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Lesson 12 NYS COMMON CORE MATHEMATICS CURRICULUM M4 PRECALCULUS AND ADVANCED TOPICS Lesson 12: Inverse Trigonometric Functions Student Outcomes ο§ Students understand that restricting a trigonometric function to a domain on which it is always increasing or always decreasing allows its inverse to be constructed. ο§ Students use inverse functions to solve trigonometric equations. Lesson Notes Students studied inverse functions in Module 3 and came to the realization that not every function has an inverse that is also a function. Students considered how to restrict the domain of a function to produce an invertible function (F-BF.B.4d). This lesson builds on that understanding of inverse functions by restricting the domains of the trigonometric functions in order to develop the inverse trigonometric functions (F-TF.B.6). In Geometry, students used arcsine, arccosine, and arctangent to find missing angles, but they did not understand inverse functions and, therefore, did not use the terminology or notation for inverse trigonometric functions. Students define the inverse trigonometric functions in this lesson. Then they use the notation sinβ1 (π₯) rather than arcsin(π₯). The focus shifts to using the inverse trigonometric functions to solve trigonometric equations (F-TF.B.7). Classwork Opening Exercise (5 minutes) Give students time to work on the Opening Exercise independently. Then have them compare answers with a partner before sharing as a class. Opening Exercise Use the graphs of the sine, cosine, and tangent functions to answer each of the following questions. a. State the domain of each function. The domain of the sine and cosine functions is the set of all real numbers. The domain of the tangent function π is the set of all real numbers π β + ππ for all integers π. π Lesson 12: Inverse Trigonometric Functions This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from PreCal-M4-TE-1.3.0-10.2015 226 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 12 NYS COMMON CORE MATHEMATICS CURRICULUM M4 PRECALCULUS AND ADVANCED TOPICS b. Would the inverse of the sine, cosine, or tangent functions also be functions? Explain. Scaffolding: None of these functions are invertible. Multiple elements of the domain are paired with a single range element. When the domain and range are exchanged to form the inverse, the result does not satisfy the definition of a function. c. If students need a review of inverse functions, use this exercise: Consider the function π(π₯) = βπ₯ β 4, which is graphed below. Graph π β1 . Find the equation of the inverse and its domain. For each function, select a suitable domain that will make the function invertible. Answers will vary, so share a variety of responses. Any answer is suitable as long as the restricted domain leaves an interval of the graph that is either always increasing or always decreasing. MP.7 Sample response: π π π = sin(π), π«: [π, ] ο§ π = cos(π), π«: [π, π ] π π π = πππ§(π), π«: [π, ] Are any of the trigonometric functions invertible? οΊ No. The inverses of the trigonometric functions are no longer functions. If necessary, remind students of the definition of an invertible function. INVERTIBLE FUNCTION: The domain of a function π can be restricted to make it invertible so that its inverse is also a function. A function is said to be invertible if its inverse is also a function. ο§ Was there only one way to restrict the domain to make each function invertible? οΊ ο§ No. There are an infinite number of ways in which we could restrict the domain of each function. We just need to erase enough of the graph to where the function is either only increasing or only decreasing. How much of the graph should we keep? οΊ We want to choose the largest subset of the domain of the function as we can (such as π(π₯) = sin(π₯)) and still have the function be continuously increasing or continuously decreasing on that interval. Ask students to share the domain restriction they chose for each of the three functions. Then, point out that while there is more than one way to do this, the convention is to use an interval that contains zero. ο§ π 2 π 2 Based on this, the convention is to restrict the domain of π(π₯) = sin(π₯) to be β β€ π₯ β€ . Does this satisfy all of our requirements? οΊ ο§ Would this same restriction work for π(π₯) = cos(π₯)? οΊ ο§ No. The graph would contain an interval of increasing and an interval of decreasing and still would not be invertible. If we want to include zero and keep the largest subset of the domain possible, what would be a logical way to restrict the domain of π(π₯) = cos(π₯)? οΊ ο§ Yes. The graph is entirely increasing. We kept as much of the domain as possible, and we included zero in the domain. Either 0 β€ π₯ β€ π or βπ β€ π₯ β€ 0 The convention is to restrict the domain of π(π₯) = cos(π₯) to 0 β€ π₯ β€ π. Lesson 12: Inverse Trigonometric Functions This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from PreCal-M4-TE-1.3.0-10.2015 227 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 12 NYS COMMON CORE MATHEMATICS CURRICULUM M4 PRECALCULUS AND ADVANCED TOPICS ο§ If we want to include zero and keep the largest subset of the domain possible, what would be a logical way to restrict the domain of π(π₯) = tan(π₯)? οΊ We could restrict the domain to values from 0 to π, but we would have to exclude π 2 from the domain. If π 2 π 2 we want to keep one branch of the graph and include 0, we should restrict the domain to β < π₯ < . ο§ π 2 π 2 The convention is to restrict the domain of π(π₯) = tan(π₯) to β < π₯ < . Example 1 (6 minutes) Allow students time to read through the example and answer part (a). Then discuss part (b) as a class. ο§ How can we find the equation of the inverse sine? Write the following on the board. π₯ = sin(π¦) ο§ Now what? We need a function that denotes that it is the inverse of the sine function. The inverse sine function is usually written as π¦ = sinβ1 (π₯). Why does this notation make sense for an inverse function? The notation π β1 (π₯) means the inverse function of π₯, so it makes sense that sinβ1 (π₯) means the inverse of sine. οΊ ο§ π 6 What is the value of sin ( )? What about sin ( 1 οΊ ο§ Both equal . 2 1 2 What is the value of sinβ1 ( )? π οΊ ο§ 5π )? 6 Why οΊ 6 π 6 and not 5π 6 ? π 2 π 2 The range of the inverse sine function is restricted to β β€ π¦ β€ , which means that while there are an infinite number of possible answers, there is only one answer that lies within this restricted interval. ο§ What is the value of sin ( οΊ ο§ 1 2 1 2 What is the value of sinβ1 (β )? οΊ ο§ β 11π )? 6 β π 6 Would it be acceptable to give the answer as οΊ No. 11π 6 11π 6 ? π is greater than . Lesson 12: 2 Inverse Trigonometric Functions This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from PreCal-M4-TE-1.3.0-10.2015 228 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 12 NYS COMMON CORE MATHEMATICS CURRICULUM M4 PRECALCULUS AND ADVANCED TOPICS ο§ Why is it important for the inverse of sine to be a function? 1 2 1 2 Otherwise there would be an infinite number of possible values of sinβ1 ( ). sinβ1 ( ) could be any οΊ 1 2 value π¦ such that sin(π¦) = . Within the restricted range, there is only one value of π¦ that satisfies the equation. Example 1 Consider the function (π) = π¬π’π§ (π), β a. π π β€πβ€ . π π Scaffolding: If students need additional practice, consider using a rapid whiteboard exchange where you present a problem such as the examples listed and students hold up their answer on a small whiteboard. In this way, you can quickly assess student understanding. State the domain and range of this function. π«: β π π β€πβ€ π π πΉ: β π β€ π β€ π b. Find the equation of the inverse function. π = π¬π’π§(π) sinβ1 ( βπ π = π¬π’π§ (π) π β2 )= 2 4 sinβ1 (β π β2 )= 2 4 cos β1 (β cos β1 ( c. State the domain and range of the inverse. sinβ1 (0) = 0 π«: β π β€ π β€ π πΉ: β π π β€πβ€ π π tanβ1 ( Exercises 1β3 (8 minutes) In these exercises, students are familiarizing themselves with the inverse trigonometric functions. Give students time to work through the exercises either individually or in pairs before sharing answers as a class. π β3 )= 3 6 π β2 )=β 2 4 3π β2 )= 2 4 π cos β1 (0) = 2 π β3 tanβ1 (β ) = β 3 6 If students are struggling, use a unit circle diagram to assist them in evaluating these expressions. Exercises 1β3 1. Write an equation for the inverse cosine function, and state its domain and range. π = ππ¨π¬ βπ (π) 2. π«: β π β€ π β€ π πΉ: π β€ π β€ π Write an equation for the inverse tangent function, and state its domain and range. π = πππ§βπ (π) Lesson 12: π«: set of all real numbers πΉ: β Inverse Trigonometric Functions This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from PreCal-M4-TE-1.3.0-10.2015 π π <π< π π 229 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. M4 Lesson 12 NYS COMMON CORE MATHEMATICS CURRICULUM PRECALCULUS AND ADVANCED TOPICS 3. Evaluate each of the following expressions without using a calculator. Use radian measures. a. βπ π¬π’π§βπ ( π ) π¬π’π§βπ (β b. π π c. βπ π ) π β π βπ ππ¨π¬ βπ ( π ) ππ¨π¬ βπ (β d. π π βπ π ) ππ π MP.7 e. π¬π’π§βπ (π) π π f. π¬π’π§βπ (βπ) π β π g. ππ¨π¬ βπ (π) h. ππ¨π¬ βπ (βπ) π i. ο§ πππ§βπ (βπ) π β π j. Scaffolding: The domain is so restricted because the input is the value of cosine, and the values of cosine range from β1 to 1. Why is the range of the inverse cosine function restricted to values from 0 to π? οΊ ο§ πππ§βπ (π) π π Why is the domain of the inverse cosine function restricted to values from β1 to 1? οΊ ο§ π The range is so restricted because we restricted the domain of the cosine function to only the values from 0 to π in order to make it an invertible function. The domain of the cosine function became the range of the inverse cosine function. What does this restriction mean in terms of evaluating an inverse trigonometric function? οΊ The answer must lie within the restricted values of the range. Pose this question to students who like a challenge: Does sin(sinβ1 (π₯)) = π₯ for all values of π₯? Yes, for all values in the domain of sinβ1 (π₯) (β1 β€ π₯ β€ 1). Does sinβ1 (sin(π₯)) = π₯ for all values of π₯? No, only for values of π₯ such π π that β β€ π₯ β€ . 2 2 Example 2 (6 minutes) Work through the examples as a class. ο§ What is the difference between solving the equation cos(π₯) = οΊ MP.6 1 2 1 1 and evaluating the expression cos β1 ( )? 2 2 When solving the equation cos(π₯) = , we are looking for all values of π₯ within the interval 1 2 1 2 0 β€ π₯ β€ 2π such that cos(π₯) = . When evaluating cos β1 ( ), we are looking for the one value within 1 2 the interval 0 β€ π¦ β€ π such that cos(π¦) = . Lesson 12: Inverse Trigonometric Functions This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from PreCal-M4-TE-1.3.0-10.2015 230 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 12 NYS COMMON CORE MATHEMATICS CURRICULUM M4 PRECALCULUS AND ADVANCED TOPICS ο§ In part (b), why do we find the inverse sine of οΊ ο§ 2 3 2 3 instead of β ? We are looking for the reference angle, which is a positive, acute measure in order to find the other solutions. When do we need a calculator to find the reference angle? οΊ π π π 6 4 3 We need a calculator when we are dealing with a value that is not a multiple of , , or or on the π₯- or π¦-axis. Example 2 Solve each trigonometric equation such that π β€ π β€ ππ . Round to three decimal places when necessary. a. π ππ¨π¬(π) β π = π ππ¨π¬(π) = π π π π Reference angle: ππ¨π¬ βπ ( ) = π π The cosine function is positive in Quadrants I and IV. π= b. ππ π and π π πβπ¬π’π§(π) + π = π π¬π’π§(π) = β π π π π Reference angle: π¬π’π§βπ ( ) = π. πππ The sine function is negative in Quadrants III and IV. π = π + π. πππ = π. πππ and π = ππ β π. πππ = π. πππ Exercises 4β8 (12 minutes) Give students time to work through the exercises either individually or in pairs. Circulate the room to ensure students understand the process of solving a trigonometric equation. For Exercises 7β8, consider using a graphing utility to either solve the equations or to check solutions calculated manually. Exercises 4β8 4. Solve each trigonometric equation such that π β€ π β€ ππ . Give answers in exact form. a. βπππ¨π¬(π) + π = π π= b. ππ ππ , π π πππ§(π) β βπ = π π= π ππ , π π Lesson 12: Inverse Trigonometric Functions This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from PreCal-M4-TE-1.3.0-10.2015 231 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 12 NYS COMMON CORE MATHEMATICS CURRICULUM M4 PRECALCULUS AND ADVANCED TOPICS c. π¬π’π§π (π) β π = π π= 5. π ππ , π π Solve each trigonometric equation such that π β€ π β€ ππ . Round answers to three decimal places. a. πβππ¨π¬(π) β π = π π = π. πππ, π. πππ b. πβcos(π) + π = π There are no solutions to this equation within the domain of the function. c. πβπ¬π’π§(π) β π = π π = π. πππ, π. πππ d. πππ§(π) = βπ. πππ π = π. πππ, π. πππ 6. A particle is moving along a straight line for π β€ π β€ ππ. The velocity of the particle at time π (in seconds) is given by π π the function π(π) = ππ¨π¬ ( π). Find the time(s) on the interval π β€ π β€ ππ where the particle is at rest (π(π) = π). The particle is at rest at π = π. π ππππππ π, π. π ππππππ π, ππ. π ππππππ π, and ππ. π ππππππ π. 7. In an amusement park, there is a small Ferris wheel, called a kiddie wheel, for toddlers. The formula π π π―(π) = ππβπ¬π’π§ (ππ (π β )) + ππ models the height π― (in feet) of the bottom-most car π minutes after the wheel begins to rotate. Once the ride starts, it lasts π minutes. a. What is the initial height of the car? π ππ. b. How long does it take for the wheel to make one full rotation? π minute c. What is the maximum height of the car? ππ ππ. d. Find the time(s) on the interval π β€ π β€ π when the car is at its maximum height. π π The car is at its maximum height when π¬π’π§ (ππ (π β )) = π, which is at π = π. π, π. π, π. π, and π. π minutes. Lesson 12: Inverse Trigonometric Functions This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from PreCal-M4-TE-1.3.0-10.2015 232 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 12 NYS COMMON CORE MATHEMATICS CURRICULUM M4 PRECALCULUS AND ADVANCED TOPICS Many animal populations fluctuate periodically. Suppose that a wolf population over an π-year period is given by 8. π π the function πΎ(π) = πππ π¬π’π§ ( π) + ππππ, where π represents the number of years since the initial population counts were made. a. Find the time(s) on the interval π β€ π β€ π such that the wolf population equals π, πππ. π = π. πππ, π. πππ The wolf population equals π, πππ after approximately π. π years and again after π. π years. b. On what time interval during the π-year period is the population below π, πππ? πΎ(π) = ππππ at π = π. πππ and π. πππ The wolf population is below π, πππ on the time interval (π. πππ, π. πππ). c. Why would an animal population be an example of a periodic phenomenon? An animal population might increase while their food source is plentiful. Then, when the population becomes too large, there is less food and the population begins to decrease. At a certain point, there are few enough animals that there is plenty of food for the entire population at which point the population begins to increase again. Closing (3 minutes) Use the following questions to summarize the lesson and check for student understanding. ο§ What does π¦ = sinβ1 (π₯) mean? οΊ ο§ π such that sin(π¦) = π₯. 2 Is cosecant the same as inverse sine? οΊ ο§ π 2 It means find the value π¦ on the interval β β€ π¦ β€ No. Cosecant is the reciprocal of sine not the inverse of sine. Suzanne says that tanβ1 (ββ3) is 5π π . When Rosanne says that it is β , Suzanne says either answer is fine 3 3 because the two rotations lie on the same spot on the unit circle. What is wrong with Suzanneβs thinking? MP.3 οΊ π 3 5π 5π because is outside of the restricted range. Because inverse 3 3 π π tangent is a function, there can only be one output value. That value must lie between β and . 2 2 tanβ1 (ββ3) = β and cannot equal Exit Ticket (5 minutes) Lesson 12: Inverse Trigonometric Functions This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from PreCal-M4-TE-1.3.0-10.2015 233 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 12 NYS COMMON CORE MATHEMATICS CURRICULUM M4 PRECALCULUS AND ADVANCED TOPICS Name Date Lesson 12: Inverse Trigonometric Functions Exit Ticket 1. State the domain and range for π(π₯) = sinβ1 (π₯), π(π₯) = cos β1 (π₯), and β(π₯) = tanβ1 (π₯). 2. Solve each trigonometric equation such that 0 β€ π₯ β€ 2π. Give answers in exact form. 3. a. 2 sin(π₯) + β3 = 0 b. tan2 (π₯) β 1 = 0 Solve the trigonometric equation such that 0 β€ π₯ β€ 2π. Round to three decimal places. β5 cos(π₯) β 2 = 0 Lesson 12: Inverse Trigonometric Functions This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from PreCal-M4-TE-1.3.0-10.2015 234 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 12 NYS COMMON CORE MATHEMATICS CURRICULUM M4 PRECALCULUS AND ADVANCED TOPICS Exit Ticket Sample Solutions 1. State the domain and range for π(π) = π¬π’π§βπ (π), π(π) = ππ¨π¬ βπ (π), and π(π) = πππ§βπ (π). For π, the domain is all real numbers π, such that βπ β€ π β€ π, and the range is all real numbers π, such that β π π β€πβ€ . π π For π, the domain is all real numbers π, such that βπ β€ π β€ π, and the range is all real numbers π, such that π β€ π β€ π . For π, the domain is all real numbers π, and the range is all real numbers π, such that β 2. Solve each trigonometric equation such that π β€ π β€ ππ . Give answers in exact form. a. π π¬π’π§(π) + βπ = π π= b. ππ ππ , π π πππ§π (π) β π = π π= 3. π π <π< . π π π ππ ππ ππ , , , π π π π Solve the trigonometric equation such that π β€ π β€ ππ . Round to three decimal places. βπ ππ¨π¬(π) β π = π π = π. πππ, π. πππ Problem Set Sample Solutions 1. Solve the following equations. Approximate values of the inverse trigonometric functions to the thousandths place, where π refers to an angle measured in radians. a. π = π ππ¨π¬(π) ππ π ± π. πππ b. c. π π π π β = π ππ¨π¬ (π β ) + π ππ π + ππ β π. πππ π ππ π β ππ + π. πππ π π = ππ¨π¬(π(π β π)) ππ π +π π Lesson 12: Inverse Trigonometric Functions This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from PreCal-M4-TE-1.3.0-10.2015 235 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 12 M4 PRECALCULUS AND ADVANCED TOPICS d. π. π = βπ. π ππ¨π¬(π π) + π. π βπ. πππ + π + ππ π π π. πππ β π + ππ π π e. π = βπ ππ¨π¬(π) β π No solutions f. π = π π¬π’π§(π) π. πππ + ππ π π β π. πππ + ππ π g. βπ = π¬π’π§ ( π (πβπ) )βπ π ππ + π h. π = π π¬π’π§(ππ + π) + π π. πππ β π + ππ π π π β π. πππ β π + ππ π π i. π π = π¬π’π§(π) π π. πππ + ππ π π β π. πππ + ππ π j. ππ¨π¬(π) = π¬π’π§(π) π= π π k. π¬π’π§(π) = πππ§(π) ππ¨π¬(π) + π π (or π. πππ + π π) π¬π’π§βπ (ππ¨π¬(π)) = ππ π ± l. π π π π πππ§(π) = π π. πππ + π π Lesson 12: Inverse Trigonometric Functions This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from PreCal-M4-TE-1.3.0-10.2015 236 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 12 NYS COMMON CORE MATHEMATICS CURRICULUM M4 PRECALCULUS AND ADVANCED TOPICS βπ = π πππ§(ππ + π) β π m. π β π + π π π π Alternatively, π.πππ β π + π π . π π = βπ. π πππ§(βπ) β π n. π. πππ + π π 2. Fill out the following tables. π βπ ππ¨π¬ βπ (π) π π¬π’π§βπ (π) ππ¨π¬ βπ (π) β π π π π π π π β βπ π β π π ππ π π π π π π π β βπ π β π π ππ π βπ π π π π π π π β π π ππ π βπ π π π π π π π π π β 3. π¬π’π§βπ (π) Let the velocity π in miles per second of a particle in a particle accelerator after π seconds be modeled by the function π = πππ§ ( a. π π π β ) on an unknown domain. ππππ π What is the π-value of the first vertical asymptote to the right of the π-axis? π = ππππ b. If the particle accelerates to ππ% of the speed of light before stopping, then what is the domain? Note: π β πππ, πππ. Round your solution to the ten-thousandths place. π. ππ β πππβπππ = πππβπππ π π π πππβπππ = πππ§ ( β ) ππππ π π π π πππ§βπ (πππβπππ) = β ππππ π ππππ π β (πππ§βπ (πππβπππ) + ) = π π π π β ππππ. ππππ So the domain is π < π β€ ππππ. ππππ. c. How close does the domain get to the vertical asymptote of the function? Very close. They are only different at the hundredths place. Lesson 12: Inverse Trigonometric Functions This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from PreCal-M4-TE-1.3.0-10.2015 237 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 12 NYS COMMON CORE MATHEMATICS CURRICULUM M4 PRECALCULUS AND ADVANCED TOPICS d. How long does it take for the particle to reach the velocity of Earth around the sun (about ππ. π miles per second)? π π π β ) ππππ π π π π πππ§βπ (ππ. π) = β ππππ π ππππ π β (πππ§βπ (ππ. π) + ) = π π π π β ππππ. πππ ππ. π = πππ§ ( It takes approximately π, πππ. πππ seconds to reach the velocity of Earth around the sun. e. What does it imply that π is negative up until π = ππππ? The particle is traveling in the opposite direction. Lesson 12: Inverse Trigonometric Functions This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from PreCal-M4-TE-1.3.0-10.2015 238 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.