Lesson Objectives

- Learn how to find reference angles
- Learn how to find trigonometric function values for non-acute angles

## How to Find Reference Angles

In this lesson, we will learn how to find trigonometric function values of non-acute angles. We will begin by discussing the concept of reference angles. Recall that quadrantal angles are angles in standard position with measures that are multiples of 90°: (90°, 180°, 270°,...). Every nonquadrantal angle in standard position will have a positive acute angle known as its reference angle. The reference angle for the angle θ is written as θ' (read as theta prime). θ' is the positive acute angle that is made by the terminal side of our angle θ and the x-axis. To see this more clearly, let's look at three diagrams, one for quadrants II, III, and IV. Note that in quadrant I, θ and θ' are the same.

Example #1: Find the reference angle for each angle.

330°

Since 330° is between 0° and 360° and lies in quadrant IV, we can find the reference angle by subtracting 360° - 330°.

360° - 330° = 30° When our angle θ is negative or has a measure that is greater than 360°, its reference angle is found by finding its coterminal angle that is between 0° and 360°. Let's look at a few examples.

Example #2: Find the reference angle for each angle.

-250°

Since -250° is negative, we first need to find a coterminal angle that is between 0° and 360°. Let's add 360° to -250°.

-250° + 360° = 110°

Now, we will find the reference angle for 110°. Since this angle lies in quadrant II, we will subtract 180° - 110°.

180° - 110° = 70° Example #3: Find the reference angle for each angle.

560°

Since 560° is greater than 360°, we first need to find a coterminal angle that is between 0° and 360°. Let's subtract 360° from 560°.

560° - 360° = 200°

Now, we will find the reference angle for 200°. Since this angle lies in quadrant III, we will subtract 200° - 180°.

200° - 180° = 20°

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

Now that we have a way to quickly obtain trigonometric function values for 30°, 45°, and 60°, let's use the concept of a reference angle to find function values of angles in quadrants II, III, and IV. We will place our 30°-60°-90° triangle on the coordinate plane. We will use our above sketch of a 150° angle to think about our sine, cosine, and tangent values of a quadrant II angle. We pick the point on the terminal side with coordinates: $$x=-\sqrt{3}$$ $$y=1$$ $$r=2$$ $$\text{sin}\hspace{.1em}θ=\frac{y}{r}=\frac{1}{2}$$ $$\text{cos}\hspace{.1em}θ=\frac{x}{r}=-\frac{\sqrt{3}}{2}$$ $$\text{tan}\hspace{.1em}θ=\frac{y}{x}=-\frac{\sqrt{3}}{3}$$ We can see that the trigonometric function values for 150° correspond to the absolute value of the trigonometric function values for the reference angle of 30°. If we want to find the trigonometric functions for a non-acute angle, we can simply find the function values for the reference angle and then apply the correct sign based on the sign rules for trigonometric functions. Let's summarize our steps below.

Example 4: Use the table above to find the exact value of each expression.

cos(-600°)

First, we want to find our reference angle. Since we have a -600° angle, we first find a coterminal angle between 0° and 360°.

-600° + 2 • 360° = -600° + 720° = 120°

Since a 120° angle lies in quadrant II, we want to subtract 180° - 120°.

180° - 120° = 60°

Now, we will find the value of cos(60°). To do this, we can reference our table above. $$\text{cos}(60°)=\frac{1}{2}$$ Lastly, let's use our sign rules to determine the correct sign. Since our angle -600° or its coterminal angle 120° lies in quadrant II, we know that cos θ is negative. Let's obtain our final answer by changing the sign: $$\text{cos}(-600°)=-\text{cos}(60°)=-\frac{1}{2}$$ Example 5: Use the table above to find the exact value of each expression.

sec(495°)

First, we want to find our reference angle. Since we have a 495° angle, we first find a coterminal angle between 0° and 360°.

495° - 360° = 135°

Since a 135° angle lies in quadrant II, we want to subtract 180° - 135°.

180° - 135° = 45°

Now, we will find the value of sec(45°). To do this, we can reference our table above. $$\text{sec}(45°)=\sqrt{2}$$ Lastly, let's use our sign rules to determine the correct sign. Since our angle 495° or its coterminal angle 135° lies in quadrant II, we know that sec θ is negative. Let's obtain our final answer by changing the sign: $$\text{sec}(495°)=-\text{sec}(45°)=-\sqrt{2}$$

### Reference Angle θ' for θ, where 0° < θ < 360°:

Quadrant | Reference Angle |
---|---|

Q I | θ' = θ |

Q II | θ' = 180° - θ |

Q III | θ' = θ - 180° |

Q IV | θ' = 360° - θ |

### Reference Angle in Quadrant II

θ' = 180° - θ### Reference Angle in Quadrant III

θ' = θ - 180°### Reference Angle in Quadrant IV

θ' = 360° - θ Let's look at an example.Example #1: Find the reference angle for each angle.

330°

Since 330° is between 0° and 360° and lies in quadrant IV, we can find the reference angle by subtracting 360° - 330°.

360° - 330° = 30° When our angle θ is negative or has a measure that is greater than 360°, its reference angle is found by finding its coterminal angle that is between 0° and 360°. Let's look at a few examples.

Example #2: Find the reference angle for each angle.

-250°

Since -250° is negative, we first need to find a coterminal angle that is between 0° and 360°. Let's add 360° to -250°.

-250° + 360° = 110°

Now, we will find the reference angle for 110°. Since this angle lies in quadrant II, we will subtract 180° - 110°.

180° - 110° = 70° Example #3: Find the reference angle for each angle.

560°

Since 560° is greater than 360°, we first need to find a coterminal angle that is between 0° and 360°. Let's subtract 360° from 560°.

560° - 360° = 200°

Now, we will find the reference angle for 200°. Since this angle lies in quadrant III, we will subtract 200° - 180°.

200° - 180° = 20°

### Trigonometric Function Values of Special Angles

Certain angles appear very frequently: 30°, 45°, and 60°. To find the six trigonometric function values for 30° and 60°, we will begin with an equilateral triangle, which is a triangle with all sides of equal length. Additionally, each angle of the triangle has a measure of 60°. We will choose a triangle such that each side has length 2. When we bisect one angle of the above equilateral triangle, we obtain two right triangles, with angles of 30°, 60°, and 90°. For convenience, let's use the right triangle on the right side. At this point, we know the hypotenuse is 2, and the side opposite the 30° angle has length 1. For the side adjacent to the 30° angle, we can find the length using the Pythagorean theorem: $$a^2 + b^2=c^2$$ Where a and b are the lengths of the legs and c is the length of the hypotenuse: $$a^2 + 1^2=2^2$$ $$a^2 + 1=4$$ $$a^2=3$$ $$a=\pm \sqrt{3}$$ Since a is a side length, we will disregard the negative square root. The side that is adjacent to the 30° angle has length $\sqrt{3}$. Now we can use our 30°-60°-90° triangle to find the values for the sine, cosine, and tangent. Let's begin with 30°: $$\text{sin}\hspace{.1em}30°=\frac{\text{opp}}{\text{hyp}}=\frac{1}{2}$$ $$\text{cos}\hspace{.1em}30°=\frac{\text{adj}}{\text{hyp}}=\frac{\sqrt{3}}{2}$$ $$\text{tan}\hspace{.1em}30°=\frac{\text{opp}}{\text{adj}}=\frac{\sqrt{3}}{3}$$ Now, we will move on to 60°: $$\text{sin}\hspace{.1em}60°=\frac{\text{opp}}{\text{hyp}}=\frac{\sqrt{3}}{2}$$ $$\text{cos}\hspace{.1em}60°=\frac{\text{adj}}{\text{hyp}}=\frac{1}{2}$$ $$\text{tan}\hspace{.1em}60°=\frac{\text{opp}}{\text{adj}}=\sqrt{3}$$ To find the trigonometric function values for a 45° angle, let's use a 45°-45°-90° triangle. This triangle is isosceles and has two sides of equal length. We will choose the lengths of the two equal sides to be 1. This gives us a hypotenuse of $\sqrt{2}$. $$\text{sin}\hspace{.1em}45°=\frac{\text{opp}}{\text{hyp}}=\frac{\sqrt{2}}{2}$$ $$\text{cos}\hspace{.1em}45°=\frac{\text{adj}}{\text{hyp}}=\frac{\sqrt{2}}{2}$$ $$\text{tan}\hspace{.1em}45°=\frac{\text{opp}}{\text{adj}}=1$$ 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° | $\frac{\sqrt{3}}{3}$ | $2$ | $\frac{2\sqrt{3}}{3}$ |

## Finding Trigonometric Function Values for Any Nonquadrantal Angle θ

- Find the reference angle θ'
- Find the trigonometric function values for reference angle θ'
- Use the sign rules to determine the correct signs for each function

Example 4: Use the table above to find the exact value of each expression.

cos(-600°)

First, we want to find our reference angle. Since we have a -600° angle, we first find a coterminal angle between 0° and 360°.

-600° + 2 • 360° = -600° + 720° = 120°

Since a 120° angle lies in quadrant II, we want to subtract 180° - 120°.

180° - 120° = 60°

Now, we will find the value of cos(60°). To do this, we can reference our table above. $$\text{cos}(60°)=\frac{1}{2}$$ Lastly, let's use our sign rules to determine the correct sign. Since our angle -600° or its coterminal angle 120° lies in quadrant II, we know that cos θ is negative. Let's obtain our final answer by changing the sign: $$\text{cos}(-600°)=-\text{cos}(60°)=-\frac{1}{2}$$ Example 5: Use the table above to find the exact value of each expression.

sec(495°)

First, we want to find our reference angle. Since we have a 495° angle, we first find a coterminal angle between 0° and 360°.

495° - 360° = 135°

Since a 135° angle lies in quadrant II, we want to subtract 180° - 135°.

180° - 135° = 45°

Now, we will find the value of sec(45°). To do this, we can reference our table above. $$\text{sec}(45°)=\sqrt{2}$$ Lastly, let's use our sign rules to determine the correct sign. Since our angle 495° or its coterminal angle 135° lies in quadrant II, we know that sec θ is negative. Let's obtain our final answer by changing the sign: $$\text{sec}(495°)=-\text{sec}(45°)=-\sqrt{2}$$

#### Skills Check:

Example #1

Find the reference angle.

-195°

Please choose the best answer.

A

85°

B

15°

C

25°

D

20°

E

-165°

Example #2

Find the exact value.

cot(-300°)

Please choose the best answer.

A

$$\text{cot}(-300°)=-1$$

B

$$\text{cot}(-300°)=1$$

C

$$\text{cot}(-300°)=2$$

D

$$\text{cot}(-300°)=\frac{\sqrt{3}}{3}$$

E

$$\text{cot}(-300°)=\frac{\sqrt{3}}{2}$$

Example #3

Find the exact value.

sec (120°)

Please choose the best answer.

A

$$\text{sec}(120°)=-1$$

B

$$\text{sec}(120°)=\frac{\sqrt{2}}{2}$$

C

$$\text{sec}(120°)=2$$

D

$$\text{sec}(120°)=-\frac{2\sqrt{3}}{3}$$

E

$$\text{sec}(120°)=-2$$

Congrats, Your Score is 100%

Better Luck Next Time, Your Score is %

Try again?

Ready for more?

Watch the Step by Step Video Lesson Take the Practice Test