Is the following a strong proof of $\frac{d}{dx} (\sin(x))$

Question

Let $y$ be a value such that $y=\sin(x)$. Then, by first principals of differentiation: $$\frac{\delta y}{\delta x} = \lim_{h\to 0} \frac{\sin(x+h)-\sin(x)}{h}$$ using the difference of angles identity, $\displaystyle \sin(A)-\sin(B)=2\cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)$, $$\frac{\delta y}{\delta x} = \lim_{h\to 0}\frac{2\cos\left(\frac{2x+h}{2}\right)\sin\left(\frac{h}{2}\right)}{h}$$ dividing top and bottom by $2$ yields, $$\frac{\delta y}{\delta x}=\lim_{h\to0}\frac{\cos\left(\frac{2x+h}{2}\right)\sin\left(\frac{h}{2}\right)}{\frac{h}{2}}$$ from here, given that $\displaystyle \lim_{\theta \to 0} \frac{\sin\theta}{\theta} = 1$, when the limit is taken, since there is no reason to take the limit "piece by piece" the entire expression is reduced to $$\frac{dy}{dx} = \cos\left(\frac{2x}{2}\right) = \cos(x)$$ short question is, is this a strong proof? Or is there another, potentially more simple proof for this scenario?