A special case of the angle-sum formulas $\cos (x+y)=\cos x \sin y -\sin x \cos y $ and $ \sin (x+y)=\sin x \cos y + \cos x \sin y$ (which is for all $x,y$), when $x=y\in (0,\pi /2)$ by elementary geometry:
In $\triangle ABC$ with $BA=CA=1 ,$
with $D$ being the mid-point of $ BC$, and $\angle BAD=x,$ we have $BD =\sin x.$ And the Cosine Formula gives $ BD^2=\frac {1}{4}BC^2=$ $=\frac {1}{4}(BA^2+CA^2-2BA\cdot CA \cos \angle BAC)=$ $=\frac {1}{2}(1-\cos 2x).$ Therefore $\sin^2 x =\frac {1-\cos 2x}{2}.$
So when $z=2x\in (0,\pi)$ we have the half-angle formulas $\sin z/2=\sqrt {(1-\cos z)/2}$ and $\cos z/2=\sqrt {1-\sin^2 z/2}=\sqrt {(1+\cos z)/2}.$
The Cosine Formula is a direct consequence of "Pythagoras". And $\sin^2 z/2+\cos^2 z/2=1$ IS "Pythagoras".
Between 21 and 22 centuries ago Archimedes used this to obtain the sin , cos, and tan of $\pi/(3\cdot 2^n)$ for $n=1,2,3,4$ starting with $\cos \pi /6=\sqrt 3\;/2,$ obtaining $3+\frac {10}{71}<48\sin \pi/48<\pi <48 \tan \pi/48<3+\frac {1}{7}.$
The general angle-sum formulas can be proved by elementary geometry. There is a very simple proof in the old classic Trigonometry, by Hobson (likely still available from Dover Publications, still a great source for cheap re-prints).
With the general angle-sum formulas, and knowing $\sin \pi /3$ and $\sin \pi/4 ,$ we can use the half-angle formulas to compute $\sin (\,A\pi/(3\cdot 2^m)+B\pi/(4\cdot 2^n)\,)$ for any $A,B\in \Bbb Z$ and any $m,n \in \Bbb Z^+\cup \{0\},$ which can be as close to any $\sin x$ as we like.
For if $x=x'+d \;$ then $|\sin x-\sin x'|=$ $=|(\sin x'\cos d+\cos x'\sin d)-\sin x'|=$ $=|(\sin x')(-1+\cos d)+\cos x' \sin d|=$ $=|(\sin x')(-2\sin^2 d/2)+\cos x' \sin d| \leq$ $\leq |\sin x'|\cdot d^2/2+|\cos x'|\cdot |\sin d|\;\leq\; d^2/2+|d|.$
Also, one of the Comments recommends the wiki article History Of trigonometry.
Given a function $f$, polynomials $p_1$ and $p_2$, and some $x_0$, we can define "better" as meaning that there is a neighborhood in which it is a better approximation. That is, if there exists $\epsilon$ such that $(|x-x_0|<\epsilon) \rightarrow (|p_1(x)-f(x)|<|p_2(x)-f(x)|)$, then near $x_0$, $p_1$ is a better approximation to $f$ than $p_2$ is.
So Taylor polynomials can, with this definition, be said to be better than any other polynomial with the same order. That is, if $f$ is analytic and $T_n$ is the $n$th order Taylor polynomial of $f$, then for all $n$th order polynomials $g$, there exists a neighborhood around $x_0$ such that $T_n$ is better than $g$.
Best Answer
I am in Israel, so I'm approximately living around Jerusalem -- even if it would take me a two hours drive to get there. For someone coming from another galaxy, I am living approximately on the sun.
Approximations are in their nature inaccurate and relative. When we say that $x$ is an approximation for $y$ we usually mean that in our context they are "pretty close". If you deal with numbers which are much larger than $10^{100^{100}}$ then $1\approx 2$ is true, and both are pretty much zero.
In terms of a sequence if we have a sequence $x_n$ which converges to $x$ then we can say that for a large enough $n$, $x_n\approx x$. It means it's close enough. In this aspect $e\approx\left(1+\frac1n\right)^n$.
In number theory we have a notion of best rational approximation (and $\frac{22}7$ is such approximation for $\pi$) meaning that if we put some constraints on the denominator, this is really the best you can get. It's a very nice theorem that the continued fraction of $r$ can be used to generate best approximations for $r$.