Lesson Objectives

- Learn how to set up the unit circle in trigonometry
- Learn how to find values of circular functions

## What is the Unit Circle?

In our algebra course, we learned how to graph circles. A unit circle is of the form: $$x^2 + y^2=1$$ This equation tells us that the center of our unit circle is at the origin (0,0) and the radius is 1: Using the unit circle to work with our trigonometric functions will lead to some interesting results. Let's first set up an angle with a terminal side in quadrant I. We will let a point that is both on the terminal side of our angle and on the unit circle be represented by (x,y): Using our unit circle, how could we find sin θ and cos θ? Let's create a right triangle: What is sin θ? $$\text{sin}\hspace{.15em}θ=\frac{y}{r}=\frac{\text{opp}}{\text{hyp}}$$ We can use our definition of y/r or opp/hyp. Either way, we think about the fact that the radius is equal to 1 since we are using the unit circle. $$r=1$$ $$\text{sin}\hspace{.15em}θ=\frac{y}{r}=y$$ What about cos θ? $$\text{cos}\hspace{.15em}θ=\frac{x}{r}=\frac{\text{adj}}{\text{hyp}}$$ Again, we can use our definition of x/r or adj/hyp. Either way, we think about the fact that the radius is equal to 1 since we are using the unit circle. $$\text{cos}\hspace{.15em}θ=\frac{x}{r}=x$$ We can now state that sin θ = y and cos θ = x. We can replace our (x,y) point with (cos θ, sin θ):

Additionally, we know the formula for the arc length on a circle: $$s=θ \cdot r$$ Suppose, we start on our unit circle at the point (1,0) and measure an arc of length s along the circle. When s is positive, this movement is measured using a counterclockwise rotation. When s is negative, this movement is measured using a clockwise rotation. Let's let the endpoint on our arc be represented with the point (x,y): When working with the unit circle, the radius r is 1. This leads to s, the arc length being equal to θ, which is our radian measure. Therefore, the trigonometric functions of angle θ measured in radians found by choosing a point (x,y) on the unit circle can be rewritten as functions of the arc length s, a real number. When shown in this way, these functions are known as circular functions.

For any real number s represented by a directed arc on the unit circle: $$\text{sin}\hspace{.15em}s=y$$ $$\text{cos}\hspace{.15em}s=x$$ $$\text{tan}\hspace{.15em}s=\frac{x}{y}, y ≠ 0$$ $$\text{csc}\hspace{.15em}s=\frac{1}{y}, y ≠ 0$$ $$\text{sec}\hspace{.15em}s=\frac{1}{x}, x ≠ 0$$ $$\text{cot}\hspace{.15em}s=\frac{x}{y}, y ≠ 0$$

The other three function values can be found using the reciprocal identities:

Let's begin with 30° ($\frac{π}{6}$), 45° ($\frac{π}{4}$), and 60° ($\frac{π}{3}$): When working with the unit circle, the x-value of the point is the cosine of the angle and the y-value of the point is the sine of the angle. For example: $$\text{sin}\hspace{.1em}\frac{π}{6}=\frac{1}{2}$$ $$\text{cos}\hspace{.1em}\frac{π}{6}=\frac{\sqrt{3}}{2}$$ So a real number of $\frac{π}{6}$ is associated with the point: $\left(\frac{\sqrt{3}}{2},\frac{1}{2}\right)$ on the unit circle.

We can use a similar thought process to come up with the coordinates for all of the points given in quadrant I. Once this is done, we can use symmetry to find the other points. The unit circle is symmetric with respect to the x-axis, the y-axis, and the origin. This means we can reflect our points across the y-axis. The y-values will stay the same and the x-values will become their opposite: Now, we will reflect all the points across the x-axis. This means the x-values will be the same and the y-values will become their opposite: By memorizing the points in quadrant I, the points for the other three quadrants can easily be obtained by reflecting across the y-axis and then reflecting across the x-axis.

Example #1: Use the unit circle above to find the exact values. $$\require{cancel}\text{sin}\hspace{.1em}\frac{3π}{4}$$ $$\text{cos}\hspace{.1em}\frac{3π}{4}$$ $$\text{tan}\hspace{.1em}\frac{3π}{4}$$ On the unit circle sin s = y and cos s = x: $$\text{sin}\hspace{.1em}\frac{3π}{4}=\frac{\sqrt{2}}{2}$$ $$\text{cos}\hspace{.1em}\frac{3π}{4}=-\frac{\sqrt{2}}{2}$$ $$\text{tan}\hspace{.1em}\frac{3π}{4}=\frac{\cancel{\sqrt{2}}}{\cancel{2}}\cdot -\frac{\cancel{2}}{\cancel{\sqrt{2}}}=-1$$ Example #2: Use the unit circle above to find the exact values. $$\text{sin}\hspace{.1em}\frac{13π}{6}$$ $$\text{sec}\hspace{.1em}\frac{13π}{6}$$ $$\text{csc}\hspace{.1em}\frac{13π}{6}$$ When our s value is larger than 360° or 2$π$ radians, we want to find a coterminal angle that is between 0° and 360° or 0 and 2$π$ radians. We can simply subtract away 2$π$ radians until we get into this range. Let's multiply by 6/6 to obtain a common denominator and then do our subtraction: $$\frac{13π}{6}- \frac{12π}{6}=\frac{π}{6}$$ On the unit circle sin s = y and cos s = x:

We can find $\frac{π}{6}$ on the unit circle. $$\text{sin}\hspace{.1em}\frac{13π}{6}=\frac{1}{2}$$ $$\text{sec}\hspace{.1em}\frac{13π}{6}=\frac{2}{\sqrt{3}}=\frac{2\sqrt{3}}{3}$$ $$\text{csc}\hspace{.1em}\frac{13π}{6}=2$$ Example #3: Use the unit circle above to find the exact values. $$\text{sin}\hspace{.1em}\left({-}\frac{23π}{4}\right)$$ $$\text{cos}\hspace{.1em}\left({-}\frac{23π}{4}\right)$$ $$\text{cot}\hspace{.1em}\left({-}\frac{23π}{4}\right)$$ When working with a negative angle, we want to find a coterminal angle that is between 0° and 360° or 0 and 2$π$ radians. We can simply add 2$π$ radians until we get into this range. Let's multiply by 4/4 to obtain a common denominator and then do our addition: $${-}\frac{23π}{4}+ \frac{8π}{4}=-\frac{15π}{4}$$ $${-}\frac{15π}{4}+ \frac{8π}{4}=-\frac{7π}{4}$$ $${-}\frac{7π}{4}+ \frac{8π}{4}=\frac{π}{4}$$ On the unit circle sin s = y and cos s = x:

We can find $\frac{π}{4}$ on the unit circle. $$\text{sin}\hspace{.1em}\left({-}\frac{23π}{4}\right)=\frac{\sqrt{2}}{2}$$ $$\text{cos}\hspace{.1em}\left({-}\frac{23π}{4}\right)=\frac{\sqrt{2}}{2}$$ $$\text{cot}\hspace{.1em}\left({-}\frac{23π}{4}\right)=1$$

### Circular Functions

Recall when working with radian measure, we use the following formula: $$θ=\frac{s}{r}$$ Where s is the arc length, and r is the radius.Additionally, we know the formula for the arc length on a circle: $$s=θ \cdot r$$ Suppose, we start on our unit circle at the point (1,0) and measure an arc of length s along the circle. When s is positive, this movement is measured using a counterclockwise rotation. When s is negative, this movement is measured using a clockwise rotation. Let's let the endpoint on our arc be represented with the point (x,y): When working with the unit circle, the radius r is 1. This leads to s, the arc length being equal to θ, which is our radian measure. Therefore, the trigonometric functions of angle θ measured in radians found by choosing a point (x,y) on the unit circle can be rewritten as functions of the arc length s, a real number. When shown in this way, these functions are known as circular functions.

For any real number s represented by a directed arc on the unit circle: $$\text{sin}\hspace{.15em}s=y$$ $$\text{cos}\hspace{.15em}s=x$$ $$\text{tan}\hspace{.15em}s=\frac{x}{y}, y ≠ 0$$ $$\text{csc}\hspace{.15em}s=\frac{1}{y}, y ≠ 0$$ $$\text{sec}\hspace{.15em}s=\frac{1}{x}, x ≠ 0$$ $$\text{cot}\hspace{.15em}s=\frac{x}{y}, y ≠ 0$$

### Trigonometric Function Values of Special Angles

We previously learned about trigonometric function values of special angles. Certain angles appear very frequently: 30°, 45°, and 60°. The function values of these special angles can be summarized using the following table:θ | sin θ | cos θ | tan θ |
---|---|---|---|

30° | $\frac{1}{2}$ | $\frac{\sqrt{3}}{2}$ | $\frac{\sqrt{3}}{3}$ |

45° | $\frac{\sqrt{2}}{2}$ | $\frac{\sqrt{2}}{2}$ | $1$ |

60° | $\frac{\sqrt{3}}{2}$ | $\frac{1}{2}$ | $\sqrt{3}$ |

θ | cot θ | sec θ | csc θ |
---|---|---|---|

30° | $\sqrt{3}$ | $\frac{2\sqrt{3}}{3}$ | $2$ |

45° | $1$ | $\sqrt{2}$ | $\sqrt{2}$ |

60° | $\frac{\sqrt{3}}{3}$ | $2$ | $\frac{2\sqrt{3}}{3}$ |

## How to Remember the Unit Circle

In most trigonometry classes, you will be required to remember the unit circle in order to quickly solve problems. This can be done quite easily given a few basic facts.Let's begin with 30° ($\frac{π}{6}$), 45° ($\frac{π}{4}$), and 60° ($\frac{π}{3}$): When working with the unit circle, the x-value of the point is the cosine of the angle and the y-value of the point is the sine of the angle. For example: $$\text{sin}\hspace{.1em}\frac{π}{6}=\frac{1}{2}$$ $$\text{cos}\hspace{.1em}\frac{π}{6}=\frac{\sqrt{3}}{2}$$ So a real number of $\frac{π}{6}$ is associated with the point: $\left(\frac{\sqrt{3}}{2},\frac{1}{2}\right)$ on the unit circle.

We can use a similar thought process to come up with the coordinates for all of the points given in quadrant I. Once this is done, we can use symmetry to find the other points. The unit circle is symmetric with respect to the x-axis, the y-axis, and the origin. This means we can reflect our points across the y-axis. The y-values will stay the same and the x-values will become their opposite: Now, we will reflect all the points across the x-axis. This means the x-values will be the same and the y-values will become their opposite: By memorizing the points in quadrant I, the points for the other three quadrants can easily be obtained by reflecting across the y-axis and then reflecting across the x-axis.

Example #1: Use the unit circle above to find the exact values. $$\require{cancel}\text{sin}\hspace{.1em}\frac{3π}{4}$$ $$\text{cos}\hspace{.1em}\frac{3π}{4}$$ $$\text{tan}\hspace{.1em}\frac{3π}{4}$$ On the unit circle sin s = y and cos s = x: $$\text{sin}\hspace{.1em}\frac{3π}{4}=\frac{\sqrt{2}}{2}$$ $$\text{cos}\hspace{.1em}\frac{3π}{4}=-\frac{\sqrt{2}}{2}$$ $$\text{tan}\hspace{.1em}\frac{3π}{4}=\frac{\cancel{\sqrt{2}}}{\cancel{2}}\cdot -\frac{\cancel{2}}{\cancel{\sqrt{2}}}=-1$$ Example #2: Use the unit circle above to find the exact values. $$\text{sin}\hspace{.1em}\frac{13π}{6}$$ $$\text{sec}\hspace{.1em}\frac{13π}{6}$$ $$\text{csc}\hspace{.1em}\frac{13π}{6}$$ When our s value is larger than 360° or 2$π$ radians, we want to find a coterminal angle that is between 0° and 360° or 0 and 2$π$ radians. We can simply subtract away 2$π$ radians until we get into this range. Let's multiply by 6/6 to obtain a common denominator and then do our subtraction: $$\frac{13π}{6}- \frac{12π}{6}=\frac{π}{6}$$ On the unit circle sin s = y and cos s = x:

We can find $\frac{π}{6}$ on the unit circle. $$\text{sin}\hspace{.1em}\frac{13π}{6}=\frac{1}{2}$$ $$\text{sec}\hspace{.1em}\frac{13π}{6}=\frac{2}{\sqrt{3}}=\frac{2\sqrt{3}}{3}$$ $$\text{csc}\hspace{.1em}\frac{13π}{6}=2$$ Example #3: Use the unit circle above to find the exact values. $$\text{sin}\hspace{.1em}\left({-}\frac{23π}{4}\right)$$ $$\text{cos}\hspace{.1em}\left({-}\frac{23π}{4}\right)$$ $$\text{cot}\hspace{.1em}\left({-}\frac{23π}{4}\right)$$ When working with a negative angle, we want to find a coterminal angle that is between 0° and 360° or 0 and 2$π$ radians. We can simply add 2$π$ radians until we get into this range. Let's multiply by 4/4 to obtain a common denominator and then do our addition: $${-}\frac{23π}{4}+ \frac{8π}{4}=-\frac{15π}{4}$$ $${-}\frac{15π}{4}+ \frac{8π}{4}=-\frac{7π}{4}$$ $${-}\frac{7π}{4}+ \frac{8π}{4}=\frac{π}{4}$$ On the unit circle sin s = y and cos s = x:

We can find $\frac{π}{4}$ on the unit circle. $$\text{sin}\hspace{.1em}\left({-}\frac{23π}{4}\right)=\frac{\sqrt{2}}{2}$$ $$\text{cos}\hspace{.1em}\left({-}\frac{23π}{4}\right)=\frac{\sqrt{2}}{2}$$ $$\text{cot}\hspace{.1em}\left({-}\frac{23π}{4}\right)=1$$

#### Skills Check:

Example #1

Find the exact value of each trigonometric function. $$\text{cos}\hspace{.15em}\left(-\frac{11π}{6}\right)$$

Please choose the best answer.

A

$$0$$

B

$$\frac{\sqrt{3}}{2}$$

C

$$\frac{1}{2}$$

D

$$-1$$

E

$$\frac{2\sqrt{3}}{3}$$

Example #2

Find the exact value of each trigonometric function. $$\text{tan}\hspace{.15em}\left(\frac{19π}{6}\right)$$

Please choose the best answer.

A

$$-\frac{3}{2}$$

B

$$-\frac{1}{2}$$

C

$$\frac{\sqrt{3}}{3}$$

D

$$0$$

E

$$2$$

Example #3

Find the exact value of each trigonometric function. $$\text{cot}\hspace{.15em}\left(\frac{17π}{3}\right)$$

Please choose the best answer.

A

$$-\frac{\sqrt{2}}{3}$$

B

$$\frac{1}{2}$$

C

$$-\frac{\sqrt{2}}{4}$$

D

$$\frac{2}{3}$$

E

$$-\frac{\sqrt{3}}{3}$$

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