Trigonometry: Degrees and Radians

Part 1 of 7
  • Lectures 27
  • Video 14 Hours
  • Skill level beginner level
  • Languages English
  • Includes Lifetime access
    30 day money back guarantee!
    Available on iOS and Android
    Certificate of Completion

How taking a course works

Discover

Find online courses made by experts from around the world.

Learn

Take your courses with you and learn anywhere, anytime.

Master

Learn and practice real-world skills and achieve your goals.

Course Description

Trigonometry: Degrees and Radians

This lesson consists of a workbook and animated solutions for the workbook.

Workbook: The workbook for Degrees and Radians can be found in the introduction. Students should make two printouts of this workbook. Label the first printout "Lecture Notes", and label the second printout "Practice".

Videos: The videos present a fully animated solution for each question in the workbook.

Strategy for Degrees and Radians:

This lesson takes four days to complete, with a time commitment of 60-75 minutes each day. Students should work through this lesson as follows:

Watch the videos and fill in your Lecture Notes workbook. (30 - 45 minutes)

In the Practice workbook, do the examples you just learned on your own, without external help. This step is required to reinforce the concepts and gain mastery. (15 - 30 minutes)

Curriculum Fit:

This lesson explores its subject matter in full depth. Teachers are encouraged to compare this lesson with their local standards and assign examples to their students accordingly. 

Calculator:

Calculator references in this lesson apply to the TI-83/83+/84 series.

What are the requirements?

  • basic algebra and trigonometry skills

What am I going to get from this course?

  • Over 27 lectures and 13.5 hours of content!
  • define degrees, radians, and revolutions
  • understand common angle terminology
  • perform conversions between degrees, radians, and revolutions
  • draw angles in standard position
  • use strategy to solve problems involving large angles, of varying types
  • for a given triangle, find the exact values of the six trigonometric ratios
  • understand the relationship between a triangle's quadrant and its trigonometric ratios
  • define arc length and use this concept to solve problems.
  • define angular speed and use this concept to solve problems

What is the target audience?

  • precalculus students
  • trigonometry students

What you get with this course?

Not for you? No problem.
30 day money back guarantee

Forever yours.
Lifetime access

Learn on the go.
Desktop, iOS and Android

Get rewarded.
Certificate of completion

Curriculum

Section 1: DAY ONE: Examples 1 - 4 (31 minutes)
05:59
Example 1: Basic Angle Terminology
  1. Define each term or phrase and draw a sample angle. In Part A, we'll define an angle in standard position.
  2. Before we can define an angle in standard position, we need to learn how to construct an angle.
  3. Draw a line in blue along the positive x-axis. This line is called the initial arm, and it's fixed in position.
  4. Now draw a second line in green that is free to rotate around the origin. This line is called the terminal arm.
  5. The rotation angle between the initial arm and the terminal arm is called the standard position angle.
  6. In Part B, we'll define positive and negative angles.
  7. An angle in standard position can be positive or negative, depending on which direction it is rotated.
  8. An angle is positive if we rotate the terminal arm counter-clockwise. In the animation, the terminal arm is rotated to positive 120°.
  9. An angle is negative if we rotate the terminal arm clockwise. In the animation, the terminal arm is rotated to negative 120°.
  10. In Part C, we'll define a reference angle.
  11. The angle formed between the terminal arm and the x-axis is called the reference angle.
  12. As an example, we'll find the reference angle of 150°.
  13. We can find the reference angle by drawing a vertical line from the terminal arm to the x-axis.
  14. The x-axis is at 180° and the angle is at 150°. Subtract the angles to get the reference angle of 30°.
  15. In Part D, we'll define co-terminal angles.
  16. If the terminal arm is rotated by a multiple of 360° in either direction, it will return to its original position. These angles are called co-terminal angles.
  17. To illustrate this, start with 60°.
  18. If we make one full rotation from this position, we get 60°+360°= 420°.
  19. We call 420° a co-terminal angle of 60° since the terminal arm occupies the same position.
  20. In Part E we'll define the principal angle.
  21. A principal angle is simply an angle that exists between 0°and 360°. We often discuss principal angles when finding the root angle of a co-terminal angle.
  22. For example, suppose we have an angle at 420°.
  23. The angle 420° has a principal angle of 60°. This is found by unwinding the terminal arm 360° back to the domain 0° ≤ θ < 360°.
  24. In Part F we'll define the general form of co-terminal angles.
  25. There are an infinite number of co-terminal angles possible for any given angle. We can represent all of these potential angles with the formula shown, where θc is the co-terminal angle, θi is the initial angle, and n is the number of full rotations.
  26. We use the positive sign when rotating counter-clockwise, and the negative sign when rotating clockwise.
  27. For an initial angle of 45°, the general form is θc = 45° + 360n when rotating counter-clockwise.
  28. The animation shows the first four positive co-terminal angles.
  29. For an initial angle of 45°, the general form is θc = 45° - 360n when rotating clockwise.
  30. The animation shows the first four negative co-terminal angles.
13:02
Example 2: Angle Types and Conversion Multipliers
  1. In this example we will learn about three angle types: Degrees, Radians, and Revolutions. In Part A, we'll define each term.
  2. One degree is defined as 1/360th of a full rotation.
  3. The diagram shows the terminal arm rotated 1°.
  4. One radian is the angle formed when the terminal arm swipes out an arc that has the same length as the terminal arm.
  5. The animation shows the terminal arm rotating 1 radian. The length of the arc is r, which is the same length as the terminal arm.
  6. One radian is approximately 57.3°.
  7. Radians are expressed as unitless quantities. If you see an angle written as 3 (without the degree symbol), it means 3 radians.
  8. Another way to visualize 1 radian is to do the following: Build an equilateral triangle with two πeces of wood and a flexible πece of metal.
  9. Now bend the triangle so the metal side pops out into an arc that corresponds to a circle with a radius of r.
  10. The angle between the πeces of wood is now 1 radian, or 57.3°.
  11. One revolution is defined as 360º, or 2π. It is one complete rotation around a circle.
  12. Note that radian angles are often expressed as multiples of π. This is done because an angle like 2π is an exact value, but its decimal equivalent 6.28 is just a rounded approximation.
  13. In Part B, use conversion multipliers to answer the questions and fill in the reference chart.
  14. A conversion multiplier is simply a fraction that is equivalent to 1. The numerator has units that we are trying to obtain, and the denominator has units that we are trying to eliminate.
  15. You've probably been using conversion multipliers for a while already. For example, to convert 3 km to metres, we perform the calculation shown.
  16. The fraction 1000m/1km is the conversion multiplier. The unwanted units, kilometres, cancel out, leaving the desired units, metres.
  17. Now we'll use conversion multipliers to convert one type of angle to a different type of angle.
  18. We can convert degrees to radians with the conversion multiplier π/180°. Note that π can be thought of as having units of radians, but this is usually left unwritten.
  19. Multiply 23° by the conversion multiplier.
  20. Use your calculator to get the numerical value 0.40. Degrees cancel, leaving units of radians. 23° is approximately 0.40 radians.
  21. Use estimation to see if this answer makes sense. 23° is about 40% of one radian (57.3°). 0.40 rad is also 40% of 1 radian. Therefore, 23° and 0.4 rad are approximately the same angle.
  22. Now we'll move on to the next conversion. We can convert degrees to revolutions with the conversion multiplier 1 rev/360°.
  23. Multiply 23° by the conversion multiplier.
  24. Use your calculator to get the numerical value 0.06. Degrees cancel, leaving units of revs. 23° is approximately 0.06 revolutions.
  25. We can use estimation to see if this answer makes sense. 23° is about one-sixteenth of a circle. 0.06 rev is about one-sixteenth of one revolution. Therefore, 23° and 0.06 rev are approximately the same angle.
  26. Now we'll move on to the next conversion. We can convert radians to degrees with the conversion multiplier 180°/π.
  27. Multiply 2.6 radians by the conversion multiplier.
  28. While radians are normally expressed without units, it's acceptable to use rad (short for radians) as a unit. This is especially helpful when using conversion multipliers.
  29. In this conversion, the radians cancel leaving units of degrees.
  30. 2.6 radians is approximately 148.97°.
  31. We can use estimation to see if this answer makes sense. If 1 radian is 57.3°, then 2.6 radians is approximately 149°. We get this with the calculation (2.6 × 57.3° ≈ 149°). Therefore, 2.6 rad and 149° are approximately the same angle.
  32. Now we'll move on to the next conversion. We can convert radians to revolutions with the conversion multiplier 1 rev/2π.
  33. Multiply 2.6 radians by the conversion multiplier.
  34. Use your calculator to get the numerical value 0.41. Radians cancel leaving units of revs.
  35. We can use estimation to see if this answer makes sense. 2.6 radians (149°) is about 40% of a circle. 0.41 revolutions is about 40% of a circle as well. Therefore, 2.6 rad and 0.41 rev are approximately the same angle.
  36. Now we'll move on to the next conversion. We can convert revolutions to degrees with the conversion multiplier 360°/1 rev.
  37. Multiply 0.75 revs by the conversion multiplier.
  38. The units of revolutions cancel out, leaving 270°.
  39. Use estimation to see if this answer makes sense. 0.75 rev is three-quarters around a circle. 270° is also three-quarters around a circle. Therefore, 0.75 rev and 270° are approximately the same angle.
  40. Now we'll move on to the next conversion. We can convert revolutions to radians with the conversion multiplier 2π/1 rev.
  41. Multiply 0.75 revs by the conversion multiplier.
  42. The units of revolutions cancel out, leaving 4.71 radians.
  43. Use estimation to see if this answer makes sense. 0.75 rev is three-quarters around a circle. If 1 radian is 57.3°, then 4.71 rad = 270°. We get this with the calculation (4.71 × 57.3° = 270°). Therefore, 0.75 rev and 4.71 rad are approximately the same angle.
  44. In Part C, contrast the decimal approximation of a radian with the exact value of a radian.
  45. When radians like 2.6, 4, or -0.712 are the result of a calculation, we call these values decimal approximations of a radian. They are not exact values since information is lost when rounding occurs.
  46. Radians like 3π, 5π/4, and -3π/2 are called exact value radians.
  47. In mathematics, exact values are considered more valuable than decimal approximations. Completing the intermediate steps of a multi-step problem using decimal approximations compounds information loss and yields unpredictable answers.
  48. Using exact values avoids this problem, since no rounding needs to occur and the final answer will always be the same.
  49. Convert 45° to a radian. In this conversion, the answer should be expressed as an approximate decimal.
  50. Multiply 45° by the conversion multiplier π/180°. Degrees cancel out, leaving 0.79 radians.
  51. Now we'll convert 45° to radians a second time. In this conversion, the answer should be expressed as an exact-value radian.
  52. Start the conversion the same way as before, using the conversion multiplier π/180°.
  53. Multiply to combine the values into one fraction. Note that degrees cancel.
  54. Reduce the fraction 45/180 to ¼. The exact value radian is π/4.
  55. You can use your calculator to help reduce fractions. Type in 45/180 (without the π) to get 0.25. Press MATH, select Frac (option 1), and press ENTER twice. The calculator will return the reduced fraction, 1/4.
06:06
Example 3: Angle Conversion Practice
  1. Convert each angle to the requested form, and round all decimals to the nearest hundredth. In Part A, we'll convert 175° to an approximate radian decimal.
  2. We'll use the conversion multiplier reference chart to assist us with our work.
  3. We are converting from degrees to radians, so multiply 175° by π/180°.
  4. Using a calculator, multiply to get 3.05 radians.
  5. In Part B, convert 210° to an exact-value radian.
  6. We'll use the conversion multiplier reference chart to assist us with our work.
  7. We are converting from degrees to radians, so multiply 210° by π/180°.
  8. Reduce the fraction to get 7π/6.
  9. In Part C, convert 120° to an exact-value revolution.
  10. We'll use the conversion multiplier reference chart to assist us with our work.
  11. We are converting from degrees to revolutions, so multiply 120° by 1 rev/360°.
  12. Reduce the fraction to get 1/3 revolutions.
  13. In Part D, convert 2.5 radians to degrees. Recall that an angle without any units is actually radians.
  14. We'll use the conversion multiplier reference chart to assist us with our work.
  15. We are converting from radians to degrees, so multiply 2.5 by 180°/π.
  16. Use your calculator to get 143.24°.
  17. In Part E, convert 3π/2 to degrees.
  18. We'll use the conversion multiplier reference chart to assist us with our work.
  19. We are converting from radians to degrees, so multiply 3π/2 by 180°/π.
  20. Use canceling and multiply to get 270°.
  21. In Part F, write 3π/2 as an approximate radian decimal.
  22. We do not need a conversion multiplier to go from an exact-value radian to an approximate radian decimal.
  23. Just evaluate the radian in your calculator! This gives an answer of 4.71 radians.
  24. In Part G, convert π/2 to an exact-value revolution.
  25. We'll use the conversion multiplier reference chart to assist us with our work.
  26. We are converting from radians to revolutions, so multiply π/2 by 1 rev/2π.
  27. Cancel and multiply to get 1/4 rev.
  28. In Part H, convert 0.5 rev to degrees.
  29. We'll use the conversion multiplier reference chart to assist us with our work.
  30. We are converting from revolutions to degrees, so multiply 0.5 rev by 360°/1 rev.
  31. Multiply to get 180°.
  32. In Part I, convert 3 rev to radians.
  33. We'll use the conversion multiplier reference chart to assist us with our work.
  34. We are converting from revolutions to radians, so multiply 3 rev by 2π/1 rev.
  35. Multiply to get 6π radians.
06:14
Example 4: Commonly Used Angles
  1. The diagram shows commonly used degrees. Find exact-value radians that correspond to each degree. When complete, memorize the diagram. In Part A, we'll find all exact value radians using a conversion multiplier.
  2. We can convert from degrees to radians using the conversion multiplier π/180°.
  3. 0° is the same as 0 radians.
  4. Multiply 30° by the conversion multiplier and reduce the fraction to get π/6.
  5. Multiply 45° by the conversion multiplier and reduce the fraction to get π/4.
  6. Multiply 60° by the conversion multiplier and reduce the fraction to get π/3.
  7. The rest of the radians can be found with similar calculations.
  8. In Part B, we'll find the exact-value radians using a shortcut.
  9. In the shortcut, we only need to know two radians: π/6 (which is 30°) and π/4 (which is 45°). In the first step, express all multiples of 30° in terms of π/6 and reduce the fraction. In the second step, express all multiples of 45° in terms of π/4 and reduce the fraction.
  10. We'll begin by finding exact-value radians that are multiples of π/6.
  11. In the first quadrant, 30° is 1π/6, 60° is 2π/6, and 90° is 3π/6. 45° is not a multiple of 30°, so ignore it for now.
  12. In the second quadrant, 120° is 4π/6, 150° is 5π/6, and 180° is 6π/6. 135° is not a multiple of 30°, so ignore it for now.
  13. In the third quadrant, 210° is 7π/6, 240° is 8π/6, and 270° is 9π/6. 225° is not a multiple of 30°, so ignore it for now.
  14. In the fourth quadrant, 300° is 10π/6, 330° is 11π/6, and 360° is 12π/6. 315° is not a multiple of 30°, so ignore it for now.
  15. Now reduce the fractions for each exact-value radian.
  16. In the first quadrant, the fractions reduce to π/6, π/3, and π/2.
  17. In the second quadrant, the fractions reduce to 2π/3, 5π/6, and π.
  18. In the third quadrant, the fractions reduce to 7π/6, 4π/3, and 3π/2.
  19. In the fourth quadrant, the fractions reduce to 5π/3, 11π/6, and 2π.
  20. Before we continue, let's clean up our work.
  21. Now we'll find exact-value radians that are multiples of π/4.
  22. In the first quadrant, 45° is 1π/4, and 90° is 2π/4.
  23. In the second quadrant, 135° is 3π/4, and 180° is 4π/4.
  24. In the third quadrant, 225° is 5π/4, and 270° is 6π/4.
  25. In the fourth quadrant, 315° is 7π/4, and 360° is 8π/4.
  26. We already have the exact value radians for 90°, 180°, 270°, and 360°, so the crossed-out radians can be ignored.
  27. Now reduce the fractions for each exact-value radian.
  28. Each radian is already in lowest terms, so we can immediately clean up our work.
  29. Even though counting radians may not seem like a shortcut at first, with practice you will gain speed. Before moving on with the lesson, practice drawing this diagram from memory a few times.
Section 2: DAY TWO: Examples 5 - 9 (38 minutes)
06:11
Example 5: Drawing Standard Position Angles
  1. Draw each of the following angles in standard position. State the reference angle. In Part A, we'll draw 210° and state its reference angle.
  2. We can draw 210° by rotating the terminal arm counter-clockwise from the positive x-axis.
  3. We can find the reference angle by drawing a vertical line from the terminal arm to the x-axis.
  4. The terminal arm is at 210°, and the negative x-axis is 180°. The reference angle is the absolute value of their difference, 30°.
  5. In Part B, draw -260° and state the reference angle.
  6. We can draw -260° by rotating the terminal arm clockwise from the positive x-axis.
  7. We can find the reference angle by drawing a vertical line from the terminal arm to the x-axis.
  8. The terminal arm is at -260°, and the negative x-axis is at -180°. The reference angle is the absolute value of their difference, 80°.
  9. In Part C, draw 5.3 radians and state the reference angle. Recall that unitless angles are radians.
  10. Convert 5.3 radians to degrees so it's easier to work with.
  11. Multiply by 180°/π to get 303.67°.
  12. 5.3 radians is approximately equal to 304°, when we round to the nearest degree.
  13. Now we'll draw 304°.
  14. The terminal arm is at 304°, and the positive x-axis is at 360°. The reference angle is the absolute value of their difference, 56°.
  15. Since the original angle was expressed as a radian, we should express the reference angle as a radian too.
  16. Multiply by π/180° to get 0.98 radians.
  17. In the diagram, replace the degree measurements we used for our work with the radian values we'll use as a final answer.
  18. In Part D, draw -5π/4 radians and state the reference angle.
  19. There are two ways to draw -5π/4.
  20. The first way is to convert to degrees using a conversion multiplier, then draw.
  21. The second way is to break -5π/4 into its component parts and use intervals to draw the angle. In this example, we'll use method 2.
  22. -5π/4 is the same thing as -5 × π/4. This means we want to rotate clockwise by five intervals of π/4, or 45°.
  23. The animation shows the rotation of the terminal arm to -5π/4.
  24. The reference angle is π/4.
  25. In Part E, draw 12π/7 and state the reference angle.
  26. The easiest way to draw 12π/7 is to convert it to degrees using a conversion multiplier. Multiply 12π/7 by 180°/π to get 309°.
  27. The animation shows the rotation of the terminal arm to 309°.
  28. We should express the reference angle as an exact-value radian since we were given an exact-value radian in the question.
  29. We know the terminal arm is at 12π/7, and the positive x-axis is at 2π. The absolute value of their difference is the reference angle.
  30. Subtract the values to get 2π/7. This is the reference angle.
06:25
Example 6: Drawing Large Angles (via the principal angle)
  1. Draw each of the following angles in standard position. State the principal and reference angles. In Part A, draw 930° and state the principal and reference angles.
  2. 930° is difficult to visualize since it is a large angle. The animation shows that multiple counter-clockwise revolutions are required to position this angle.
  3. If we find the principal angle, we can just draw that. Start at 930°, then subtract 360° twice to get the principal angle of 210°.
  4. Note that we can think of this calculation as unwinding our way back to the principal angle.
  5. The terminal arm is at 210° and the negative x-axis is at 180°. The absolute value of their difference is the reference angle, 30°.
  6. In Part B, draw -855° and state the principal and reference angles.
  7. -855° is difficult to visualize since it is a large angle. The animation shows that multiple clockwise revolutions are required to position this angle.
  8. If we find the principal angle, we can just draw that. Start at -855°, then add 360° three times to get the principal angle of 225°.
  9. Note that we can think of this calculation as winding-up towards the principal angle.
  10. The terminal arm is at 225° and the negative x-axis is at 180°. The absolute value of their difference is the reference angle, 45°.
  11. In Part C, draw 9 (radians) and state the principal and reference angles.
  12. 9 radians is difficult to draw because i) it's in radian decimal form, and ii) it's a large angle that is more than one revolution.
  13. Let's convert this to degrees. Multiply 9 radians by 180°/π to get 516°.
  14. Now find the principal angle. Start at 516°, then subtract 360° to get the principal angle of 156°.
  15. The animation shows the terminal arm rotating to 156°.
  16. The reference angle is the absolute value of (156° - 180°), which is 24°.
  17. Since the original angle was expressed as an approximate radian, we should convert the principal and reference angles to radians.
  18. The principal angle, 156°, can be multiplied by π/180° to get 2.72 rad.
  19. The reference angle, 24°, can be multiplied by π/180° to get 0.42 rad.
  20. In Part D, draw -10π/3 radians and state the principal and reference angles.
  21. There are two ways we could draw this angle: i) Convert to degrees, or ii) Count intervals of π/3. In this example, we'll use option ii), counting intervals.
  22. We know that π/3 = 60°, so -10π/3 is the same as -10×(60°), or -10×(π/3).
  23. The means we can count out 60° (or π/3) ten times clockwise to arrive at the angle.
  24. In the animation, we rotate the terminal arm clockwise by 10 intervals of π/3. This is -600°.
  25. Start at -600°, then add 360° two times to get the principal angle of 120°.
  26. The terminal arm is at 120°, and the negative x-axis is at 180°. The absolute value of their difference is the reference angle, 60°.
  27. Since the original angle was an exact radian, our answers should be too.
  28. The principal angle, 120°, can be multiplied by π/180° to get 2π/3.
  29. The reference angle, 60°, can be multiplied by π/180° to get π/3.
10:31
Example 7: Finding Co-Terminal Angles
  1. For each angle, find all co-terminal angles within the stated domain. In Part A, find the co-terminal angles of 60° between -360° and 1080°.
  2. Begin with the terminal arm at 60°.
  3. Find the positive co-terminal angles by adding 360° until we are outside the allowed domain.
  4. Both 420° and 780° are within the allowed domain, but 1140° is too large.
  5. Return the terminal arm to 60°.
  6. Find the negative co-terminal angles by subtracting 360° until we are outside the allowed domain.
  7. -300° falls within the allowed domain, but -660° does not.
  8. The co-terminal angles of 60° that exist within the specified domain are listed below. 60° is greyed out from the list since an angle can't be co-terminal with itself.
  9. In Part B, find the co-terminal angles of -495° between -1080° and 720°.
  10. Begin by finding the principal angle of -495°. Rotate twice counter-clockwise to get the principal angle of 225°.
  11. Position the terminal arm at this angle.
  12. Find the positive co-terminal angles by adding 360° until we are outside the allowed domain.
  13. 585° falls within the specified domain, but 945° does not.
  14. Return the terminal arm to 225°.
  15. Find the negative co-terminal angles by subtracting 360° until we are outside the allowed domain.
  16. -135°, -495°, and -855° are within the specified domain, but -1215° is not.
  17. The co-terminal angles of -495° that exist within the specified domain are listed below. -495° is greyed out from the list since an angle can't be co-terminal with itself. Note that 225° is both the principal angle and a co-terminal angle of -495°.
  18. In Part C, we'll find the co-terminal angles of 11.78 radians between -2π and 4π.
  19. Let's convert 11.78 radians to degrees so we can visualize it more clearly. Multiply 11.78 radians by the conversion multiplier 180°/π to get 675°.
  20. Now find the principal angle. 675° - 360° = 315°.
  21. It may also be helpful to convert the domain to degrees. -2π times 180°/π is -360°. 4π times 180°/π is 720°.
  22. Rewrite the domain from radians to degrees.
  23. Now that we have re-written all the radians as degrees, store them for use.
  24. Begin by positioning the terminal arm at 315°.
  25. Find the positive co-terminal angles by adding 360° until we are outside the allowed domain.
  26. 675° falls within the specified domain, but 1035° does not.
  27. Return the terminal arm to 315°.
  28. Find the negative co-terminal angles by subtracting 360° until we are outside the allowed domain.
  29. -45° is within the specified domain, but -405° is not.
  30. The co-terminal angles of 675° that exist within the specified domain are listed below. 675° is greyed out from the list since an angle can't be co-terminal with itself. Note that 315° is both the principal angle and a co-terminal angle of 675°.
  31. The original angle was an approximate radian, so we should express the answer in radian form too.
  32. Convert each co-terminal angle to radians using the conversion multiplier π/180°.
  33. For our final answer, rewrite the co-terminal angles from degrees to approximate radian decimals.
  34. In Part D, we'll find the co-terminal angles of 8π/3 between -13π/2 and 37π/5.
  35. Let's convert 8π/3 to degrees so we can visualize it more clearly. Multiply 8π/3 by the conversion multiplier 180°/π to get 480°.
  36. Now find the principal angle: 480° - 360° = 120°.
  37. It may also be helpful to convert the domain to degrees. -13 π/2 × 180°/π gives us -1170°. 37π/5 × 180°/π gives us 1332°.
  38. Rewrite the domain from radians to degrees.
  39. Now that we have re-written all the radians as degrees, store them for use.
  40. Begin by positioning the terminal arm at 120°.
  41. Find the positive co-terminal angles by adding 360° until we are outside the allowed domain.
  42. 480°, 840°, and 1200° exist within the specified domain, but 1560° does not.
  43. Return the terminal arm to 120°.
  44. Find the negative co-terminal angles by subtracting 360° until we are outside the allowed domain.
  45. -240°, -600°, and -960° exist within the specified domain, but -1320° does not.
  46. The co-terminal angles of 480° that exist within the specified domain are listed below. 480° is greyed out from the list since an angle can't be co-terminal with itself. In this example, 120° is both the principal angle and a co-terminal angle of 480°.
  47. The original angle was expressed as an exact radian (8π/3), so we should convert our answer to this form as well.
  48. Convert each degree to an exact-value radian using the conversion multiplier π/180°. This is our final answer.
06:11
Example 8: Finding the Principal Angle (with estimation)
  1. For each angle, use estimation to find the principal angle. In Part A, use estimation to find the principal angle of 1893°.
  2. Let's outline some rules for finding the principal angle of a large angle.
  3. If we are finding the principal angle of a large positive angle, we want a multiple of 360° or 2π that is less positive than the original angle.
  4. For example: the principal angle of 900° is 900° - 2(360°) = 180°. Notice that the angle subtracted (720°) is less positive than 900°.
  5. If we are finding the principal angle of a large negative angle, we want a multiple of -360° or -2π that is more negative than the original angle.
  6. For example: the principal angle of -1300° is -1300° - 4(-360°) = 140°. Notice that the angle subtracted (-1440°) is more negative than -1300°.
  7. In both cases, we subtract the multiple of 360° or 2π from the original angle. Following these rules will ensure that the result is in fact the principal angle.
  8. Using estimation and guess-and-check, find the multiple of 360° that is closest to and less positive than 1893°.
  9. 360° × 5 = 1800°.
  10. Now subtract: 1893° - 1800° = 93°.
  11. Here is a tip: 1893°/360° = 5.26, so there are five full revolutions of 360° before we get to 1893°.
  12. In Part B, find the principal angle of -437.24 rad.
  13. Using estimation and guess-and-check, find the multiple of -2π that is closest to and more negative than -437.24 rad.
  14. -2π × 70 = -439.82 rad.
  15. Now subtract: -437.24 rad - (-439.82 rad) = 2.58 rad.
  16. Here is a tip: -437.24 rad / -2π = 69.6, so 70 full revolutions of -2π are required for an angle more negative than -437.24.
  17. In Part C, find the principal angle of 912π/15.
  18. We can find the answer to this quickly by splitting the numerator into a multiple of 15 and its remainder.
  19. This gives us 900π/15 + 12π/15.
  20. This reduces to 60π + 4π/5.
  21. The principal angle is 4π/5.
  22. In Part D, find the principal angle of 95π/6.
  23. We can find the answer to this quickly by splitting the numerator into a multiple of 6 and its remainder.
  24. This gives us 90π/6 + 5π/6.
  25. This reduces to 15π + 5π/6.
  26. Watch Out! 15π is not a multiple of 2π! Let's try this again.
  27. This time try 84π/6 + 11π/6.
  28. This reduces to 14π + 11π/6.
  29. The principal angle is 11π/6.
06:34
Example 9: General Form of Co-Terminal Angles
  1. Use the general form of co-terminal angles to find the specified angle. In Part A, we have a principal angle of 300°, and we want to find the co-terminal angle after 3 rotations counter-clockwise.
  2. Recall that the general form of co-terminal angles is θc = θi ± n(360°) in degrees, or θc = θi ± n(2π) in radians.
  3. The initial angle is 300°, and there are 3 rotatations counter-clockwise. We want to add, since the angle is becoming more positive.
  4. The co-terminal angle after 3 rotations counter-clockwise is 1380°.
  5. In Part B, we have a principal angle of 2π/5, and we want to find the co-terminal angle 14 after rotations clockwise.
  6. Start with the general form of co-terminal angles for radians, θc = θi ± n(2π).
  7. The initial angle is 2π/5, and there are 14 rotations clockwise. We want to subtract, since the angle is becoming more negative.
  8. Simplify the expression.
  9. Get a common denominator of 5.
  10. Subtract to get -138π/5. This is the co-terminal angle after 14 rotations clockwise.
  11. In Part C, how many rotations are required to find the principal angle of -4300°? State the principal angle.
  12. We have an initial angle of -4300°, and we want to rotate the terminal arm counter-clockwise until we have a positive angle. Set the equation equal to zero and solve for n.This will tell us how many revolutions are required to get an angle of zero.
  13. Isolate the term with n on the left side of the equation.
  14. Divide both sides by 360°.
  15. This gives us n = 11.94.
  16. It takes 11.94 counter-clockwise revolutions to get to zero.
  17. Keep going to 12 full revolutions to return the terminal arm to its original position.
  18. The principal angle is -4300° + (12)(360°) = 20°.
  19. In Part D, how many rotations are required to find the principal angle of 32π/3? State the principal angle.
  20. We have an initial angle of 32π/3, and we want to rotate the terminal arm clockwise as many times as possible without getting a negative angle. Set the equation equal to zero and solve for n. This will tell us how many revolutions are required to get an angle of zero.
  21. Isolate the term with n on the right side of the equation.
  22. Multiply both sides of the equation by 3 to eliminate the fraction.
  23. Divide both sides of the equation by 6π to isolate n.
  24. This gives us n = 5.33.
  25. It takes 5.33 clockwise revolutions to get to zero.
  26. Backtrack to 5 revolutions to return the terminal arm to its original position.
  27. Now find the principal angle using 32π/3 – n(2π). Start with the original angle, and subtract 2π n times to get the principal angle.
  28. Rewrite the expression and plug in 5 revolutions for n. We need to unwind the terminal arm by 5 full revolutions to get the principal angle.
  29. Simplify the expression.
  30. Get a common denominator of 3.
  31. Subtract to get 2π/3. The is the principal angle as an exact-value radian.
Section 3: DAY THREE: Examples 10 - 13 (36 minutes)
08:47
Example 10: Six Trigonometric Ratios
  1. In addition to the three primary trigonometric ratios (sinθ, cosθ, and tanθ), there are three reciprocal ratios (cscθ, secθ, and cotθ). Given a triangle with side lengths of x and y, and a hypotenuse of length r, the six trigonometric ratios are shown. In Part A, if the point P(-5, 12) exists on the terminal arm of an angle θ in standard position, determine the exact values of all six trigonometric ratios. State the reference angle and the standard position angle.
  2. Use the point P(-5, 12) to draw the terminal arm.
  3. Create a triangle by connecting the terminal arm to the x-axis. The reference angle is inside the triangle as shown.
  4. Use the coordinates of the point to label the sides of the triangle. The adjacent side is -5, and the opposite side is 12.
  5. We need the hypotenuse before we can continue. Use the Pythagorean theorem to find this value.
  6. The Pythagorean theorem is x2 + y2 = r2.
  7. Replace x with the adjacent side -5, and replace y with the opposite side 12.
  8. This gives us 169 = r2
  9. Square root both sides to get r = ±13.
  10. The hypotenuse is always positive, so we'll use r = 13.
  11. Now state the six trigonometric ratios. sinθ = 12/13.
  12. cosθ = -5/13
  13. tanθ = 12/-5
  14. cscθ = 13/12. This is the reciprocal of sinθ.
  15. secθ = 13/-5. This is the reciprocal of cosθ.
  16. cotθ = 12/-5. This is the reciprocal of tanθ.
  17. Now we'll use one of the trig ratios to find the reference angle θ. In this example, we'll choose sinθ.
  18. We can get θ by itself by taking the inverse sine of both sides.
  19. Use the sine inverse feature of your calculator to get θ = 67.38°. This is the reference angle. Make sure your calculator is in DEGREE mode when evaluating the angle. (We will learn more about calculator modes in the Trigonometric Equations lessons.)
  20. Subtract the reference angle from 180° to get the principal angle. (180° - 67.38° = 112.62°)
  21. It's possible to get different answers for θ depending on which trigonometric ratio you use.
  22. If we find θ using cosθ = -5/13, we get θ = 112.62°.
  23. It's up to you, and your knowledge of the terminal arm position, to properly deduce if the angle output is the reference angle or some other angle.
  24. Now we'll move on to Part B. If the point P(2, -3) exists on the terminal arm of an angle θ in standard position, determine the exact values of all six trigonometric ratios. State the reference angle and the standard position angle.
  25. Use the point P(2, -3) to draw the terminal arm.
  26. Create a triangle by connecting the terminal arm to the x-axis. The reference angle is inside the triangle as shown.
  27. Use the coordinates of the point to label the sides of the triangle. The adjacent side is 2, and the opposite side is -3.
  28. We need the hypotenuse before we can continue. Use the Pythagorean theorem to find this value.
  29. Replace x with the adjacent side, 2, and replace y with the opposite side, -3.
  30. This gives us 13 = r2.
  31. Square root both sides to get r = ±root(13).
  32. The hypotenuse is always positive, so use r = + root(13).
  33. Now state the six trigonometric ratios. sinθ = -3/root(13).
  34. cosθ = 2/root(13).
  35. tanθ = -3/2.
  36. cscθ = root(13)/-3. This is the reciprocal of sinθ.
  37. secθ = root(13)/2. This is the reciprocal of cosθ.
  38. cotθ = 2/-3. This is the reciprocal of tanθ.
  39. Both sinθ and cosθ have radicals in the denominator. It is considered good form to remove the radicals from the denominator using the process known as rationalizing the denominator. For sinθ, multiply the numerator and denominator by root(13) to get -3root(13)/13. For cosθ, multiply the numerator and denominator by root(13) to get 2root(13)/13.
  40. Now we'll use one of the trig ratios to find the reference angle θ. We'll use the cosine ratio, cosθ = 2root(13)/13.
  41. Take the inverse cosine of both sides to get θ by itself.
  42. Use the inverse cosine feature of your calculator to get θ = 56.31°. This is the reference angle.
  43. Subtract the reference angle from 360° to get the principal angle. (360° - 56.31° = 303.69°)
10:29
Example 11: Signs of Trigonometric Ratios (by quadrant)
  1. Determine the sign of each trigonometric ratio in each quadrant. In Part A, we'll find the sign of sinθ in each quadrant.
  2. Recall that sinθ = y/r, or opposite over hypotenuse.
  3. In quadrant I, the opposite side is 4, and the hypotenuse is 5.
  4. sinθ is positive in quadrant I, since both the opposite side and hypotenuse are positive.
  5. In quadrant II, the opposite side is 4, and the hypotenuse is 5.
  6. sinθ is positive in quadrant II, since both the opposite side and hypotenuse are positive.
  7. In quadrant III, the opposite side is -4, and the hypotenuse is 5.
  8. sinθ is negative in quadrant III, since the opposite side is negative and hypotenuse is positive.
  9. In quadrant IV, the opposite side is -4, and the hypotenuse is 5.
  10. sinθ is negative in quadrant IV, since the opposite side is negative and the hypotenuse is positive.
  11. General Rule: sinθ is positive in quadrants where the y-values are positive. sinθ is negative in quadrants where the y-values are negative.
  12. In Part B, we'll find the sign of cosθ in each quadrant.
  13. Recall that cosθ = x/r, or adjacent over hypotenuse.
  14. In quadrant I, the adjacent side is 3, and the hypotenuse is 5.
  15. cosθ is positive in quadrant I, since both the adjacent side and hypotenuse are positive.
  16. In quadrant II, the adjacent side is -3, and the hypotenuse is 5.
  17. cosθ is negative in quadrant II, since the adjacent side is negative and the hypotenuse is positive.
  18. In quadrant III, the adjacent side is -3, and the hypotenuse is 5.
  19. cosθ is negative in quadrant III, since the adjacent side is negative and the hypotenuse is positive.
  20. In quadrant IV, the adjacent side is 3, and the hypotenuse is 5.
  21. cosθ is positive in quadrant IV, since both the adjacent side and hypotenuse are positive.
  22. General Rule: cosθ is positive in quadrants where the x-values are positive. cosθ is negative in quadrants where the x-values are negative.
  23. In Part C, we'll find the sign of tanθ in each quadrant.
  24. Recall that tanθ = y/x, or opposite over adjacent.
  25. In quadrant I, the opposite side is 4, and the adjacent side is 3.
  26. tanθ is positive in quadrant I, since both the adjacent side and the opposite side are positive.
  27. In quadrant II, the opposite side is 4, and the adjacent side is -3.
  28. tanθ is negative in quadrant II, since the opposite side is positive and the adjacent side is negative.
  29. In quadrant III, the opposite side is -4, and the adjacent side is -3.
  30. tanθ is positive in quadrant III, since the opposite side is negative and the adjacent side is negative.
  31. In quadrant IV, the opposite side is -4, and the adjacent side is 3.
  32. tanθ is negative in quadrant IV since the opposite side is negative and the adjacent side is positive.
  33. General Rule: tanθ is positive in quadrants where the x and y-values have the same sign. tanθ is negative in quadrants where the x and y-values have different signs.
  34. In Part D, we'll find the sign of cscθ in each quadrant.
  35. Recall that cscθ = r/y, or hypotenuse over opposite. This is the reciprocal of sinθ.
  36. In quadrant I, the hypotenuse is 5 and the opposite side is 4.
  37. cscθ is positive in quadrant I, since both the hypotenuse and the opposite side are positive.
  38. On your own, verify that quadrant II is positive, quadrant III is negative, and quadrant IV is negative.
  39. General Rule: cscθ is positive in the same quadrants that sinθ is positive. (where the y-values are positive). cscθ is negative in the same quadrants that sinθ is negative. (where the y-values are negative)
  40. In Part E, we'll find the sign of secθ in each quadrant.
  41. Recall that secθ = r/x, or hypotenuse over adjacent. This is the reciprocal of cosθ.
  42. In quadrant I, the hypotenuse is 5 and the adjacent side is 3.
  43. secθ is positive in quadrant I, since both the hypotenuse and the adjacent side are positive.
  44. On your own, verify that quadrant II is negative, quadrant III is negative, and quadrant IV is positive.
  45. General Rule: secθ is positive in the same quadrants that cosθ is positive. (where the x-values are positive). secθ is negative in the same quadrants that cosθ is negative. (where the x-values are negative)
  46. In Part F, we'll find the sign of cotθ in each quadrant.
  47. Recall that cotθ = x/y, or adjacent over opposite. This is the reciprocal of tanθ.
  48. In quadrant I, the adjacent is 3 and the opposite is 4.
  49. cotθ is positive in quadrant I, since both the adjacent and opposite sides are positive.
  50. On your own, verify that quadrant II is negative, quadrant III is positive, and quadrant IV is negative.
  51. General Rule: cotθ is positive in the same quadrants that tanθ is positive. (where the x and y-values have the same sign.) cotθ is negative in the same quadrants that tanθ is negative. (where the x and y-values have different signs.)
  52. Now we'll move on to Part G. How do the quadrant signs of the reciprocal trigonometric ratios (cscθ, secθ, and cotθ) compare to the quadrant signs of the primary trigonometric ratios (sinθ, cosθ, and tanθ)?
  53. sinθ and its reciprocal, cscθ, share the same quadrant signs.
  54. cosθ and its reciprocal, secθ, share the same quadrant signs.
  55. tanθ and its reciprocal, cotθ, share the same quadrant signs.
06:20
Example 12: What Quadrant is the Angle In?
  1. Given the following conditions, find the quadrant(s) where the angle θ could potentially exist.
  2. We'll begin with part i)
  3. sinθ is negative in quadrants where the y-value is negative.
  4. The angle θ could exist in either quadrant III or quadrant IV.
  5. Now we'll move on to part ii)
  6. cosθ is positive in quadrants where the x-value is positive.
  7. The angle θ could exist in either quadrant I or quadrant IV.
  8. Now we'll move on to part iii)
  9. tanθ is positive in quadrants where the x and y-values have the same sign.
  10. The angle θ could exist in either quadrant I or quadrant III.
  11. Now we'll move on to Part B.
  12. We'll begin with part i)
  13. sinθ is positive in quadrants where the y-value is positive. (Quadrants I & II)
  14. cosθ is positive in quadrants where the x-value is positive. (Quadrants I & IV)
  15. There is overlap in Quadrant I. That is where the angle is located.
  16. Now we'll move on to part ii)
  17. secθ is positive in the same quadrants where cosθ is positive. (Quadrants I & IV)
  18. tanθ is negative in quadrants where the x and y-values have different signs. (Quadrants II & IV)
  19. There is overlap in Quadrant IV. That is where the angle is located.
  20. Now we'll move on to part iii)
  21. cscθ is negative in the same quadrants where sinθ is negative. (Quadrants III & IV)
  22. cotθ is positive in the same quadrants where tanθ is positive. (Quadrants I & III)
  23. There is overlap in Quadrant III. That is where the angle is located.
  24. Now we'll move on to Part C.
  25. We'll begin with part i)
  26. sinθ is negative in quadrants where the y-value is negative. (Quadrants III & IV)
  27. Note: For the purpose of finding quadrants, cscθ = 1/2 is the same as cscθ > 0. This is because we are not trying to get an actual value, we just need to know if cscθ is a positive or negative quantity.
  28. cscθ = 1/2 means that cscθ > 0. cscθ is positive in the same quadrants where sinθ is positive. (Quadrants I & II)
  29. No angle satisfies these conditions, so there is no solution.
  30. Now we'll move on to part ii)
  31. cosθ = -root(3)/2 is the same as cosθ < 0. cosθ is negative in quadrants where the x-value is negative. (Quadrants II & III)
  32. cscθ is negative in the same quadrants where sinθ is negative. (Quadrants III & IV)
  33. There is overlap in Quadrant III. That is where the angle is located.
  34. Now we'll move on to part iii)
  35. secθ is positive in the same quadrants where cosθ is positive. (Quadrants I & IV)
  36. tanθ = 1 means that tanθ > 0. tanθ is positive in quadrants where the x and y-values have the same sign. (Quadrants I & III)
  37. There is overlap in Quadrant I. That is where the angle is located.
10:33
Example 13: Finding Trigonometric Ratios (with a domain restriction)
  1. Given one trigonometric ratio, find the exact values of the other five trigonometric ratios. State the reference angle and the standard position angle, to the nearest hundredth of a radian. In Part A, we have cosθ = -12/13, and the angle θ must be between π and 3π/2.
  2. The domain tells us that the angle we are trying to find must be in quadrant III. (between 180° and 270°)
  3. The cosine ratio can be used to draw the sides of a triangle.
  4. Recall that cosθ is adjacent over hypotenuse. The adjacent value is -12, and the hypotenuse (which is always positive) is 13.
  5. Use the Pythagorean theorem to find the opposite side. The Pythagorean theorem is x2 + y2 = r2.
  6. The adjacent side is -12, so plug this in for x. The hypotenuse is 13, so plug this in for r.
  7. This gives us 144 + y2 = 169.
  8. Subtract 144 from each side of the equation to isolate y2. This gives us y2 = 25.
  9. Square root each side of the equation to get y = ±5.
  10. Choose y = -5, since y-values are negative in quadrant III.
  11. Now that we know all three sides of the triangle, we can state the six trigonometric ratios.
  12. sinθ = -5/13
  13. cosθ = -12/13
  14. tanθ = -5/-12 = 5/12
  15. cscθ = 13/-5. This is the reciprocal of sinθ.
  16. secθ = 13/-12. This is the reciprocal of cosθ.
  17. cotθ = -12/-5 = 12/5. This is the reciprocal of tanθ.
  18. Now we need to find the reference angle. Which trigonometric ratio should we πck to get the reference angle?
  19. Try to select a trigonometric ratio that is positive when finding the reference angle. This will avoid getting angles that are negative or larger than 90°. Recall that reference angles are always between 0° and 90°. (0°≤ θr ≤ 90°)
  20. In this example, we'll use tanθ = 5/12.
  21. Isolate θ by taking the inverse tan of each side.
  22. This gives us the reference angle, θ = 22.62°.
  23. The standard position angle is found by adding the reference angle to 180°. This gives us 202.62°.
  24. The question asks for angles to the nearest hundredth of a radian, so convert each degree to a radian decimal.
  25. Multiply 22.62° by the conversion multiplier π/180° to get 0.39 rad.
  26. And finally, multiply 202.62° by the conversion multiplier π/180° to get 3.54 rad.
  27. In Part B, we have cscθ = 7/3, and the angle θ must be between π/2 and π.
  28. The domain tells us that the angle we are trying to find must be in quadrant II. (between 90° and 180°)
  29. The cosecant ratio can be used to draw the sides of a triangle.
  30. Recall that cscθ is hypotenuse over opposite. The hypotenuse (which is always positive) is 7, and the opposite value is 3.
  31. Use the Pythagorean theorem to find the opposite side. The Pythagorean theorem is x2 + y2 = r2.
  32. The opposite side is 3. Plug this in for y. The hypotenuse is 7. Plug this in for r.
  33. This gives us x2 + 9 = 49.
  34. Subtract 9 from both sides of the equation to isolate x2. This gives us x2 = 40.
  35. Square root both sides to get x = ±root(40)
  36. root(40) can be simplified to a mixed radical. 40 is the same as 4 × 10.
  37. The square root of 4 is 2, and root(10) can`t be simplified further.
  38. Since x-values are negative in quadrant II, choose the negative sign.
  39. The value of x is -2×root(10).
  40. Now that we know all three sides of the triangle, we can state the six trigonometric ratios.
  41. sinθ = 3/7.
  42. cosθ = -2root(10)/7
  43. tanθ = 3/-2root(10)
  44. cscθ = 7/3. This is the reciprocal of sinθ.
  45. secθ = 7/-2root(10). This is the reciprocal of cosθ.
  46. cotθ = -2root(10)/3. This is the reciprocal of tanθ.
  47. We should rationalize the denominator for tanθ and secθ. For tanθ, multiply the numerator and denominator by root(10) to get 3root(10) /-20. For secθ, multiply the numerator and denominator by root(10) to get 7root(10) /-20.
  48. Now we need to find the reference angle. Which trigonometric ratio should we πck to get the reference angle?
  49. Try to select a trigonometric ratio that is positive when finding the reference angle. In this example, we`ll use sinθ = 3/7.
  50. Get θ by itself by taking the inverse sine of each side.
  51. This gives us the reference angle, θ = 25.38°.
  52. The standard position angle is found by subtracting the reference angle from 180°. This gives us 154.62°.
  53. The question asks for angles to the nearest hundredth of a radian, so convert each degree to a radian decimal.
  54. Multiply 25.38° by the conversion multiplier π/180° to get 0.44 rad.
  55. And finally, multiply 154.62° by the conversion multiplier π/180° to get 2.70 rad.
Section 4: DAY FOUR: Examples 14 - 18 (41 minutes)
09:56
Example 14: Finding Trigonometric Ratios (by finding the quadrant first)
  1. Given one trigonometric ratio, find the exact values of the other five trigonometric ratios. State the reference angle and the standard position angle, to the nearest hundredth of a degree. In Part A, we have secθ = 5/4, and sinθ < 0.
  2. The first thing we need to do is determine which quadrant the angle exists in.
  3. secθ = 5/4, which means that secθ is positive.
  4. Since secθ is the reciprocal of cosθ, secθ is positive in the same quadrants that cosθ is positive.
  5. cosθ is positive in the quadrants where the x-values are positive. This occurs in quadrants I & IV.
  6. sinθ is negative in the quadrants where the y-values are negative. This occurs in quadrants III & IV.
  7. There is overlap in quadrant IV. This is where the angle exists.
  8. Recall that secθ = hypotenuse over adjacent. If secθ = 5/4, the hypotenuse is 5 and the adjacent is 4.
  9. Use this information to draw a triangle in quadrant IV.
  10. Use the Pythagorean theorem to find the opposite side. The Pythagorean theorem is x2 + y2 = r2.
  11. The adjacent side is 4. Put this in for x. The hypotenuse is 5. Put this in for r.
  12. This gives us 16 + y2 = 25.
  13. Subtract 16 from both sides of the equation to get y2 = 9.
  14. Square root both sides to get y = ±3.
  15. Since the y-value is negative in quadrant IV, use -3 for the triangle.
  16. Now that we have all three sides of the triangle, we can state the six trigonometric ratios.
  17. sinθ = -3/5.
  18. cosθ = 4/5.
  19. tanθ = -3/4.
  20. cscθ = 5/-3. This is the reciprocal of sinθ.
  21. secθ = 5/4. This is the reciprocal of cosθ.
  22. cotθ = 4/-3. This is the reciprocal of tanθ.
  23. Now we need to find the reference angle. Use a positive trigonometric ratio to get the reference angle.
  24. In this example, we'll use cosθ = 4/5.
  25. Get θ by itself by taking the inverse cosine of each side.
  26. The reference angle is 36.87°.
  27. Position the reference angle within the triangle.
  28. The standard position angle is found by subtracting the reference angle from 360°. This gives us 323.13°.
  29. In Part B, we have tanθ = -2/3, and secθ > 0.
  30. The first thing we need to do is determine which quadrant the angle exists in.
  31. tanθ = -2/3, which means that tanθ is negative.
  32. tanθ is negative in quadrants where the x and y-values have opposite signs. This happens in quadrants II & IV.
  33. secθ is positive in the same quadrants where cosθ is positive.
  34. cosθ is positive in quadrants where the x-values are positive. This occurs in quadrants I & IV.
  35. There is overlap in quadrant IV. This is where the angle exists.
  36. Recall that tanθ = opposite/adjacent. If tanθ = -(2/3), and the angle exists in quadrant IV, the opposite side is -2 and the adjacent side is 3.
  37. Use this information to draw a triangle in quadrant IV.
  38. Use the Pythagorean theorem to find the opposite side. The Pythagorean theorem is x2 + y2 = r2.
  39. The adjacent side is 3. Put this in for x. The opposite side is -2. Put this in for y.
  40. This gives us 9 + 4 = r2.
  41. Add to get 13 = r2.
  42. Square root both sides to get r = ±root(13).
  43. The hypotenuse is always positive, so choose r = +root(13).
  44. Now that we have all three sides of the triangle, we can state the six trigonometric ratios.
  45. sinθ = -2/root(13).
  46. cosθ = 3/root(13).
  47. tanθ = -2/3.
  48. cscθ = root(13)/-2. This is the reciprocal of sinθ.
  49. secθ = root(13)/3. This is the reciprocal of cosθ.
  50. cotθ = 3/-2. This is the reciprocal of tanθ.
  51. We should rationalize the denominator for sinθ and cosθ. For sinθ, multiply the numerator and denominator by root(13). This gives us -2root(13)/13. For cosθ, multiply the numerator and denominator by root(13). This gives us 3root(13)/13.
  52. Now we need to find the reference angle. Use a positive trigonometric ratio to get the reference angle.
  53. In this example, we'll use cosθ = 3root(13)/13.
  54. Get θ by itself by taking the inverse cosine of each side.
  55. This gives us the reference angle, 33.69°.
  56. Position the reference angle within the triangle.
  57. The standard position angle is found by subtracting the reference angle from 360°, to get 326.31°.
07:04
Example 15: Unexpected Calculator Results (when finding the angle)
  1. When you solve a trigonometric equation in your calculator, the answer you get for θ can seem unexpected. Complete the following chart to learn how the calculator processes your attempt to solve for θ.
  2. When we try to solve for θ in an equation like cosθ = -3/5, the calculator has to guess which quadrant the triangle is in.
  3. A calculator has a preferential order in which it tries to place a triangle.
  4. If it's possible to place a triangle in quadrant I, the calculator will always choose this quadrant first.
  5. If quadrant I is not an option, the calculator will try to place the triangle in quadrant IV.
  6. If neither quadrant I nor quadrant IV are options, the calculator will place the triangle in quadrant II.
  7. The calculator will never place the triangle in quadrant III.
  8. Now we'll complete the chart.
  9. If the angle θ could exist in either quadrant I or II,
  10. The calculator always picks quadrant I.
  11. If the angle θ could exist in either quadrant I or III,
  12. The calculator always picks quadrant I.
  13. If the angle θ could exist in either quadrant I or IV,
  14. The calculator always picks quadrant I.
  15. If the angle θ could exist in either quadrant II or III,
  16. The calculator always picks quadrant II.
  17. Note that the calculator returns quadrant II angles as a standard position angle greater than 90°.
  18. If the angle θ could exist in either quadrant II or IV,
  19. The calculator always picks quadrant IV.
  20. Note that the calculator returns quadrant IV angles as a negative standard position angle.
  21. If the angle θ could exist in either quadrant III or IV,
  22. The calculator always picks quadrant IV.
  23. Now we'll move on to Part B. Given the point P(-4, 3), Mark tries to find the reference angle using a sine ratio, Jordan tries to find it using a cosine ratio, and Dylan tries to find it using a tangent ratio. Why does each person get a different result from their calculator?
  24. Given the point P(-4, 3), we can construct a triangle in the second quadrant.
  25. Use the Pythagorean theorem to find the hypotenuse of the triangle, which is 5. Label all three sides of the triangle.
  26. We can now write the three trigonometric ratios: sinθ = 3/5, cosθ = -4/5, tanθ = 3/-4
  27. Now we'll inspect Mark's calculation.
  28. Even though we know the triangle is in quadrant II (we have the picture), the calculator has no way of knowing the triangle is in quadrant II.
  29. The calculator will find the angle using the chart we completed in part (a). There is no further need for the triangle diagram above, since the calculator can't use it.
  30. Since sinθ > 0, the angle could potentially exist in quadrants I or II.
  31. When the calculator has to choose between quadrants I & II, it always picks quadrant I.
  32. Use the inverse sine feature of your calculator to get θ = 36.87°.
  33. Now we'll inspect Jordan's calculation.
  34. Since cosθ < 0, the angle could potentially exist in quadrants II or III.
  35. When the calculator has to choose between quadrants II & III, it always picks quadrant II.
  36. Use the inverse cosine feature of your calculator to get θ = 143.13°. Note that the calculator returns quadrant II angles as a standard position angle greater than 90°.
  37. Now we'll inspect Dylan's calculation.
  38. Since tanθ < 0, the angle could potentially exist in quadrants II or IV.
  39. When the calculator has to choose between quadrants II & IV, it always picks quadrant IV.
  40. Use the inverse tan feature of your calculator to get θ = -36.87°. Note that the calculator returns quadrant IV angles as a negative standard position angle.
05:40
Example 16: Arc Length
  1. The formula for arc length is a = rθ, where a is is the arc length, θ is the central angle in radians, and r is the radius of the circle. The radius and arc length must have the same units. In Part A, derive the formula for arc length, a = rθ.
  2. The terminal arm must be rotated by an angle of 2pi to swipe out a complete circle.
  3. If a circle sector is formed with a central angle of θ (where 0 < θ < 2pi), then the percentage of the circle occupied by the sector is θ/2pi.
  4. We know that the circumference of a complete circle is C = 2πr.
  5. The arc length can be found by multiplying the circumference by the sector percentage. This gives us a = 2πr × θ/2pi, which is equal to rθ. This completes our derivation of the arc length formula.
  6. in Part B, solve for a, to the nearest hundredth.
  7. Before we can find the arc length, we need to convert the angle from degrees to radians.
  8. Convert 153° to radians with the conversion multiplier pi/180°. This gives us 2.67 radians.
  9. Now use the arc length formula to find the arc length.
  10. Plug in 5 cm for the radius, and 2.67 radians for θ.
  11. The arc length is 13.35 cm.
  12. In Part C, solve for θ. Express your answer as a degree, to the nearest hundredth.
  13. Use the arc length formula to solve for θ.
  14. Rearrange the formula to get θ = a/r.
  15. The arc length is 6 cm, and the radius is 3 cm.
  16. Divide to get 2 radians.
  17. Convert 2 radians to a degree using the conversion multiplier 180°/pi. The angle is 114.59°.
  18. In Part D, solve for r, to the nearest hundredth.
  19. Use the arc length formula to solve for r.
  20. Rearrange the formula to get r = a/θ.
  21. The arc length is 1.23pi cm, and the angle is pi/2.
  22. To divide fractions, multiply the numerator by the reciprocal of the denominator.
  23. This gives us r = 2.46 cm.
  24. In Part E, solve for n. Express your answer as an exact-value radian.
  25. Use the arc length formula to solve for the angle corresponding to the arc of 5pi cm.
  26. Rearrange the arc length formula to get θ = a/r.
  27. The arc length is 5pi cm, and the radius is 6 cm.
  28. Centimetres cancel out, leaving us with 5pi/6 radians.
  29. Note that the angle we just found, 5pi/6, is not the angle we need. Subtract 5pi/6 from 2pi to get the angle n.
  30. Write 2pi - 5pi/6.
  31. Get a common denominator of 6.
  32. Subtract to get 7pi/6. This is the value of angle n.
06:11
Example 17: Area of a Circle Sector
  1. In this example we will learn how to calculate the area of a circle sector. In Part A, derive the formula for the area of a circle sector, A = r2θ/2.
  2. We can begin the derivation by using the area of a circle, A = πr2.
  3. If the complete circle is 2π, and the sector angle is θ, then the percentage of the circle occupied by the sector is θ/2π.
  4. Multiply the area of the full circle by the sector percentage to get the area of the sector.
  5. Cancel the π's to get A = r2θ/2.
  6. In Part B, find the area of the shaded region.
  7. The formula for the area of a circle sector is A = r2θ/2.
  8. Plug in the radius and sector angle.
  9. 42 = 16.
  10. Simplify.
  11. Reduce 8/6 to 4/3 and multiply to get 28π/3. We have found an area, so the units are in square centimetres.
  12. In Part C, find the area of the shaded region.
  13. Before we can find the area, we need to find the radian angle of the shaded region.
  14. 360° - 240° = 120°.
  15. 120° is 2π/3 as an exact value radian.
  16. The formula for the area of a circle sector is A = r2θ/2.
  17. Plug in the radius and sector angle.
  18. 32 = 9.
  19. Cancel to get 6π/2.
  20. The area is 3π cm2.
  21. In Part D, find the area of the shaded region.
  22. Before we can find the area, we need to convert the sector angles to radians.
  23. 60° is π/3 as an exact value radian.
  24. 120° is 2π/3 as an exact value radian.
  25. The formula for the area of a circle sector is A = r2θ/2.
  26. Add the sector areas together to get the total area.
  27. The area of a sector is r2θ/2. The first sector has an angle of θ1, and the second sector has an angle of θ2.
  28. Factor out r2/2.
  29. Plug in the radius and sector angles.
  30. Simplify.
  31. The area is 81π/2 cm2.
  32. In Part E, find the area of the shaded region.
  33. We can get the area of the shaded region by subtracting the inner sector from the outer sector.
  34. The shaded area is the difference between the total sector area and the inner sector area.
  35. The area of a sector is r2θ/2. Note that the sector angle θ is identical for both sectors, so it can be factored out.
  36. Factor out θ/2.
  37. Plug in the sector angle and radius values.
  38. Divide the fraction and square the values in the brackets.
  39. Simplify.
  40. The area of the shaded region is 15π cm2.
09:21
Example 18: Problem Solving (angular speed)
  1. The formula for angular speed is omega = ∆θ/∆T, where Omega is the angular speed, ∆θ is the change in angle, and ∆T is the change in time. Calculate the requested quantity in each scenario. Round all decimals to the nearest hundredth. In Part A, a bicycle wheel makes 100 complete revolutions in one minute. Calculate the angular speed in degrees per second.
  2. Angular speed is just like regular speed, except we use angles as our distance, instead of metres.
  3. Common units for angular speed are: rev/s, degrees/s, and radians/s.
  4. The animation shows an object moving in a circle with an angular speed of 1 rev/s.
  5. If we double the original angular speed, we have 2 rev/s.
  6. If we halve the original angular speed, we have 0.5 rev/s.
  7. Now we'll answer the question in Part A.
  8. We want the answer to be expressed in degrees per second, so convert revolutions to degrees, and convert minutes to seconds.
  9. Convert 100 rev to degrees using the conversion multiplier 360°/1 rev. This gives us 36000°.
  10. Convert 1 minute to seconds using the conversion multiplier 60 s/1 min. This gives us 60 seconds.
  11. Now we'll use the formula for angular speed.
  12. Plug in 36000° for the change in angle, and plug in 60 s for the change in time.
  13. This gives us 600°/s.
  14. In Part B, a Ferris wheel rotates 1020° in 4.5 minutes. Calculate the angular speed in radians per second.
  15. We want the answer to be expressed in radians per second, so convert degrees to to radians, and convert minutes to seconds.
  16. Convert 1020° to radians using the conversion multiplier π/180°. This gives us 17π/3.
  17. Convert 4.5 minutes to seconds using the conversion multiplier 60 s/1 min. This gives us 270 seconds.
  18. Now we'll use the formula for angular speed.
  19. Plug in 17π/3 for the change in angle, and 270 seconds for the change in time.
  20. This gives us 0.07 rad/s.
  21. In Part C, the moon orbits Earth once every 27 days. Calculate the angular speed in revolutions per second. If the average distance from the Earth to the moon is 384 400 km, how far does the moon travel in one second?
  22. The moon orbiting the Earth once is equivalent to 1 rev.
  23. We need rev/s, so convert 27 days to seconds.
  24. We can use a sequence of conversion multipliers to convert 27 days to seconds. There are 24 hours in one day, there are 60 minutes in 1 hour, and there are 60 seconds in 1 minute.
  25. This gives us 2332800 seconds.
  26. Now we'll use the formula for angular speed.
  27. plug in 1 rev for the change in angle, and 2332800 s for the change in time.
  28. This gives us a value with units of revolutions per second.
  29. In one second, the moon makes 4.29×10-7 revolutions around Earth.
  30. In one full revolution, the moon travels the circumference of a circle with a radius of 384 400 km.
  31. The circumference of a circle is C = 2πr.
  32. Plug in 384 400 km for the radius.
  33. Multiply to get a circumference of 2.42 × 106 km.
  34. To get the distance the moon travels in one second, multiply the orbital circumference by the number of revolutions in one second.
  35. This gives us 1.04 km as our answer.
  36. In Part D, a cooling fan rotates with an angular speed of 4200 rpm. What is the speed in rps?
  37. Rewrite 4200 rpm as a fraction.
  38. We need to convert minutes to seconds. Multiply by the conversion multiplier 1 min /60 s.
  39. This gives us 70 rps.
  40. In Part E, a bike is ridden at a speed of 20 km/h, and each wheel has a diameter of 68 cm. Calculate the angular speed of one of the bicycle wheels and express the answer using revolutions per second.
  41. Calculate the circumference of a wheel. Since a wheel is just a circle, we can use the circumference formula C = 2πr.
  42. If the diameter is 0.68 m, the radius is half this value. Plug in 0.34 m for r.
  43. Multiply to get the circumference, 2.14 m.
  44. Next, convert the speed to m/s.
  45. We can use a sequence of conversion multipliers to convert km/h to m/s. There are 1000 m in a kilometre, and 1 hour has 3600 s.
  46. This gives us 5.56 m/s.
  47. If the wheel travels 5.56 m in one second, and the circumference of the wheel is 2.14 m, use a conversion multiplier to get the number of revolutions in one second.
  48. 1 revolution is the same as 2.14 m.
  49. This gives us 2.60 rev/s.
02:48
Example 19: Problem Solving (orbiting satellite)
  1. A satellite orbiting Earth 340 km above the surface makes one complete revolution every 90 minutes. The radius of Earth is approximately 6370 km. In Part A, calculate the angular speed of the satellite. Express your answer as an exact value, in radians/second.
  2. When the satellite revolves once around Earth, it travels an angle of 2π radians.
  3. The time for one revolution is 90 minutes, or 5400 seconds. (90 min × 60 s/min)
  4. Now we'll use the formula for angular speed.
  5. Plug in 2π for the change in angle, and 5400 seconds for the change in time. This reduces to π/2700 radians per second, which is our exact-value answer for the angular speed.
  6. Now we'll move on to Part B. How many kilometres does the satellite travel in one minute? Round your answer to the nearest hundredth of a kilometre.
  7. In part (a), we found the angular speed of the satellite, π/2700 radians per second.
  8. In one second, the satellite travels π/2700 rad. In 60 seconds (one minute), the satellite travels 60×(π/2700) rad, or π/45 rad when reduced.
  9. If the radius of the Earth is 6370 km, and the distance from the surface to the satellite is 340 km, then the distance from the centre of the Earth to the satellite is: 6370 km + 340 km = 6710 km.
  10. Now we'll use the arc length formula.
  11. Plug in 6710 km for r, and π/45 rad for θ. In one minute, the satellite travels approximately 468.45 km.
Section 5: Other Lesson Materials
Workbook
22 pages
00:50
Degrees and Radians: Introduction
  1. Welcome to Degrees and Radians. This topic will take four days to complete.
  2. In Day 1 we will define degrees, radians, and revolutions. We will also learn how to convert between angle types.
  3. In Day 2 we will draw standard position angles and study co-terminal angles.
  4. In Day 3 we will learn about the six trigonometric ratios.
  5. In Day 4 we will learn about arc length. We will also solve problems with degrees and radians.
01:09
Degrees and Radians: Summary
  1. You have now completed Degrees and Radians. Over the past four days you have learned:
  2. the definitions of degrees, radians, and revolutions.
  3. how to use conversion multipliers.
  4. how to draw standard position angles.
  5. how to find the exact values of the six trigonometric ratios for a given angle.
  6. the relationship between a triangle's quadrant and the sign of its trigonometric ratios.
  7. the definition of arc length.
  8. how to apply degrees, radians, and revolutions to real-world contexts.
Degrees and Radians
11 questions
Section 6: Full Course PDF Files for Math 30-1 and Pre-Calculus 12 Workbook
Formula Sheet
2 pages
Condensed Workbook (No Whitespace)
117 pages
Expanded Workbook (Book 1 - Includes Whitespace)
257 pages
Expanded Workbook (Book 2 - Includes Whitespace)
230 pages
Answer Key
39 pages

Instructor Biography

Barry Mabillard , Math Animator

I am an independent developer of math websites aligned to the Canadian mathematics curriculum. (www.math30.ca and www.math10.ca). These animated tutorials can be used for Pre-Calculus 12 (BC), Math 30-1 (Alberta), Pre-Calculus 30 (Saskatchewan), and Pre-Calculus 40S (Manitoba). Students in other areas may find the content useful as a supplement to their local mathematics curriculum.

Join the biggest student community

5,900,000

Hours of video content

22,000,000

Course Enrollments

6,500,000

Students

Reviews

Average Rating
4.8
Details
  1. 5 Stars
    19
  2. 4 Stars
    4
  3. 3 Stars
    1
  4. 2 Stars
    0
  5. 1 Stars
    0
    • Arkadiy Deliev

    Simply Great

    A really great course with understandable material and a good instructor. Worth learning from!

    • Therese Bertrand

    Very thorough

    I am at a new job, teaching high school Trigonometry. The school had no Trigonometry text so I ventured online to see what was available. I really liked the way that this course was broken down. I have gained ideas from the lessons taught here to use with my students. I appreciate the step by step and very thorough instruction. I also appreciate the opportunity to download all of the information and the opportunity to take this class at no cost.

    • Oscar Layne

    Good

    The teacher does everything step by step it is hard to get lost how he breaks everything down

    • James Mohart

    Great Course

    Very specific and too the point making the lecture very understandable.

    • J_a_spooner@yahoo.com

    A good intro course.

Show more reviews
Ready to start learning?
Enroll for free now