Purely algebraic proof of the trigonometric inequalities While calculating various limits of trigonometric functions, one must resort to the squeeze theorem which is founded on the inequalities $$1 > \frac{\sin x}{x} > \cos x$$ for some "small" $x$. These inequalities are, however, always (to my knowledge) established geometrically by drawing various triangles and circles where one sees that they hold.
Is there a purely algebraic proof of this inequalities, using only the properties of trigonometric functions and not relying on the underlying geometry?
 A: Both:
$$\frac{\sin x}{x}\leq 1,\qquad\frac{\tan x}{x}\geq 1$$
can be seen as convexity inequalities. The first one follows from the fact that $\frac{d}{dx}\sin x = \cos x$ is decreasing on $[0,\pi]$ and the second one follows from the fact that $\frac{d}{dx}\tan x=\frac{1}{\cos^2 x}$ is increasing over the same interval.
A: Are you happy to start from:
$\sin x=x-\frac {x^3}{3!}+\frac{x^5}{5!}-...$
$\frac{\sin x}x=1-\frac {x^2}{3!}+\frac{x^4}{5!}-...$
$\cos x=1-\frac {x}{2!}+\frac{x^4}{4!}-...$
$\cos x-\frac{\sin x}x=-x^2\left(\frac 1{2!}-\frac 1{3!}\right)+x^4\left(\frac 1 {4!}-\frac 1{5!}\right)+...$
A: I don't know if it is "algebraic", but you might start from the observation that the function
$$
\theta\in \mathbb{R}\mapsto e^{i\theta}\in \mathbb{C}
$$
is Lipschitz with constant $1$. This can be proved via the fundamental theorem of calculus:
$$
\left\lvert e^{i\theta+h}-e^{i\theta}\right\rvert=\left\lvert e^{i\theta}\int_0^h ie^{i\, h'}\, dh'\right\rvert\le\lvert h\rvert.$$
So
$$
\left\lvert\frac{e^{i\theta}-1}{\theta}\right\rvert^2\le 1, $$
and applying Euler's formula one obtains
$$
\frac{(\cos \theta -1)^2}{\theta^2}+\frac{\sin^2\theta}{\theta^2}\le 1.
$$
In particular, 
$$
\left\lvert \frac{\sin \theta}{\theta}\right\rvert\le 1,\qquad\left\lvert\frac{1-\cos\theta}{\theta}\right\rvert\le 1$$ for all $\theta\in \mathbb{R}$. 
A: On some suitable interval around $x$ $$1\geq \cos(x) \geq \cos(x) - x \sin(x)\geq 0.$$ Now integrate on this interval to find for $x\geq 0$ $$x \geq \sin(x) \geq x \cos(x)$$ and the opposite inequalities for $x\leq 0$.
A: Trigonometric identities and other such properties of trig functions are independent of whether angles are measured in radians, degrees, or what have you.  But the squeeze theorem for $\sin x\over x$ and other such relations do depend on $x$ being measured in radians, so the underlying geometry is inescapable.
