Derivative of cos(x) from the limit definition

Start with $\frac{f(x+h)-f(x)}{h} = \frac{\cos(x+h) - \cos(x)}{h}$. Substitute the angle addition formula: $\cos(x+h) = \cos(x)\cos(h) - \sin(x)\sin(h)$, so the numerator becomes $\cos(x)\cos(h) - \sin(x)\sin(h) - \cos(x)$. Grouping the $\cos(x)$ terms gives $\cos(x)[\cos(h)-1] - \sin(x)\sin(h)$.

Divide by $h$ to write the difference quotient as $\cos(x)\cdot\frac{\cos(h)-1}{h} - \sin(x)\cdot\frac{\sin(h)}{h}$. Taking the limit as $h \to 0$, use the known limits $\lim_{h\to 0}\frac{\cos(h)-1}{h}=0$ and $\lim_{h\to 0}\frac{\sin(h)}{h}=1$. This yields $f'(x) = \cos(x)\cdot 0 - \sin(x)\cdot 1 = -\sin(x)$.