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# Inverses of Trigonometric Functions

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Your instructor gives you a trigonometric function, $f(x) = 3\sin (x) + 5$ , and asks you to find the inverse. You are all set to start manipulating the equation, when you realize that you don't know just how to do this. Your instructor suggests that you try finding the inverse through graphing instead.

Are you able to do this?

Keep reading, and by the end of this Concept, you'll be able to find the inverse of trig function and others through graphing instead of algebra.

### Guidance

In other Concepts, two different ways to find the inverse of a function were discussed: graphing and algebra. However, when finding the inverse of trig functions, it is easy to find the inverse of a trig function through graphing. Consider the graph of a sine function shown here:

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.

The inverse of $f(x)$ is 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 Concept 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.

#### Example A

Establish an alternative domain that makes $y=sin(x)$ a one to one function.

Solution: Any number of possible solutions can be given, but the important point is that the function must pass the "horizontal line test" and the "vertical line test". This means that a horizontal line drawn through the graph will intersect the function in only one place, and a vertical line drawn through the graph will intersect the function in only one place.

For the sine curve, this means that the function can't "turn over" or "go in the other direction", since then it couldn't pass the horizontal line test. So any part of the function that starts at the bottom of the "y" values and stops at the top of the "y" values will work. (Any value that starts at the top of the "y" values and stops at the bottom of the "y" values will work as well.

In this example, you can see that the function starts at $\frac{\pi}{2}$ and stops at $\frac{3\pi}{2}$ .

#### Example B

Find the range of the function given in Example A

Solution: You can see that the function still has the same "y" range of values, since the function $y = \sin x$ moves up and down between -1 and 1. Therefore, the range is $-1 \le y \le 1$ .

#### Example C

Find the domain and range of the inverse of the function given in Example A.

Solution: Since the domain of the inverse function is the range of the original function and the range of the inverse function is the domain of the original function, you only have to take the "x" and "y" values of the original function and reverse them to get the domain and range range of the inverse function.

Therefore, the domain of $y = \sin^{-1} x$ as described in Example A is $-1 \le x \le 1$ and the range is $\frac{\pi}{2} \le y \le \frac{3\pi}{2}$ .

### Vocabulary

Inverse Function: An inverse function is a function that undoes another function.

### Guided Practice

1. 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.

2. 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.

3. Graph $y = 2\sin^{-1}(2x)$

Solutions:

1.

$y = \frac{1}{2} \cos^{-1} (3x+1)$ 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.

2.

$y = 3-\tan^{-1} (x-2)$ 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 $-\tan^{-1}$ .

3.

### Concept Problem Solution

To find the inverse of this function through graphing, first restrict the domain of the function so that it is one to one. A graph of $f(x) = 3\sin (x) + 5$ , restricted so that the domain is $-\frac{\pi}{2}$ to $\frac{\pi}{2}$ looks like this:

If you apply the inverse reflection principle, you can see that the inverse of this function looks like this:

### Practice

1. Why does the domain of a trigonometric function have to be restricted in order to find its inverse function?
2. If the domain of $f(x)=\cos(x)$ is $[0,\pi]$ , what is the domain and range of $f^{-1}(x)$ ?.
3. If the domain of $g(x)=\sin(x)$ is $[-\frac{\pi}{2},\frac{\pi}{2}]$ , what is the domain and range of $g^{-1}(x)$ ?.
4. Establish an alternative domain that makes $y=\cos(x)$ a function.
5. What is the domain and range of the inverse of the function from the previous problem.
6. Establish an alternative domain that makes $y=\tan(x)$ a function.
7. What is the domain and range of the inverse of the function from the previous problem.

Sketch a graph of each function. Use the domains presented in this concept.

1. $y=2\sin^{-1}(3x-1)$
2. $y=-3+\cos^{-1}(2x)$
3. $y=1+2\tan^{-1}(x+2)$
4. $y=4\sin^{-1}(x-4)$
5. $y=2+\cos^{-1}(x+3)$
6. $y=1+\cos^{-1}(2x-3)$
7. $y=-3+\tan^{-1}(3x+1)$
8. $y=-1+2\sin^{-1}(x+5)$