**Proof.**Let f(x) = \cos(x). Then

\begin{align*} f'(x) &= \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h} \\ &= \lim_{h \rightarrow 0} \frac{\cos(x + h) - \cos(x)}{h} \end{align*}

\begin{equation*} \cos(x + y) = \cos(x)\cos(y) - \sin(x)\sin(y) \end{equation*}

\begin{equation*} \cos(x) - \cos(y) = -2\sin(\frac{x + y}{2})\sin(\frac{x - y}{2}) \end{equation*}

\begin{align*} \lim_{h \rightarrow 0} \frac{-2\sin(\frac{2x + h}{2})\sin(\frac{h}{2})}{2} &= \lim_{h \rightarrow 0} -\sin(\frac{2x + h}{2}) \lim_{h \rightarrow 0} \frac{2\sin(\frac{h}{2})}{h} \\ &= \lim_{h \rightarrow 0} -\sin(\frac{2x + h}{2}) \lim_{h \rightarrow 0} \frac{\sin(\frac{h}{2})}{h/2} \end{align*}

\begin{align*} \lim_{h \rightarrow 0} -\sin(\frac{2x + h}{2}) = -\sin(2x/2) = -\sin(x). \end{align*}