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.

Here is a proof that uses the fact, once in $(4)$ and twice in $(5)$, that 
$$
\lim_{x\to0}\frac{\sin(x)}{x}=1\tag{1}
$$
A geometric proof of $(1)$ can be found in this answer.

Answer Using the Duplication Formulas for $\boldsymbol{\sin(x)}$ and $\boldsymbol{\cos(x)}$
Using the duplication formulas for $\sin(x)$ and $\cos(x)$, we get
$$
\begin{align}
\frac{2\sin(x/2)-\sin(x)}{x^3}
&=\frac{2\sin(x/2)-2\sin(x/2)\cos(x/2)}{x^3}\\
&=\frac{2\sin(x/2)}x\,\frac{1-\cos(x/2)}{x^2}\\
&=\frac{2\sin(x/2)}x\,\frac{2\sin^2(x/4)}{x^2}\tag{2}
\end{align}
$$
Substituting $x\mapsto\frac{x}{2^k}$ in $(2)$ gives
$$
\frac{2^{k+1}\sin\left(\frac{x}{2^{k+1}}\right)-2^k\sin\left(\frac{x}{2^k}\right)}{x^3}=4^{-k}\frac{2\sin\left(\frac{x}{2^{k+1}}\right)}{\frac{x}{2^k}}\,\frac{2\sin^2\left(\frac{x}{2^{k+2}}\right)}{\frac{x^2}{4^k}}\tag{3}
$$
Summing in $k$, the left side of $(3)$ telescopes:
$$
\frac{x-\sin(x)}{x^3}=\sum_{k=0}^\infty4^{-k}\frac{2\sin\left(\frac{x}{2^{k+1}}\right)}{\frac{x}{2^k}}\,\frac{2\sin^2\left(\frac{x}{2^{k+2}}\right)}{\frac{x^2}{4^k}}\tag{4}
$$
Since the terms of the right side of $(4)$ are dominated by $\frac18\cdot4^{-k}$, Dominated Convergence yields
$$
\begin{align}
\lim_{x\to0}\frac{x-\sin(x)}{x^3}
&=\sum_{k=0}^\infty4^{-k}\lim_{x\to0}\frac{2\sin\left(\frac{x}{2^{k+1}}\right)}{\frac{x}{2^k}}\,\lim_{x\to0}\frac{2\sin^2\left(\frac{x}{2^{k+2}}\right)}{\frac{x^2}{4^k}}\\
&=\frac43\cdot1\cdot\frac18\\[6pt]
&=\frac16\tag{5}
\end{align}
$$

Answer Inspired by Equation $\boldsymbol{(2)}$ from Ricardo Cruz's Answer
Recall the triplication formula for $\sin(x)$:
$$
\sin(3x)=3\sin(x)-4\sin^3(x)\tag{6}
$$
Substituting $x\mapsto\frac{x}{3^{k+1}}$ gives
$$
\frac{3^{k+1}\sin\left(\frac{x}{3^{k+1}}\right)-3^k\sin\left(\frac{x}{3^k}\right)}{x^3}=4\cdot9^{-k}\frac{\sin^3\left(\frac{x}{3^{k+1}}\right)}{\frac{x^3}{27^k}}\tag{7}
$$
Summing in $k$, the left side of $(7)$ telescopes:
$$
\frac{x-\sin(x)}{x^3}=\sum_{k=0}^\infty4\cdot9^{-k}\frac{\sin^3\left(\frac{x}{3^{k+1}}\right)}{\frac{x^3}{27^k}}\tag{8}
$$
Since the terms of the right side of $(8)$ are dominated by $\frac4{27}\cdot9^{-k}$, Dominated Convergence yields
$$
\begin{align}
\lim_{x\to0}\frac{x-\sin(x)}{x^3}
&=\sum_{k=0}^\infty4\cdot9^{-k}\lim_{x\to0}\frac{\sin^3\left(\frac{x}{3^{k+1}}\right)}{\frac{x^3}{27^k}}\\
&=4\cdot\frac98\cdot\frac1{27}\\[6pt]
&=\frac16\tag{9}
\end{align}
$$
