Derivatives of Trigonometric functions
The Zebra Danio and its escape response
UBC Calculus Online Course Notes

Inverse Trigonometric Functions

You probably remember from high school that there are many occassions in which an angle is specified by giving the value of one of the trigonometric functions at this angle. For instance, you might know that

\[ 
\tan(\phi) = \frac{1}{\sqrt{3}} 
 \]

but what you'd really like to know is the value of $  \phi  $ .

This is similar to when you want to know what value of $  x 
 $ has $  x^2 = 3  $ . In that case, you use an inverse function---namely, the square root function---which undoes the operation of the squaring function. In the same way, we will want to build an inverse tangent function which undoes the operation of the tangent function.



The Inverse Tangent Function

Let's begin by thinking about the graph of $  \tan(t) 
 $ . If we want to solve $  \tan(\phi) = 
\frac{1}{\sqrt{3}} $ , we may draw a horizontal line $ 
\frac{1}{\sqrt{3}}  $ units above the $  t  $ axis and choose one of the points which lies on the intersection of the graph and the horizontal line.

From this demonstration, you can see that, as you vary the horizontal line, there are always lots of solutions. However, there is always a unique solution between $  -\frac \pi 2  $ and $  \frac \pi 2  $ . For this reason, we have a well-defined function if define the inverse tangent function by saying

$  y = \tan^{-1}(x)  $ if $  y  $ is the value between $  -\frac \pi 2 $ and $  \frac \pi 2 $ such that $  \tan(y) 
= x $ .

This is similar to the square root function: there are two values which satisfy $  x^2 = 4  $ but we agree, by convention, that the square root of 4 is the positive value.

Here are some famous values of the inverse tangent function:

$  x  $ $  \tan^{-1}(x)  $
$  -\sqrt{3}  $ $  -\frac \pi 3  $
-1 $   -\frac \pi 4  $
$  -1/\sqrt{3}  $ $ -\frac \pi 6  $
0 0
$  1/\sqrt{3}  $ $  \frac \pi 6  $
1 $  \frac \pi 4  $
$  \sqrt{3}  $ $ \frac \pi 3  $

In fact, we can sketch the graph of the inverse tangent as below


The derivative of the inverse tangent

In our usual way, we are able to compute the derivative of the inverse tangent function using implicit differentation. Let's begin:

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

Here we have used the trigonometric identity that $  \sec^2(y) = 1 + \tan^2(y)  $ . This is just another guise for the most famous trigonometric identity: $  \cos^2(y) + 
\sin^2(y) = 1  $ .

Our conclusion is that

\[  \frac{d}{dx} \tan^{-1}(x) = \frac 1{1+x^2} 
 \]


Other inverse trigonometric functions

In the same way, we can build up the inverse sine and cosine functions:

$  y = \sin^{-1}(x)  $ if $  y  $ is the value between $ 
-\frac \pi 2 
 $ and $  \frac \pi 2  $ such that $  \sin(y) = x  $ .

We can understand the graph of the inverse sine function

Also,

$  y = \cos^{-1}(x)  $ if $  y  $ is the value between $  0  $ and $  \pi  $ such that $  \cos(y) = x  $ .

Notice that the domain of both of these functions is restricted: if $  |x| > 1 $ , there is no angle $  y  $ such that $  \sin(y) = x $ . This means that we require $  |x| \leq 1 
 $ in both of these functions.

You should verify for yourself the following relationships:

 \begin{eqnarray*} 
\frac{d}{dx} \sin^{-1}(x)  & = & \frac{1}{\sqrt{1-x^2}} \\ 
\frac{d}{dx} \cos^{-1}(x)  & = & -\frac{1}{\sqrt{1-x^2}} 
\end{eqnarray*}