Statement
The derivative of $ \sin(x) $ is:
$$ \tfrac{d}{dx}\big(\sin(x)\big) = \cos(x) $$
Proof
Take the definition of the derivative:
$$ \tfrac{d}{dx}\big(\sin(x)\big) = \lim_{h \to 0} \left(\frac{\sin(x + h) - \sin(x)}{h}\right) $$
Use the angle sum formula for sine to expand $ \sin(x + h) $:
$$ \tfrac{d}{dx}\big(\sin(x)\big) = \lim_{h \to 0} \left(\frac{\sin(x)\cos(h) + \cos(x)\sin(h) - \sin(x)}{h}\right) $$
Factor out $ \sin(x) $.
$$ \tfrac{d}{dx}\big(\sin(x)\big) = \lim_{h \to 0} \left(\frac{\sin(x)\big(\cos(h) - 1\big) + \cos(x)\sin(h)}{h}\right) $$
Split the limit.
$$ \tfrac{d}{dx}\big(\sin(x)\big) = \lim_{h \to 0} \left(\frac{\sin(x)\big(\cos(h) - 1\big)}{h}\right) + \lim_{h \to 0} \left(\frac{\cos(x)\sin(h)}{h}\right) $$
Factor out all factors without $ h $.
$$ \tfrac{d}{dx}\big(\sin(x)\big) = \sin(x) \cdot \lim_{h \to 0} \left(\frac{\cos(h) - 1}{h}\right) + \cos(x) \cdot \lim_{h \to 0} \left(\frac{\sin(h)}{h}\right) $$
Evaluate these limits. Use that the second limit is equal to $ 1 $ and $ \cos(0) = 1 $.
$$\begin{aligned}\tfrac{d}{dx}\big(\sin(x)\big) &= \sin(x) \cdot \lim_{h \to 0} \left(\frac{1 - 1}{h}\right) + \cos(x) \cdot 1 \newline&= \sin(x) \cdot 0 + \cos(x) \cdot 1 \newline&= \cos(x)\end{aligned}$$