Is it possible to deduce that $\sin(x)< x$ for $0x$? It is not difficult to individually prove that in the domain $0<x<\pi$, we have 
$$\tan x > x \quad(A)$$
and 
$$\sin x < x \quad (B)$$
However, I was wondering if it was possible to get result $(B)$ for free from $(A)$, or possibly visa versa.

If we restrict our attention to the interval $\left(0,\frac{\pi}{2}\right)$, the inequality
$$\tan(x)>x\tag{A'} $$
implies, by termwise integration, that over the same interval
$$ -\log\cos(x) > \frac{x^2}{2}\tag{A'1}$$
$$ \cos(x)<e^{-x^2/2}<1\tag{A'2} $$
hold. By applying termwise integration again, we may easily check that $(A')$ implies 
$$ \sin(x) < x \tag{B'} $$
so $(A')\to(B')$. With the same approach, $(B')$ implies
$$ \cos(x) > 1-\frac{x^2}{2}\tag{B'1}$$
$$ \sin(x) > x-\frac{x^3}{6}\tag{B'2}$$
$$ \cos(x) < 1-\frac{x^2}{2}+\frac{x^4}{24}\tag{B'3}$$
so from $(B'2)$ and $(B'3)$ we have:
$$ \tan(x) > \frac{x-\frac{x^3}{6}}{1-\frac{x^2}{2}+\frac{x^4}{24}}>x \tag{B'4}$$
since $\frac{\pi}{2}<\sqrt{8}$. It follows that $(B')\to (A')$, so $(A')$ and $(B')$ are equivalent.
