Archimedes used the fact that
$\sin(x)< x< \tan(x)$ when $0< x < \cfrac{\pi}2$
to prove that the perimeter of a polygon inscribed in a circle is less than the circumference of the circle. Similarly, he used this to show that the perimeter of a polygon circumscribed about a circle exceeds the circumference of the circle. I want to get some idea as to how Archimedes proved the above result.
Consider the following triangles in the unit circle:
We know the following: $\overline{XY} =\overline{XZ} = 1$, $\overline{XV} = \cos\theta$, $\overline{ZV} = \sin\theta$, $\overline{WY} = \tan\theta$. Now, the area of the triangle $\Delta XZY$ is obviously $\frac{1}{2}\sin\theta$. The area of the sector of the unit circle containing $\Delta XZY$ is $\frac{1}{2}\theta$. Finally, the area of $\Delta XWY$ is $\frac{1}{2}\tan\theta$. Since $\Delta XWY$ contains the sector, which contains $\Delta XZY$, we have: $$\frac{1}{2}\sin\theta < \frac{1}{2}\theta < \frac{1}{2}\tan\theta$$ It is easy to go from this to $$\sin\theta < \theta < \tan\theta$$