We begin our study of trigonometry by considering right triangles and acute angles measured in degrees. An acute angle is an angle with measure greater than $0°$ and less than $90°$ . Greek letters such as $\alpha$ (alpha), $\beta$ (beta), $\gamma$ (gamma), $\theta$ (theta), and $\varphi$ (phi) are often used to denote an angle. Consider a right triangle with one of its acute angles labeled $\theta$ . The side opposite the right angle is called the hypotenuse. The other sides of the triangle are referenced by their position relative to the acute angle $\theta$ . One side is opposite $\theta$ and one is adjacent to $\theta$ .

The lengths of the sides of the triangle are used to define the six trigonometric ratios:

The sine of $\theta$ is the length of the side opposite $\theta$ divided by the length of the hypotenuse (see Fig. 1):

The ratio depends on the measure of angle $\theta$ , and thus the ratio is a function of $\theta$ . The notation $\text{sin }\theta$ actually means $\text{sin }\left(\theta \right)$ , where sin, or sine, is the name of the function.

The cosine of $\theta$ is the length of the side adjacent to $\theta$ divided by the length of the hypotenuse (see Fig. 2):

The tangent of $\theta$ is the length of the side opposite $\theta$ divided by the length of the side adjacent to $\theta$ (see Fig. 3):

The six trigonometric ratios, or trigonometric functions, are defined as follows. Here the domain of each function is the set of acute angles. Later in this chapter, the domain will be extended first to the set of all angles, or rotations, and then to the set of real numbers.

Example 1

In the right triangle shown at left, find the six trigonometric function values of (a) $\theta$ and (b) $\alpha$ .

Solution

We use the definitions.

1. a)

   $\begin{array}{ccc}\text{sin }\theta ={\displaystyle \frac{\text{opp}}{\text{hyp}}}={\displaystyle \frac{12}{13}},\hfill & \text{csc }\theta ={\displaystyle \frac{\text{hyp}}{\text{opp}}}={\displaystyle \frac{13}{12}},\hfill & \hfill \\ \text{cos }\theta ={\displaystyle \frac{\text{adj}}{\text{hyp}}}={\displaystyle \frac{5}{13}},\hfill & \text{sec }\theta ={\displaystyle \frac{\text{hyp}}{\text{adj}}}={\displaystyle \frac{13}{5}},\hfill & \mathbf{\text{The references to opposite, adjacent, and hypotenuse are relative to}}\mathit{\theta }.\hfill \\ \text{tan }\theta ={\displaystyle \frac{\text{opp}}{\text{adj}}}={\displaystyle \frac{12}{5}},\hfill & \text{cot }\theta ={\displaystyle \frac{\text{adj}}{\text{opp}}}={\displaystyle \frac{5}{12}}\hfill & \hfill \end{array}$
2. b)
   $$\begin{array}{ccc}\text{sin }\alpha ={\displaystyle \frac{\text{opp}}{\text{hyp}}}={\displaystyle \frac{5}{13}},\hfill & \text{csc }\alpha ={\displaystyle \frac{\text{hyp}}{\text{opp}}}={\displaystyle \frac{13}{5}},\hfill & \hfill \\ \text{cos }\alpha ={\displaystyle \frac{\text{adj}}{\text{hyp}}}={\displaystyle \frac{12}{13}},\hfill & \text{sec }\alpha ={\displaystyle \frac{\text{hyp}}{\text{adj}}}={\displaystyle \frac{13}{12}},\hfill & \mathbf{\text{The references to opposite, adjacent, and hypotenuse are relative to}}\mathit{\alpha }.\hfill \\ \text{tan }\alpha ={\displaystyle \frac{\text{opp}}{\text{adj}}}={\displaystyle \frac{5}{12}},\hfill & \text{cot }\alpha ={\displaystyle \frac{\text{adj}}{\text{opp}}}={\displaystyle \frac{12}{5}}\hfill & \hfill \end{array}$$

Now Try Exercise 1.

In Example 1(a), we note that the value of $\text{csc }\theta ,{\displaystyle \frac{13}{12}}$ . is the reciprocal of

${\displaystyle \frac{12}{13}}$ . the value of $\text{sin }\theta$ . Likewise, we see the same reciprocal relationship between

the values of $\text{sec }\theta$ and $\text{cos }\theta$ and between the values of $\text{cot }\theta$ and $\text{tan }\theta$ . For any angle, the cosecant, secant, and cotangent function values are the reciprocals of the sine, cosine, and tangent function values, respectively.

If we know the values of the sine, cosine, and tangent functions of an angle, we can use these reciprocal relationships to find the values of the cosecant, secant, and cotangent functions of that angle.

Triangles are said to be similar if their corresponding angles have the same measure. In similar triangles, the lengths of corresponding sides are in the same ratio. The right triangles shown below are similar. Note that the corresponding angles are equal, and the length of each side of the second triangle is four times the length of the corresponding side of the first triangle.

Let’s observe the sine, cosine, and tangent values of $\beta$ in each triangle. Can we expect corresponding function values to be the same?

For the two triangles, the corresponding values of $\text{sin}\beta$, $\text{cos}\beta$ , and $\text{tan }\beta$ are the same. The lengths of the sides are proportional—thus the ratios are the same. This must be the case because in order for the sine, cosine, and tangent to be functions, there must be only one output (the ratio) for each input (the angle $\beta$ ).

We can find the other five trigonometric function values of an acute angle when one of the function-value ratios is known.

Example 4

If $\text{sin }\beta ={\displaystyle \frac{6}{7}}$ and $\beta$ is an acute angle, find the other five trigonometric function values of $\beta$ .

Solution

We know from the definition of the sine function that the ratio

$${\displaystyle \frac{6}{7}}\text{is}{\displaystyle \frac{\text{opp}}{\text{hyp}}}.$$

Using this information, let’s consider a right triangle in which the hypotenuse has length 7 and the side opposite $\beta$ has length 6. To find the length of the side adjacent to $\beta$ , we use the Pythagorean equation:

$$\begin{array}{lll}\hfill a^2+b^2& =\hfill & c^2\hfill \\ \hfill a^2+6^2& =\hfill & 7^2\hfill \\ \hfill a^2+36& =\hfill & 49\hfill \\ \hfill a^2& =\hfill & 49-36=13\hfill \\ \hfill a& =\hfill & \sqrt{13}.\hfill \end{array}$$

We now use the lengths of the three sides to find the other five ratios:

$$\begin{array}{cc}\text{sin }\beta ={\displaystyle \frac{6}{7}},\hfill & \text{csc }\beta ={\displaystyle \frac{7}{6}},\hfill \\ \text{cos }\beta ={\displaystyle \frac{\sqrt{13}}{7}},\hfill & \text{sec }\beta ={\displaystyle \frac{7}{\sqrt{13}}}\text{,}\text{or}{\displaystyle \frac{7\sqrt{13}}{13}},\hfill \\ \text{tan }\beta ={\displaystyle \frac{6}{\sqrt{13}}}\text{,}\text{or}{\displaystyle \frac{6\sqrt{13}}{13}},\hfill & \text{cot }\beta ={\displaystyle \frac{\sqrt{13}}{6}}.\hfill \end{array}$$

Now Try Exercise 9.

In Examples 1 and 4, we found the trigonometric function values of an acute angle of a right triangle when the lengths of the three sides were known. In most situations, we are asked to find the function values when the measure of the acute angle is given. For certain special angles such as $30°$ , $45°$ , and $60°$ , which are frequently seen in applications, we can use geometry to determine the function values.

A right triangle with a $45°$ angle actually has two $45°$ angles. Thus the triangle is isosceles, and the legs are the same length. Let’s consider such a triangle whose legs have length 1. Then we can find the length of its hypotenuse, $c$ , using the Pythagorean equation as follows:

Such a triangle is shown below. From this diagram, we can easily determine the trigonometric function values of $45°$ .

It is sufficient to find only the function values of the sine, cosine, and tangent, since the other three function values are their reciprocals.

It is also possible to determine the function values of $30°$ and $60°$ . A right triangle with $30°$ and $60°$ acute angles is half of an equilateral triangle, as shown in the following figure. Thus if we choose an equilateral triangle whose sides have length 2 and take half of it, we obtain a right triangle that has a hypotenuse of length 2 and a leg of length 1. The other leg has length a, which can be found as follows:

We can now determine the function values of $30°$ and $60°$ :

Since we will often use the function values of $30°$, $45°$ . and $60°$ , either the triangles that yield them or the values themselves should be memorized.

Let’s now use what we have learned about trigonometric functions of special angles to solve problems. We will consider such applications in greater detail in Section 6.2.

Example 5 Height of a Fireworks Display.

Trajectories for fireworks involve variables such as the launch angle, the launch velocity, and the size of the shell. Using physics to calculate the trajectory, fireworks technicians know exactly the path of the shell. Launch angles vary from $45°$ to $90°$ . For every inch of the diameter of the shell, a firework can travel about 100 ft vertically and 70 ft horizontally. These distances can vary depending on air resistance. (Sources: thinkquest.org, Pyrotechnico; www.dispatch.com, Aaron Harden) A 6-in. shell launched at an angle of $60°$ travels a horizontal distance of 390 ft. Approximate the height of the fireworks display. Round the answer to the nearest foot.

Solution

We begin with a diagram of the situation. We know the measure of an acute angle and the length of the adjacent side.

Since we want to determine the length of the opposite side, we can use either the tangent ratio or the cotangent ratio. In this case, we use the tangent ratio:

$$\begin{array}{lllll}\hfill \tan {60}^{°}& =\hfill & {\displaystyle \frac{\text{opp}}{\text{adj}}}\hfill & =\hfill & {\displaystyle \frac{h}{390}}\hfill \\ \hfill 390\cdot \tan {60}^{°}& =\hfill & h\hfill & \hfill & \mathbf{\text{Multiplying by}}\mathbf{390}\mathbf{\text{on both sides}}\hfill \\ \hfill 390\cdot \sqrt{3}& =\hfill & h\hfill & \hfill & \mathbf{\text{Substituting}}\mathbf{tan}{60}^{°}=\sqrt{3}\hfill \\ \hfill 675& \approx \hfill & h.\hfill & \hfill & \hfill \end{array}$$

The fireworks display is approximately 675 ft high.

Now Try Exercise 29.

Historically, the measure of an angle has been expressed in degrees, minutes, and seconds. One minute, denoted $1\prime$ , is such that $60\prime =1°$ , or $1\prime ={\displaystyle \frac{1}{60}}\cdot \left(1°\right)$ . One second, denoted $1"$ , is such that $60"=1\prime$ , or $1"={\displaystyle \frac{1}{60}}\cdot \left(1\prime \right)$ . Then 61 degrees, 27 minutes, 4 seconds could be written as $61°27\prime 4\text{″}$ . This $D^{°}M\text{'}S"$ form was common before the widespread use of calculators. Now the preferred notation is to express fraction parts of degrees in decimal degree form. For example, $61°27\prime 4^{″}\approx 61.45°$ in decimal degree form. Although the $\text{D}°\text{M}\prime {\text{S}}^{″}$ notation is still widely used in navigation, we will most often use the decimal form in this text.

Most calculators can convert $5°42\prime 30^{″}$ notation to decimal degree notation and vice versa. Procedures among calculators vary.

So far we have measured angles using degrees. Another useful unit for angle measure is the radian, which we will study in Section 6.4. Most calculators work with either degrees or radians. Be sure to use whichever mode is appropriate. In this section, we use the DEGREE mode.

Keep in mind the difference between an exact answer and an approximation. For example,

But using a calculator, you get an answer like

Calculators generally provide values only of the sine, cosine, and tangent functions. You can find values of the cosecant, secant, and cotangent by taking reciprocals of the sine, cosine, and tangent functions, respectively.

We can use a calculator to find an angle for which we know a trigonometric function value.

Example 9

Find the acute angle, to the nearest tenth of a degree, whose sine value is approximately 0.20113—that is, given $\text{sin}\theta =0.20113$, find $\theta$ .

Solution

The quickest way to find the angle with a calculator is to use an inverse function key. (We first studied inverse functions in Section 5.1 and will consider inverse trigonometric functions in Section 7.4.) First check to be sure that your calculator is in DEGREE mode. Usually two keys must be pressed in sequence. For this example, if we press

we find that the acute angle whose sine is 0.20113 is approximately $11.60304613°$, or $11.6°$ .

Now Try Exercise 75.

Example 10 Ladder Safety.

A window-washing crew has purchased new 30-ft extension ladders. The manufacturer states that the safest placement on a wall is to extend the ladder to 25 ft and to position the base 6.5 ft from the wall (Source: R. D. Werner Co., Inc.). What angle does the ladder make with the ground in this position?

Solution

We make a drawing and then use the most convenient trigonometric function. Because we know the length of the side adjacent to $\theta$ and the length of the hypotenuse, we choose the cosine function.

From the definition of the cosine function, we have

$$\cos \theta ={\displaystyle \frac{\text{adj}}{\text{hyp}}}={\displaystyle \frac{6.5\text{ ft}}{25\text{ ft}}}=\mathrm{0.26.}$$

Using a calculator, we find the acute angle whose cosine is 0.26:

Thus when the ladder is in its safest position, it makes an angle of about $75°$ with the ground.

Two angles are complementary whenever the sum of their measures is $90°$ . Each is the complement of the other. In a right triangle, the acute angles are complementary, because the sum of all three angle measures is $180°$ and the right angle accounts for $90°$ of this total. Thus if one acute angle of a right triangle is $\theta$, the other is $90°-\theta$ .

The six trigonometric function values of each of the acute angles in the right triangle below are listed on the right. Note that $53°$ and $37°$ are complementary angles because $53°+37°=90°$ .

For these angles, we note that

The sine of an angle is also the cosine of the angle’s complement. Similarly, the tangent of an angle is the cotangent of the angle’s complement, and the secant of an angle is the cosecant of the angle’s complement. These pairs of functions are called cofunctions. A list of cofunction identities follows.