
% koblitz 124 notes---Chapter 9

\chapter{\index{ Derivative of Trig Functions} Derivative of Trig Functions }


\section{The Derivative of \boldmath $\mbox{sin}\, x$}

To find the \index{ slope of the tangent line} slope of the tangent line to the curve $y=f(x)=\sin x$, we must
go back to the definition of the derivative as the limit of the difference
quotient: 
\[
\frac{dy}{ dx}
   =\lim_{\Delta x\longrightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}
   =\lim_{\Delta x\longrightarrow 0}\frac{\sin(x+\Delta x)-\sin x}{\Delta x}.
\]
Recall the procedure used to find the derivative
of the much simpler function $x^2$: expand $(x+\Delta x)^2$, simplify
algebraically, and then take the limit as $\Delta x$ gets closer and closer
to 0.  We do essentially the same thing for $\sin x$, but we need the
following three basic facts in order to do this:

\smallskip
\begin{align}
\text{Fact~1.~}&\sin(\alpha+\beta)=\sin\alpha\cos\beta+
\cos\alpha\sin\beta;
\notag\\
\text{Fact~2.~}&\lim_{\theta\longrightarrow 0}\frac{\sin\theta}{\theta}=1;
\notag\\
\text{Fact~3.~}&
\lim_{\theta\longrightarrow 0}\frac{1-\cos\theta}{\theta}=0.
\notag
\end{align}

\smallskip

Fact (1) was proved in the last section (Problem 4 of the homework).  Facts
(2) and (3) can be seen from the diagram below, where $\theta$ is a small
angle in radians.  By the definition of the radian measurement of an angle,
$\theta$ is the distance along the circumference from $A$ to $C$.
Meanwhile, $\sin\theta=\overline{AB}$ and $1-\cos\theta=\overline{BC}$.

%--- figure JA ---------------------
%DIA Diagram on p. 47.

\begin{center}
\begin{psfigure}{4in}{1in}{figJA.ps}
\puttext{-20,-15}{$\theta$}
\puttext{0,15}{$1$}
\puttext{110,35}{$A$}
\puttext{100,-35}{$B$}
\puttext{115,-35}{$C$}
\puttext{-125,-25}{$O$}
\puttext{80,0}{$\sin(\theta)$}
\puttext{120,0}{$\theta$}
\end{psfigure}\\
\end{center}
%-----------------------------------

\noindent
Thus, Fact (2) says that although the line $\overline{AB}$ is smaller than
the arc $AC$, the ratio of the two lengths gets closer and closer to 1 as
$\theta$ gets smaller.  Fact (3) says that the ratio of the line
$\overline{BC}$ to the arc $AC$ gets closer and closer to 0 as $\theta$
gets smaller.  We are now ready to simplify the above difference quotient.

\smallskip
We have:
\begin{align}
\frac{\sin(x+\Delta x)-\sin x}{\Delta x}
&=\frac{\sin x\cos(\Delta x)+\cos x\sin(\Delta x)-\sin x}{\Delta x}
\notag\\
\intertext{(by Fact (1) with $\alpha=x$ and $\beta=\Delta x$)}
&=\frac{\cos x\sin(\Delta x)-\sin x(1-\cos\Delta x)}{\Delta x}
\notag\\
\intertext{(putting $\sin x\cos(\Delta x)$~last on top and factoring
out $-\sin x$)} 
&=\cos x\frac{\sin(\Delta x)}{\Delta x}-\sin x\frac{1-\cos(\Delta x)}{
\Delta x} \notag\\
&\longrightarrow (\cos x)\cdot 1-(\sin x)\cdot 0=\cos x
\qquad\text{as}\qquad \Delta x\longrightarrow 0.
\notag
\end{align}
The last step comes from Facts (2) and (3) with $\theta$ replaced by
$\Delta x$.  We therefore conclude the following basic derivative formula:
\[
\frac{d}{ dx}\sin x=\cos x.
\]



% Example 1
\begin{example}

You know the exact value $\sin(30^\circ)=0.5$.  Find $\sin(31^\circ)$ using
the \index{ tangent line approximation} tangent line approximation, and compare with the exact value (found by
calculator).

\smallskip

In the tangent line approximation we choose $f(x)=\sin x$, $x=30^\circ=\pi/6$
and $\Delta x=1^\circ=\pi/180$.  Here it was important that we changed
$\Delta x$ from degrees to radians.  By the tangent line approximation,
\begin{align*}
   \sin(31^\circ)&=f(\frac{\pi}{ 6}+\frac{\pi}{ 180})
    \approx 0.5+\frac{\pi}{ 180} f'(\frac{\pi}{ 6})\\
           &=0.5+\frac{\pi}{ 180}\cos(\frac{\pi}{ 6})
           =0.5+\frac{\pi}{ 180} \cdot \frac{\sqrt{3}}{2}=0.515115;
\end{align*}
the exact value is $0.515038\dots$.

\end{example}


When we have a derivative formula for a new function, we can combine it
with other rules to find the derivative of a more complicated function that
is built up from the basic function.



% Example 2
\begin{example}

Find the derivative of $\sin^2(\sqrt{x})$. 

\smallskip

The notation $\sin^2$(something) means $\Bigl(\sin($something$)\Bigr)^2$.
To evaluate this derivative, we have to use the chain rule twice: 
\begin{align*}
\frac{d}{ dx}\Bigl(\sin\sqrt{x}\Bigr)^2
 &=
   2\sin\sqrt{x}\>\frac{d}{ dx}
    \sin\sqrt{x}=2\sin\sqrt{x}\cos\sqrt{x}\>\frac{d}{ dx}\sqrt{x}\\
 &=
   2\sin\sqrt{x}\cos\sqrt{x}\>(1/2\sqrt{x})
  =
   \frac{\sin\sqrt{x}\cos\sqrt{x} }{ \sqrt{x}}.
\end{align*}

\end{example}


\section{Other \index{ Trig Derivatives} Trig Derivatives }  All of the other trig functions can be
expressed in terms of the sine, and so their derivatives can easily be
calculated.  The most important ones are:
\begin{align}
\frac{d}{dx}\cos x&=\frac{d}{dx}\sin(x+\frac{\pi}{ 2})=
\cos(x+\frac{\pi}{ 2})=\sin((x+\frac{\pi}{ 2})+\frac{\pi}{ 2})=
\sin(x+\pi)=-\sin x;
\notag\\
\frac{d}{dx}\tan x&=
\frac{d}{dx}\frac{\sin x}{\cos x}
=\frac{\disp \cos x\frac{d}{dx}\sin x -\sin x\frac{d}{dx}\cos x}{\cos^2x}
=\frac{\cos^2x+\sin^2x}{\cos^2x}=
\frac{1}{\cos^2x}=\sec^2x;
\notag\\
\frac{d}{dx}\sec x&=\frac{d}{dx}\Bigl(\cos x\Bigr)^{-1}=
(-1)\Bigl(\cos x\Bigr)^{-2}(-\sin x)=\frac{\sin x}{\cos^2x}=\tan x\sec x.
\notag
\end{align}



