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\begin{document}


\title{On-line Math 21}


\email{david.johnson@lehigh.edu}


\urladdr{http://www.lehigh.edu/\textasciitilde{}dlj0}

\maketitle

\subsection{Trigonometric functions}


\subsubsection{Derivatives of \protect\( \sin (x)\protect \) and \protect\( \cos (x)\protect \).}

\begin{thm*}
The functions \( \sin x \) and \( \cos x \) have the following derivatives:
\begin{eqnarray*}
(\sin x)' & = & \cos x\\
(\cos x)' & = & -\sin x
\end{eqnarray*}

\end{thm*}
\begin{proof}
For the first one, we really have to work out the details:  \begin{eqnarray*}
(\sin x)' & = & \lim _{h\rightarrow 0}\frac{\sin (x+h)-\sin (x)}{h}\\
 & = & \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}\\
 & = & \sin (x)\lim _{h\rightarrow 0}\frac{\cos (h)-1}{h}+\cos (x)\lim _{h\rightarrow 0}\frac{\sin (h)}{h}\\
 & = & \sin (x)\cdot 0+\cos (x)\cdot 1=\cos x.
\end{eqnarray*}
For the second, you could do exactly that same thing, with the other sum rule
for \( \cos (x+h) \). But, there is another way. Several, in fact. If you can
believe that \( \cos x \) is differentiable, then one way to find the derivative
is to look at the equation: \( \cos ^{2}x+\sin ^{2}x=1, \) that is, \[
(\cos x)^{2}+(\sin x)^{2}=1.\]
 Now, from the examples of the product rule, we figured out that, for any  function
\( f(x) \), \[
((f(x))^{2})'=2f(x)f'(x).\]
 Differentiate both sides of the equation \( (\cos x)^{2}+(\sin x)^{2}=1 \),
using this rule. The derivative of the right hand side is easy, it's 0. Since
the function on the right (1) is the same as the function on the left, then
the derivative of the function on the right is the same as the derivative of
the  function on the left: \begin{eqnarray*}
\left( (\cos x)^{2}+(\sin x)^{2}\right) ' & = & 1'\\
2(\cos x)(\cos x)'+2(\sin x)(\sin x)' & = & 0\\
2(\cos x)(\cos x)'+2(\sin x)(\cos x) & = & 0.
\end{eqnarray*}
 I didn't change the \( (\cos x)' \) term, because we don't yet know what that
is.  But, dividing by 2, putting the second term on the other side of the equation
 (with a -- sign), and dividing out by \( \cos x \), we have: \( (\cos x)'=-\sin x, \)
as claimed. \end{proof}

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