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