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 $\arcsin$ .

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

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

Figure 12: Graphs of sine and arcsine.

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 $\arccos$ .

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

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 $\arctan$ .

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

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

Figure 14: Graphs of tan and arctan.

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:

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

   
   
   Therefore

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

   
   
   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:

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

   
   

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

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

   
   
   Therefore

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

   
   
   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:

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

   
   

We will have an important usage of these identities in Chapter 6.