Prove $\sin x < x < \tan x$ for $0 < x < \pi/2$ algebraically I know there are some proofs showing that, for  $0 < x < \pi/2$:
$\sin x < x < \tan x $
with the use of the unit circle, however, can I prove this algebraically in some way as well?

Since
$\sin' = \cos
$ and
$\cos' = -\sin$
and
$0 < \cos < 1$
for $0 < x < \pi/2$,
$\sin(x)
=\int_0^x \cos(t) dt
\lt x
$.
Since
$\tan'(x)
=\sec^2(x)
=\dfrac1{\cos^2(x)}
$,
$\tan(x)
=\int_0^x\dfrac1{\cos^2(t)}dt
\gt\int_0^x 1dt
=x
$.
