In his book Visual Complex Analysis, Tristran Needham uses Newton's method of infinitesimals to derive the derivative $\tan(\theta)' = 1 + \tan^2(\theta)$ by using the following picture (where $T = \tan(\theta)$):
If $d\theta$ "goes to" zero, the black triangle becomes similar to the shaded triangle which yields $\frac{dT}{L d\theta} = \frac{L}{1} \Rightarrow \frac{dT}{d\theta} = L^2 = 1 + T^2$.
If I try to apply the same logic to $\sin(\theta)$ I don't get the correct result. My thinking: use the unit hypothenuse instead of the unit base. Then $L \rightarrow 1$, $1 \rightarrow C = \cos(\theta)$, $T \rightarrow S = \sin(\theta)$, $dT \rightarrow dS$ and $L \cdot d\theta \rightarrow d\theta$. But this gives the wrong result $\frac{dS}{d\theta} = \frac{1}{C}$.
Where's my error here? Please don't simply respond by objecting that the method is not rigorous.
It is the wrong picture for $\sin\theta.$
The equivalent picture would be something like...