Review the concept of an inverse function.
Define inverse trigonometric functions.
Learn how to simplify expressions involving both trigonometric and inverse trigonometric functions.
The inverse to a given function reverses the action of this function. In other words, the inverse function undoes whatever the function does. In this section, we recall the formal definition of an inverse function, state the necessary conditions for an inverse function to exist, and use this to define inverse trigonometric functions. We then discuss how to simlify expressions involving both trigonometric and inverse trigonometric functions.
Given a function with domain and range , its inverse function (if it exists) is the function with domain and range such that
It follows that, for a function and its inverse ,
These are commonly referred to as cancellation equations .
Note that the -1 is not used as an exponent here, that is, .
The figure below shows the relationship between the domain and range of and the domain and range of .
Recall that a function has exactly one output for each input. Therefore, to define an inverse function, we need to map each input to exactly one output. For example, let’s try to find the inverse function for . Solving the equation for we obtain . This formula does not describe as a function of , because there are two values of corresponding to every . So the problem with trying to find an inverse function for is that two inputs can be sent to the same output. In contrast, there is no such issue with because, for this function, each input is sent to a different output. A function that sends each input to a different output is called a one-to-one function.
We say that is a one-to-one function if whenever .
One way to determine whether a function is one-to-one is by looking at its graph. If a function is one-to-one, then no two inputs can be sent to the same output. Therefore, if we draw a horizontal line anywhere in the -plane it cannot intersect the graph more than once. This establishes the horizontal line test to determine whether a function is one-to-one.
Horizontal Line Test
A function is one-to-one if and only if every horizontal line intersects the graph of no more than once.
Do not confuse the horizontal line test with the vertical line test that determines whether a given curve is the graph of a function.
Determining Whether a Function Is One-to-One
For each of the following functions, use the horizontal line test to determine whether it is one-to-one.
Since the horizontal line for any integer intersects the graph more than once, this function is not one-to-one. (The function represented by this graph is called the floor function, it is denoted by , and is equal to the biggest integer that does not exceed .)
Since every horizontal line intersects the graph at most once, this function is one-to-one.
Is the function graphed in the following image one-to-one?
Graphing Inverse Functions
We now discuss the relationship between the graph of a function and the graph of its inverse. Consider the graph of shown below and a point on the graph. Since , we have that . Therefore, when we graph , the point is on the graph. As a result, the graph of is a reflection of the graph of about the line .
Sketching Graphs of Inverse Functions
For the graph of in the following image, sketch a graph of by sketching the line and using symmetry. Identify the domain and range of .
Reflect the graph about the line . The domain of coincides with the range of , which is , and the range of coincides with the domain of , which is . In fact, , , that can be verified by solving for and interchanging the variables and in the formula obtained.
Sketch the graph of and the graph of its inverse using the symmetry property of inverse functions.
As we have seen, does not have an inverse function because it is not one-to-one. However, we can choose a subset of the domain of such that the function still has the same range over this subset and is one-to-one on it. Such subset is called a restricted domain. By restricting the domain of , we can define a new function such that the domain of is the restricted domain of and for all in the domain of . Then we can define an inverse function for on that domain. For example, since is one-to-one on the interval and the function still takes all non-negative values (which is the range of ) when changes in this interval, we can define a new function such that the domain of is and for all in its domain. Since is a one-to-one function, it has an inverse function, given by the formula . On the other hand, the function is also one-to-one and takes all non-negative values on the domain . Therefore, we could also define a new function such that the domain of is and for all in the domain of . Then is a one-to-one function and must also have an inverse. Its inverse is given by the formula .
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. Let us start with the sine function. The sine function is one-to-one and takes all values from its range on an infinite number of intervals, but the standard convention is to restrict the domain to the interval . By doing so, we define the inverse sine function on the domain such that for any in the interval , the inverse sine function tells us which angle in the interval satisfies . 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.
The inverse sine function, denoted or , and the inverse cosine function, denoted or are defined on the domain as follows:
The inverse tangent function, denoted or , and inverse cotangent function, denoted or , are defined on the domain as follows:
The inverse cosecant function, denoted or , and inverse secant function, denoted or , are defined on the domain as follows:
Finding Domains of Functions Involving Inverse Trigonometric Functions
Determine the domain of the function .
Since the domain of inverse sine is , we have that must satisfy . We solve each of the inequalities separately.
Zeroes of the right hand side are -2 and -1, and they divide the -axis into subintervals , , and . Using sample points or any other appropriate method, we determine that the inequality is satisfied for , where the endpoints of the intervals are included since the inequality is not strict. Likewise,
Since must satisfy both inequalities, our domain is going to be the intersection of and , which is .
Find the domain of the function .
To graph the inverse trigonometric functions, we use the graphs of the trigonometric functions restricted to the domains specified earlier and reflect the graphs about the line to obtain the curves shown below.
When evaluating an inverse trigonometric function, the output is an angle. For example, to evaluate , we need to find an angle such that . Clearly, many angles have this property. However, given the definition of , we need the angle that not only solves this equation, but also lies in the interval We conclude that .
We now consider a composition of a trigonometric function and its inverse. For example, consider the two expressions and . For the first one, we simplify as follows: .
For the second one, we have .
The inverse function is supposed to “undo” the original function, so why isn’t ? Recalling our definition of inverse functions, a function and its inverse satisfy the conditions for all in the domain of and for all in the domain of , so what happened here? The issue is that the inverse sine function, is the inverse of the restricted sine function defined on the domain . Therefore, for in the interval , it is true that . However, for values of outside this interval, the equation does not hold, even though is defined for all real numbers .
What about ? Does that have a similar issue? The answer is no . Since the domain of is the interval , we conclude that if and the expression is not defined for other values of . To summarize,
Likewise, for the cosine function, and .
Similar properties hold for the other trigonometric functions and their inverses.
Evaluating Expressions Involving Inverse Trigonometric Functions
Evaluate each of the following expressions.
Evaluating is equivalent to finding the angle such that and . Using the unit circle, we see that the angle satisfies these two conditions. Therefore, .
Since is not equal to a cotangent of a standard angle, we cannot evaluate without a calculator. However, since the functions and are mutually inverse and is in the domain of which is , we conclude that .
Let . Then the angle we are looking for should satisfy and . We have that
Since sine function is -periodic, . However, , which means that does not belong to the interval , and hence it is not our answer yet. To “bring” the angle into the desired interval, we can use either the unit circle or the properties of the sine function, in this case, . We have
, and hence, . Because , we obtain that .
Let . Then , , and we need to find . Note that since , we can even say that . We can solve this question either using right triangle trigonometry, or applying an appropriate trigonometric formula. With the geometric approach, we draw a right triangle with acute angle , opposite side 3 and adjacent side 4, as . Then, by the Pythagorean theorem, the hypotenuse has length . It follows that . Alternatively, we can use the formula (that follows from the Pythagorean trigonometric identity by expressing and in terms of and and bringing the right hand side to the common denominator). Since secant is positive on the interval , we obtain that
Note that since , we have that , and so is well-defined. If , then , and we are looking for . Because , we have that , and hence is well-defined. Using a triangle diagram (, so we take adjacent side to be , hypotenuse to be , and then, applying the Pythagoren theorem, we obtain that the opposite side is going to be ), we find that . Alternatively, and which follows from the Pythagorean trigonometric identity since on , implying that . It is an easy, yet useful, exercise to verify that the two answers we obtained are actually the same.
Key Concepts and Equations
For a function to have an inverse, the function must be one-to-one. Given the graph of a function, we can determine whether the function is one-to-one by using the horizontal line test.
If a function is not one-to-one, we can restrict the domain to a smaller domain where the function is one-to-one and then define the inverse of the function on the smaller domain.
For a function and its inverse for all in the domain of and for all in the domain of .
Since the trigonometric functions are periodic, we need to restrict their domains to define the inverse trigonometric functions.
The graph of a function and its inverse are symmetric about the line .
For the following exercises, find the domains of the given functions.
For the following exercises, evaluate the functions, when possible. If the function is undefined, explain why.
- inverse function
- for a function , the inverse function is defined if is one-to-one by whenever
- one-to-one function
- a function is one-to-one if whenever
- horizontal line test
- a function is one-to-one if and only if every horizontal line intersects the graph of at most once
- restricted domain
- a subset of the domain of a function
- inverse trigonometric functions
- the inverses of the trigonometric functions are defined on restricted domains where they are one-to-one functions