# 1.3: Basic Trigonometric Functions

**At Grade**Created by: CK-12

## Learning Objectives

- Find the values of the six trigonometric functions for angles in right triangles.

## Introduction

Consider a situation in which you are building a ramp for wheelchair access to a building. If the ramp must have a height of 8 feet, and the angle of the ramp must be about

Solving this kind of problem requires trigonometry. 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.

## 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

Since all right triangles with the same acute angles are similar, this function 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 over the hypotenuse of the triangle. Using this triangle, we have:

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:

There are a few important things to note about the way we write these functions. First, keep in mind that the abbreviations *sine*, with a long

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

**Solution:**

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

If

If we use triangle

Also notice that the tangent function is the same as the slope of the hypotenuse.

**Example 2:** Find

**Solution:**

Using triangle

Using triangle

An easy way to remember the ratios of the sine, cosine, and tangent functions is SOH-CAH-TOA.

## Secant, Cosecant, and Cotangent Functions

We can define three more functions also based on a right triangle. They are the reciprocals of sine, cosine and tangent.

If

If

If

**Example 3:** Find the secant, cosecant, and cotangent of angle

**Solution:**

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

\begin{align*}5^2 + 12^2 & = H^2\\ 25 + 144 & = H^2\\ 169 & = H^2\\ H & = 13\end{align*}

Now we can find the secant, cosecant, and cotangent of angle \begin{align*}B\end{align*}:

\begin{align*}\sec B & = \frac{\text{hypotenuse}}{\text{adjacent side}} = \frac{13}{12}\\ \csc B & = \frac{\text{hypotenuse}}{\text{opposite side}} = \frac{13}{5}\\ \cot B & = \frac{\text{adjacent side}}{\text{opposite side}} = \frac{12}{5}\end{align*}

## Points to Consider

- Do you notice any similarities between the sine of one angle and the cosine of the other, in the same triangle?

## Review Questions

- Find the values of the six trig functions of angle \begin{align*}A\end{align*}.
- Consider triangle \begin{align*}VET\end{align*} below. Find the length of the hypotenuse and values of the six trig functions of angle \begin{align*}T\end{align*}.
- Consider the right triangle below.
- Find the hypotenuse.
- Find the six trigonometric functions of \begin{align*}\angle{X}\end{align*}.
- Find the six trigonometric functions of \begin{align*}\angle{Z}\end{align*}.

- Looking back at #3, are any functions of \begin{align*}\angle{X}\end{align*} equal to any of the functions of \begin{align*}\angle{Z}\end{align*}? If so, which ones? Do you think this could be generalized for ANY pair of acute angles in the same right triangle (also called complements)?
- Consider an isosceles right triangle with legs of length 2. Find the sine, cosine and tangent of both acute angles.
- Consider an isosceles right triangle with legs of length \begin{align*}x\end{align*}. Find the sine, cosine and tangent of both acute angles. Write down any similarities or patterns you notice with #5.
- Consider a \begin{align*}30-60-90\end{align*} triangle with hypotenuse of length 10. Find the sine, cosine and tangent of both acute angles.
- Consider a \begin{align*}30-60-90\end{align*} triangle with short leg of length \begin{align*}x\end{align*}. Find the sine, cosine and tangent of both acute angles. Write down any similarities or patterns you notice with #7.
- Consider a right triangle, \begin{align*}ABC\end{align*}. If \begin{align*}\sin A = \frac{9}{41}\end{align*}, find the length of the third side.

## Review Answers

- \begin{align*}\sin A = \frac{9}{15} = \frac{3}{5}, \cos A = \frac{12}{15} = \frac{4}{5}, \tan A = \frac{9}{12} = \frac{3}{4}, \csc A = \frac{15}{9} = \frac{5}{3}, \sec A = \frac{15}{12} = \frac{5}{4}, \cot A = \frac{12}{9} = \frac{4}{3}\end{align*}
- The hypotenuse is \begin{align*}17 \left ( \sqrt{15^2 + 8^2} = \sqrt{225 + 64} = \sqrt{289} = 17 \right ).\end{align*} \begin{align*}\sin T = \frac{15}{17}, \cos T = \frac{8}{17}, \tan T = \frac{15}{8}, \csc T = \frac{17}{15}, \sec T = \frac{17}{8}, \cot T = \frac{8}{15}\end{align*}
- The hypotenuse is \begin{align*}13 \left ( \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \right ).\end{align*}
- \begin{align*}\sin X = \frac{12}{13}, \cos X = \frac{5}{13}, \tan X = \frac{12}{5}, \csc X = \frac{13}{12}, \sec X = \frac{13}{5}, \cot X = \frac{5}{12}\end{align*}
- \begin{align*}\sin Z = \frac{5}{13}, \cos Z = \frac{12}{13}, \tan Z = \frac{5}{12}, \csc Z = \frac{13}{5}, \sec Z = \frac{13}{12}, \cot Z = \frac{12}{5}\end{align*}

- From #3, we can conclude that \begin{align*}\sin X = \cos Z, \cos X = \sin Z, \tan X = \cot Z, \cot X = \tan Z, \csc X = \sec Z\end{align*} and \begin{align*}\sec X = \csc Z\end{align*}. Yes, this can be generalized for all complements.
- The hypotenuse is \begin{align*}2\sqrt{2}\end{align*}. Each angle is \begin{align*}45^\circ\end{align*}, so the sine, cosine, and tangent are the same for both angles. \begin{align*}\sin 45^\circ = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}, \cos 45^\circ = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}, \tan 45^\circ = \frac{2}{2} = 1\end{align*}
- If the legs are length \begin{align*}x\end{align*}, then the hypotenuse is \begin{align*}x\sqrt{2}\end{align*}. For \begin{align*}45^\circ\end{align*}, the sine, cosine, and tangent are: \begin{align*}\sin 45^\circ = \frac{x}{x\sqrt{2}} = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}, \cos 45^\circ = \frac{x}{x\sqrt{2}} = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}, \tan 45^\circ = \frac{x}{x} = 1\end{align*} This tells us that regardless of the length of the sides of an isosceles right triangle, the sine, cosine and tangent of \begin{align*}45^\circ\end{align*} are always the same.
- If the hypotenuse is 10, then the short leg is 5 and the long leg is \begin{align*}5\sqrt{3}\end{align*}. Recall, that \begin{align*}30^\circ\end{align*} is opposite the short side, or 5, and \begin{align*}60^\circ\end{align*} is opposite the long side, or \begin{align*}5\sqrt{3}\end{align*}. \begin{align*}\sin 30^\circ & = \frac{5}{10} = \frac{1}{2}, \cos 30^\circ = \frac{5\sqrt{3}}{10} = \frac{\sqrt{3}}{2}, \tan 30^\circ = \frac{5}{5\sqrt{3}} = \frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{3}\\ \sin 60^\circ & = \frac{5\sqrt{3}}{10} = \frac{\sqrt{3}}{2}, \cos 60^\circ = \frac{5}{10} = \frac{1}{2}, \tan 60^\circ = \frac{5\sqrt{3}}{5} = \sqrt{3}\end{align*}
- If the short leg is \begin{align*}x\end{align*}, then the long leg is \begin{align*}x\sqrt{3}\end{align*} and the hypotenuse is \begin{align*}2x\end{align*}. \begin{align*}30^\circ\end{align*} is opposite the short side, or \begin{align*}x\end{align*}, and \begin{align*}60^\circ\end{align*} is opposite the long side, or \begin{align*}x\sqrt{3}\end{align*}. \begin{align*}\sin 30^\circ & = \frac{x}{2x} = \frac{1}{2}, \cos 30^\circ = \frac{x\sqrt{3}}{2x} = \frac{\sqrt{3}}{2}, \tan 30^\circ = \frac{x}{x\sqrt{3}} = \frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{3}\\ \sin 60^\circ & = \frac{x\sqrt{3}}{2x} = \frac{\sqrt{3}}{2}, \cos 60^\circ = \frac{x}{2x} = \frac{1}{2}, \tan 60^\circ = \frac{x\sqrt{3}}{x} = \sqrt{3}\end{align*} This tells us that regardless of the length of the sides of a \begin{align*}30-60-90\end{align*} triangle, the sine, cosine and tangent of \begin{align*}30^\circ\end{align*} and \begin{align*}60^\circ\end{align*} are always the same. Also, \begin{align*}\sin 30^\circ = \cos 60^\circ\end{align*} and \begin{align*}\cos 30^\circ = \sin 60^\circ\end{align*}.
- If \begin{align*}\sin A = \frac{9}{41}\end{align*}, then the opposite side is \begin{align*}9x\end{align*} (some multiple of 9) and the hypotenuse is \begin{align*}41x\end{align*}. Therefore, working with the Pythagorean Theorem would give us the length of the other leg. Also, we could notice that this is a Pythagorean Triple and the other leg is \begin{align*}40x\end{align*}.

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