an arc of a circle has a length is dependent on the fact that the function $f(x) = \sqrt{1 - x^{2}}$ is monotone on $[0, 1]$. Thus I remove the geometrical illustration here (i.e. I remove the drawing and you will have to spend extra effort to visualize / understand what follows) and present the classical proof.
Let $\mathcal{C}$ be the set of points $(x, y)$ in plane which satisfy $$x^{2} + y^{2} = 1\tag{1}$$ and consider a part of it namely the smaller part from point $A = (1, 0)$ to $B = (0, 1)$ and let $P = (p, q)$ be an intermediate point on this arc $AB$. Let $C = (p, 0)$ and clearly we can see that the length of line segment $PC$ i.e. $q$ is less than the length of line segment $AP$ i.e. $\sqrt{(p - 1)^{2} + q^{2}}$.
Now the calculation of arc length $L$ of arc $AP$ is somewhat tedious. If is defined as the supremum of lengths of a polygonal arcs with points lying on arc $AP$ such that $A$ is the starting point of the polygonal arcs and the final point is $P$. Clearly line segment $AP$ of length $\sqrt{(p - 1)^{2} + q^{2}}$ is also one such polygonal arc and hence by definition of supremum we have $$L \geq \sqrt{(p - 1)^{2} + q^{2}} > q$$ By the definition of circular functions the coordinates of point $P$ are $(\cos L, \sin L)$ and hence $p = \cos L, q = \sin L$. We have thus proved that $$\sin L < L\tag{2}$$ for values of $L$ such that the the point $P$ lies in the smaller part of the curve $\mathcal{C}$ between $A$ and $B$. This means that $L > 0$ and $$L < \int_{0}^{1}\frac{dx}{\sqrt{1 - x^{2}}}$$ where the integral on right is the length of the whole arc $AB$ based on formula of arc-length in terms of integral. If we define circular functions on the basis of arc-length (as done above) then the constant $\pi$ is defined to be twice the above integral i.e. $$\pi = 2\int_{0}^{1}\frac{dx}{\sqrt{1 - x^{2}}}$$ Thus we have finally proved that $\sin L < L$ for $0 < L < \pi/2$.

Contrary to what many believe the definition of circular functions via the use of circle $\mathcal{C}$ defined by $x^{2} + y^{2} = 1$ is fully rigorous. This is also the traditional approach used in textbooks of trigonometry. However the fundamental point in this approach is to justify one of the following two things and define the circular functions accordingly:

An arc of a circle has a well defined length. If $A = (1, 0)$ is a fixed point on circle $\mathcal{C}$ and $P$ is a point on same curve $\mathcal{C}$ with length of arc $AP$ as $x$ then the coordinates of the point $P$ are defined to be $(\cos x, \sin x)$.
A sector of a circle has a well defined area. If $A = (1, 0)$ is a fixed point on circle $\mathcal{C}$ and $P$ is a point on same curve $\mathcal{C}$ with area of sector corresponding to arc $AP$ as $x/2$ then the coordinates of the point $P$ are defined to be $(\cos x, \sin x)$.

The traditional presentation of circular functions uses one of the two above definitions (the one on area is simpler to execute) but typically the justifications of length and area are ommitted because they require the notion of integral. Moreover it can be easily established via integral calculus that both the above definitions are equivalent. For more details see my blog post.