Inverse Trigonometric Functions.

Let $ f$ be the function defined on the interval $ \left[ \frac {\pi}{2},\frac {\pi}{2} \right]$ by $ f(x)=\sin x$ . The function $ f$ is continuous and strictly increasing on $ \left[ \frac {\pi}{2},\frac {\pi}{2} \right]$ , and $ f \left( \left[ \frac {\pi}{2},\frac {\pi}{2} \right] \right) = [-1,1]$ . Therefore $ f$ is invertible. Its inverse is called arcsinus and denoted $ \arcsin$ .

\begin{displaymath}\begin{cases}y=\arcsin x  -1 \leq x \leq 1 \end{cases} \Lon...
...in y  - \frac {\pi}{2} \leq y \leq \frac {\pi}{2} \end{cases}\end{displaymath}    

For example: $ \arcsin 0 = 0$ ; $ \arcsin \frac 12 = \frac {\pi}{6}$ .

Figure 12: Graphs of sine and arcsine.
\begin{figure}\centering
\mbox{
\epsfig{file=Sin-arcsin.eps, height=5cm}
}\end{figure}

Remark 5.7.7   For any $ x \in [-1,1]$ , we have $ \sin (\arcsin x)=x$ , but generally $ \arcsin (\sin x) \neq x$ . For example, $ \arcsin (\sin 137 \pi ) = \arcsin 0 =0$ .

Let $ g$ be the function defined on the interval $ [0, \pi ]$ by $ g(x)=\cos x$ . The function $ g$ is continuous and strictly decreasing on $ [0, \pi ]$ , and $ g ([0, \pi ]) = [-1,1]$ . Therefore $ g$ is invertible. Its inverse is called arccosinus and denoted $ \arccos$ .

\begin{displaymath}\begin{cases}y=\arccos x  -1 \leq x \leq 1 \end{cases} \Lon...
...tarrow \begin{cases}x = \cos y  0 \leq y \leq \pi \end{cases}\end{displaymath}    

For example: $ \arccos 0 = \frac {\pi}{2}$ ; $ \arccos \frac 12 = \frac {\pi}{3}$ .

Figure 13: Graphs of cosine and arccosine.

Remark 5.7.8   For any $ x \in [-1,1]$ , we have $ \cos (\arccos x)=x$ , but generally $ \arccos (\cos x) \neq x$ . For example, $ \arccos (\cos 137 \pi ) = \arccos (-1) =\pi$ .

Let $ h$ be the function defined on the interval $ \left( \frac {\pi}{2},\frac {\pi}{2} \right)$ by $ h(x)=\tan x$ . The function $ h$ is continuous and strictly increasing on $ \left( \frac {\pi}{2},\frac {\pi}{2} \right)$ , and $ h \left( \left( \frac {\pi}{2},\frac {\pi}{2} \right) \right)) = \mathbb{R}$ . Therefore $ h$ is invertible. Its inverse is called arctangent and denoted $ \arctan$ .

\begin{displaymath}\begin{cases}y=\arctan x  x \in \mathbb{R} \end{cases} \Lon...
...tan y  -\frac {\pi}{2} \leq y \leq \frac {\pi}{2} \end{cases}\end{displaymath}    

For example: $ \arctan 0 = 0$ ; $ \arctan 1 = \frac {\pi}{4}$ .

Figure 14: Graphs of tan and arctan.
\begin{figure}\centering
\mbox{\epsfig{file=TangentsInverseGraphs.eps, height=5cm}}\end{figure}

Remark 5.7.9   For any $ x \in \mathbb{R}$ , we have $ \tan (\arctan x)=x$ , but generally $ \arctan (\tan x) \neq x$ . For example, $ \arctan (\tan 137 \pi ) = \arctan 0 =0$ .

The inverse trigonometric functions verify many interesting identities; let us see some of them as examples.

Example 5.7.10   We wish to simplify the expressions $ \cos (\arcsin x)$ and $ \sin (\arccos x)$ , for any $ x \in [-1,1]$ .
  1. For any $ x \in [-1,1]$ , the following identity holds:

    $\displaystyle \sin^2 (\arccos x) + \cos^2 (\arccos x) =1$    

    Therefore

    $\displaystyle \sin^2 (\arccos x)=1-\cos^2 (\arccos x)=1-x^2.$    

    It follows that $ \sin (\arccos x)=\pm \sqrt{1-x^2}$ . As $ \arccos x $ belongs to the interval $ ([0, \pi ]$ , we know that $ \sin (\arccos x) \geq 0$ . Finally we have:

    $\displaystyle \sin (\arccos x)=\sqrt{1-x^2}$ (5.1)

  2. For any $ x \in [-1,1]$ , the following identity holds:

    $\displaystyle \sin^2 (\arcsin x) + \cos^2 (\arcsin x) =1$    

    Therefore

    $\displaystyle \cos^2 (\arcsin x)=1-\sin^2 (\arcsin x)=1-x^2.$    

    It follows that $ \cos (\arcsin x)=\pm \sqrt{1-x^2}$ . As $ \arcsin x $ belongs to the interval $ ([-\pi /2, \pi /2 ]$ , we know that $ \cos (\arcsin x) \geq 0$ . Finally we have:

    $\displaystyle \cos (\arcsin x)=\sqrt{1-x^2}$ (5.2)

We will have an important usage of these identities in Chapter 6.
Noah Dana-Picard 2007-12-28