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

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

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

\begin{align*} &\lim_{h \rightarrow 0} \frac{\sin(x)\cos(h) + \cos(x)\sin(h) - \sin(x)}{h} = \\ &\lim_{h \rightarrow 0} \frac{\sin(x)(\cos(h) - 1) + \cos(x)\sin(h)}{h} = \\ &\lim_{h \rightarrow 0} \frac{\sin(x)(\cos(h) - 1)}{h} + \lim_{h \rightarrow 0} \frac{\cos(x)\sin(h)}{h} = \\ &\lim_{h \rightarrow 0} \sin(x) \lim_{h \rightarrow 0}\frac{(\cos(h) - 1)}{h} + \lim_{h \rightarrow 0} \cos(x) \lim_{h \rightarrow 0}\frac{\sin(h)}{h} \end{align*}

\begin{equation*} \sin(x) \cdot 0 + \cos(x) \cdot 1 = \cos(x). \end{equation*}

\begin{equation*} f'(x) = \cos(x) \end{equation*}