Single Variable Calculus, Chapter 3, 3.4, Section 3.4, Problem 20

Using the definition of derivative, prove that if $f(x) = \cos x$, then $f'(x) = - \sin x$

$f(x + h) = \cos x$

From the formula of addition of two angles of cosine




$
\begin{equation}
\begin{aligned}

\cos (A + B) =& \cos A \cos B - \sin A \sin B
\\
\\
\cos (x + h) =& \cos x \cos h - \sin x \sin h
\\
\\
f'(x) =& \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}
&& \text{}
\\
\\
f'(x) =& \lim_{h \to 0} \frac{\cos (x + h) - \cos x}{h}
&& \text{}
\\
\\
f'(x) =& \lim_{h \to 0} \frac{\cos x \cos h - \sin x \sin h - \cos x}{h}
&& \text{Group the equation}
\\
\\
f'(x) =& \lim_{h \to 0} \cos x \left( \frac{\cos h - 1}{h} \right) - \sin x \left( \frac{\sin h}{h} \right)
&& \text{Apply the Trigonometric Identity}
\\
\\
f'(x) =& \lim_{h \to 0} \cos x (0) - \sin x (1)
&& \text{Simplify the equation}
\\
\\
f'(x) =& - \sin x

\end{aligned}
\end{equation}
$