Inverse Trigonometric Functions

Learning Outcomes

Evaluate inverse trigonometric functions

The six basic trigonometric functions are periodic, and therefore they are not one-to-one. However, if we restrict the domain of a trigonometric function to an interval where it is one-to-one, we can define its inverse. Consider the sine function. The sine function is one-to-one on an infinite number of intervals, but the standard convention is to restrict the domain to the interval $[-\frac{\pi}{2},\frac{\pi}{2}]$ . By doing so, we define the inverse sine function on the domain $[-1,1]$ such that for any $x$ in the interval $[-1,1]$ , the inverse sine function tells us which angle $\theta$ in the interval $[-\frac{\pi}{2},\frac{\pi}{2}]$ satisfies $\sin \theta =x$ . Similarly, we can restrict the domains of the other trigonometric functions to define inverse trigonometric functions, which are functions that tell us which angle in a certain interval has a specified trigonometric value.

Definition

The inverse sine function, denoted $\sin^{-1}$ or arcsin, and the inverse cosine function, denoted $\cos^{-1}$ or arccos, are defined on the domain $D=\{x|-1 \le x \le 1\}$ as follows:

\begin{matrix} {sin}^{- 1} (x) = y if and only if sin (y) = x and - \frac{π}{2} \leq y \leq \frac{π}{2}; {cos}^{- 1} (x) = y if and only if cos (y) = x and 0 \leq y \leq π \end{matrix}

$\begin{array}{c}\sin^{-1}(x)=y \,\, \text{if and only if} \, \sin (y)=x \, \text{and} \, -\frac{\pi}{2} \le y \le \frac{\pi}{2};\hfill \\ \cos^{-1}(x)=y \,\, \text{if and only if} \, \cos (y)=x \, \text{and} \, 0 \le y \le \pi \hfill \end{array}$

The inverse tangent function, denoted $\tan^{-1}$ or arctan, and inverse cotangent function, denoted $\cot^{-1}$ or arccot, are defined on the domain [latex]D=\{x|-\infty

[latex]\begin{array}{c}\tan^{-1}(x)=y \,\, \text{if and only if} \, \tan (y)=x \, \text{and} \, -\frac{\pi}{2}

The inverse cosecant function, denoted $\csc^{-1}$ or arccsc, and inverse secant function, denoted $\sec^{-1}$ or arcsec, are defined on the domain $D=\{x| \, |x| \ge 1\}$ as follows:

\begin{matrix} {csc}^{- 1} (x) = y if and only if csc (y) = x and - \frac{π}{2} \leq y \leq \frac{π}{2}, y \neq 0; {sec}^{- 1} (x) = y if and only if sec (y) = x and 0 \leq y \leq π, y \neq \frac{π}{2} \end{matrix}

$\begin{array}{c}\csc^{-1}(x)=y \,\, \text{if and only if} \, \csc (y)=x \, \text{and} \, -\frac{\pi}{2} \le y \le \frac{\pi}{2}, \, y\ne 0;\hfill \\ \sec^{-1}(x)=y \,\, \text{if and only if} \, \sec (y)=x \, \text{and} \, 0 \le y \le \pi, \, y \ne \frac{\pi}{2}\hfill \end{array}$

To graph the inverse trigonometric functions, we use the graphs of the trigonometric functions restricted to the domains defined earlier and reflect the graphs about the line $y=x$ (Figure 16).

An image of six graphs. The first graph is of the function “f(x) = sin inverse(x)”, which is an increasing curve function. The function starts at the point (-1, -(pi/2)) and increases until it ends at the point (1, (pi/2)). The x intercept and y intercept are at the origin. The second graph is of the function “f(x) = cos inverse (x)”, which is a decreasing curved function. The function starts at the point (-1, pi) and decreases until it ends at the point (1, 0). The x intercept is at the point (1, 0). The y intercept is at the point (0, (pi/2)). The third graph is of the function f(x) = tan inverse (x)”, which is an increasing curve function. The function starts close to the horizontal line “y = -(pi/2)” and increases until it comes close the “y = (pi/2)”. The function never intersects either of these lines, it always stays between them - they are horizontal asymptotes. The x intercept and y intercept are both at the origin. The fourth graph is of the function “f(x) = cot inverse (x)”, which is a decreasing curved function. The function starts slightly below the horizontal line “y = pi” and decreases until it gets close the x axis. The function never intersects either of these lines, it always stays between them - they are horizontal asymptotes. The fifth graph is of the function “f(x) = csc inverse (x)”, a decreasing curved function. The function starts slightly below the x axis, then decreases until it hits a closed circle point at (-1, -(pi/2)). The function then picks up again at the point (1, (pi/2)), where is begins to decrease and approach the x axis, without ever touching the x axis. There is a horizontal asymptote at the x axis. The sixth graph is of the function “f(x) = sec inverse (x)”, an increasing curved function. The function starts slightly above the horizontal line “y = (pi/2)”, then increases until it hits a closed circle point at (-1, pi). The function then picks up again at the point (1, 0), where is begins to increase and approach the horizontal line “y = (pi/2)”, without ever touching the line. There is a horizontal asymptote at the “y = (pi/2)”.

Figure 16. The graph of each of the inverse trigonometric functions is a reflection about the line $y=x$ of the corresponding restricted trigonometric function.

Interactive

Go to the following site for more comparisons of functions and their inverses.

When evaluating an inverse trigonometric function, the output is an angle. For example, to evaluate $\cos^{-1}(\frac{1}{2})$ , we need to find an angle $\theta$ such that $\cos \theta =\frac{1}{2}$ . Clearly, many angles have this property. However, given the definition of $\cos^{-1}$ , we need the angle $\theta$ that not only solves this equation, but also lies in the interval $[0,\pi]$ . We conclude that $\cos^{-1}(\frac{1}{2})=\frac{\pi}{3}$ . Review the following table of common sine and cosine values using reference angles, if necessary.

Recall: Commonly encountered angles in first quadrant of the unit circle

Angle	0	$\frac{\pi }{6}$ , or 30°	$\frac{\pi }{4}$ , or 45°	$\frac{\pi }{3}$ , or 60°	$\frac{\pi }{2}$ , or 90°
Cosine	1	$\frac{\sqrt{3}}{2}$	$\frac{\sqrt{2}}{2}$	$\frac{1}{2}$	0
Sine	0	$\frac{1}{2}$	$\frac{\sqrt{2}}{2}$	$\frac{\sqrt{3}}{2}$	1

We now consider a composition of a trigonometric function and its inverse. For example, consider the two expressions $\sin (\sin^{-1}(\frac{\sqrt{2}}{2}))$ and $\sin^{-1}(\sin(\pi))$ . For the first one, we simplify as follows:

sin ({sin}^{- 1} (\frac{\sqrt{2}}{2})) = sin (\frac{π}{4}) = \frac{\sqrt{2}}{2}

$\sin (\sin^{-1}(\frac{\sqrt{2}}{2}))= \sin (\frac{\pi}{4})=\frac{\sqrt{2}}{2}$

For the second one, we have

{sin}^{- 1} (sin (π)) = {sin}^{- 1} (0) = 0

$\sin^{-1}( \sin (\pi))=\sin^{-1}(0)=0$

The inverse function is supposed to “undo” the original function, so why isn’t $\sin^{-1}(\sin (\pi))=\pi$ ? Recalling our definition of inverse functions, a function $f$ and its inverse $f^{-1}$ satisfy the conditions $f(f^{-1}(y))=y$ for all $y$ in the domain of $f^{-1}$ and $f^{-1}(f(x))=x$ for all $x$ in the domain of $f$ , so what happened here? The issue is that the inverse sine function, $\sin^{-1}$ , is the inverse of the restricted sine function defined on the domain $[-\frac{\pi}{2},\frac{\pi}{2}]$ . Therefore, for $x$ in the interval $[-\frac{\pi}{2},\frac{\pi}{2}]$ , it is true that $\sin^{-1}(\sin x)=x$ . However, for values of $x$ outside this interval, the equation does not hold, even though $\sin^{-1}(\sin x)$ is defined for all real numbers $x$ .

What about $\sin (\sin^{-1}y)$ ? Does that have a similar issue? The answer is no. Since the domain of $\sin^{-1}$ is the interval $[-1,1]$ , we conclude that $\sin (\sin^{-1}y)=y$ if $-1 \le y \le 1$ and the expression is not defined for other values of $y$ . To summarize,

sin ({sin}^{- 1} y) = y if - 1 \leq y \leq 1

$\sin (\sin^{-1}y)=y \, \text{ if } \, -1 \le y \le 1$

and

{sin}^{- 1} (sin x) = x if - \frac{π}{2} \leq x \leq \frac{π}{2}

$\sin^{-1}( \sin x)=x \, \text{ if } \, -\frac{\pi}{2} \le x \le \frac{\pi}{2}$

Similarly, for the cosine function,

cos ({cos}^{- 1} y) = y if - 1 \leq y \leq 1

$\cos (\cos^{-1}y)=y \, \text{ if } \, -1 \le y \le 1$

and

{cos}^{- 1} (cos x) = x if 0 \leq x \leq π

$\cos^{-1}( \cos x)=x \, \text{ if } 0 \le x \le \pi$

Similar properties hold for the other trigonometric functions and their inverses.

Example: Evaluating Expressions Involving Inverse Trigonometric Functions

Evaluate each of the following expressions.

$\sin^{-1}\left(-\frac{\sqrt{3}}{2}\right)$
$\tan \left(\tan^{-1}\left(-\frac{1}{\sqrt{3}}\right)\right)$
$\cos^{-1}\left( \cos \left(\frac{5\pi}{4}\right)\right)$
$\sin^{-1}\left( \cos \left(\frac{2\pi}{3}\right)\right)$

Show Solution

Evaluating $\sin^{-1}\left(−\frac{\sqrt{3}}{2}\right)$ is equivalent to finding the angle $\theta$ such that $\sin \theta =−\frac{\sqrt{3}}{2}$ and $−\frac{\pi}{2} \le \theta \le \frac{\pi}{2}$ . The angle $\theta =−\frac{\pi}{3}$ satisfies these two conditions. Therefore, $\sin^{-1}\left(−\frac{\sqrt{3}}{2}\right)=−\frac{\pi}{3}$ .
First we use the fact that $\tan^{-1}\left(-\frac{1}{\sqrt{3}}\right)=−\frac{\pi}{6}$ . Then $\tan \left(\frac{\pi}{6}\right)=-\frac{1}{\sqrt{3}}$ . Therefore, $\tan \left(\tan^{-1}\left(-\frac{1}{\sqrt{3}}\right)\right)=-\frac{1}{\sqrt{3}}$ .
To evaluate $\cos^{-1}\left( \cos \left(\frac{5\pi}{4}\right)\right)$ , first use the fact that $\cos \left(\frac{5\pi}{4}\right)=−\frac{\sqrt{2}}{2}$ . Then we need to find the angle $\theta$ such that $\cos (\theta )=−\frac{\sqrt{2}}{2}$ and $0 \le \theta \le \pi$ . Since $\frac{3\pi}{4}$ satisfies both these conditions, we have $\cos \left(\cos^{-1}\left(\frac{5\pi}{4}\right)\right)= \cos \left(\cos^{-1}\left(−\frac{\sqrt{2}}{2}\right)\right)=\frac{3\pi}{4}$ .
Since $\cos \left(\frac{2\pi}{3}\right)=-\frac{1}{2}$ , we need to evaluate $\sin^{-1}\left(-\frac{1}{2}\right)$ . That is, we need to find the angle $\theta$ such that $\sin (\theta )=-\frac{1}{2}$ and $−\frac{\pi}{2} \le \theta \le \frac{\pi}{2}$ . Since $−\frac{\pi}{6}$ satisfies both these conditions, we can conclude that $\sin^{-1}\left( \cos \left(\frac{2\pi}{3}\right)\right)=\sin^{-1}\left(-\frac{1}{2}\right)=−\frac{\pi}{6}$ .

Watch the following video to see the worked solution to Example: Evaluating Expressions Involving Inverse Trigonometric Functions

Closed Captioning and Transcript Information for Video

Activity: The Maximum Value of a Function

In many areas of science, engineering, and mathematics, it is useful to know the maximum value a function can obtain, even if we don’t know its exact value at a given instant. For instance, if we have a function describing the strength of a roof beam, we would want to know the maximum weight the beam can support without breaking. If we have a function that describes the speed of a train, we would want to know its maximum speed before it jumps off the rails. Safe design often depends on knowing maximum values.

This project describes a simple example of a function with a maximum value that depends on two equation coefficients. We will see that maximum values can depend on several factors other than the independent variable $x$ .

Consider the graph in Figure 17 of the function $y= \sin x + \cos x$ . Describe its overall shape. Is it periodic? How do you know?

Figure 17. The graph of $y= \sin x + \cos x$ .

Using a graphing calculator or other graphing device, estimate the $x$ – and $y$ -values of the maximum point for the graph (the first such point where $x>0$ ). It may be helpful to express the $x$ -value as a multiple of $\pi$ .
Now consider other graphs of the form $y=A \sin x + B \cos x$ for various values of $A$ and $B$ . Sketch the graph when $A = 2$ and $B = 1$ , and find the $x$ – and $y$ -values for the maximum point. (Remember to express the $x$ -value as a multiple of $\pi$ , if possible.) Has it moved?
Repeat for $A = 1, \, B = 2$ . Is there any relationship to what you found in part (2)?
Complete the following table, adding a few choices of your own for $A$ and $B$ :

$A$ $B$ $x$ $y$ $A$ $B$ $x$ $y$

0 1 $\sqrt{3}$ 1

1 0 1 $\sqrt{3}$

1 1 12 5

1 2 5 12

2 1

2 2

3 4

4 3
Try to figure out the formula for the $y$ -values.
The formula for the $x$ -values is a little harder. The most helpful points from the table are $(1,1), \, (1,\sqrt{3}), \, (\sqrt{3},1)$ . (Hint: Consider inverse trigonometric functions.)
If you found formulas for parts (5) and (6), show that they work together. That is, substitute the $x$ -value formula you found into $y=A \sin x + B \cos x$ and simplify it to arrive at the $y$ -value formula you found.

Module 1: Functions and Graphs