Applying the basic definition of the derivative and the previously mentioned two formulas give:

$$\begin{align*} \frac{d}{dx} \sin{x} &=\ \lim_{\Delta x\to 0}\frac{\sin(x+\Delta x) -\sin{x}}{\Delta x}\\ &=\ \lim_{\Delta x\to 0}\frac{2\sin\frac{\Delta x}{2}\cos\left(x+\frac{\Delta x}{2}\right)}{\Delta x}\\ &=\ \lim_{\Delta x\to 0}\frac{\sin\frac{\Delta x}{2}}{\frac{\Delta x}{2}}\cos\left(x+\frac{\Delta x}{2}\right)\\ &=\ \cos{x} \end{align*}$$

$$\therefore\quad\boxed{\frac{d}{dx} \sin{x}=\cos{x}}$$