Is there a geometric method to show $\sin x \sim x- \frac{x^3}{6}$ I found a geometric method to show $$\text{when}\; x\to 0 \space , \space \cos x\sim 1-\frac{x^2}{2}$$ like below :

Suppose $R_{circle}=1 \to \overline{AB} =2$ in $\Delta AMB$ we have $$\overline{AM}^2=\overline{AB}\cdot\overline{AH} \tag{*}$$and 
$$\overline{AH}=\overline{OA}-\overline{OH}=1-\cos x$$ when $x$ is very small
w.r.t. * we can conclude $$x^2 \approx 2(1-\cos x) \\\to \cos x \sim 1- \frac{x^2}{2}$$
Now I have two question :
$\bf{1}:$ Is there other idea(s) to prove (except Taylor series) $x\to 0  , \space \cos x\sim 1-\frac{x^2}{2}\\$    
$\bf{2}:$ How can show $\sin x \sim x- \frac{x^3}{6}$ with a geometric concept ?
Thanks in advance.

Let assume that for small $x$, $$\sin x = x  + a x^2 + bx^3 + O(x^4)\tag{1}$$

In this trigonometric circle we can compute $BC$ in two ways:
First, considering the $ABC$ triangle:
$$BC= \sqrt{AB^2+AC^2}=\sqrt{\sin^2 x+\left(1-\sqrt{1-\sin^2 x}\right)^2}=\\= x+ax^2+\left(b+\frac18\right)x^3+O(x^4) \tag{2}$$ 
[*]
Second, bisecting the angle $x$: 
$$BC = 2 \sin \frac{x}{2}=x+\frac{a}{2} x^2 + \frac{b}{4}x^3 + O(x^4)\tag{3}$$ 
Equating (2) and (3) we get $a=0$ and $b=-\frac{1}{6}$, that is: $\sin x = x - \frac16 x^3 + O(x^4)$
I'm not sure if this a geometric proof, though... 

[*] To prove (2) we only need to know the expansion, for $t\to 0$: $$\sqrt{1+t}=1+\frac{t}{2}+\cdots \tag{4}$$
(here and in the following expansions the dots means a remainder $O(t^{k+1})$ where $k$ is the highest exponent in the preceeding terms).
Then 
$$\sqrt{1-t^2}=1-\frac{t^2}{2}+\cdots \\
(1-\sqrt{1-t^2})^2=\frac{t^4}{4}+\cdots\\
t^2+(1-\sqrt{1-t^2})^2=t^2(1+\frac{t^2}{4}+\dots)\\
\sqrt{t^2+(1-\sqrt{1-t^2})^2}=t (1+\frac{t^2}{8}+\cdots)
=t +\frac{t^3}{8}+\cdots
$$
Letting $t=x+ax^2+bx^3+O(x^4)$ that gives
$$ x+ax^2+bx^3+\frac{x^3}{8}+O(x^4)$$
With some more work (I've verified it with Maxima), doing higher order expansions we can get more terms ($\sin x = x -\frac16 x^3 + \frac{1}{120}x^5 +\cdots$).
