Here's a simple proof,
not original by me, using calculus.
It can easily be extended
to show that the power series
for sine and cosine are enveloping,
that is successive terms
bracket the function.
Start with
this definition of
since and cosine:
$\sin' = \cos
$,
$\cos' = -\sin
$,
$\sin(0) = 0$,
$\cos(0) = 1$.
These imply
$\sin^2+\cos^2 = 1$.
For small $t$,
$1 \ge \cos(t)
\ge 0
$
so
$\sin(x)
=\int_0^x \cos(t)dt
\le x
$.
Therefore
$1-\cos(x)
=\int_0^x \sin(t) dt
\le \int_0^x t dt
= \frac{t^2}{2}
$
so
$\cos(t)
\ge 1-\frac{t^2}{2}
$.
Therefore
$\sin(x)
=\int_0^x \cos(t)dt
\ge\int_0^x (1-\frac{t^2}{2})dt
=x-\frac{x^3}{6}
$.
So we already have
$x-\frac{x^3}{6}
\le \sin(x)
\le x
$.