Inverse Functions: Learn It 4

Lumen Learning

Inverse Functions: Learn It 4

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.

inverse trigonometric functions

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 $D=\{x|-\infty < x < \infty \}$ as follows:

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

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

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}

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

Graphs of Inverse Trigonometric Functions

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.

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 $\cos^{-1}(\frac{1}{2})=\frac{\pi}{3}$ .

Knowing the common values of sine and cosine for key angles can simplify the process of evaluating inverse trigonometric functions, making it a smoother and quicker task.

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	$0$	$\frac{1}{2}$	$\frac{\sqrt{2}}{2}$	$\frac{\sqrt{3}}{2}$	$1$

Compositions of Inverse Trigonometric Functions

When working with inverse trigonometric functions, it’s crucial to understand how composition works. For instance, when we compose $\sin$ and its inverse $\sin^{-1}$ , such as in $\sin{(\sin^{-1}(y))}$ , we’re essentially undoing the sine function, which should give us the original input, $y$ . However, this holds true only if $y$ falls within the range of $\sin^{-1}$ , which is $[1,1]$ . So $\sin{(\sin^{-1}(y))}=y$ for $-1 \le y \le 1$ . Conversely, when we consider $\sin^{-1}({\sin(x)})$ , the result is $x$ only if $x$ is within the restricted domain of $\sin^{-1}$ , which is $[-\frac{\pi}{2},\frac{\pi}{2}]$ . The same principle applies to $\cos$ 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:

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

For the second one, we have

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

This is because the value of $π$ falls outside the restricted range of the inverse sine function, which is $[-\frac{\pi}{2},\frac{\pi}{2}]$ .

To summarize,

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

and

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

Similarly, for the cosine function,

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

and

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

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

How to: Composing Inverse Trig Functions

Check the Range: Ensure the value inside the inverse function is within the inverse function’s range. For $\sin^{-1}$ , the value must be between $-1$ and $1$ .
Apply the Function: Perform the composition by applying the inverse function first.
Reverse the Process: Apply the original trigonometric function to the result.
Restrict the Range: Remember that for $\sin{(\sin^{-1}({x}))}$ and $\cos{(\cos^{-1}({x}))}$ , the original $x$ is retrieved only if it’s in the principal range of the inverse function.
Verify: Plug the result back into the original function to confirm the outcome.

Evaluate each of the following expressions.

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

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 this example.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this video using this link (opens in new window).

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

$B$ :

$A$	$B$	$A$	$B$
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.