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.

Not a complete answer, I am afraid. Approximating the arc $AB$ as a line segment leads to the correct approximation of $\cos(x) \sim 1 - \frac{x^2}{2}$ but leads to the inaccurate result for the corresponding sine as $ \sin(x) \sim x - \frac{x^3}{8}$. Nevertheless, I post my approach, since it has not been covered in the preceding answers.   

In the above diagram, we have $OA=OB=1$ and $\angle BOA = x$
Let the co-ordinates of point $B$ be $(h,k)$
Since $B$ lies on the unit circle, $$h^2 + k^2 =1\tag{1}$$ 
The equation of the line $AB$ is given as $y^*=m(x^*-1)$ and since B lies on the line $AB$, we have $$ k=m(h-1\tag{2})$$ where $m$ is the slope of $AB$.
From $(1)$ and $(2)$, we have 
$$\Rightarrow h^2 + m^2(h-1)^2=1 $$
$$\Rightarrow  h^2(1+m^2) - h(2m^2) + (m^2 -1)=0 $$
which gives two values of $h$, of which one value $h=1$ is for the point $A$
$$\Rightarrow  h=\frac{m^2 -1}{m^2 +1}, \; k =\frac{-2m}{m^2 +1} \tag{3}$$
The length of the line segment $AB$ is assumed to be $x$, hence 
$$\Rightarrow (h-1)^2 + k^2 =x^2 \tag{4}$$
Using $(2)$ and $(4)$, we get
$$ \Rightarrow (1+m^2)(h-1)^2 =x^2$$
$$\Rightarrow x=|h-1|\sqrt{1+m^2}=(1-h)\sqrt{1+m^2}$$
From $(3)$, we have 
$$ \Rightarrow x=\frac{2}{\sqrt{1+m^2}}$$
$$ \Rightarrow 1+m^2 = \frac{4}{x^2}$$
$$ \Rightarrow m = \sqrt{\frac{4-x^2}{x^2}} \tag{5}$$
Now, using $(5)$ and $(3)$, 
$$ h=OD=\frac{4-2x^2}{4}, \; k=DB= \frac{x\sqrt{4-x^2}}{2}$$
Thus $OD = \cos(x) \sim 1-\frac{x^2}{2}$ and $DB= \sin(x) \sim x(1-\frac{x^2}{4})^{1/2}$
This gives $\sin(x) \sim x - \frac{x^3}{8}$, which isn't quite correct.
