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Lesson 7 NYS COMMON CORE MATHEMATICS CURRICULUM M2 ALGEBRA II Lesson 7: Secant and the Co-Functions Student Outcomes ο§ Students define the secant function and the co-functions in terms of points on the unit circle. They relate the names for these functions to the geometric relationships among lines, angles, and right triangles in a unit circle diagram. ο§ Students use reciprocal relationships to relate the trigonometric functions to each other and use these relationships to evaluate trigonometric functions at multiples of 30, 45, and 60 degrees. Lesson Notes The geometry of the unit circle and its related triangles provide a clue as to how the different reciprocal functions got their names. This lesson draws out the connections among tangent and secant lines of a circle, angle relationships, and the trigonometric functions. The names for the various trigonometric functions make more sense to students when viewed through the lens of geometric figures, providing students with an opportunity to practice MP.7. Students make sense of the domain and range of these functions and use the definitions to evaluate the trigonometric functions for rotations that are multiples of 30, 45, and 60 degrees. The relevant vocabulary upon which this lesson is based appears below. SECANT FUNCTION (description): The secant function, sec: {π₯ β β | π₯ β 90 + 180π for all integers π} β β can be defined as follows: Let π be any real number such that π β 90 + 180π for all integers π. In the Cartesian plane, rotate the initial ray by π degrees about the origin. Intersect the resulting terminal ray with the unit circle to get a point (π₯π , π¦π ). The value of sec(π°) is 1 π₯π . COSECANT FUNCTION (description): The cosecant function, csc: {π₯ β β | π₯ β 180π for all integers π} β β can be defined as follows: Let π be any real number such that π β 180π for all integers π. In the Cartesian plane, rotate the initial ray by π degrees about the origin. Intersect the resulting terminal ray with the unit circle to get a point (π₯π , π¦π ). The value of csc(π°) is 1 π¦π . COTANGENT FUNCTION (description): The cotangent function, cot: {π₯ β β | π₯ β 180π for all integers π} β β can be defined as follows: Let π be any real number such that π β 180π for all integers π. In the Cartesian plane, rotate the initial ray by π degrees about the origin. Intersect the resulting terminal ray with the unit circle to get a point (π₯π , π¦π ). The value of cot(π°) is Lesson 7: π₯π π¦π Secant and the Co-Functions This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M2-TE-1.3.0-08.2015 . 99 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 7 NYS COMMON CORE MATHEMATICS CURRICULUM M2 ALGEBRA II Classwork Opening Exercise (5 minutes) Give students a short time to work independently on this Opening Exercise (about 2 minutes). Lead a whole-class discussion afterward to reinforce the reasons why the different segments have the given measures. If students have difficulty naming these segments in terms of trigonometric functions they have already studied, remind them of the conclusions of the previous few lessons. Opening Exercise Find the length of each segment below in terms of the value of a trigonometric function. πΆπΈ = π·πΈ = πΉπΊ = Debrief this exercise with a short discussion. For the purposes of this section, limit the values of π to be between 0 and 90. Make sure that each student has labeled the proper line segments on his paper as sin(π°), cos(π°), tan(π°), and sec(π°) before moving on to Example 1. ο§ Why is ππ = sin(π°)? Why is ππ = cos(π°)? οΊ ο§ The values of the sine and cosine functions correspond to the π¦- and π₯-coordinates of a point on the unit circle where the terminal ray intersects the circle after a rotation of π degrees about the origin. Why is π π = tan(π°)? οΊ In Lesson 5, we used similar triangles to show that Μ Μ Μ Μ π π had length ππ ππ = sin(π°) . cos(π°) This quotient is tan(π°). Lesson 7: Secant and the Co-Functions This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M2-TE-1.3.0-08.2015 100 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 7 NYS COMMON CORE MATHEMATICS CURRICULUM M2 ALGEBRA II Scaffolding: Since there are ways to calculate lengths for nearly every line segment in this diagram using the length of the radius, or the cosine, sine, or tangent functions, it makes sense to find the length of Μ Μ Μ Μ ππ , the line segment on the terminal ray that intersects the tangent line. ο§ What do you call a line that intersects a circle at more than one point? οΊ ο§ ο§ Have students use differentcolored highlighters or pencils to mark proportional segments. Consider writing out the proportional relationships using the segment names first. It is called a secant line. Μ Μ Μ Μ lies on the line tangent to In Lesson 6, we saw that π π = tan(π°), where π π the unit circle at (1,0), which helped to explain how this trigonometric function got its name. Letβs introduce a new function sec(π°), the secant of π, to be the length of Μ Μ Μ Μ ππ since this segment is on the secant line that contains the terminal ray. Then the secant of π is sec(π°) = ππ . For example, ππ ππ = ππ ππ . ο§ Provide a numeric example for students to work first. Example 1 (5 minutes) Direct students to label the appropriate segments on their papers if they have not already done so. Reproduce the Opening Exercise diagram on chart paper, and label the segments on the diagram in a different color than the rest of the diagram so they are easy to see. Refer to this chart often in the next section of the lesson. ο§ For more advanced students, skip the Opening Exercise and start by stating that a new function, sec( π°) = ππ , is being introduced since Μ Μ Μ Μ ππ is on the secant line that passes through the center of the circle. Example 1 Use similar triangles to find the value of π¬ππ(π½°) in terms of one other trigonometric function. Approach 1: By similar triangles, πΆπΉ πΆπΊ = πΆπ· , so π¬ππ(π½°) πΆπΈ π πΆπ· π¬ππ(π½°) = π ; thus, π¬ππ(π½°) = ππ¨π¬(π½°) π . ππ¨π¬(π½°) Approach 2: By similar triangles πΆπΉ = , so πΆπΊ πΆπΈ π thus, π¬ππ(π½°) = . ππ¨π¬(π½°) π = πππ§(π½°) π¬π’π§(π½°) = π¬π’π§(π½°)/ππ¨π¬(π½°) π¬π’π§(π½°) = π¬π’π§(π½°) β π ππ¨π¬(π½°) π¬π’π§(π½°) = π ; ππ¨π¬(π½°) Exercise 1 (5 minutes) Following the same technique as in the previous lesson with the tangent function, use this working definition of the secant function to extend it outside of the first quadrant. Present the definition of secant, and then ask students to answer the following questions with a partner or in writing. Circulate around the classroom to informally assess understanding and provide guidance. Lead a whole-class discussion based on these questions after giving individuals or partners a few minutes to record their thoughts. Encourage students to revise what they wrote as the discussion progresses. Start a bulleted list of the main points of this discussion on the board. Encourage students to draw a picture representing both the circle description of the secant function and its relationship to cos(π°) to assist them in answering the question. A series of guided questions follows that help in scaffolding this discussion. Lesson 7: Secant and the Co-Functions This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M2-TE-1.3.0-08.2015 101 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 7 NYS COMMON CORE MATHEMATICS CURRICULUM M2 ALGEBRA II Exercise 1 The definition of the secant function is offered below. Answer the questions to better understand this definition and the domain and range of this function. Be prepared to discuss your responses with others in your class. Let π½ be any real number. In the Cartesian plane, rotate the nonnegative π-axis by π½ degrees about the origin. Intersect this new ray with the unit circle to get a point (ππ½ , ππ½ ). If ππ½ β π, then the value of π¬ππ(π½°) is π¬ππ(π½°) is undefined. π ππ½ . Otherwise, In terms of the cosine function, π¬ππ(π½°) = π ππ¨π¬(π½°) for ππ¨π¬(π½°) β π. a. What is the domain of the secant function? The domain of the secant function is all real numbers π½ such that π½ β ππ + ππππ, for all integers π. b. The domains of the secant and tangent functions are the same. Why? Both the tangent and secant functions can be written as rational expressions with ππ¨π¬(π½°) in the denominator. That is, πππ§(π½°) = π¬π’π§(π½°) π , and π¬ππ(π½°) = . Since the restricted values in their domains ππ¨π¬(π½°) ππ¨π¬(π½°) occur when the denominator is zero, it makes sense that they have the same domain. MP.3 c. What is the range of the secant function? How is this range related to the range of the cosine function? Since π¬ππ(π½°) = π , and βπ β€ ππ¨π¬(π½°) β€ π, we see that either π¬ππ(π½°) β₯ π, or π¬ππ(π½°) β€ βπ. ππ¨π¬(π½°) Thus, the range of the secant function is (ββ, βπ] βͺ [π, β). d. Is the secant function a periodic function? If so, what is its period? Yes, its period is πππ. Discussion (5 minutes) Use these questions to scaffold the discussion to debrief the preceding exercise. ο§ What are the values of sec(π°) when the terminal ray is horizontal? When the terminal ray is vertical? οΊ When the terminal ray is horizontal, it will coincide with the positive or negative π₯-axis, and sec(π°) = οΊ 1 1 = 1, or sec(π°) = = β1. 1 β1 When the terminal ray is vertical, the π₯-coordinate of point π is zero, so sec(π°) = 1 , which is 0 undefined. If we use the geometric interpretation of sec(π°) as the length of the segment from the origin to the intersection of the terminal ray and the tangent line to the circle at (1,0), then sec(π°) is undefined because the secant line containing the terminal ray and the tangent line will be parallel and will not intersect. Lesson 7: Secant and the Co-Functions This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M2-TE-1.3.0-08.2015 102 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 7 NYS COMMON CORE MATHEMATICS CURRICULUM M2 ALGEBRA II ο§ Name several values of π for which sec(π°) is undefined. Explain your reasoning. οΊ ο§ What is the domain of the secant function? οΊ ο§ The secant of π is undefined when cos(π°) = 0 or when the terminal ray intersects the unit circle at the π¦-axis. Some values of π that meet these conditions include 90°, 270°, and 450°. Answers will vary but should be equivalent to all real numbers except 90 + 180π, where π is an integer. Students may also use set-builder notation such as {π β β| cos(π°) β 0}. The domains of the tangent and secant functions are the same. Why? οΊ Approach 1 (circle approach): ο§ οΊ The tangent and secant segments are defined based on their intersection, so at times when they are parallel, these segments do not exist. Approach 2 (working definition approach): ο§ Both the tangent and secant functions are defined as rational expressions with cos(π) in the denominator. That is, tan(π°) = sin(π°) 1 , and sec(π°) = . Since the restricted values in cos(π°) cos(π°) their domains occur when the denominator is zero, it makes sense that they have the same domain. ο§ How do the values of the secant and cosine functions vary with each other? As cos(π°) gets larger, what happens to the value of sec(π°)? As cos(π°) gets smaller but stays positive, what happens to the value of sec(π°)? What about when cos(π°) < 0? οΊ ο§ What is the smallest positive value of sec(π°)? Where does this occur? οΊ ο§ The largest negative value of sec(π°) = β1 and occurs when cos(π°) = β1; that is, when π = 180 + 360π, for π an integer. What is the range of the secant function? οΊ ο§ The smallest positive value of sec(π°) is 1 and occurs when cos(π°) = 1, that is, when π = 360π, for π an integer. What is the largest negative value of sec(π°)? Where does this occur? οΊ ο§ We know that sec(π°) and cos(π°) are reciprocals of each other, so as the magnitude of one gets larger, the magnitude of the other gets smaller and vice versa. As cos(π°) increases, sec(π°) gets closer to 1. As cos(π°) decreases but stays positive, sec(π°) increases without bound. When cos(π°) is negative, as cos(π°) increases but stays negative, sec(π°) decreases without bound. As cos(π°) decreases, sec(π°) gets closer to β1. All real numbers outside of the interval (β1,1), including β1 and 1. Is the secant function a periodic function? If so, what is its period? οΊ The secant function is periodic, and the period is 360°. Exercise 2 (5 minutes) These questions get students thinking about the origin of the names of the co-functions. They start with the diagram from the Opening Exercise and ask students how the diagram below compares to it. Then, the point of Example 2 is to introduce the sine, secant, and tangent ratios of the complement to π and justify why we name these functions cosine, cosecant, and cotangent. Lesson 7: Secant and the Co-Functions This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M2-TE-1.3.0-08.2015 103 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 7 NYS COMMON CORE MATHEMATICS CURRICULUM M2 ALGEBRA II Exercise 2 In the diagram, the horizontal blue line is tangent to the unit circle at (π, π). Scaffolding: a. How does this diagram compare to the one given in the Opening Exercise? It is basically the same diagram with different angles and lengths marked. We can consider a rotation through π· degrees measured clockwise from the π-axis instead of a rotation through π½ degrees measured counterclockwise from the π-axis. It is essentially a reflection of the diagram from the Opening Exercise across the diagonal line with equation π = π. b. ο§ Students struggling to see the complementary relationships among the trigonometric functions may benefit from drawing the unit circle on a transparency or patty paper and reflecting across the diagonal line π¦ = π₯ so that the π₯-axis and π¦-axis are interchanged. ο§ Students may need to be shown where sin(π°) is on the diagram as well. What is the relationship between π· and π½? Angles π»πΆπ· and πΊπΆπ· are complements, so π· + π½ = ππ. MP.7 c. Which segment in the figure has length π¬π’π§(π½°)? Which segment has length ππ¨π¬(π½°)? πΆπΌ = π¬π’π§(π½°), and πΌπ· = ππ¨π¬(π½°). d. Which segment in the figure has length π¬π’π§(π·°)? Which segment has length ππ¨π¬(π·°)? πΌπ· = π¬π’π§(π·°), and πΆπΌ = ππ¨π¬(π½°). e. How can you write π¬π’π§(π½°) and ππ¨π¬(π½°) in terms of the trigonometric functions of π·? π¬π’π§(π½°) = ππ¨π¬(π·°), and ππ¨π¬(π½°) = π¬π’π§(π·°). Briefly review the solutions to these exercises before moving on. Reinforce the result of part (e) that the cosine of an acute angle is the same as the sine of its complement. Ask students to consider whether the trigonometric ratios of the complement of an angle might be related to the original angle in a similar fashion. Record student responses to the exercises and any other predictions or thoughts on another sheet of chart paper. Lesson 7: Secant and the Co-Functions This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M2-TE-1.3.0-08.2015 104 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 7 NYS COMMON CORE MATHEMATICS CURRICULUM M2 ALGEBRA II Example 2 (5 minutes) Use similar triangles to show that if π + π½ = 90, then sec(π½°) = 1 1 and tan(π½°) = . Depending on studentsβ sin(π°) tan(π°) level, provide additional scaffolding or just have groups work independently on this. Circulate around the classroom as they work. Example 2 The horizontal blue line is tangent to the circle at (π, π). a. If two angles are complements with measures π· and π½ as shown in the diagram, use similar triangles to show that π¬ππ(π·°) = π . π¬π’π§(π½°) Consider triangles β³ πΆπ·πΌ and β³ πΆπΉπ». Then π¬π’π§(π½°) = ππ¨π¬(π·°), we can write πΆπΉ πΆπ· = π»πΉ , so π¬ππ(π·°) πΌπ· π = πππ§(π·°) . Since ππ¨π¬(π½°) = π¬π’π§(π·°) and ππ¨π¬(π½°) π¬π’π§(π·°) ( ) ππ¨π¬(π·°) π¬ππ(π·°) = ππ¨π¬(π½°) ππ¨π¬(π½°) ( ) π¬π’π§(π½°) = ππ¨π¬(π½°) = b. π . π¬π’π§(π½°) If two angles are complements with measures π· and π½ as shown in the diagram, use similar triangles to show that πππ§(π·°) = π . πππ§(π½°) Using triangles β³ πΆπ·πΌ and β³ πΆπΉπ», we have then π πππ§(π½°) Lesson 7: = πΉπ» πΆπ» ππ¨π¬(π½°) π ; so, πππ§(π·°) = . πππ§(π½°) π¬π’π§(π½°) = πΌπ· . Then πΆπΌ Secant and the Co-Functions This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M2-TE-1.3.0-08.2015 πππ§(π·°) π = ππ¨π¬(π½°) π¬π’π§(π½°) . Since πππ§(π½°) = , ππ¨π¬(π½°) π¬π’π§(π½°) 105 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 7 NYS COMMON CORE MATHEMATICS CURRICULUM M2 ALGEBRA II Discussion (7 minutes) The reciprocal functions 1 sin(π°) and 1 tan(π°) are called the cosecant and cotangent functions. The cosine function has already been defined in terms of a point on the unit circle. The same can be done with the cosecant and cotangent functions. All three functions introduced in this lesson are defined below. Scaffolding: 1 For each of the reciprocal functions, use the facts that sec(π°) = , To help students keep track of cos(π°) 1 1 csc(π°) = , and cot(π°) = to extend the definitions of these functions sin(π°) tan(π°) beyond the first quadrant and to help with the following ideas. Have students work in pairs to answer the following questions before sharing their conclusions with the class. the reciprocal definitions, mention that every trigonometric function has a co-function. Discussion Definitions of the cosecant and cotangent functions are offered below. Answer the questions to better understand the definitions and the domains and ranges of these functions. Be prepared to discuss your responses with others in your class. Let π½ be any real number such that π½ β ππππ, for all integers π. In the Cartesian plane, rotate the initial ray by π½ degrees about the origin. Intersect the resulting terminal ray with the unit circle to get a point (ππ½ , ππ½ ). The value of ππ¬π(π½°) is The value of ππ¨π(π½°) is π ππ½ ππ½ ππ½ . . The secant, cosecant, and cotangent functions are often referred to as reciprocal functions. Why do you think these functions are so named? These functions are the reciprocals of the three main trigonometric functions: π¬ππ(π½°) = π π π , ππ¬π(π½°) = , and ππ¨π(π½°) = . ππ¨π¬(π½°) π¬π’π§(π½°) πππ§(π½°) Why are the domains of these functions restricted? We restrict the domain to prevent division by zero. We restrict the domain because the geometric shapes defining the functions must make sense (i.e., based on the intersection of distinct parallel lines). The domains of the cosecant and cotangent functions are the same. Why? Both the cosecant and cotangent functions are equal to rational expressions that have π¬π’π§(π½°) in the denominator. What is the range of the cosecant function? How is this range related to the range of the sine function? The sine function has range [βπ, π], so the cosecant function has range (ββ, βπ] βͺ [π, β). The two ranges only intersect at βπ and at π. Lesson 7: Secant and the Co-Functions This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M2-TE-1.3.0-08.2015 106 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 7 NYS COMMON CORE MATHEMATICS CURRICULUM M2 ALGEBRA II What is the range of the cotangent function? How is this range related to the range of the tangent function? The range of the cotangent function is all real numbers. This is the same as the range of the tangent function. ο§ Why are the secant, cosecant, and cotangent functions called reciprocal functions? οΊ Each one is a reciprocal of cosine, sine, or tangent according to their definitions. For example, csc(π°) is equal to ο§ The range of the cosecant function is all real numbers except between β1 and 1; that is, the range of the cosecant function is (ββ, β1] βͺ [1, β), which is the same as the range of the secant function. No, sec(π°) and csc(π°) cannot be between 0 and 1, but cot(π°) can. Both the secant and cosecant functions are reciprocals of functions that range from β1 to 1, while the cotangent function is the reciprocal of a function that ranges across all real numbers. Whenever tan(π°) > 1, 0 < cot(π°) < 1. What is the value of cot(90°)? οΊ ο§ The greatest negative value of the cosecant function is β1. Can sec(π°) or csc(π°) be a number between 0 and 1? Can cot(π°)? Explain why or why not. οΊ ο§ As with the secant function, the smallest positive value of the cosecant function is 1. This is because the cosecant function is the reciprocal of the sine function, which has a maximum value of 1. Also, the secant and cosecant functions have values based on the length of line segments from the center of the circle to the intersection with the tangent line. Thus, when positive, they must always be greater than or equal to the radius of the circle. What is the range of the cosecant function? What other trigonometric function has this range? οΊ ο§ The cosecant and cotangent functions share the same domain. Both functions can be written as a rational expression with sin(π°) in the denominator, so whenever sin(π°) = 0, they are undefined. The domain is all real numbers except π such that sin(π°) = 0. More specifically, the domain is {π β β|π β 180π, for all integers π}. What is the greatest negative value of the cosecant function? οΊ ο§ , the reciprocal of sin(π°). What is the smallest positive value of the cosecant function? οΊ ο§ π¦π Which two of the reciprocal functions share the same domain? Why? What is their domain? οΊ ο§ 1 cot(90°) = cos(90°) 0 = = 0. sin(90°) 1 How does the range of the cotangent function compare to the range of the tangent function? Why? οΊ The ranges of the tangent and cotangent functions are the same. When tan(π°) is close to 0, cot(π°) is far from 0, either positive or negative depending on the sign of tan(π°). When tan(π°) is far from 0, then cot(π°) is close to 0. Finally, when one function is undefined, the other is 0. Closing (4 minutes) ο§ The secant and cosecant functions are reciprocals of the cosine and sine functions, respectively. The tangent and cotangent functions are also reciprocals of each other. It is helpful to draw the circle diagram used to define the tangent and secant functions to ensure mistakes are not made with the relationships derived from similar triangles. Lesson 7: Secant and the Co-Functions This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M2-TE-1.3.0-08.2015 107 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 7 NYS COMMON CORE MATHEMATICS CURRICULUM M2 ALGEBRA II ο§ This is a good time to summarize all six trigonometric functions. With a partner, in writing, or as a class, have students summarize the definitions for sine, cosine, tangent, and the three reciprocal functions along with their domains. Use this as an opportunity to check for any gaps in understanding. Use the following summary as a model: Let π½ be any real number. In the Cartesian plane, rotate the initial ray by π½ degrees about the origin. Intersect the resulting terminal ray with the unit circle to get a point (ππ½ , ππ½ ). Then: Function Value For any π½ such that β¦ Formula Sine ππ½ π½ is a real number Cosine ππ½ π½ is a real number Tangent ππ½ ππ½ π½ β ππ + ππππ, for all integers π πππ§(π½°) = π¬π’π§(π½°) ππ¨π¬(π½°) Secant π ππ½ π½ β ππ + ππππ, for all integers π π¬ππ(π½°) = π ππ¨π¬(π½°) Cosecant π ππ½ π½ β ππππ, for all integers π ππ¬π(π½°) = π π¬π’π§(π½°) Cotangent ππ½ ππ½ π½ β ππππ, for all integers π ππ¨π(π½°) = ππ¨π¬(π½°) π¬π’π§(π½°) Exit Ticket (4 minutes) Lesson 7: Secant and the Co-Functions This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M2-TE-1.3.0-08.2015 108 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 7 NYS COMMON CORE MATHEMATICS CURRICULUM M2 ALGEBRA II Name Date Lesson 7: Secant and the Co-Functions Exit Ticket Consider the following diagram, where segment π΄π΅ is tangent to the circle at π·. Right triangles π΅π΄π, π΅ππ·, ππ΄π·, and ππ·πΆ are similar. Identify each length π΄π·, ππ΄, ππ΅, and π΅π· as one of the following: tan(π°), cot(π°), sec(π°), and csc(π°). Lesson 7: Secant and the Co-Functions This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M2-TE-1.3.0-08.2015 109 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 7 NYS COMMON CORE MATHEMATICS CURRICULUM M2 ALGEBRA II Exit Ticket Sample Solutions Consider the following diagram, where segment π¨π© is tangent to the circle at π«. Right triangles π©π¨πΆ, π©πΆπ«, πΆπ¨π«, and πΆπ«πͺ are similar. Identify each length π¨π«, πΆπ¨, πΆπ©, and π©π« as one of the following: πππ§(π½°), ππ¨π(π½°), π¬ππ(π½°), and ππ¬π(π½°). Since β³ πΆπ«πͺ ~ β³ πΆπ«π¨, Since β³ πΆπ«πͺ ~ β³ πΆπ«π¨, Since β³ πΆπ«πͺ ~ β³ π©π¨πΆ, Since β³ πΆπ«πͺ ~ β³ π©πΆπ«, Lesson 7: π¨π« πΆπ« πΆπ¨ πΆπ« πΆπ© πΆπ¨ π©π« πΆπ« = = = = πͺπ« πΆπͺ πΆπ« πΆπͺ πΆπͺ πͺπ« πΆπͺ πͺπ« so so so so π¨π« π πΆπ¨ π = = π¬π’π§(π½°) ππ¨π¬(π½°) π and thus πΆπ¨ = π¬ππ(π½°). ππ¨π¬(π½°) πΆπ© π¬ππ(π½°) π©π« π = and thus π¨π« = πππ§(π½°). = ππ¨π¬(π½°) π¬π’π§(π½°) ππ¨π¬(π½°) π¬π’π§(π½°) and thus πΆπ¨ = π¬ππ(π½°) ππ¨π¬(π½°) π = = ππ¬π(π½°). π¬π’π§(π½°) π¬π’π§(π½°) and thus π©π« = ππ¨π(π½°). Secant and the Co-Functions This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M2-TE-1.3.0-08.2015 110 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 7 NYS COMMON CORE MATHEMATICS CURRICULUM M2 ALGEBRA II Problem Set Sample Solutions 1. Use the reciprocal interpretations of π¬ππ(π½°), ππ¬π(π½°), and ππ¨π(π½°) and the unit circle provided to complete the table. π½, in degrees π¬ππ(π½°) ππ¬π(π½°) ππ¨π(π½°) π π Undefined Undefined ππ πβπ π π βπ ππ βπ βπ π ππ π πβπ π βπ π ππ Undefined π π πππ βπ πβπ π πππ βπ Undefined Undefined πππ ββπ ββπ π πππ βπ πππ Undefined βπ π πππ βπ ββπ βπ πππ πβπ π βπ ββπ Lesson 7: Secant and the Co-Functions This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M2-TE-1.3.0-08.2015 β πβπ π β βπ π βπ π 111 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 7 NYS COMMON CORE MATHEMATICS CURRICULUM M2 ALGEBRA II 2. 3. Find the following values from the information given. a. π¬ππ(π½°); ππ¨π¬(π½°) = π. π b. ππ¬π(π½°); π¬π’π§(π½°) = βπ. ππ c. ππ¨π(π½°); πππ§(π½°) = ππππ d. π¬ππ(π½°); ππ¨π¬(π½°) = βπ. π e. ππ¬π(π½°); π¬π’π§(π½°) = π f. ππ¨π(π½°); πππ§(π½°) = βπ. ππππ π π ππ = = π. π π/ππ π ππ¬π(π½°) = π π = = βππ βπ. ππ βπ/ππ ππ¨π(π½°) = π π = πππ§(π½°) ππππ π¬ππ(π½°) = π π ππ = =β βπ. π βπ/ππ π ππ¬π(π½°) = π π = , but this is undefined. π¬π’π§(π½°) π ππ¨π(π½°) = π π πππππ = =β = βππππ βπ. ππππ βπ/πππππ π Choose three π½ values from the table in Problem π for which π¬ππ(π½°), ππ¬π(π½°), and πππ§(π½°) are defined and not π¬ππ(π½°) zero. Show that for these values of π½, For π½ = ππ°, πππ§(π½°) = βπ π , and For π½ = ππ°, πππ§(π½°) = π, and = πππ§(π½°). ππ¬π(π½°) π¬ππ(π½°) = ππ¬π(π½°) πβπ/π π = π¬ππ(π½°) βπ . Thus, for π½ = ππ, = πππ§(π½°). π ππ¬π(π½°) π¬ππ(π½°) βπ π¬ππ(π½°) = = π. Thus, for π½ = ππ, = πππ§(π½°). ππ¬π(π½°) βπ ππ¬π(π½°) For π½ = ππ°, πππ§(π½°) = βπ, and 4. π¬ππ(π½°) = π¬ππ(π½°) ππ¬π(π½°) = π πβπ π = π βπ = βπ. Thus, for π½ = ππ, π¬ππ(π½°) = πππ§(π½°). ππ¬π(π½°) Find the value of π¬ππ(π½°)ππ¨π¬(π½°) for the following values of π½. a. π½ = πππ π π We know that ππ¨π¬(πππ°) = β , so π¬ππ(πππ°) = βπ, and then π¬ππ(πππ°)ππ¨π¬(πππ°) = π. b. π½ = πππ We know that ππ¨π¬(πππ°) = β c. π , so sππ(πππ°) = ββπ, and then π¬ππ(πππ°)ππ¨π¬(πππ°) = π. π½ = πππ We know that ππ¨π¬(πππ°) = d. βπ βπ π , so π¬ππ(πππ°) = π , and then π¬ππ(πππ°)ππ¨π¬(πππ°) = π. βπ Explain the reasons for the pattern you see in your responses to parts (a)β(c). If ππ¨π¬(π½°) β π, then π¬ππ(π½°) = Lesson 7: π π , so we know that π¬ππ(π½°) ππ¨π¬(π½°) = β ππ¨π¬(π½°) = π. ππ¨π¬(π½°) ππ¨π¬(π½°) Secant and the Co-Functions This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M2-TE-1.3.0-08.2015 112 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 7 NYS COMMON CORE MATHEMATICS CURRICULUM M2 ALGEBRA II 5. Draw a diagram representing the two values of π½ between π and πππ so that π¬π’π§(π½°) = β πππ§(π½°), π¬ππ(π½°), and ππ¬π(π½°) for each value of π½. 6. πππ§(πππ°) = βπ, π¬ππ(πππ°) = βπ, and ππ¬π(πππ°) = β πβπ . π πππ§(πππ°) = ββπ, π¬ππ(πππ°) = π, and ππ¬π(πππ°) = β πβπ . π π βπ π . Find the values of π Find the value of (π¬ππ(π½°)) β (πππ§(π½°)) when π½ = πππ. π π π (π¬ππ(πππ°)) β (πππ§(πππ°)) = (βπ) β ππ = π 7. π π Find the value of (ππ¬π(π½°)) β (ππ¨π(π½°)) when π½ = πππ. π π π (ππ¬π(π½°)) β (ππ¨π(π½°)) = (βπ)π β (ββπ) = π β π = π Extension: 8. Using the formulas π¬ππ(π½°) = π¬ππ(π½°) π π , ππ¬π(π½°) = , and ππ¨π(π½°) = πππ§π(π½°), show that ππ¬π(π½°) = πππ§(π½°), ππ¨π¬(π½°) π¬π’π§(π½°) where these functions are defined and not zero. π π¬ππ(π½°) ππ¨π¬(π½°) π π¬π’π§(π½°) = = β π¬π’π§(π½°) = = πππ§(π½°) π ππ¬π(π½°) ππ¨π¬(π½°) ππ¨π¬(π½°) π¬π’π§(π½°) 9. Tara showed that π¬ππ(π½°) ππ¬π(π½°) = πππ§(π½°), for values of π½ for which the functions are defined and ππ¬π(π½°) β π, and then concluded that π¬ππ(π½°) = π¬π’π§(π½°) and ππ¬π(π½°) = ππ¨π¬(π½°). Explain what is wrong with her reasoning. Just because For instance, π¨ π© π π πͺ = = , this does not mean that π¨ = πͺ and π© = π«. It only means that they are equivalent fractions. π« π . Substituting any value of π½ verifies that Tara is incorrect. Also, since the only values of π½ in ππ the range of both secant and sine are β π and π, we would have either π¬ππ(π½°) = π¬π’π§(π½°) = βπ or π¬ππ(π½°) = π¬π’π§(π½°) = π. Since the value of the sine function is either β π ππ π, the value of the cosine function would be zero. Since π¬ππ(π½°) = π , we would have ππ¨π¬(π½°) = π or ππ¨π¬(π½°) = βπ. Since ππ¨π¬(π½°) cannot be ππ¨π¬(π½°) simultaneously zero and nonzero, it is impossible to have π¬ππ(π½°) = π¬π’π§(π½°) and ππ¬π(π½°) = ππ¨π¬(π½°). Lesson 7: Secant and the Co-Functions This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M2-TE-1.3.0-08.2015 113 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 7 M2 ALGEBRA II 10. From Lesson 6, Ren remembered that the tangent function is odd, meaning that βπππ§(π½°) = πππ§(βπ½°) for all π½ in the domain of the tangent function. He concluded because of the relationship between the secant function, cosecant function, and tangent function developed in Problem 9, it is impossible for both the secant and the cosecant functions to be odd. Explain why he is correct. If we assume that both the secant and cosecant functions are odd, then the tangent function could not be odd. That is, we would get πππ§(βπ½°) = π¬ππ(βπ½°) βπ¬ππ(π½°) π¬ππ(π½°) = = = πππ§(π½°), ππ¬π(βπ½°) βππ¬π(π½°) ππ¬π(π½°) but that would contradict βπππ§(π½°) = πππ§(π½°). Thus, it is impossible for both the secant and cosecant functions to be odd. Lesson 7: Secant and the Co-Functions This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M2-TE-1.3.0-08.2015 114 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.