# High School Trigonometry/Defining Trigonometric Functions

Consider a situation in which you are building a ramp so that people in wheelchairs can access a building. If the ramp must have a height of 8 feet, and the angle of the ramp must be about 5°, how long must the ramp be?

Solving this kind of problem requires trigonometry. Recall that in the first lesson, you learned that the word trigonometry comes from two words meaning triangle and measure. In this lesson we will define six trigonometric functions. For each of these functions, the elements of the domain are angles. We will define these functions in two ways: first, using right triangles, and second, using angles of rotation. Once we have defined these functions, we will be able to solve problems like the one above.

## Learning Objectives

• Find the values of the six trig functions for angles in right triangles.
• Find the values of the six trig functions for angles of rotation.
• Work with angles in the unit circle.

## The Sine, Cosine, and Tangent Functions

The first three trigonometric functions we will work with are the sine, cosine, and tangent functions. As noted above, the elements of the domains of these functions are angles. We can define these functions in terms of a right triangle: The elements of the range of the functions are particular ratios of sides of triangles.

We define the sine function as follows: For an acute angle x in a right triangle, is the sin x ratio of the side opposite of the angle to the hypotenuse of the triangle. For example, in the triangle shown above, we have:

${\displaystyle \sin(A)={\frac {a}{c}}}$
${\displaystyle \sin(B)={\frac {b}{c}}}$

Since all right triangles with the same acute angle are similar, this function is will produce the same ratio, no matter which triangle is used. Thus, it is a well defined function.

Similarly, the cosine of an angle is defined as the ratio of the side adjacent (next to) the angle to the hypotenuse of the triangle. In the triangle above, we have:

${\displaystyle \cos(A)={\frac {b}{c}}}$
${\displaystyle \cos(B)={\frac {a}{c}}}$

Finally, the tangent of an angle is defined as the ratio of the side opposite the angle to the side adjacent to the angle. In the triangle above, we have:

${\displaystyle \tan(A)={\frac {a}{b}}}$
${\displaystyle \tan(B)={\frac {b}{a}}}$

There are a few important things to note about the way we write these functions. First, keep in mind that the abbreviations sin(x), cos(x), and tan(x) are just like f(x). They simplify stand for specific kinds of functions. Second, be careful about how you pronounce the names of the functions. When we write sin x it is still pronounced sine, with a long "i". When we write cos x, we still say co-sine. And when we write tan x, we still say tangent. (Sometimes casually people say "cos" and "tan", however, it shouldn't be surprising that "sin" is always pronounced "sine"!)

We can use these definitions to find the sine, cosine, and tangent values for angles in a right triangle.

 Example 1 Find the sine, cosine, and tangent of angle A: Solution: ${\displaystyle \sin A={\frac {\text{opposite side}}{\text{hypotenuse}}}={\frac {4}{5}}}$ ${\displaystyle \cos A={\frac {\text{adjacent side}}{\text{hypotenuse}}}={\frac {3}{5}}}$ ${\displaystyle \tan A={\frac {\text{opposite side}}{\text{adjacent side}}}={\frac {4}{3}}}$

One of the reasons that these functions will help us solve problems is that these ratios will always be the same, as long as the angles are the same. Consider for example, a triangle similar to triangle ABC.

If CP has length 3, then side AP of triangle NAP is 6. Because NAP is similar to ABC, side NP has length 8. This means the hypotenuse AN has length 10. (We can show this either using the proportions from the similar triangles, or by using the Pythagorean Theorem.)

If we use triangle NAP to find the sine, cosine, and tangent of angle A, we get:

${\displaystyle \sin A={\frac {\text{opposite side}}{\text{hypotenuse}}}={\frac {8}{10}}={\frac {4}{5}}}$
${\displaystyle \cos A={\frac {\text{adjacent side}}{\text{hypotenuse}}}={\frac {6}{10}}={\frac {3}{5}}}$
${\displaystyle \tan A={\frac {\text{opposite side}}{\text{adjacent side}}}={\frac {8}{6}}={\frac {4}{3}}}$
 Example 2 Find sin(B) using triangle ABC and triangle NAP. Solution: Using triangle ABC: sin B = ${\displaystyle {\tfrac {3}{5}}}$ Using triangle NAP: sin B = ${\displaystyle {\tfrac {6}{10}}}$ = ${\displaystyle {\tfrac {3}{5}}}$

## Secant, Cosecant, and Cotangent Functions

We can define three more functions also based on a right triangle.

Table 1.9
Function name Definition Example
Secant ${\displaystyle {\tfrac {\text{hypotenuse}}{\text{adjacent side}}}}$ In triangle ABC, sec A = ${\displaystyle {\tfrac {c}{b}}}$
Cosecant ${\displaystyle {\tfrac {\text{hypotenuse}}{\text{opposite side}}}}$ In triangle ABC, csc A = ${\displaystyle {\tfrac {c}{a}}}$
Cotangent ${\displaystyle {\tfrac {\text{adjacent side}}{\text{opposite side}}}}$ In triangle ABC, cot A = ${\displaystyle {\tfrac {b}{a}}}$
Example 3

Find the secant, cosecant, and cotangent of angle B.

Solution:

First, we must find the length of the hypotenuse. We can do this using the Pythagorean Theorem:

 52 + 122 = H2 25 + 144 = H2 169 = H2 H = 13

Now we can find the secant, cosecant, and cotangent of angle B:

${\displaystyle \sec(B)={\frac {\text{hypotenuse}}{\text{adjacent side}}}={\frac {13}{12}}}$
${\displaystyle \csc(B)={\frac {\text{hypotenuse}}{\text{opposite side}}}={\frac {13}{5}}}$
${\displaystyle \cot(B)={\frac {\text{adjacent side}}{\text{opposite side}}}={\frac {12}{5}}}$

## Trigonometric Functions of Angles in Standard Position

Above, we defined the six trigonometric functions for angles in right triangles. We can also define the same functions in terms of angles of rotation. Consider an angle in standard position, whose terminal side intersects a circle of radius r. We can think of the radius as the hypotenuse of a right triangle:

The point (x, y) where the terminal side of the angle intersects the circle tells us the lengths of the two legs of the triangle. Now, we can define the trigonometric functions in terms of x, y, and r:

 ${\displaystyle \cos \theta ={\frac {x}{r}}}$ ${\displaystyle \sec \theta ={\frac {r}{x}}}$ ${\displaystyle \sin \theta ={\frac {y}{r}}}$ ${\displaystyle \csc \theta ={\frac {r}{y}}}$ ${\displaystyle \tan \theta ={\frac {y}{x}}}$ ${\displaystyle \cot \theta ={\frac {x}{y}}}$

Now we can extend these functions to include non-acute angles.

Example 4

The point (−3, 4) is a point on the terminal side of an angle in standard position. Determine the values of the six trigonometric functions of the angle.

Solution:

Notice that the angle is more than 90 degrees, and that the terminal side of the angle lies in the second quadrant. This will influence the signs of the trigonometric functions.

 ${\displaystyle \cos \theta ={\frac {-3}{5}}}$ ${\displaystyle \sec \theta ={\frac {5}{-3}}}$ ${\displaystyle \sin \theta ={\frac {4}{5}}}$ ${\displaystyle \csc \theta ={\frac {5}{4}}}$ ${\displaystyle \tan \theta ={\frac {4}{-3}}}$ ${\displaystyle \cot \theta ={\frac {-3}{4}}}$

Notice that the value of r depends on the coordinates of the given point. You can always find the value of r using the Pythagorean Theorem. However, often we look at angles in a circle with radius 1. As you will see next, doing this allows us to simplify the definitions of the functions.

## The Unit Circle

Consider an angle in standard position, such that the point (x, y) on the terminal side of the angle is a point on a circle with radius 1.

This circle is called the unit circle. With r = 1, we can define the trigonometric functions in the unit circle:

 ${\displaystyle \cos \theta ={\frac {x}{r}}={\frac {x}{1}}=x}$ ${\displaystyle \sec \theta ={\frac {r}{x}}={\frac {1}{x}}}$ ${\displaystyle \sin \theta ={\frac {y}{r}}={\frac {y}{1}}=y}$ ${\displaystyle \csc \theta ={\frac {r}{y}}={\frac {1}{y}}}$ ${\displaystyle \tan \theta ={\frac {y}{x}}}$ ${\displaystyle \cot \theta ={\frac {x}{y}}}$

Notice that in the unit circle, the sine and cosine of an angle are the x and y coordinates of the point on the terminal side of the angle. Now we can find the values of the trigonometric functions of any angle of rotation, even the quadrantal angles, which are not angles in triangles.

We can use the figure above to determine values of the trig functions for the quadrantal angles. For example, sin(90°) = y = 1.

 Example 5 Use the unit circle above to find each value: a. cos 90° b. cot 180° c. sec 0° Solution: a. cos 90° = 0 The ordered pair for this angle is (0, 1). The cosine value is the x coordinate, 0. b. cot 180° is undefined The ordered pair for this angle is (-1, 0). The ratio ${\displaystyle {\tfrac {x}{y}}}$ is ${\displaystyle {\tfrac {-1}{0}}}$, which is undefined. c. sec 0° = 1 The ordered pair for this angle is (1, 0). The ratio is ${\displaystyle {\tfrac {1}{x}}}$ is ${\displaystyle {\tfrac {1}{1}}}$ = 1.

There are several important angles in the unit circle that you will work with extensively in your study of trigonometry: 30°, 45°, and 60°. To find the values of the trigonometric functions of these angles, we need to know the ordered pairs. Let's begin with 30°.

The terminal side of the angle intersects the unit circle at the point ${\displaystyle {\tfrac {\sqrt {3}}{2}},{\tfrac {1}{2}}}$. (You will prove this in one of the review exercises.). Therefore we can find the values of any of the trig functions of 30°. For example, the cosine value is the x-coordinate, so cos (30°) = ${\displaystyle {\tfrac {\sqrt {3}}{2}}}$. Because the coordinates are fractions, we have to do a bit more work in order to find the tangent value:

${\displaystyle \tan(30^{\circ })={\frac {y}{x}}={\frac {\frac {1}{2}}{\frac {\sqrt {3}}{2}}}={\frac {1}{2}}\cdot {\frac {2}{\sqrt {3}}}={\frac {1}{\sqrt {3}}}}$

In the review exercises you will find the values of the remaining four trig functions of this angle. The table below summarizes the ordered pairs for 30°, 45°, and 60° on the unit circle.

Table 1.10
Angle x coordinate y coordinate
30° ${\displaystyle {\tfrac {\sqrt {3}}{2}}}$ ${\displaystyle {\tfrac {1}{2}}}$
45° ${\displaystyle {\tfrac {\sqrt {2}}{2}}}$ ${\displaystyle {\tfrac {\sqrt {2}}{2}}}$
60° ${\displaystyle {\tfrac {1}{2}}}$ ${\displaystyle {\tfrac {\sqrt {3}}{2}}}$

We can use these values to find the values of any of the six trig functions of these angles.

 Example 6 Find the value of each function. a. cos (45°) b. sin (60°) c. tan (45°) Solution: a. cos (45°) = ${\displaystyle {\tfrac {\sqrt {2}}{2}}}$ The cosine value is the x coordinate of the point. b. sin (60°) = ${\displaystyle {\tfrac {\sqrt {3}}{2}}}$ The sine value is the y coordinate of the point. c. tan (45°) = 1 The tangent value is the ratio of the y coordinate to the x coordinate. Because the x and y coordinates are the same for this angle, the tangent ratio is 1.

## Lesson Summary

In this chapter we have defined the six trigonometric functions. First we defined the functions for angles in right triangles, and then we defined them for angles of rotation. We considered angles formed when the terminal side of an angle intersected a circle of radius r, and then we focused in on the unit circle, which has radius 1. The unit circle will be used extensively throughout the remainder of the chapter.

## Points to Consider

• How is the Pythagorean Theorem useful in trigonometry?
• How can some values of the trig functions be negative? How is it that some are undefined?
• Why is the unit circle and the trig functions defined on it useful, even when the hypotenuses of triangles in the problem are not 1?

## Review Questions

1. Find the values of the six trig functions of angle A.
2. Consider triangle VET below.
(a) Find the length of the hypotenuse.
(b) Find the values of the six trig functions of angle T.
3. The point (3, −4) is a point on the terminal side of an angle θ in standard position.
(a) Determine the radius of the circle.
(b) Determine the values of the six trigonometric functions of the angle.
ii. The values are:
4. The point (−5, −12) is a point on the terminal side of an angle θ in standard position.
(a) Determine the radius of the circle.
(b) Determine the values of the six trigonometric functions of the angle.
ii. The values are:
5. The terminal side of the angle 270° intersects the unit circle at (0, −1). Use this ordered pair to find the six trig functions of 270°.
6. In the lesson you learned that the terminal side of the angle 30° intersects the unit circle at the point ${\displaystyle \left({\tfrac {\sqrt {3}}{2}},{\tfrac {1}{2}}\right)}$. Here you will prove that this is true.
(a) Explain why triangle ABD is an equiangular triangle. What is the measure of angle DAB?
(b) What is the length of BD? How do you know?
(c) What is the length of BC and CD? How do you know?
(d) Now explain why the ordered pair is ${\displaystyle \left({\tfrac {\sqrt {3}}{2}},{\tfrac {1}{2}}\right)}$
(e) Why does this tell you that the ordered pair for 60° is ${\displaystyle \left({\tfrac {1}{2}},{\tfrac {\sqrt {3}}{2}}\right)}$
7. In the lesson you learned that the terminal side of the angle 45° is ${\displaystyle \left({\tfrac {\sqrt {2}}{2}},{\tfrac {\sqrt {2}}{2}}\right)}$. Use the figure below and the Pythagorean Theorem to show that this is true.
8. State the values of the six trig functions of 60°.
9. In what quadrants will an angle in standard position have a positive tangent value? Explain your thinking.
10. Sketch the angle 150° on the unit circle. How is this angle related to 30°? What do you think the ordered pair is?

1. Trigonometric

2.  ${\displaystyle \cos A={\frac {12}{15}}={\frac {4}{5}}}$ ${\displaystyle \sec A={\frac {15}{12}}={\frac {5}{4}}}$ ${\displaystyle \sin A={\frac {9}{15}}={\frac {3}{5}}}$ ${\displaystyle \csc A={\frac {15}{9}}={\frac {5}{3}}}$ ${\displaystyle \tan A={\frac {9}{12}}={\frac {3}{4}}}$ ${\displaystyle \cot A={\frac {12}{9}}={\frac {4}{3}}}$
3. The length of the hypotenuse is 17.
 ${\displaystyle \cos T={\frac {8}{17}}}$ ${\displaystyle \sec T={\frac {17}{8}}}$ ${\displaystyle \sin T={\frac {15}{17}}}$ ${\displaystyle \csc T={\frac {17}{15}}}$ ${\displaystyle \tan T={\frac {15}{8}}}$ ${\displaystyle \cot T={\frac {8}{15}}}$

4.  ${\displaystyle \cos \theta ={\frac {3}{5}}}$ ${\displaystyle \sec \theta ={\frac {5}{3}}}$ ${\displaystyle \sin \theta ={\frac {-4}{5}}}$ ${\displaystyle \csc \theta ={\frac {5}{-4}}}$ ${\displaystyle \tan \theta ={\frac {-4}{3}}}$ ${\displaystyle \cot \theta ={\frac {3}{-4}}}$

5.  ${\displaystyle \cos \theta ={\frac {-5}{13}}}$ ${\displaystyle \sec \theta ={\frac {13}{-5}}}$ ${\displaystyle \sin \theta ={\frac {-12}{13}}}$ ${\displaystyle \csc \theta ={\frac {13}{-12}}}$ ${\displaystyle \tan \theta ={\frac {-12}{-5}}={\frac {12}{5}}}$ ${\displaystyle \cot \theta ={\frac {-5}{-12}}={\frac {5}{12}}}$

6.  ${\displaystyle \cos 270^{\circ }=0}$ ${\displaystyle \sec 270^{\circ }={\text{undefined}}}$ ${\displaystyle \sin 270^{\circ }=-1}$ ${\displaystyle \csc 270^{\circ }={\frac {1}{-1}}=-1}$ ${\displaystyle \tan 270^{\circ }={\text{undefined}}}$ ${\displaystyle \cot 270^{\circ }=0}$
7.
(a) The triangle is equiangular because all three angles measure 60 degrees. Angle DAB measures 60 degrees because it is the sum of two 30 degree angles.
(b) BD has length 1 because it is one side of an equiangular, and hence equilateral triangle.
(c) BC and CD each have length ${\displaystyle {\tfrac {1}{2}}}$, as they are each half of BD. This is the case because triangle ABC and ADC are congruent.
(d) We can use the Pythagorean Theorem to show that the length of AC is ${\displaystyle {\tfrac {\sqrt {3}}{2}}}$. If we place angle BAC as an angle in standard position, then AC and BC correspond to the x and y coordinates where the terminal side of the angle intersect the unit circle. Therefore the ordered pair is ${\displaystyle \left({\tfrac {\sqrt {3}}{2}},{\tfrac {1}{2}}\right)}$.
(e) If we draw the angle 60° in standard position, we will also obtain a 30 – 60 – 90 triangle, but the side lengths will be interchanged. So the ordered pair for 60° is ${\displaystyle \left({\tfrac {1}{2}},{\tfrac {\sqrt {3}}{2}}\right)}$.
8.  ${\displaystyle n^{2}+n^{2}=1^{2}\,\!}$ ${\displaystyle 2n^{2}=1\,\!}$ ${\displaystyle n^{2}={\tfrac {1}{2}}}$ ${\displaystyle n=\pm {\sqrt {\tfrac {1}{2}}}}$ ${\displaystyle n=\pm {\tfrac {1}{\sqrt {2}}}}$ ${\displaystyle n=\pm {\tfrac {1}{\sqrt {2}}}\cdot {\tfrac {\sqrt {2}}{\sqrt {2}}}=\pm {\tfrac {\sqrt {2}}{2}}}$
Because the angle is in the first quadrant, the x and y coordinates are positive.

9.  ${\displaystyle \cos 60^{\circ }={\frac {1}{2}}}$ ${\displaystyle \sec 60^{\circ }={\frac {1}{\frac {1}{2}}}=2}$ ${\displaystyle \sin 60^{\circ }={\frac {\sqrt {3}}{2}}}$ ${\displaystyle \csc 60^{\circ }={\frac {1}{\frac {\sqrt {3}}{2}}}={\frac {2}{\sqrt {3}}}={\frac {2{\sqrt {3}}}{3}}}$ ${\displaystyle \tan 60^{\circ }={\frac {\frac {\sqrt {3}}{2}}{\frac {1}{2}}}={\sqrt {3}}}$ ${\displaystyle \cot 60^{\circ }={\frac {\frac {1}{2}}{\frac {\sqrt {3}}{2}}}={\frac {1}{\sqrt {3}}}\ {\text{or}}\ {\frac {\sqrt {3}}{3}}}$
10. An angle in the first quadrant, as the tangent is the ratio of two positive numbers. An angle in the third quadrant, as the tangent in the ratio of two negative numbers, which will be positive.
11. The terminal side of the angle is a reflection of the terminal side of 30°. From this, students should see that the ordered pair is ${\displaystyle \left(-{\tfrac {\sqrt {3}}{2}},{\tfrac {1}{2}}\right)}$.

## Vocabulary

A side adjacent to an angle is the side next to the angle. In a right triangle, it is the leg that is next to the angle.
hypotenuse
The hypotenuse is the longest side in a right triangle, opposite the right angle.
leg
The legs of a right triangle are the two shorter sides.
Pythagorean Theorem
The Pythagorean theorem states the relationship among the sides of a right triangle: Leg 12 + Leg 22 = Hypotenuse2