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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 inverse trig functions}


\subsection*{Inverse Trigonometric Functions.}

For someone uncomfortable with trigonometric functions and all their identities,
the thought of having to worry about \textit{inverse}-trigonometric functions
may be a little unnerving. But don't let it be. These functions are simply additions
to our ``zoo'' of available functions. There actually are some inverse-trigonometric
identities, but they are not that useful and we will not worry about them.

So why study inverse trigonometric functions at all? Well, we want to be able
to integrate and differentiate just about every ``normal'' function we can.
I'm not considering inverse trig functions as ``normal'' in that sense. What
we will see today, though, is that the derivatives of these inverse trig functions
are algebraic, sometimes even rational functions, so in order to antidifferentiate
those functions, we need to talk about inverse trig functions.


\subsubsection*{The Inverse sine }

\( \sin ^{-1} \) is the inverse of \( \sin  \). By saying that, we have to
mean that, if you restrict the domain of \( \sin (x) \) to the right interval
so that it is one-to-one, then there is an inverse, which we call \( \sin ^{-1}(x) \).
It is also called \( \arcsin (x) \) or \( \mathrm{Arcsin}(x) \). Maple understands
\( \arcsin (x) \). Technically, \( \sin ^{-1}(x) \) is defined by: 
\[
\sin ^{-1}(x):=y\Leftrightarrow y\in [-\pi /2,\pi /2],\, \, \mathrm{and}\, \sin (y)=x.\]


To find the derivative of \( \sin ^{-1}(x) \), we do as we did before: define
\( y \) to be \( y=\sin ^{-1}(x) \). Then we want \( dy/dx \). By definition,
\( \sin (y)=x \), and so: 
\begin{eqnarray*}
\sin (y) & = & x.\, \, \mathrm{Now},\, \mathrm{differentiate}\, implicitly:\\
\cos (y)\frac{dy}{dx} & = & 1.\, \, \mathrm{Solve}\, \mathrm{for}\, \, \frac{dy}{dx}:\\
\frac{dy}{dx} & = & \frac{1}{\cos (y)}.
\end{eqnarray*}


Here it is a little complicated to solve for \( dy/dx \) in terms of \( x \)
directly.

\vspace{0.3cm}
{\par\centering \includegraphics{right-triangle.eps} \par}
\vspace{0.3cm}

Think of a right triangle, with angle \( y \). If \( \sin y=x \), then we
can choose our scale so that the hypotenuse is 1 (we divide by the hypotenuse
to find \( \sin y \), so we do want to have it be 1 to simplify the arithmetic).
Then, the opposite side to \( y \) has length \( x \), since 
\[
\sin y=opp/hyp=x/1=x.\]
 The formula above has \( \cos x \), however. You see that from the same picture,
\[
\cos (y)=\sqrt{1-x^{2}}.\]
 So, substitute that for \( \cos y \) and you have: 
\[
(\sin ^{-1}(x))'=\frac{1}{\sqrt{1-x^{2}}}.\]



\subsubsection*{The Inverse cosine }

You can think about how you define the inverse cosine, just as for \( \sin ^{-1} \).
The only really new thing is the restriction on \( y \), which can't be the
same as for \( \sin ^{-1} \) since it isn't one-to-one there. 
\[
\cos ^{-1}(x):=y\Leftrightarrow y\in [0,\pi ]\, \, \mathrm{and}\, \, \cos (y)=x.\]
 If you were to go through the calculations again, you would have that: 
\[
(\cos ^{-1}(x))'=\frac{-1}{\sqrt{1-x^{2}}}.\]
 Since the derivatives are almost the same (they differ by just the sign), we
don't worry too much about \( \cos ^{-1}x \). Actually, there is an \textit{inverse-trigonometric
identity} behind this.


\subsubsection*{The Inverse tangent }

This one is perhaps easier. We start out as before:
\[
\tan ^{-1}(x):=y\iff y\in (-\pi /2,\pi /2)and\tan (y)=x.\]


And: 
\begin{eqnarray*}
\tan ^{-1}(x) & := & y\\
x & = & \tan y\\
1 & = & \sec ^{2}y\frac{dy}{dx}\\
\cos ^{2}y & = & \frac{dy}{dx}.
\end{eqnarray*}


Now, look at the right triangle again, this time with the side opposite \( y \)
having length (by scaling) \( x \), and the adjacent side \( 1 \).

\vspace{0.3cm}
{\par\centering \includegraphics{right-triangle2.eps} \par}
\vspace{0.3cm}

That puts the denominator in the equation \( \tan (y)=opp/adj \) to 1, so we
can easily manipulate it. Now consider the formula above. Since \( \cos (y)=adj/hyp \),
\[
\cos (y)=1/\sqrt{1+x^{2}}.\]
 Plugging in to the equation above, and we're done: 
\[
\left( \tan ^{-1}(x)\right) '=\frac{1}{1+x^{2}}.\]



\subsubsection*{The Inverse secant }

Three more to go. Actually, though, after this one the rest of the inverse trig
functions will not give any new integration (antidifferentiation) formulas.
The reason is the same as for the inverse cosine. So, this is the last one.
\[
\sec ^{-1}(x):=y\Leftrightarrow y\in [0,\pi /2)\, \textrm{or}\, y\in [\pi ,3\pi /2),\, \, \mathrm{and}\, \, \sec (y)=x.\]
 You should note that the domain here for \( \sec ^{-1}(x) \) is a bit nonstandard
in the choice of domain.

And: 
\begin{eqnarray*}
\sec ^{-1}(x) & := & y\\
x & = & \sec y\\
1 & = & \sec y\tan y\frac{dy}{dx}\\
\frac{1}{\sec y\tan y} & = & \frac{dy}{dx}.
\end{eqnarray*}


Here it is a little messier. For the right triangle with standard angle \( y \),

\vspace{0.3cm}
{\par\centering \includegraphics{right-triangle3.eps} \par}
\vspace{0.3cm}

scaled so that the adjacent side is 1 (the denominator of \( \sec y \)), and
hypotenuse \( x \), then \( \sec y=x \) and 
\[
\tan y=\sqrt{x^{2}-1},\]
 so 
\[
(\sec ^{-1}(x))'=\frac{1}{x\sqrt{x^{2}-1}}.\]


I must confess. I cheated. I described these pictures, with the right triangles
and all, forgetting that the domain of these functions include some \( x<0 \).
But there the geometry is invalid. You actually only have to see how the signs
of the trig functions change when \( x<0 \) to see how to extend them. But
that is usually unneeded, since the signs won't change. For \( \sec ^{-1}x \),
however, it matters, since 
\[
\tan y=-\sqrt{x^{2}-1}\]
 for \( y\in (\pi /2,\pi ] \). But then also \( \sec y=x \) is negative, so
multiplying by \( -1 \) just adds in absolute value signs. For the other piece
of the domain, the absolute value sign changes nothing. so, really: 
\[
(\sec ^{-1}(x))'=\frac{1}{|x|\sqrt{x^{2}-1}}.\]


\end{document}
