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Lesson 4 NYS COMMON CORE MATHEMATICS CURRICULUM M4 PRECALCULUS AND ADVANCED TOPICS Lesson 4: Addition and Subtraction Formulas Student Outcomes ο§ Students use the addition formulas to derive the double-angle formulas for sine, cosine, and tangent and evaluate trigonometric expressions for specific values of π. ο§ Students use the double-angle formulas to derive the half-angle formulas for sine, cosine, and tangent and evaluate trigonometric expressions for specific values of π. Lesson Notes In the previous lesson, the students applied an analytic method to prove the subtraction formula for cosine. They then derived the remaining addition and subtraction formulas using their understanding of the periodicity and symmetry of the sine, cosine, and tangent functions. In this lesson, students continue to apply their understanding of the properties of the trigonometric functions to determine double and half-angle formulas for sine, cosine, and tangent, which they apply to evaluate trigonometric functions for specific input values. Classwork Opening (5 minutes) ο§ Letβs return one more time to our carousel model of the unit circle. We know that a riderβs position after a rotation π is (π₯π , π¦π ), where π₯π = cos(π) and π¦π = sin(π). In the last lesson, we developed the sum and difference formulas for sine and cosine; that is, if we know sin(πΌ), sin(π½), cos(πΌ), and cos(π½), then we can find the values of sin(πΌ + π½), sin(πΌ β π½), cos(πΌ + π½), and cos(πΌ β π½). Students should reflect on the following prompts. After a minute, they should share their reflections with a partner. Select several students to share their ideas; ask them to explain a strategy or display a procedure on the board. ο§ π We know the position of the rider after rotating by π = . How can we use the addition formulas to find her position after rotating by οΊ MP.3 & MP.7 ο§ 2π 3 3 ? Answers will vary but might include using the formulas we have derived in previous lessons; e.g., we could rewrite sin(2π) as sin(π + π) and apply the formula for the sine of a sum to find the position of a rider after a rotation of 2π. π We know the position of the rider after rotating by π = . How can we determine his position after rotating 4 π by ? 8 οΊ Answers will vary but might include applying the angle sum formulas in reverse after writing sin(2π) as π π π 4 8 8 sin(π + π), using this particular angle measure, sin ( ) = sin ( + ). Lesson 4: Date: Addition and Subtraction Formulas 4/30/17 © 2015 Common Core, Inc. Some rights reserved. commoncore.org 60 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 4 NYS COMMON CORE MATHEMATICS CURRICULUM M4 PRECALCULUS AND ADVANCED TOPICS Exercise 1 (7 minutes) Scaffolding: In this exercise, students derive the double-angle formula for sine using the addition formula. Allow students to struggle with this task before explaining that they apply the sum formulas with 2π = π + π. This provides an opportunity for them to further develop their mathematical reasoning skills and to practice looking for structure. Students later apply the double-angle formulas to determine the half-angle formulas, which allows them to evaluate trigonometric functions for a larger number of input values. Encourage advanced students to think about ways to write these formulas in terms of sine or cosine only. The exercise should be completed in pairs and then discussed in a whole-class setting after a few minutes. Be sure to identify these formulas as the double-angle formulas for sine and cosine. Exercises 1. Derive formulas for the following: a. π¬π’π§(ππ½) π¬π’π§(ππ½) = π¬π’π§(π½ + π½) = π¬π’π§(π½)ππ¨π¬(π½) + ππ¨π¬(π½)π¬π’π§(π½) = ππ¬π’π§(π½)ππ¨π¬(π½) MP.7 b. ππ¨π¬(ππ½) ππ¨π¬(ππ½) = ππ¨π¬(π½ + π½) = ππ¨π¬(π½)ππ¨π¬(π½) β π¬π’π§(π½)π¬π’π§(π½) = ππ¨π¬ π (π½) β π¬π’π§π (π½) ο§ Why was it helpful to rewrite 2π as (π + π) when deriving the double-angle formula? οΊ ο§ While we donβt know how to find the value of sin(2π) directly, we do know how to evaluate the sine of the sum (π + π) using the formula sin(π + π) = sin(π)cos(π) + cos(π)sin(π) = 2sin(π)cos(π). Our formula for calculating cos(2π) is written in terms of cos 2 (π) and sin2 (π). Is there a way to write this formula in terms of the cosine function only? Explain. οΊ Yes. Since cos 2 (π) + sin2 (π) = 1, we can substitute sin2 (π) = 1 β cos 2 (π). Then cos 2 (π) β sin2 (π) = cos 2 (π) β (1 β cos 2 (π)) = 2cos 2 (π) β 1. ο§ Can we write cos(2π) in terms of the sine function only? Explain. οΊ Yes. Since cos 2 (π) + sin2 (π) = 1, we can substitute cos 2 (π) = 1 β sin2 (π). Then cos 2 (π) β sin2 (π) = (1 β sin2 (π)) β sin2 (π) = 1 β 2sin2 (π). Exercises 2β3 (10 minutes) Students should complete the exercises in pairs or small groups. Each pair or small group should verify one of the identities in Exercise 1. After a few minutes, volunteers could display their solutions. Other students should be allowed MP.3 to offer alternative approaches or critiques. Exercise 2 could be completed in pairs or as part of a teacher-led discussion. The discussion of Exercise 2 could include a verification, using the unit circle, of the signs on the coordinates of the riderβs position. Lesson 4: Date: Addition and Subtraction Formulas 4/30/17 © 2015 Common Core, Inc. Some rights reserved. commoncore.org 61 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 4 NYS COMMON CORE MATHEMATICS CURRICULUM M4 PRECALCULUS AND ADVANCED TOPICS 2. Use the double-angle formulas for sine and cosine to verify these identities: a. πππ§(ππ½) = ππππ§(π½) πβπππ§π (π½) πππ§(ππ½) = π¬π’π§(ππ½) ππ¨π¬(ππ½) ο§ Point students to a visual representation of the definition of the tangent π ππ¬π’π§(π½)ππ¨π¬(π½) ππ¨π¬ π (π½) =( ( ) ) π ππ¨π¬ π (π½) β π¬π’π§π (π½) π (π½) ππ¨π¬ ππ¬π’π§(π½)ππ¨π¬(π½) ππ¨π¬ π (π½) = ππ¨π¬ π (π½) β π¬π’π§π (π½) ππ¨π¬ π (π½) as tan(π) = ππ¬π’π§(π½)ππ¨π¬(π½) ππ¨π¬(π½)ππ¨π¬(π½) ππ¨π¬ π (π½) π¬π’π§π (π½) β ππ¨π¬ π (π½) ππ¨π¬ π (π½) ππππ§(π½) = π β πππ§π (π½) π¬π’π§π (π½) = πβππ¨π¬(ππ½) π . ο§ Cue students to use a procedure similar to that used in Exercise 1 to verify the identity in Exercise 2 part (c). π β ππ¨π¬(ππ½) π β (π β ππ¬π’π§π (π½)) ππ¬π’π§π (π½) = = = π¬π’π§π (π½) π π π c. sin(π) cos(π) ο§ Cue students to look at the structure of the right side of the identity in Exercise 2 part (a) to help them simplify the expression on the left side. For instance, ask the students, βWhat term could we divide our expression by so the first term in the denominator reduces to 1?β = b. Scaffolding: π¬π’π§(ππ½) = βππ¬π’π§π (π½) + ππ¬π’π§(π½) π¬π’π§(ππ½) = π¬π’π§(ππ½ + π½) = π¬π’π§(ππ½)ππ¨π¬(π½) + ππ¨π¬(ππ½)π¬π’π§(π½) = ππ¬π’π§(π½)(ππ¨π¬(π½))(ππ¨π¬(π½)) + (π β ππ¬π’π§π (π½))(π¬π’π§(π½)) = ππ¬π’π§(π½)ππ¨π¬π (π½) + π¬π’π§(π½) β ππ¬π’π§π (π½) = ππ¬π’π§(π½)(π β π¬π’π§π (π½)) + π¬π’π§(π½) β ππ¬π’π§π (π½) = ππ¬π’π§(π½) β ππ¬π’π§π (π½) + π¬π’π§(π½) β ππ¬π’π§π (π½) = βππ¬π’π§π (π½) + ππ¬π’π§(π½) 3. MP.2 Suppose that the position of a rider on the unit circle carousel is (π. π, βπ. π) for a rotation π½. What is the position of the rider after rotation by ππ½? πππ½ = ππ¨π¬(ππ½) = ππ¨π¬ π (π½) β π¬π’π§π (π½) = π. ππ β (βπ. π)π = π. ππ β π. ππ = π. ππ πππ½ = π¬π’π§(ππ½) = ππ¬π’π§(π½)ππ¨π¬(π½) = π(βπ. π)(π. π) = βπ. ππ The riderβs position is (π. ππ, βπ. ππ). Lesson 4: Date: Addition and Subtraction Formulas 4/30/17 © 2015 Common Core, Inc. Some rights reserved. commoncore.org 62 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 4 NYS COMMON CORE MATHEMATICS CURRICULUM M4 PRECALCULUS AND ADVANCED TOPICS Discussion (5 minutes) In this example, students use the double-angle formula for cosine to derive the half-angle formula for cosine. The halfangle formula provides students with an efficient means of evaluating trigonometric functions for a wider variety of inputs. The example should be completed as part of a teacher-led discussion. ο§ What are some of the limitations of the formulas we have derived so far in evaluating trigonometric expressions for specific inputs? οΊ π π π π 2 6 4 3 Since we only determined the exact values of trigonometric functions for multiples of , , , and we are limited in the values of π for which we can evaluate trigonometric expressions. In other words, we can only evaluate sin(π), cos(π), and tan(π) for values of π that can be found by adding or subtracting π π π π multiples of , , , and . 6 4 3 ο§ 2 π π In the Opening, we thought about how we could find sin ( ) and cos ( ) since we know the values of sine and 8 π 8 π cosine at . Do we have a way to evaluate sine and cosine at ? 4 οΊ ο§ Why might it be helpful to derive half-angle formulas for sine, cosine, and tangent? οΊ ο§ 8 No, not yet. Answers will vary but should address that we could evaluate the trigonometric functions for more values of π. Recall that the double-angle formula for cosine is cos(2π) = 2cos 2 (π) β 1. How can we use this formula to π find a value for cos ( ) ? 8 οΊ Allow students to work with a partner and struggle with this question for a minute or two before revealing the answer. π π We can use the double-angle formula with cos ( ) = cos (2 β ), which gives 4 π π 4 8 2 8 π β2 8 2 cos ( ) = cos (2 β ) = 2 cos ( ) β 1. Then we can solve ο§ This gives π cos 2 ( ) 8 1 β2 = ( 2 2 π π 8 8 = 2 cos 2 ( ) β 1 for cos ( ). + 1). How do we know whether or not we need the positive or negative square π root when we solve for cos ( ) ? 8 οΊ π It will depend on the quadrant in which the initial ray lands after rotating by . Cosine is positive in π 8 Quadrants I and IV and negative in Quadrants II and III. Since rotation by terminates in Quadrant I, 8 π the value of cos ( ) is positive. 8 Exercises 4β7 (10 minutes) Students should complete the exercises in pairs, first working on the problems independently and then verifying their solutions with a partner. At an appropriate time, selected students should explain their solutions, including how they determined the signs for the position coordinates in Exercise 6 and the signs for the evaluated trigonometric functions in Exercise 7. As time permits, additional students should be encouraged to share alternative approaches to solving the π problems; e.g., there could be multiple ways to evaluate cos ( ) in Exercise 7 part (b). Note: if calculators are available, 12 students could use them to verify their solutions and approximate the values in Exercises 6 and 7. If they are not available, answers could be left in their exact form. Lesson 4: Date: Addition and Subtraction Formulas 4/30/17 © 2015 Common Core, Inc. Some rights reserved. commoncore.org 63 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 4 NYS COMMON CORE MATHEMATICS CURRICULUM M4 PRECALCULUS AND ADVANCED TOPICS 4. ππ¨π¬(π½)+π . π π½ π Use the double-angle formula for cosine to establish the identity ππ¨π¬ ( ) = ±β π½ π π½ π π½ π Since π½ = π ( ), the double-angle formula gives ππ¨π¬ (π ( )) = πππ¨π¬ π ( ) β π. Then we have π½ ππ¨π¬(π½) = πππ¨π¬ π ( ) β π π π½ π π + ππ¨π¬(π½) = πππ¨π¬ ( ) π π + ππ¨π¬(π½) π½ π = ππ¨π¬ ( ) π π ο§ Have advanced students derive a formula for the one-third angle for sine instead of verifying the identities in Exercise 5. ο§ Cue students in Exercise 7 by asking them, βIs there a way to use the addition or subtraction formulas to π½ ππ¨π¬(π½) + π ππ¨π¬ ( ) = ±β π π 5. Scaffolding: π evaluate cos ( )?β 12 Use the double-angle formulas to verify these identities: a. π½ π πβππ¨π¬(π½) π π¬π’π§ ( ) = ±β π½ π π½ π π½ π Since π½ = π ( ), the double-ange formulas give ππ¨π¬ (π ( )) = π β ππ¬π’π§π ( ). Then we have π½ π ππ¨π¬(π½) = π β ππ¬π’π§π ( ) π½ π β ππ¨π¬(π½) = ππ¬π’π§π ( ) π π β ππ¨π¬(π½) π½ = π¬π’π§π ( ) π π π½ π β ππ¨π¬(π½) π¬π’π§ ( ) = ±β π π b. π½ π πβππ¨π¬(π½) π+ππ¨π¬(π½) πππ§ ( ) = ±β π β ππ¨π¬(π½) π½ π¬π’π§ ( ) ±β π½ π β ππ¨π¬(π½) π π πππ§ ( ) = = = ±β π½ π π + ππ¨π¬(π½) ππ¨π¬ ( ) ±βππ¨π¬(π½) + π π π Lesson 4: Date: Addition and Subtraction Formulas 4/30/17 © 2015 Common Core, Inc. Some rights reserved. commoncore.org 64 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 4 NYS COMMON CORE MATHEMATICS CURRICULUM M4 PRECALCULUS AND ADVANCED TOPICS 6. MP.2 The position of a rider on the unit circle carousel is (π. π, βπ. π) after a rotation by π½ where π β€ π½ < ππ . What is π½ the position of the rider after rotation by ? π Given that ππ¨π¬(π½) is positive and π¬π’π§(π½) is negative, the rider is located in Quadrant IV after rotation by π½, so ππ π½ ππ π < π½ π½ π½ < ππ . This means that < < π , which is in Quadrant II, so ππ¨π¬ ( ) is negative and π¬π’π§ ( ) is positive. π π π π π½ ππ¨π¬(π½) + π π. π + π ππ½ = ππ¨π¬ ( ) = ±β = ββ β βπ. ππ π π π π π½ π β ππ¨π¬(π½) π β π. π ππ½ = π¬π’π§ ( ) = ±β =β β π. ππ π π π π π½ The riderβs position is approximately (βπ. ππ, π. ππ) after rotation by . π 7. Evaluate the following trigonometric expressions. a. π¬π’π§ ( ππ ) π π ππ ππ βπ π β (βππ¨π¬ ( )) ππ π βπ + π βπ β ππ¨π¬ ( π ) β π π¬π’π§ ( ) = π¬π’π§ ( ) = = = β π. ππ π π π π π b. πππ§ ( π ) ππ πππ§ ( π π π β ππ¨π¬ ( ) π ππ ) = πππ§ ( ππ ) = β π ππ π π + ππ¨π¬ ( ) ππ ππ¨π¬ ( π π π π π π π π βπ βπ βπ βπ + βπ ) = ππ¨π¬ ( β ) = ππ¨π¬ ( ) ππ¨π¬ ( ) + π¬π’π§ ( ) π¬π’π§ ( ) = ( ) + ( )= ππ π π π π π π π π π π π πππ§ ( βπ + βπ πβ π π β βπ β βπ π )=β =β β π. ππ ππ + π + βπ + βπ βπ βπ π+ π Lesson 4: Date: Addition and Subtraction Formulas 4/30/17 © 2015 Common Core, Inc. Some rights reserved. commoncore.org 65 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 4 NYS COMMON CORE MATHEMATICS CURRICULUM M4 PRECALCULUS AND ADVANCED TOPICS Closing (3 minutes) Have students respond in writing to one or more of the prompts below: ο§ Write the double-angle formulas that we studied in this lesson for sin(2π), cos(2π), and tan(2π). ο§ Write the half-angle formulas that we studied in this lesson for sin ( ), cos ( ), and tan ( ). ο§ How can our understanding of the trigonometric functions help us determine the position coordinates for a carousel rider on our unit circle model regardless of the size of the rotation π? Share your thoughts with a partner. ο§ Answers will vary. Possible acceptable responses could include: π π π 2 2 2 π π π 6 4 3 οΊ We have found the exact position coordinates for rotations , , and . οΊ Our understanding of the unit circle allows us to determine the position coordinates for rotations that π are multiples of . οΊ The double- and triple-angle formulas enable us to find position coordinates for rotations that are π π π multiples of , , and . οΊ The half-angle and addition/subtraction formulas enable us to find position coordinates for many other rotational values. 2 6 4 3 Lesson Summary The double-angle and half-angle formulas for sine, cosine, and tangent are summarized below. For all real numbers π½ for which the expressions are defined, π¬π’π§(ππ½) = ππ¬π’π§(π½)ππ¨π¬(π½) ππ¨π¬(ππ½) = ππ¨π¬ π (π½) β π¬π’π§π (π½) = πππ¨π¬ π (π½) β π = π β ππ¬π’π§π (π½) ππππ§(π½) πππ§(ππ½) = π β πππ§π (π½) π½ π β ππ¨π¬(π½) π¬π’π§ ( ) = ±β π π π ππ¨π¬(π½) + π ππ¨π¬ ( ) = ±β π π π½ π β ππ¨π¬(π½) πππ§ ( ) = ±β π π + ππ¨π¬(π½) Exit Ticket (5 minutes) Lesson 4: Date: Addition and Subtraction Formulas 4/30/17 © 2015 Common Core, Inc. Some rights reserved. commoncore.org 66 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 4 NYS COMMON CORE MATHEMATICS CURRICULUM M4 PRECALCULUS AND ADVANCED TOPICS Name Date Lesson 4: Addition and Subtraction Formulas Exit Ticket 1. Show that cos(3π) = 4cos 3 (π) β 3cos(π). 2. Evaluate cos ( ) using the half-angle formula, and then verify your solution using a different formula. 7π 12 Lesson 4: Date: Addition and Subtraction Formulas 4/30/17 © 2015 Common Core, Inc. Some rights reserved. commoncore.org 67 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 4 NYS COMMON CORE MATHEMATICS CURRICULUM M4 PRECALCULUS AND ADVANCED TOPICS Exit Ticket Sample Solutions 1. Show that ππ¨π¬(ππ½) = πππ¨π¬ π (π½) β πππ¨π¬(π½). ππ¨π¬(ππ½) = ππ¨π¬(ππ½ + π½) = ππ¨π¬(ππ½)ππ¨π¬(π½) β π¬π’π§(ππ½)π¬π’π§(π½) = (πππ¨π¬ π (π½) β π)(ππ¨π¬(π½)) β ππ¬π’π§(π½)ππ¨π¬(π½)(π¬π’π§(π½)) = πππ¨π¬ π (π½) β ππ¨π¬(π½) β ππ¬π’π§π (π½)ππ¨π¬(π½) = πππ¨π¬ π (π½) β ππ¨π¬(π½) β π(π β ππ¨π¬ π (π½))(ππ¨π¬(π½)) = πππ¨π¬ π (π½) β ππ¨π¬(π½) β πππ¨π¬(π½) + πππ¨π¬ π (π½) = πππ¨π¬ π (π½) β πππ¨π¬(π½) 2. Evaluate ππ¨π¬ ( ππ ) using the half-angle formula, and then verify your solution using a different formula. ππ ππ ππ ββπ +π ππ β βππ¨π¬ ( π ) + π ππ¨π¬ ( ) = ππ¨π¬ ( π ) = β =β π β βπ. ππ ππ π π π ππ π π π π π π βπ π βπ βπ βπ β βπ ππ¨π¬ ( ) = ππ¨π¬ ( + ) = ππ¨π¬ ( ) ππ¨π¬ ( ) β π¬π’π§ ( ) π¬π’π§ ( ) = ( )β β βπ. ππ ( )β ππ π π π π π π π π π π π Problem Set Sample Solutions 1. Evaluate the following trigonometric expressions. a. π π π π ππ¬π’π§ ( ) ππ¨π¬ ( ) π π π¬π’π§ ( ) = b. π π π π c. d. π¬π’π§ ( π π ) ππ¨π¬ (ππ) π π (π¬π’π§ ( π)) = π ππ¬π’π§ (β ππ ππ ) ππ¨π¬ (β ) ππ ππ ππ¬π’π§ ( βππ π π ) = βππ¬π’π§ ( ) = βπ ( ) = βπ π π π ππ¨π¬ π ( ππ ππ ) β π¬π’π§π ( ) π π ππ¨π¬ ( e. π ππ βπ βπ ππ π ) = βππ¨π¬ ( ) = β π π π πππ¨π¬ π ( π π π )βπ ππ ππ¨π¬ ( ) = βπ Lesson 4: Date: π Addition and Subtraction Formulas 4/30/17 © 2015 Common Core, Inc. Some rights reserved. commoncore.org 68 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 4 NYS COMMON CORE MATHEMATICS CURRICULUM M4 PRECALCULUS AND ADVANCED TOPICS f. π π π β ππ¬π’π§π (β ) ππ¨π¬ ( g. βπ βπ )= π π ππ¨π¬ π (β π π πππ )βπ ππ (πππ¨π¬π (β πππ ππ π π βπππ ) β π) β π = π (ππ¨π¬ ( π π π π π π βπ π )) β π = π (ππ¨π¬ ( π)) β π = π ( π ) β π = βπ π β π π π π π πβπππ§π ( ) π ππππ§( ) h. π π πππ§ ( ) = π ππ ) ππ ππ πβπππ§π (β ) ππ ππππ§(β i. πππ§ (β j. βπ ππ π π ) = πππ§ ( ) = = π π βπ π π π ππ¨π¬ π ( ) π π ππ¨π¬ π ( ) = k. π+ππ¨π¬(π π+ ππ π βπ π) = = + π π π π β π π ππ¨π¬ ( ) Rotation by π½ = π π terminates in Quadrant I; therefore, ππ¨π¬ ( ) has a positive value. π π π+ππ¨π¬(π π βπ βπ+βπ π) =β + = π π π π π π ππ¨π¬ ( ) = β l. ππ¨π¬ (β ππ ) π Rotation by π½ = β ππ¨π¬ (β m. ππ ππ terminates in Quadrant III; therefore, ππ¨π¬ (β ) has a negative value. π π βπ+βπ π+ππ¨π¬(βππ π+ππ¨π¬(π ππ π βπ π) π) ) = ββ = ββ = ββ + =β π π π π π π π¬π’π§π ( π ) ππ π¬π’π§π ( πβππ¨π¬( π ) πβ π π π βπ )= = = β ππ π π π π π Lesson 4: Date: βπ Addition and Subtraction Formulas 4/30/17 © 2015 Common Core, Inc. Some rights reserved. commoncore.org 69 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 4 NYS COMMON CORE MATHEMATICS CURRICULUM M4 PRECALCULUS AND ADVANCED TOPICS n. π¬π’π§ ( π ) ππ Rotation by π½ = π¬π’π§ ( o. π π terminates in Quadrant I; therefore, π¬π’π§ ( ) has a positive value. ππ ππ β β ) πβππ¨π¬(π π π = βπ + βπ = π+ π )=β ππ π π π π π¬π’π§ (β ππ ) ππ Rotation by π½ = β π¬π’π§ (β p. ππ ππ terminates in Quadrant IV; therefore π¬π’π§ (β ) has a negative value. ππ ππ β πβππ¨π¬(βππ β ) π+ππ¨π¬(π ππ π) π = ββπ + βπ = β π+ π ) = ββ = ββ ππ π π π π π π π πππ§ ( ) Rotation by π½ = π π π π terminates in Quadrant I; therefore, πππ§ ( ) has a positive value. π π πβππ¨π¬(π πβ π πββπ β π) = β βππ = β = π β πβπ β π. πππ ) π+ππ¨π¬(π π+βπ π+ π π β πππ§ ( ) = β π π π π π π π πβππ¨π¬(π πβ π πββπ π) = βππ = = βπ β π β π. πππ π βπ π¬π’π§( π ) β πππ§ ( ) = πππ§ ( π ) = q. π π¬π’π§(π βπ π) π = β π. πππ π = π+ππ¨π¬(π ) π+βπ π+βπ π β πππ§ ( ) = πππ§ ( π ) = π π πππ§ ( π ) ππ πππ§ ( π ) πβππ¨π¬(π π π = βπβ π = βπββπ = βπ β πβπ β π. πππ )=β βπ ππ ) π+ππ¨π¬(π π+βπ π+ π π πππ§ ( π π ) π¬π’π§(π π π π ) = πππ§ ( π ) = = πβπ = = π β βπ β π. πππ π ππ π π+ππ¨π¬(π ) π+ π+βπ π πππ§ ( π π ) πβππ¨π¬(π π π = πβ π = π β ) = πππ§ ( π ) = βπ β π. πππ π ππ π ) π¬π’π§(π π β β π Lesson 4: Date: Addition and Subtraction Formulas 4/30/17 © 2015 Common Core, Inc. Some rights reserved. commoncore.org 70 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 4 NYS COMMON CORE MATHEMATICS CURRICULUM M4 PRECALCULUS AND ADVANCED TOPICS r. πππ§ (β ππ ) π Rotation by π½ = β πππ§ (β ππ ππ terminates in Quadrant IV; therefore, πππ§ (β ) has a negative value. π π β πβππ¨π¬(βππ π+ππ¨π¬(π π+ ππ ππ π+βπ π) π) ) = ββ = ββ = ββ = ββ = ββπ + πβπ β βπ. πππ π β π πβππ¨π¬(π ) πββπ π+ππ¨π¬(βππ ) πβ π π π πππ§ (β ππ ππ βπ π¬π’π§ (β ) β β ππ π π = ββπ β βπ. πππ ) = πππ§ ( π ) = = ππ π π π + ππ¨π¬ (β ) π β βπ π β βπ π π πππ§ (β πβππ¨π¬(β π ) π+ π π+βπ βππ ππ ) = πππ§ ( π ) = = βππ = = ββπ β π β βπ. πππ π π ββπ π¬π’π§(βππ ) β ππ π 2. β π Show that π¬π’π§(ππ) = ππ¬π’π§(π)ππ¨π¬ π (π) β π¬π’π§π (π). (Hint: Use π¬π’π§(ππ) = ππ¬π’π§(π)ππ¨π¬(π) and the sine sum formula.) π¬π’π§(ππ) = π¬π’π§(π + (ππ)) = π¬π’π§(π)ππ¨π¬(ππ) + ππ¨π¬(π)π¬π’π§(ππ) = π¬π’π§(π)[ππ¨π¬ π (π) β π¬π’π§π (π)] + ππ¨π¬(π)[ππ¬π’π§(π)ππ¨π¬(π)] = π¬π’π§(π)ππ¨π¬π (π) β π¬π’π§π (π) + ππ¬π’π§(π)ππ¨π¬ π (π) = ππ¬π’π§(π)ππ¨π¬π (π) β π¬π’π§π (π) 3. Show that ππ¨π¬(ππ) = ππ¨π¬ π (π) β ππ¬π’π§π (π)ππ¨π¬(π). (Hint: Use ππ¨π¬(ππ) = ππ¨π¬ π (π) β π¬π’π§π (π) and the cosine sum formula.) ππ¨π¬(ππ) = ππ¨π¬(π + (ππ)) = ππ¨π¬(π)ππ¨π¬(ππ) β π¬π’π§(π)π¬π’π§(ππ) = ππ¨π¬(π)[ππ¨π¬ π (π) β π¬π’π§π (π)] β π¬π’π§(π)[ππ¬π’π§(π)ππ¨π¬(π)] = ππ¨π¬ π (π) β ππ¨π¬(π)π¬π’π§π (π) β πππ¨π¬(π)π¬π’π§π (π) = ππ¨π¬ π (π) β πππ¨π¬(π)π¬π’π§π (π) 4. Use ππ¨π¬(ππ) = ππ¨π¬ π (π) β π¬π’π§π (π) to establish the following formulas. a. ππ¨π¬ π (π) = π+ππ¨π¬(ππ) π ππ¨π¬(ππ) = ππ¨π¬ π (π) β π¬π’π§π (π) = ππ¨π¬ π (π) β (π β ππ¨π¬ π (π)) = πππ¨π¬ π (π) β π Therefore, ππ¨π¬ π (π) = b. π¬π’π§π (π) = π+ππ¨π¬(ππ) . π πβππ¨π¬(ππ) . π ππ¨π¬(ππ) = ππ¨π¬ π (π) β π¬π’π§π (π) = (π β π¬π’π§π (π)) β π¬π’π§π (π) = π β ππ¬π’π§π (π) Therefore, π¬π’π§π (π) = Lesson 4: Date: πβππ¨π¬(ππ) . π Addition and Subtraction Formulas 4/30/17 © 2015 Common Core, Inc. Some rights reserved. commoncore.org 71 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 4 NYS COMMON CORE MATHEMATICS CURRICULUM M4 PRECALCULUS AND ADVANCED TOPICS 5. π½ π Jamia says that because sine is an odd function, π¬π’π§ ( ) is always negative if π½ is negative. That is, she says that for πβππ¨π¬(π½) . Is she correct? Explain how you know. π π½ π negative values of π¬π’π§ ( ) = ββ Jamia is not correct. Consider π½ = β Thus, π¬π’π§ (β 6. ππ ) is positive. π π½ ππ ππ ππ . In this case, = β , and rotation by β terminates in Quadrant II. π π π π π π π π ) is using the difference formula for sine since = β . Fred ππ ππ π π π says that there is another way to calculate π¬π’π§ ( ). Who is correct, and why? ππ Ginger says that the only way to calculate π¬π’π§ ( Fred is correct. We can use the half-angle formula with π½ = 7. π π to calculate π¬π’π§ ( ). π ππ π½ π Henry says that by repeatedly applying the half-angle formula for sine we can create a formula for π¬π’π§ ( ) for any positive integer π. Is he correct? Explain how you know. π½ π ) for positive integers π. There is π Henry is not correct. Repeating this process will only give us formulas for π¬π’π§ ( π½ π no way to derive a formula for quantities such as π¬π’π§ ( ) using this method. Lesson 4: Date: Addition and Subtraction Formulas 4/30/17 © 2015 Common Core, Inc. Some rights reserved. commoncore.org 72 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.