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# 4.2: Graphing Inverse Trigonometric Functions

Created by: CK-12

## Learning Objectives

• Understand the meaning of restricted domain as it applies to the inverses of the six trigonometric functions.
• Apply the domain, range and quadrants of the six inverse trigonometric functions to evaluate expressions.

## Finding the Inverse by Mapping

Determining an inverse function algebraically can be both involved and difficult, so it is useful to know how to map $f$ to $f^{-1}$. The graph of $f$ can be used to produce the graph of $f^{-1}$ by applying the inverse reflection principle:

The points $(a, b)$ and $(b, a)$ in the coordinate plane are symmetric with respect to the line $y = x$.

The points $(a, b)$ and $(b, a)$ are reflections of each other across the line $y = x$.

Example 1: Find the inverse of $f(x) = \frac{1}{x-5}$ mapping.

Solution: From the last section, we know that the inverse of this function is $y = \frac{5x+1}{x}$. To find the inverse by mapping, pick several points on $f(x)$, reflect them using the reflection principle and plot. Note: The coordinates of some of the points are rounded.

A: (4, -1)

B: (4.8, -5)

C: (2, -0.3)

D: (0, -0.2)

E: (5.3, 3.3)

F: (6, 1)

G: (8, 0.3)

H: (11, 0.2)

Now, take these eight points, switch the $x$ and $y$ and plot $(y, x)$. Connect them to make the inverse function.

$A^{-1}: \ (-1, 4)$

$B^{-1}: \ (-5, 4.8)$

$C^{-1}: \ (-0.3, 2)$

$D^{-1}: \ (-0.2, 0)$

$E^{-1}: \ (3.3, 5.3)$

$F^{-1}: \ (1, 6)$

$G^{-1}: \ (0.3, 8)$

$H^{-1}: \ (0.2, 11)$

Not all functions have inverses that are one-to-one. However, the inverse can be modified to a one-to-one function if a “restricted domain” is applied to the inverse function.

Example 2: Find the inverse of $f(x) = x^2-4$.

Solution: Let’s use the graphic approach for this one. The function is graphed in blue and its inverse is red.

Clearly, the inverse relation is not a function because it does not pass the vertical line test. This is because all parabolas fail the horizontal line test. To “make” the inverse a function, we restrict the domain of the original function. For parabolas, this is fairly simple. To find the inverse of this function algebraically, we get $f^{-1}(x) =\sqrt{x+4}$. Technically, however, the inverse is $\pm \sqrt{x+4}$ because the square root of any number could be positive or negative. So, the inverse of $f(x) = x^2-4$ is both parts of the square root equation, $\sqrt{x+4}$ and $-\sqrt{x+4}$. $\sqrt{x+4}$ will yield the top portion of the horizontal parabola and $-\sqrt{x+4}$ will yield the bottom half. Be careful, because if you just graph $f^{-1}(x) = \sqrt{x+4}$ in your graphing calculator, it will only graph the top portion of the inverse.

This technique of sectioning the inverse is applied to finding the inverse of trigonometric functions because it is periodic.

## Finding the Inverse of the Trigonometric Functions

In order to consider the inverse of this function, we need to restrict the domain so that we have a section of the graph that is one-to-one. If the domain of $f$ is restricted to $-\frac{\pi}{2} \le x \le \frac{\pi}{2}$ a new function $f(x) = \sin x, -\frac{\pi}{2} \le x \le \frac{\pi}{2}$. is defined. This new function is one-to-one and takes on all the values that the function $f(x) = \sin x$ takes on. Since the restricted domain is smaller, $f(x) = \sin x, -\frac{\pi}{2} \le x \le \frac{\pi}{2}$ takes on all values once and only once.

In the previous lesson the inverse of $f(x)$ was represented by the symbol $f^{-1}(x)$, and $y = f^{-1}(x) \Leftrightarrow f(y) = x$. The inverse of $\sin x, -\frac{\pi}{2} \le x \le \frac{\pi}{2}$ will be written as $\sin^{-1} x$. or $\arcsin x$.

$\begin{Bmatrix}y = \sin^{-1} x\\\quad or\\y = \arcsin x\end{Bmatrix} \Leftrightarrow \sin y = x$

In this lesson we will use both $\sin^{-1} x$ and $\arcsin x$ and both are read as “the inverse sine of $x$” or “the number between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ whose sine is $x$.”

The graph of $y = \sin^{-1} x$ is obtained by applying the inverse reflection principle and reflecting the graph of $y=\sin x, -\frac{\pi}{2} \le x \le \frac{\pi}{2}$ in the line $y = x$. The domain of $y = \sin x$ becomes the range of $y = \sin^{-1} x$, and hence the range of $y = \sin x$ becomes the domain of $y = \sin^{-1} x$.

Another way to view these graphs is to construct them on separate grids. If the domain of $y = \sin x$ is restricted to the interval $\left [ -\frac{\pi}{2}, \frac{\pi}{2} \right ]$, the result is a restricted one-to one function. The inverse sine function $y = \sin^{-1} x$ is the inverse of the restricted section of the sine function.

The domain of $y = \sin x$ is $\left [ -\frac{\pi}{2}, \frac{\pi}{2} \right ]$ and the range is [-1, 1].

The restriction of $y = \sin x$ is a one-to-one function and it has an inverse that is shown below.

The domain of $y = \sin^{-1}$ is [-1, 1] and the range is $\left [ -\frac{\pi}{2}, \frac{\pi}{2} \right ]$.

The inverse functions for cosine and tangent are defined by following the same process as was applied for the inverse sine function. However, in order to create one-to-one functions, different intervals are used. The cosine function is restricted to the interval $0 \le x \le \pi$ and the new function becomes $y = \cos x, 0 \le x \le \pi$. The inverse reflection principle is then applied to this graph as it is reflected in the line $y = x$ The result is the graph of $y = \cos^{-1} x$ (also expressed as $y = \arccos x$).

Again, construct these graphs on separate grids to determine the domain and range. If the domain of $y = \cos x$ is restricted to the interval $[0, \pi]$, the result is a restricted one-to one function. The inverse cosine function $y = \cos^{-1} x$ is the inverse of the restricted section of the cosine function.

The domain of $y = \cos x$ is $[0, \pi]$ and the range is [-1, 1].

The restriction of $y = \cos x$ is a one-to-one function and it has an inverse that is shown below.

The statements $y = \cos x$ and $x = \cos y$ are equivalent for $y-$values in the restricted domain $[0, \pi]$ and $x-$values between -1 and 1.

The domain of $y = \cos^{-1} x$ is [-1, 1] and the range is $[0, \pi]$.

The tangent function is restricted to the interval $-\frac{\pi}{2} < x < \frac{\pi}{2}$ and the new function becomes $y = \tan x, -\frac{\pi}{2} < x < \frac{\pi}{2}$. The inverse reflection principle is then applied to this graph as it is reflected in the line $y = x$. The result is the graph of $y = \tan^{-1} x$ (also expressed as $y = \arctan x$).

Graphing the two functions separately will help us to determine the domain and range. If the domain of $y = \tan x$ is restricted to the interval $\left [ -\frac{\pi}{2}, \frac{\pi}{2} \right ]$, the result is a restricted one-to one function. The inverse tangent function $y = \tan^{-1} x$ is the inverse of the restricted section of the tangent function.

The domain of $y = \tan x$ is $\left [ -\frac{\pi}{2}, \frac{\pi}{2} \right ]$ and the range is $[-\infty, \infty]$.

The restriction of $y = \tan x$ is a one-to-one function and it has an inverse that is shown below.

The statements $y = \tan x$ and $x = \tan y$ are equivalent for $y-$values in the restricted domain $\left [ -\frac{\pi}{2}, \frac{\pi}{2} \right ]$ and $x-$values between -4 and +4.

The domain of $y = \tan^{-1} x$ is $[-\infty, \infty]$ and the range is $\left [ -\frac{\pi}{2}, \frac{\pi}{2} \right ]$.

The above information can be readily used to evaluate inverse trigonometric functions without the use of a calculator. These calculations are done by applying the restricted domain functions to the unit circle. To summarize:

Restricted Domain Function Inverse Trigonometric Function Domain Range Quadrants
$y = \sin x$ $\left [ -\frac{\pi}{2}, \frac{\pi}{2} \right ]$ [-1, 1] 1 AND 4

$y = \arcsin x$

$y = \sin^{-1} x$

[-1, 1] $\left [ -\frac{\pi}{2}, \frac{\pi}{2} \right ]$
$y = \cos x$ $[0, \pi]$ [-1, 1] 1 AND 2

$y = \arccos x$

$y = \cos^{-1} x$

[-1, 1] $[0, \pi]$
$y = \tan x$ $\left ( -\frac{\pi}{2}, \frac{\pi}{2} \right )$ $(-\infty, \infty)$ 1 AND 4

$y = \arctan x$

$y = \tan^{-1}x$

$(-\infty, \infty)$ $\left ( -\frac{\pi}{2}, \frac{\pi}{2} \right )$

Now that the three trigonometric functions and their inverses have been summarized, let’s take a look at the graphs of these inverse trigonometric functions.

## Points to Consider

• What are the restricted domains for the inverse relations of the trigonometric functions?
• Can the values of the special angles of the unit circle be applied to the inverse trigonometric functions?

## Review Questions

Study each of the following graphs and answer these questions:

(a) Is the graphed relation a function?

(b) Does the relation have an inverse that is a function?

Find the inverse of the following functions using the mapping principle.

1. $f(x) = x^2 + 2x -15$
2. $y = 1 + 2\sin x$
3. Sketch a graph of $y = \frac{1}{2} \cos^{-1} (3x+1)$. Sketch $y = \cos^{-1} x$ on the same set of axes and compare how the two differ.
4. Sketch a graph of $y = 3-\tan^{-1} (x-2)$. Sketch $y = \tan^{-1} x$ on the same set of axes and compare how the two differ.
5. Graph $y = 2\sin^{-1}(2x)$
6. Graph $y =4 + \cos^{-1} \frac{1}{3}x$
7. Remember that sine and cosine are out of phase with each other, $\sin x = \cos \left ( x-\frac{\pi}{2} \right )$. Find the inverse of $y = \cos \left ( x-\frac{\pi}{2} \right )$. Is the inverse of $y = \cos \left (x-\frac{\pi}{2} \right )$ the same as $y = \sin^{-1}x$? Why or why not?

1. The graph represents a one-to-one function. It passes both a vertical and a horizontal line test. The inverse would be a function.
2. The graph represents a function, but is not one-to-one because it does not pass the horizontal line test. Therefore, it does not have an inverse that is a function.
3. The graph does not represent a one-to-one function. It fails a vertical line test. However, its inverse would be a function.
4. By selecting 4-5 points and switching the $x$ and $y$ values, you will get the red graph below.
5. By selecting 4-5 points and switching the $x$ and $y$ values, you will get the red graph below.
6. $y = \frac{1}{2} \cos^{-1} (3x+1)$ is in blue and $y =\cos^{-1}(x)$ is in red. Notice that $y = \frac{1}{2} \cos^{-1}(3x+1)$ has half the amplitude and is shifted over -1. The 3 seems to narrow the graph.
7. $y = 3-\tan^{-1} (x-2)$ is in blue and $y = \tan^{-1} x$ is in red. $y = 3-\tan^{-1} (x-2)$ is shifted up 3 and to the right 2 (as indicated by point $C$, the “center”) and is flipped because of the $-\tan^{-1}$.
8. $y & = \cos \left( x-\frac{\pi}{2} \right )\\x & = \cos \left( y-\frac{\pi}{2} \right )\\\cos^{-1} x & = y-\frac{\pi}{2}\\\frac{\pi}{2} + \cos^{-1} x & = y$ $\sin^{-1} x \ne \frac{\pi}{2} + \cos^{-1} x$, graphing the two equations will illustrate that the two are not the same. This is because of the restricted domain on the inverses. Since the functions are periodic, there is a phase shift of cosine that, when the inverse is found, is equal to sine inverse.

Feb 23, 2012

Dec 13, 2013