**Trigonometric functions** are used to model many phenomena, including sound waves, vibrations of strings, alternating electrical current, and the motion of pendulums. In fact, almost any repetitive, or cyclical, motion can be modeled by some combination of trigonometric functions. In this section, we define the six basic trigonometric functions and look at some of the main identities involving these functions.

# Radian Measure

To use trigonometric functions, we first must understand how to measure the angles. Although we can use both radians and degrees, **radians** are a more natural measurement because they are related directly to the unit circle, a circle with radius 1. The radian measure of an angle is defined as follows. Given an angle , let be the length of the corresponding arc on the unit circle ((Figure)). We say the angle corresponding to the arc of length 1 has radian measure 1.

Since an angle of 360° corresponds to the circumference of a circle, or an arc of length , we conclude that an angle with a degree measure of 360° has a radian measure of . Similarly, we see that 180° is equivalent to radians. (Figure) shows the relationship between common degree and radian values.

Degrees | Radians | Degrees | Radians |
---|---|---|---|

0 | 0 | 120 | |

30 | 135 | ||

45 | 150 | ||

60 | 180 | ||

90 |

### Converting between Radians and Degrees

- Express 225° using radians.
- Express rad using degrees.

#### Solution

Use the fact that 180° is equivalent to radians as a conversion factor: .

- rad
- rad =

Express 210° using radians. Express rad using degrees.

#### Solution

rad; 330°

#### Hint

radians is equal to .

# The Six Basic Trigonometric Functions

Trigonometric functions allow us to use angle measures, in radians or degrees, to find the coordinates of a point on any circle—not only on a unit circle—or to find an angle given a point on a circle. They also define the relationship among the sides and angles of a triangle.

To define the trigonometric functions, first consider the unit circle centered at the origin and a point on the unit circle. Let be an angle with an initial side that lies along the positive -axis and with a terminal side that is the line segment . An angle in this position is said to be in *standard position* ((Figure)). We can then define the values of the six trigonometric functions for in terms of the coordinates and .

### Definition

Let be a point on the unit circle centered at the origin . Let be an angle with an initial side along the positive -axis and a terminal side given by the line segment . The trigonometric functions are then defined as

If , then and are undefined. If , then and are undefined.

We can see that for a point on a circle of radius with a corresponding angle , the coordinates and satisfy.

The values of the other trigonometric functions can be expressed in terms of , and ((Figure)).

(Figure) shows the values of sine and cosine at the major angles in the first quadrant. From this table, we can determine the values of sine and cosine at the corresponding angles in the other quadrants. The values of the other trigonometric functions are calculated easily from the values of and .

0 | 0 | 1 |

1 | 0 |

### Evaluating Trigonometric Functions

Evaluate each of the following expressions.

#### Solution

- On the unit circle, the angle corresponds to the point . Therefore, .
- An angle corresponds to a revolution in the negative direction, as shown. Therefore, .
- An angle . Therefore, this angle corresponds to more than one revolution, as shown. Knowing the fact that an angle of corresponds to the point , we can conclude that .

Evaluate and .

#### Solution

#### Hint

Look at angles on the unit circle.

As mentioned earlier, the ratios of the side lengths of a right triangle can be expressed in terms of the trigonometric functions evaluated at either of the acute angles of the triangle. Let be one of the acute angles. Let be the length of the adjacent leg, be the length of the opposite leg, and be the length of the hypotenuse. By inscribing the triangle into a circle of radius , as shown in (Figure), we see that , and satisfy the following relationships with :

### Constructing a Wooden Ramp

A wooden ramp is to be built with one end on the ground and the other end at the top of a short staircase. If the top of the staircase is 4 ft from the ground and the angle between the ground and the ramp is to be , how long does the ramp need to be?

#### Solution

Let denote the length of the ramp. In the following image, we see that needs to satisfy the equation . Solving this equation for , we see that ft.

A house painter wants to lean a 20-ft ladder against a house. If the angle between the base of the ladder and the ground is to be , how far from the house should she place the base of the ladder?

#### Solution

10 ft

#### Hint

Draw a right triangle with hypotenuse 20 ft.

# Trigonometric Identities

A **trigonometric identity** is an equation involving trigonometric functions that is true for all angles for which the functions are defined. We can use the identities to help us solve or simplify equations. The main trigonometric identities are listed next.

### Rule: Trigonometric Identities

**Reciprocal identities**

**Pythagorean identities**

**Addition and subtraction formulas**

**Double-angle formulas**

### Solving Trigonometric Equations

For each of the following equations, use a trigonometric identity to find all solutions.

[reveal-answer q=”501288″]Show Answer[/reveal-answer][hidden-answer a=”501288″]a. Using the double-angle formula for , we see that is a solution of

if and only if

,

which is true if and only if

.

To solve this equation, it is important to note that we need to factor the left-hand side and not divide both sides of the equation by . The problem with dividing by is that it is possible that is zero. In fact, if we did divide both sides of the equation by , we would miss some of the solutions of the original equation. Factoring the left-hand side of the equation, we see that is a solution of this equation if and only if

.

Since when

,

and when

, or ,

we conclude that the set of solutions to this equation is

, and .

b. Using the double-angle formula for and the reciprocal identity for , the equation can be written as

.

To solve this equation, we multiply both sides by to eliminate the denominator, and say that if satisfies this equation, then satisfies the equation

.

However, we need to be a little careful here. Even if satisfies this new equation, it may not satisfy the original equation because, to satisfy the original equation, we would need to be able to divide both sides of the equation by . However, if , we cannot divide both sides of the equation by . Therefore, it is possible that we may arrive at extraneous solutions. So, at the end, it is important to check for extraneous solutions. Returning to the equation, it is important that we factor out of both terms on the left-hand side instead of dividing both sides of the equation by . Factoring the left-hand side of the equation, we can rewrite this equation as

.

Therefore, the solutions are given by the angles such that or . The solutions of the first equation are . The solutions of the second equation are . After checking for extraneous solutions, the set of solutions to the equation is

and .

Find all solutions to the equation .

#### Solution

for

#### Hint

Use the double-angle formula for cosine.

### Proving a Trigonometric Identity

Prove the trigonometric identity .

#### Solution

We start with the identity.

Dividing both sides of this equation by , we obtain.

Since and , we conclude that.

Prove the trigonometric identity .

#### Hint

Divide both sides of the identity by .

# Graphs and Periods of the Trigonometric Functions

We have seen that as we travel around the unit circle, the values of the trigonometric functions repeat. We can see this pattern in the graphs of the functions. Let be a point on the unit circle and let be the corresponding angle. Since the angle and correspond to the same point , the values of the trigonometric functions at and at are the same. Consequently, the trigonometric functions are **periodic functions**. The period of a function is defined to be the smallest positive value such that for all values in the domain of . The sine, cosine, secant, and cosecant functions have a period of . Since the tangent and cotangent functions repeat on an interval of length , their period is ((Figure)).

Just as with algebraic functions, we can apply transformations to trigonometric functions. In particular, consider the following function:.

In (Figure), the constant causes a horizontal or phase shift. The factor changes the period. This transformed sine function will have a period . The factor results in a vertical stretch by a factor of . We say is the “amplitude of .” The constant causes a vertical shift.

Notice in (Figure) that the graph of is the graph of shifted to the left units. Therefore, we can write . Similarly, we can view the graph of as the graph of shifted right units, and state that .

A shifted sine curve arises naturally when graphing the number of hours of daylight in a given location as a function of the day of the year. For example, suppose a city reports that June 21 is the longest day of the year with 15.7 hours and December 21 is the shortest day of the year with 8.3 hours. It can be shown that the function

is a model for the number of hours of daylight as a function of day of the year ((Figure)).

### Sketching the Graph of a Transformed Sine Curve

Sketch a graph of .

#### Solution

This graph is a phase shift of to the right by units, followed by a horizontal compression by a factor of 2, a vertical stretch by a factor of 3, and then a vertical shift by 1 unit. The period of is .

Describe the relationship between the graph of and the graph of .

#### Solution

To graph , the graph of needs to be compressed horizontally by a factor of 4, then stretched vertically by a factor of 3, then shifted down 5 units. The function will have a period of and an amplitude of 3.

#### Hint

The graph of can be sketched using the graph of and a sequence of three transformations.

### Key Concepts

- Radian measure is defined such that the angle associated with the arc of length 1 on the unit circle has radian measure 1. An angle with a degree measure of 180° has a radian measure of rad.
- For acute angles , the values of the trigonometric functions are defined as ratios of two sides of a right triangle in which one of the acute angles is .
- For a general angle , let be a point on a circle of radius corresponding to this angle . The trigonometric functions can be written as ratios involving , and .
- The trigonometric functions are periodic. The sine, cosine, secant, and cosecant functions have period . The tangent and cotangent functions have period .

# Key Equations

**Generalized sine function**

An **inverse function** reverses the operation done by a particular function. In other words, whatever a function does, the inverse function undoes it. In this section, we define an inverse function formally and state the necessary conditions for an inverse function to exist. We examine how to find an inverse function and study the relationship between the graph of a function and the graph of its inverse. Then we apply these ideas to define and discuss properties of the inverse trigonometric functions.

# Existence of an Inverse Function

We begin with an example. Given a function and an output , we are often interested in finding what value or values were mapped to by . For example, consider the function . Since any output , we can solve this equation for to find that the input is . This equation defines as a function of . Denoting this function as , and writing , we see that for any in the domain of . Thus, this new function, , “undid” what the original function did. A function with this property is called the inverse function of the original function.

### Definition

Given a function with domain and range , its inverse function (if it exists) is the function with domain and range such that if . In other words, for a function and its inverse , for all in , and for all in .

Note that is read as “f inverse.” Here, the -1 is not used as an exponent and . (Figure) 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 arrive at the equation . This equation does not describe as a function of because there are two solutions to this equation for every is that two inputs are sent to the same output for each output discussed earlier did not have this problem. For that function, each input was sent to a different output. A function that sends each input to a *different* output is called a** one-to-one function.**

### Definition

We say a is a one-to-one function if when .

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, according to the **horizontal line test**, it cannot intersect the graph more than once. We note that the horizontal line test is different from the vertical line test. The vertical line test determines whether a graph is the graph of a function. The horizontal line test determines whether a function is one-to-one ((Figure)).

### Rule: Horizontal Line Test

A function is one-to-one if and only if every horizontal line intersects the graph of no more than once.

### 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.

#### Solution

- Since the horizontal line for any integer intersects the graph more than once, this function is not one-to-one.
- Since every horizontal line intersects the graph once (at most), this function is one-to-one.

Is the function graphed in the following image one-to-one?

#### Solution

No.

#### Hint

Use the horizontal line test.

# Finding a Function’s Inverse

We can now consider one-to-one functions and show how to find their inverses. Recall that a function maps elements in the domain of to elements in the range of . The inverse function maps each element from the range of back to its corresponding element from the domain of . Therefore, to find the inverse function of a one-to-one function , given any in the range of , we need to determine which in the domain of satisfies . Since is one-to-one, there is exactly one such value . We can find that value by solving the equation for . Doing so, we are able to write as a function of where the domain of this function is the range of and the range of this new function is the domain of . Consequently, this function is the inverse of , and we write . Since we typically use the variable to denote the independent variable and to denote the dependent variable, we often interchange the roles of and , and write . Representing the inverse function in this way is also helpful later when we graph a function and its inverse on the same axes.

### Problem-Solving Strategy: Finding an Inverse Function

- Solve the equation for .
- Interchange the variables and and write .

### Finding an Inverse Function

Find the inverse for the function . State the domain and range of the inverse function. Verify that .

#### Solution

Follow the steps outlined in the strategy.

Step 1. If , then and .

Step 2. Rewrite as and let .

Therefore, .

Since the domain of is , the range of is . Since the range of is , the domain of is .

You can verify that by writing.

Note that for to be the inverse of , both and for all in the domain of the inside function.

Find the inverse of the function . State the domain and range of the inverse function.

#### Solution

. The domain of is . The range of is .

#### Hint

Use the (Note) for finding inverse functions.

## Graphing Inverse Functions

Let’s consider the relationship between the graph of a function and the graph of its inverse. Consider the graph of shown in (Figure) and a point on the graph. Since , then . 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 .

#### Solution

Reflect the graph about the line . The domain of is . The range of is . By using the preceding strategy for finding inverse functions, we can verify that the inverse function is , as shown in the graph.

Sketch the graph of and the graph of its inverse using the symmetry property of inverse functions.

#### Solution

#### Hint

The graphs are symmetric about the line .

## Restricting Domains

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 is one-to-one. This 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 , 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 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 ((Figure)).

### Restricting the Domain

Consider the function .

- Sketch the graph of and use the horizontal line test to show that is not one-to-one.
- Show that is one-to-one on the restricted domain . Determine the domain and range for the inverse of on this restricted domain and find a formula for .

#### Solution

- The graph of is the graph of shifted left 1 unit. Since there exists a horizontal line intersecting the graph more than once, is not one-to-one.
- On the interval is one-to-one.

The domain and range of are given by the range and domain of , respectively. Therefore, the domain of is and the range of is . To find a formula for , solve the equation for . If , then . Since we are restricting the domain to the interval where , we need . Therefore, . Interchanging and , we write and conclude that .

Consider restricted to the domain . Verify that is one-to-one on this domain. Determine the domain and range of the inverse of and find a formula for .

#### Solution

The domain of is . The range of is . The inverse function is given by the formula .

#### Hint

The domain and range of is given by the range and domain of , respectively. To find , solve for .

# 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 ((Figure)). 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 . 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 i**nverse 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 or arcsin, and the inverse cosine function, denoted or arccos, are defined on the domain as follows:

The inverse tangent function, denoted or arctan, and inverse cotangent function, denoted or arccot, are defined on the domain as follows:

The inverse cosecant function, denoted or arccsc, and inverse secant function, denoted or arcsec, are defined on the domain as follows:

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 ((Figure)).

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 , 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,

and.

Similarly, 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.

#### Solution

- Evaluating is equivalent to finding the angle such that and . The angle satisfies these two conditions. Therefore, .
- First we use the fact that . Then . Therefore, .
- To evaluate , first use the fact that . Then we need to find the angle such that and . Since satisfies both these conditions, we have .
- Since , we need to evaluate . That is, we need to find the angle such that and . Since satisfies both these conditions, we can conclude that .

### 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 .

- Consider the graph in (Figure) of the function . Describe its overall shape. Is it periodic? How do you know?
**Figure 6.**The graph of .Using a graphing calculator or other graphing device, estimate the – and -values of the maximum point for the graph (the first such point where -value as a multiple of . - Now consider other graphs of the form for various values of and . Sketch the graph when and , and find the – and -values for the maximum point. (Remember to express the -value as a multiple of , if possible.) Has it moved?
- Repeat for . Is there any relationship to what you found in part (2)?
- Complete the following table, adding a few choices of your own for and :

011101111251251221223443 - Try to figure out the formula for the -values.
- The formula for the -values is a little harder. The most helpful points from the table are . (
*Hint*:*Consider inverse trigonometric functions.)* - If you found formulas for parts (5) and (6), show that they work together. That is, substitute the -value formula you found into and simplify it to arrive at the -value formula you found.

### Key Concepts

- 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 .

# Key Equations

**Inverse functions**

for all in , and for all in .[/latex]