I'm attempting to show that
$$ (ds)^2 = (dr)^2 + (r d\theta)^2 $$
Through a direct geometric proof rather than the easier substitution of $x= r\cos(\theta), y = r\sin(\theta)$ into $(ds)^2 = (dx)^2 + (dy)^2$
So I considered the diagram below:
Which by application of the law of cosines should suggest that
$$ (ds)^2 = (r)^2 + (r+dr)^2 - 2(r)(r+dr)\cos(d\theta) $$
But this is incorrect since after simplification I end up wtih
$$ (ds)^2 = (dr)^2$$
and am entirely missing my $d\theta$ dependence.
Where am I going wrong in setting up the diagram?
$\require{cancel}$
Expand $\cos ({\rm d}\theta)$ up to second order
$$ \cos ({\rm d}\theta) \approx 1 - \frac{({\rm d}\theta)^2}{2!} \tag{1} $$
Which results in
\begin{eqnarray} ({\rm d}s)^2 &=& (r + {\rm d}r)^2 + (r)^2 - 2 r(r + {\rm d}r) \cos ({\rm d}\theta) \\ &=& r^2 + 2r{\rm d}r + r^2 + ({\rm d}r)^2 - 2r^2 \cos ({\rm d}\theta) - 2r{\rm d}r \cos ({\rm d}\theta) \\ &\stackrel{(1)}{=}& \color{blue}{\cancel{2r^2}} + \color{red}{\cancel{2r{\rm d}r}} + ({\rm d}r)^2 - 2r^2\left(\color{blue}{\cancel{1}} - \frac{({\rm d}\theta)^2}{2!} \right) - 2r{\rm d}r\left(\color{red}{\cancel{1}} - \frac{({\rm d}\theta)^2}{2!} \right) \\ &=& ({\rm d}r)^2 + (r{\rm d}\theta)^2 - \color{orange}{r{\rm d}r ({\rm d}\theta)^2} \\ &\approx& ({\rm d}r)^2 + (r{\rm d}\theta)^2 \end{eqnarray}
Where I dropped the orange term because it is third order in the approximation