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Lesson 15 NYS COMMON CORE MATHEMATICS CURRICULUM M2 ALGEBRA II Lesson 15: What Is a Trigonometric Identity? Student Outcomes ο§ Students prove the Pythagorean identity sin2 (π₯) + cos 2 (π₯) = 1. ο§ Students extend trigonometric identities to the real line, with attention to domain and range. ο§ Students use the Pythagorean identity to find sin(π), cos(π), or tan(π), given sin(π), cos(π), or tan(π) and the quadrant of the terminal ray of the rotation. Lesson Notes The lesson begins with an example that develops and proves the Pythagorean identity for all real numbers. An equivalent form of the Pythagorean identity is developed, and students observe that there are special values for which the resulting functions are not defined; therefore, there are values for which the identity does not hold. Students then examine the domains for several identities and use the Pythagorean identity to find one function in terms of another in a given quadrant. Classwork Opening (5 minutes): The Pythagorean Identity Lessons 4 and 5 extended the definitions of the sine and cosine functions so that sin(π) and cos(π) are defined for all real numbers π. ο§ What is the equation of the unit circle centered at the origin? οΊ ο§ Recall that this equation is a special case of the Pythagorean theorem. Given the number π, there is a unique point π on the unit circle that results from rotating the positive π₯-axis through π radians around the origin. What are the coordinates of π? οΊ ο§ The coordinates of π are (π₯π , π¦π ), where π₯π = cos(π), and π¦π = sin(π). How can you combine this information to get a formula involving sin(π) and cos(π)? οΊ ο§ The unit circle has equation π₯ 2 + π¦ 2 = 1. Replace π₯ and π¦ in the equation of the unit circle by the coordinates of π. That gives sin2 (π) + cos 2 (π) = 1, where π is any real number. Notice that we use the notation sin2 (π) in place of (sin(π))2 . Both are correct, but the first is notationally simpler. Notice also that neither is the same expression as sin(π 2 ). Lesson 15: What Is a Trigonometric Identity? 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 Scaffolding: ο§ English language learners may need support such as choral recitation or a graphic organizer for learning the word identity. ο§ An alternative, more challenging Opening might be, βProve that sin2 (π) + cos 2 (π) = 1, for all real values of π.β (MP.3) 270 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 15 NYS COMMON CORE MATHEMATICS CURRICULUM M2 ALGEBRA II ο§ The equation sin2 (π) + cos 2(π) = 1 is true for all real numbers π and is an identity. The functions on either side of the equal sign are equivalent for every value of π. They have the same domain, the same range, and the same rule of assignment. You saw some polynomial identities in Module 1, and weβve developed some identities for sine, cosine, and tangent observed from graphs in this module. The identity we just proved is a trigonometric identity, and it is called the Pythagorean identity because it is another important consequence of the Pythagorean theorem. Example 1 (8 minutes): Another Identity? This example gets into the issue of what an identity is. Students should work in pairs to answer the questions. ο§ Scaffolding: Divide both sides of the Pythagorean identity by cos 2(π). What happens to the identity? οΊ The equation becomes tan2 (π) + 1 = 1 , which we can restate as cos2 (π) tan2 (π) + 1 = sec 2 (π). This appears to be another identity. ο§ π 2 οΊ ο§ 3π ? Why? 2 Both tan(π) and sec(π) are undefined at these values of π. That happens because cos(π) = 0 for those values of π, and we cannot divide by zero. Therefore, we can no longer say that the equation is true for all real numbers π. ο§ Model the substitution process, if needed, showing explicitly how the equation becomes tan2 (π) + 1 = 1 . cos2(π) How do we need to modify our claim about what looks like a new identity? οΊ ο§ π 2 What happens when π = β , π = , and π = ο§ Circulate to identify student pairs who might be prepared to share their results and to assist any students having trouble. We need to say that tan2 (π) + 1 = sec 2 (π) is a trigonometric identity for all real numbers π such that the functions π(π) = tan2 (π) + 1 and π(π) = sec 2 (π) are defined. In some cases, the functions are defined, and in other cases, they are not defined. For which values of π are the functions π(π) = tan2 (π) + 1 and π(π) = sec 2 (π) defined? οΊ The function π(π) = tan2 (π) + 1 is defined for π β οΊ The function π(π) = sec 2 (π) is defined for π β π + ππ, for all integers π. 2 π + ππ, for all integers π. 2 ο§ What is the range of each of the functions π(π) = tan2 (π) + 1 and π(π) = sec 2 (π)? ο§ The two functions have the same domain and the same range, and they are equivalent. ο§ Therefore, tan2 (π) + 1 = sec 2 (π) is a trigonometric identity for all real numbers π such that π β οΊ The ranges of the functions are all real numbers π(π) β₯ 1 and π(π) β₯ 1. π + ππ, for 2 all integers π. For any trigonometric identity, we need to specify not only the two functions that are equivalent, but also the values for which the identity is true. Be sure that the discussion clarifies that any equation is not automatically an identity. The equation needs to involve the equivalence of two functions and include the specification of their identical domains. Lesson 15: What Is a Trigonometric Identity? 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 271 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 15 NYS COMMON CORE MATHEMATICS CURRICULUM M2 ALGEBRA II Exercises 1β3 (25 minutes) Students should work on these exercises individually and then share their results either in a group or with the whole class. If some students are struggling, they should be encouraged to work together with the teacher while the others work individually. Exercises 1β3 1. Recall the Pythagorean identity π¬π’π§π (π½) + ππ¨π¬ π (π½) = π, where π½ is any real number. a. π π Find π¬π’π§(π), given ππ¨π¬(π) = , for β π < π < π. π π π π π ππ π = , and π¬π’π§(π) = β , or ππ ππ π π π π π¬π’π§(π) = . Since β < π < π, we know that π¬π’π§(π) is negative; therefore, π¬π’π§(π) = β . π π π From the Pythagorean identity, π¬π’π§π (π) + ( ) = π. So, π¬π’π§π (π) = π β b. Find πππ§(π), given ππ¨π¬(π) = β π π , for < π < π . ππ π π π ππ πππ ππ ) = π. So, π¬π’π§π (π) = π β = , and π¬π’π§(π) = β , ππ πππ πππ ππ π ππ ππ or π¬π’π§(π) = . Since < π < π , we know that π¬π’π§(π) is positive; therefore, π¬π’π§(π) = . ππ ππ π From the Pythagorean identity, π¬π’π§π (π) + (β Therefore, πππ§(π) = c. ππ π¬π’π§(π) ππ = ππ = β . ππ¨π¬(π) β π π ππ Write πππ§(π) in terms of ππ¨π¬(π), for π < π < ππ . π From the Pythagorean identity, π¬π’π§π (π) = π β ππ¨π¬ π (π). Therefore, either π¬π’π§(π) = ββπ β ππ¨π¬ π (π), or π¬π’π§(π) = βπ β ππ¨π¬ π (π). Since π¬π’π§(π) is negative for π < π < πππ§(π) = 2. ππ , π¬π’π§(π) = ββπ β ππ¨π¬ π (π). Because π ββπβππ¨π¬π(π) π¬π’π§(π) , we have πππ§(π) = . ππ¨π¬(π) ππ¨π¬(π) Use the Pythagorean identity to do the following: a. Rewrite the expression ππ¨π¬(π½) π¬π’π§π (π½) β ππ¨π¬(π½) in terms of a single trigonometric function. State the resulting identity. ππ¨π¬(π½) π¬π’π§π (π½) β ππ¨π¬(π½) = ππ¨π¬(π½)(π¬π’π§π (π½) β π) = ππ¨π¬(π½)(β ππ¨π¬ π (π½)) = β ππ¨π¬ π (π½) Therefore, ππ¨π¬(π½) π¬π’π§π (π½) β ππ¨π¬(π½) = β ππ¨π¬ π (π½) for all real numbers π½. b. Rewrite the expression (π β ππ¨π¬ π (π½)) ππ¬π(π½) in terms of a single trigonometric function. State the resulting identity. (π β ππ¨π¬ π (π½)) ππ¬π(π½) = π¬π’π§π (π½) ππ¬π(π½) π = π¬π’π§π (π½) π¬π’π§(π½) = π¬π’π§(π½) Therefore, (π β ππ¨π¬ π (π½)) ππ¬π(π½) = π¬π’π§(π½) for π½ β ππ , for all integers π. Lesson 15: What Is a Trigonometric Identity? 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 272 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 15 NYS COMMON CORE MATHEMATICS CURRICULUM M2 ALGEBRA II c. Find all solutions to the equation π π¬π’π§π (π½) = π + ππ¨π¬(π½) in the interval (π, ππ). Draw a unit circle that shows the solutions. π π¬π’π§π (π½) = π + ππ¨π¬(π½) π(π β ππ¨π¬ π ( π½)) = π + ππ¨π¬(π½) π β πππ¨π¬ π (π½) = π + ππ¨π¬ π½ β πππ¨π¬ π (π½) β ππ¨π¬(π½) = π πππ¨π¬ π (π½) + ππ¨π¬(π½) = π ππ¨π¬(π½) (π ππ¨π¬(π½) + π) = π π π Therefore, ππ¨π¬(π½) = π, or ππ¨π¬(π½) = β . See the unit circle on the right, which shows the four π π points where ππ¨π¬(π½) = π, or ππ¨π¬(π½) = β . π π In the interval (π, ππ ), ππ¨π¬(π½) = π only if π½ = , or π½= ππ π ππ ππ . Also, ππ¨π¬(π½) = β only if π½ = , or π½ = . π π π π π ππ ππ ππ , , and . π π π π Therefore, the solutions of the equations in the interval (π, ππ ) are , 3. Which of the following statements are identities? If a statement is an identity, specify the values of π where the equation holds. a. π¬π’π§(π + ππ ) = π¬π’π§(π) where the functions on both sides are defined. This is an identity defined for all real numbers. b. π¬ππ(π) = π where the functions on both sides are defined. This is not an identity. The functions are not equivalent for all real numbers. For example, although π π π¬ππ(π) = π, π¬ππ ( ) = βπ. The functions have equal values only when π is an integer multiple of ππ . Additionally, the ranges are different. The range of π(π) = π¬ππ(π) is all real numbers π such that π β€ βπ or π β₯ π, whereas the range of π(π) = π is the single number π. c. π¬π’π§(βπ) = π¬π’π§(π) where the functions on both sides are defined. This is not an identity; this statement is only true when π¬π’π§(π) = π, which happens only at integer multiples of π . d. π + πππ§π (π) = π¬ππ π (π) where the functions on both sides are defined. This is an identity. The functions on either side are defined for π½ β e. π + ππ , for all integers π. π π π π¬π’π§ ( β π) = ππ¨π¬(π) where the functions on both sides are defined. This is an identity defined for all real numbers. Lesson 15: What Is a Trigonometric Identity? 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 273 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 15 NYS COMMON CORE MATHEMATICS CURRICULUM M2 ALGEBRA II f. π¬π’π§π (π) = πππ§π (π) for all real π. This is not an identity. The equation π¬π’π§π (π) = πππ§π (π) is only true where π¬π’π§π (π) = π π¬π’π§ (π) , so ππ¨π¬π(π) ππ¨π¬ π (π) = π, and then ππ¨π¬(π) = π, or ππ¨π¬(π) = βπ, which gives π = π π, for all integers π. For all other π values of π, the functions on the two sides are not equal. Moreover, πππ§π (π) is defined only for π½ β + ππ , π for all integers π, whereas π¬π’π§π (π) is defined for all real numbers. π π βπ Another argument for why this statement is not an identity is that π¬π’π§π ( ) = ( π π π π π π π π ) = , but πππ§π ( ) = ππ = π, and π β ; therefore, the statement is not true for all values of π. Closing (2 minutes) ο§ One trigonometric equation is cos(π₯) = οΊ 5 . Explain why this equation is not a trigonometric identity. 13 The functions on each side of the equal sign have the same domain. The left side, π(π₯) = cos(π₯), is defined for all real π₯. The right side, π(π₯) = οΊ 5 , is also defined for all real π₯. 13 The two functions are not, however, equivalent. The left side is a trigonometric function that equals only sometimes. For example, if π₯ = 0, then cos(π₯) = 1. The right side, in contrast, is a constant function. The functions π(π₯) and π(π₯) are not equal on all values for which they are defined. οΊ 5 5 is the single number π¦ = . This is further evidence that the two functions are different. 13 13 The graphs of the two functions show how the functions are different. The graph of π(π₯) = cos(π₯) is periodic, whereas the graph of π(π₯) = οΊ 13 The range of π(π₯) = cos(π₯) is the set of all real numbers π¦ such that β1 β€ π¦ β€ 1, whereas the range of π(π₯) = οΊ 5 5 is a horizontal line. See the graphs below. 13 Because the two functions are not equivalent wherever they are defined, the equation is not an identity. Lesson Summary The Pythagorean identity: π¬π’π§π (π½) + ππ¨π¬ π (π½) = π for all real numbers π½. Exit Ticket (5 minutes) Lesson 15: What Is a Trigonometric Identity? 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 274 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 15 NYS COMMON CORE MATHEMATICS CURRICULUM M2 ALGEBRA II Name Date Lesson 15: What Is a Trigonometric Identity? Exit Ticket April claims that 1 + cos2 (π) 1 = 2 is an identity for all real numbers π that follows from the Pythagorean identity. sin2(π) sin (π) a. For which values of π are the two functions π(π) = 1 + b. Show that the equation 1 + c. Is April correct? Explain why or why not. d. Write the equation 1 + cos2(π) 1 and π(π) = 2 defined? sin2(π) sin (π) cos2(π) 1 = 2 follows from the Pythagorean identity. sin2(π) sin (π) cos2 (π) 1 = 2 in terms of other trigonometric functions, and state the resulting 2 sin (π) sin (π) identity. Lesson 15: What Is a Trigonometric Identity? 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 275 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 15 NYS COMMON CORE MATHEMATICS CURRICULUM M2 ALGEBRA II Exit Ticket Sample Solutions April claims that π + a. ππ¨π¬π(π½) π = is an identity for all real numbers π½ that follows from the Pythagorean identity. π π π¬π’π§ (π½) π¬π’π§ (π½) For which values of π½ are the two functions π(π½) = π + ππ¨π¬π(π½) π and π(π½) = defined? π π π¬π’π§ (π½) π¬π’π§ (π½) Both functions contain π¬π’π§(π½) in the denominator, so they are undefined if π¬π’π§(π½) = π. Thus, the two functions π and π are defined when π½ β ππ , for all integers π. b. Show that the equation π + ππ¨π¬π(π½) π = follows from the Pythagorean identity. π π π¬π’π§ (π½) π¬π’π§ (π½) By the Pythagorean identity, π¬π’π§π (π½) + ππ¨π¬ π (π½) = π. If π¬π’π§(π½) β π, then, π¬π’π§π (π½) ππ¨π¬ π (π½) π + = π¬π’π§π (π½) π¬π’π§π (π½) π¬π’π§π (π½) π+ c. ππ¨π¬ π (π½) π = . π¬π’π§π (π½) π¬π’π§π (π½) Is April correct? Explain why or why not. No. While Aprilβs equation does follow from the Pythagorean identity, it is not valid for all real numbers π½. For example, if π½ = π , then both sides of the equation are undefined. In order to divide by π¬π’π§π (π½), we need to be sure that we are not dividing by zero. d. Write the equation π + ππ¨π¬π(π½) π = in terms of other trigonometric functions, and state the resulting π π π¬π’π§ (π½) π¬π’π§ (π½) identity. Because ππ¨π(π½) = ππ¨π¬(π½) π and ππ¬π(π½) = , we can rewrite the equation as π¬π’π§(π½) π¬π’π§(π½) π+ ππ¨π¬ π (π½) π = π¬π’π§π(π½) π¬π’π§π (π½) π π ππ¨π¬(π½) π π+( ) ) =( π¬π’π§(π½) π¬π’π§(π½) π + ππ¨π π (π½) = ππ¬π π (π½). Thus, π + ππ¨π π (π½) = ππ¬π π (π½), where π½ β ππ , for all integers π. Lesson 15: What Is a Trigonometric Identity? 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 276 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 15 NYS COMMON CORE MATHEMATICS CURRICULUM M2 ALGEBRA II Problem Set Sample Solutions Problems are intended to give students practice in distinguishing trigonometric identities from other trigonometric equations, in distinguishing identities defined for all real numbers from those that are defined on a subset of the real numbers, and in using the Pythagorean identity and given information to find values of trigonometric functions. 1. Which of the following statements are trigonometric identities? Graph the functions on each side of the equation. a. πππ§(π) = π¬π’π§(π) where the functions on both sides are defined. ππ¨π¬(π) This is an identity that is defined for π β b. π + ππ , for all integers π. See the identical graphs above. π ππ¨π¬ π (π) = π + π¬π’π§(π) where the functions on both sides are defined. This is not an identity. For example, when π = π, the left side of the equation is π, and the right side is also π. π But when π = , the left side is π, and the right side is π. The graphs below are clearly different. π c. π π ππ¨π¬ ( β π) = π¬π’π§(π) where the functions on both sides are defined. This is an identity that is defined for all real numbers π. See the identical graphs below. Lesson 15: What Is a Trigonometric Identity? 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 277 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 15 NYS COMMON CORE MATHEMATICS CURRICULUM M2 ALGEBRA II 2. Determine the domain of the following trigonometric identities: a. ππ¨π¬(π) where the functions on both sides are defined. π¬π’π§(π) ππ¨π(π) = This identity is defined only for π β ππ , for all integers π. b. ππ¨π¬(βπ) = ππ¨π¬ (π) where the functions on both sides are defined. This identity is defined for all real numbers π. c. π¬ππ(π) = π where the functions on both sides are defined. ππ¨π¬(π) This identity is defined for π β 3. π + ππ , for all integers π. π Rewrite π¬π’π§(π)ππ¨π¬π (π) β π¬π’π§(π) as an expression containing a single term. π¬π’π§(π) β π¬π’π§(π)ππ¨π¬ π (π) = π¬π’π§(π)(π β ππ¨π¬ π (π)) = π¬π’π§(π)π¬π’π§π (π) = π¬π’π§π (π) 4. Suppose π < π½ < ππ¨π¬(π½) = 5. βπ π If ππ¨π¬(π½) = β π βπ Either π¬π’π§(π½) = 6. π π and π¬π’π§(π½) = . What is the value of ππ¨π¬(π½)? π βπ , what are possible values of π¬π’π§(π½)? π , or π¬π’π§(π½) = β βπ π βπ Use the Pythagorean identity π¬π’π§π (π½) + ππ¨π¬ π (π½) = π, where π½ is any real number, to find the following: a. ππ¨π¬(π½), given π¬π’π§(π½) = π π , for < π½ < π . ππ π π π ππ πππ ππ ) . So, ππ¨π¬ π (π½) = π β = , and ππ¨π¬(π½) = , or ππ πππ πππ ππ ππ ππ ππ¨π¬(π½) = β . In the second quadrant, ππ¨π¬(π½) is negative, so ππ¨π¬(π½) = β . ππ ππ From the Pythagorean identity, ππ¨π¬ π (π½) = π β ( b. πππ§(π), given ππ¨π¬(π) = β π βπ , for π < π < ππ . π From the Pythagorean identity, π¬π’π§π (π) = π β ( π βπ π ) . So, π¬π’π§π (π) = Because π¬π’π§(π) is negative in the third quadrant, πππ§(π) = βπ β β Lesson 15: What Is a Trigonometric Identity? 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 π βπ π π π , and π¬π’π§(π) = , or π¬π’π§(π) = β . π βπ βπ = π. π 278 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 15 NYS COMMON CORE MATHEMATICS CURRICULUM M2 ALGEBRA II 7. The three identities below are all called Pythagorean identities. The second and third follow from the first, as you saw in Example 1 and the Exit Ticket. a. For which values of π½ are each of these identities defined? i. π¬π’π§π (π½) + ππ¨π¬ π (π½) = π, where the functions on both sides are defined. Defined for any real number π½. ii. πππ§π (π½) + π = π¬ππ π(π½), where the functions on both sides are defined. Defined for real numbers π½ such that π½ β iii. π + ππ , for all integers π. π π + ππ¨π π (π½) = ππ¬π π(π½), where the functions on both sides are defined. Defined for real numbers π½ such that π½ β ππ , for all integers π. b. For which of the three identities is π in the domain of validity? Identities i and ii c. For which of the three identities is π π in the domain of validity? Identities i and iii d. For which of the three identities is β π in the domain of validity? π Identities i, ii, and iii Lesson 15: What Is a Trigonometric Identity? 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 279 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
