Chapter Trigonometric Functions 1.1 Therefore, the –570° and 150° angles are coterminal All angles that are coterminal with 150° are 150° + n ⋅ 360°, where n is an integer Angles 1.1 Practice Exercises complement: 90º − 67° = 23º supplement: 180º − 67° = 113º a α = 37 + 12 ′ + 37 ′′ + β = + 24 + 15′ + 45′′ α + β = = = 61 + 27 ′ + 82 ′′ 61 + 27 ′ + (60 + 22 )′′ 61° + 28′ + 22 ′′ = 61°28′ 22′′ In the figure, OP is the diagonal of the × square PQRO So OP bisects the right angle QOR Therefore θ = 180° + ⋅ 90° = 225° Since there are infinitely many angles with OP as the terminal side, and those angles differ by 360°, θ = 225° + n ⋅ 360°, n any integer b α = 37 + 12 ′ + 37 ′′ = 37° + 11′ + 1′ + 37 ′′ = 37° + 11′ + 60′′ + 37 ′′ = 37° + 11′ + 97 ′′ = 36° + 1° + 11′ + 97 ′′ = 36° + 60′ + 11′ + 97 ′′ = 36° + 71′ + 97 ′′ α = 36 + 71′ + 97 ′′ − β = − ( 24° + 15′ + 45′′ ) α −β = 12 + 56′ + 52′′ = 12 56′52′′ 22 ⎞ ⎛ + 13°9′22′′ = ⎜13 + ⎟ ° ≈ 13.16° ⎝ 60 3600 ⎠ 41.275° = 41° + 0.275(60′) = 41° + 16.5′ = 41° + 16′ + 0.5(60′′ ) = 41°16′ 30′′ Angles and are supplements, and angles and are congruent, so angles and are supplements Therefore, m∠2 + m∠7 = 180° ⇒ (6 x − 3) ° + (8 x + 43) ° = 180° ⇒ (14 x + 40) ° = 180° ⇒ 14 x = 140° ⇒ x = 10° a Since 765° is a positive angle greater than 720° = (360°)(2), subtract 720° to get an angle between 0° and 360° 765° – 720° = 45° Therefore, the 765° and 45° angles are coterminal All angles that are coterminal with 45° are 45° + n ⋅ 360°, where n is an integer b Adding 2(360°) = 720° to –570° gives an angle between 0° and 360° –570° + 720° = 150° Thus, m∠2 = (6 ⋅ 10 − 3) ° = 57° and m∠7 = (8 ⋅ 10 + 43) ° = 123° 1.1 A Exercises: Basic Skills and Concepts The degree measure of one complete revolution is 360° The sum of two complementary angles is 90° An angle is in standard position if the initial side is the positive x-axis and the vertex is at the origin Copyright © 2011 Pearson Education Inc Publishing as Addison-Wesley Chapter Trigonometric Functions For any integer n, an angle of θ ° + n ⋅ 360° has the same terminal side as the angle of θ degrees 20 a α +β = True False An angle is standard position is quadrantal if its terminal side lies on a coordinate axis complement: 90° – 47° = 43° supplement: 180° – 47° = 133° b 21 a 10 complement: none because the measure of the angle is greater than 90° supplement: 180° – 160° = 20° = 22 a 11 complement: none because the measure of the angle is greater than 90°; supplement: none because the measure of the angle is greater than 180° 12 complement: none because the measure of the angle is negative; supplement: none because the measure of the angle is negative 15 180° − θ 18 For an angle θ to be its own supplement, we must have θ = 180° − θ ⇒ 2θ = 180° ⇒ θ = 90° 19 a b 23 a b α = 34 + 12′ − β = − 27 + 5′ α −β = + ′ = 7°7 ′ α = 35 + 43′ + β = + 15 + 35′ 50 + + 18′ = 51 + 18′ 51 18′ α = 35° + 43′ − β = − 15° + 35′ α + β = 20° + 8′ = 20°8′ α = 15 + 38′ + β = + 13 + 45′ α +β = = = 28 + 83′ = 28 + 60 ′ + 23′ 28 + + 23′ = 29 + 23′ 29 23′ b α = 15° + 38′ = 14° + 1° + 38′ = 14° + 60′ + 38′ = 14° + 98′ α = 14 + 98′ − β = − 13 + 45′ α = 34 + 12′ + β = + 27 + 5′ α + β = 61 + 17 ′ = 61°17 ′ α = 47° + 54′ − β = − 12° + 14′ α − β = 35° + 40′ = 35°40′ = = 16 180° − θ 17 For an angle θ to be its own complement, we must have θ = 90° − θ ⇒ 2θ = 90° ⇒ θ = 45° 59 + 68′ = 59° + 60 ′ + 8′ 59 + + 8′ = 60 + 8′ = 60 8′ α + β = 50 + 78′ = 50 + 60 ′ + 18′ 13 90° − θ 14 An obtuse angle does not have a complement 41 + 25′ = 41 25′ α = 47 + 54′ + β = + 12 + 14′ α +β = b 87 + 49′ = 87°49′ α = 64 + 37′ − β = − 23 + 12′ α −β = complement: 90° – 75° = 15° supplement: 180° – 75° = 105° complement: none because the measure of the angle is greater than 90° supplement: 180° – 120° = 60° α = 64 + 37 ′ + β = + 23 + 12′ α −β = 24 a + 53′ = 53′ α = 28 + 42′ + β = + 16 + 56′ α +β = = = 44 + 98′ = 44 + 60 ′ + 38′ 44 + + 38′ = 45 + 38′ 45 38′ Copyright © 2011 Pearson Education Inc Publishing as Addison-Wesley Section 1.1 Angles b α = 28° + 42′ = 27° + 1° + 42′ = 27° + 60′ + 42′ = 27° + 102′ b α = 89 + 45′ + 40′′ = 89 + 44′ + 1′ + 40′′ = 89 + 44 ′ + 60′′ + 40 ′′ = 89 + 44 ′ + 100 ′′ α = 27 + 102′ − β = − 16 + 56′ α −β = 25 a α = 89 + 44′ + 100′′ − β = − 56 + 35′ + 46′′ 11 + 46′ = 11 46′ α = 70 + 12′ + 15′′ + β = + 54 + 18′ α −β = 29 a α + β = 124 + 30′ + 15′′ = 124 30′ 15′′ = = = α = 16 + 15′ + 12′′ + β = + 12 + 23′ α +β = 28 + 38′ + 12′′ = 28 38′ 12′′ b α = 16 + 15′ + 12′′ = 15° + 1° + 15′ + 12′′ = 15° + 60′ + 15′ + 12′′ = 15° + 75′ + 12′′ α = 15° + 75′ + 12′′ − β = − ( 12° + 23′ ) 3° + 52′ + 12′′ = 3°52′12′′ α −β = 27 a b 30 a α = 89 + 45′ + 40′′ + β = + 56 + 35′ + 46′′ α + β = 145 + 80′ + 86′′ = = = 145 + 60′ + 20′ + 60′′ + 26′′ 145 + + 20′ + 1′ + 26′′ 146 + 21′ + 26′′ = 146 21′ 26′′ 33° + 9′ + 54′′ = 33°9′ 54′′ 407 407 407 408 + 90′ + 100′′ + 60′ + 30′ + 60′′ + 40′′ + + 30′ + 1′ + 40′′ + 31′ + 40′′ = 408 31′ 40′′ β = 220° + 34′ + 67′′ = 219° + 1° + 34′ + 67 ′′ = 219° + 60′ + 34′ + 67 ′′ = 219° + 94′ + 67′′ α = 187° + 56′ + 93′′ − β = − ( 219° + 94′ + 67′′ ) α = 240 + 35′ + 48′′ + β = + 335 + 6′ + 54′′ α +β = = = = α + β = 20 + 42′ + 58′′ = 20 42′ 58′′ 28 a ) α − β = − ( 32° + 38′ + 34′′ ) = −32°38′ 34′′ α = 12 + 15′ + 22′′ + β = + + 27 ′ + 36′′ b α = 12 + 15′ + 22′′ = 12° + 14′ + 1′ + 22′′ = 12° + 14′ + 60′′ + 22′′ = 12° + 14′ + 82′′ = 11° + 1° + 14′ + 82′′ = 11° + 60′ + 14′ + 82′′ = 11° + 74′ + 82′′ α = 11° + 74′ + 82′′ − β = − ( 8° + 27 ′ + 36′′ ) 3° + 47 ′ + 46′′ = 3°47 ′46′′ α −β = ( α = 187 + 56′ + 33′′ + β = + 220 + 34′ + 67 ′′ α +β = b α = 70 + 12′ + 15′′ = 69° + 1° + 12′ + 15′′ = 69° + 60′ + 12′ + 15′′ = 69° + 72′ + 15′′ α = 69° + 72′ + 15′′ − β = − ( 54° + 18′ ) α − β = 15° + 54′ + 15′′ = 15°54′ 15′′ 26 a b 575 575 575 575 + 41′ + 102′′ + 41′ + 60′′ + 42′′ + 41′ + 1′ + 42′′ + 42′ + 42′′ = 575 42′ 42′′ β = 335 + 6′ + 54′′ = 334° + 1° + 6′ + 54′′ = 334° + 60′ + 6′ + 54′′ = 334° + 66′ + 54′′ α = 240° + 35′ + 48′′ − β = − (334° + 66′ + 54′′ ) α − β = − ( 94° + 31′ + 6′′ ) = −94°31′ 6′′ 31 45 ⎞ ⎛ 70°45′ = ⎜ 70 + ⎟ ° = 70.75° ⎝ 60 ⎠ 38 ⎞ ⎛ 32 38°38′ = ⎜ 38 + ⎟ ° ≈ 38.63° ⎝ 60 ⎠ 33 42 30 ⎞ ⎛ 23°42′30′′ = ⎜ 23 + + ⎟ ° ≈ 23.71° ⎝ 60 3600 ⎠ 34 50 50 ⎞ ⎛ + 45°50′50′′ = ⎜ 45 + ⎟ ° ≈ 45.85° ⎝ 60 3600 ⎠ Copyright © 2011 Pearson Education Inc Publishing as Addison-Wesley Chapter Trigonometric Functions 35 42 57 ⎞ ⎛ −15°42′ 57 ′′ = − ⎜15 + + ⎟ ° ≈ −15.72° ⎝ 60 3600 ⎠ 36 18 13 ⎞ ⎛ −70°18′ 13′′ = − ⎜ 70 + + ⎟ ° ≈ −70.30° ⎝ 60 3600 ⎠ 37 27.32° = 27° + 0.32(60′ ) = 27° + 19.2′ = 27° + 19′ + 0.2(60′′ ) = 27°19′12′′ 38 120.64° = 120° + 0.64(60′ ) = 120° + 38.4′ = 120° + 38′ + 0.4(60′′ ) = 120°38′ 24′′ 46 Quadrant II 47 39 13.347° = 13° + 0.347(60′ ) = 13° + 20.82′ = 13° + 20′ + 0.82(60′′ ) = 13°20′ 49′′ 40 110.433° = 110° + 0.433(60′) = 110° + 25.98′ = 110° + 25′ + 0.98(60′′) = 110°25′ 59′′ 41 19.0511° = 19° + 0.0511(60′) = 19° + 3.066′ = 19° + 3′ + 0.066(60′′) = 19°3′ 4′′ Quadrant IV 48 42 82.7272° = 82° + 0.7272(60′ ) = 82° + 43.632′ = 82° + 43′ + 0.632(60′′ ) = 82°43′ 38′′ 43 Quadrant III 49 Quadrant II 44 Quadrant II 50 Quadrant IV 45 Quadrant I Quadrant I 51 Since 400° is a positive angle greater than 360°, subtract 360° to get an angle between 0° and 360°: 400° – 360° = 40° Therefore, the 400° and 40° angles are coterminal All angles that are coterminal with 40° are 40° + n ⋅ 360°, where n is an integer Copyright © 2011 Pearson Education Inc Publishing as Addison-Wesley Section 1.1 Angles 52 Since 700° is a positive angle greater than han 360°, subtract 360° to get an angle between 0° and 360°: 700° – 360° = 340° Therefore, the 700° and 340° angles are coterminal All angles that are coterminal with 340° are 340° + n ⋅ 360°, where n is an integer 53 Since 1785° is a positive angle greater than 1440° = (360°)(4), subtract 1440° to get an angle between 0° and 360° 1785° – 1440° = 345° Therefore, the 1785° and 345° angles are coterminal All angles that are coterminal with 345° are 345° + n ⋅ 360°, where n is an integer 54 Since 2064° is a positive angle greater than 1800° = (360°)(5), subtract 1800° to get an angle between 0° and 360° 2064° – 1800° = 264° Therefore, the 2064° and 264° angles are coterminal All angles that are coterminal with 264° are 264° + n ⋅ 360°, where n is an integer 55 Adding 360° to –50° gives an angle between 0° and 360°: –50° + 360° = 310° Therefore, the –50° and 310° angles are coterminal All angles that are coterminal with 310° are 310° + n ⋅ 360°, where n is an integer 56 Adding 360° to –225° gives an angle between 0° and 360°: –225° + 360° = 135° Therefore, the –225° and 135° angles are coterminal All angles that are coterminal with 135° are 135° + n ⋅ 360°, where n is an integer 57 Adding 2(360°) = 720° to –400° gives an angle between 0° and 360°: –400° + 720° = 320° Therefore, the –400° and 320° angles are coterminal All angles that are coterminal with 320° are 320° + n ⋅ 360°, where n is an integer 58 Adding 2(360°) = 720° to –700° gives an angle between 0° and 360°: –700° + 720° = 20° Therefore, the –700° and 20° angles are coterminal All angles that are coterminal with 20° are 20° + n ⋅ 360°, where n is an integer For exercises 59–66, review the explanations in Example and Practice Problem 59 (3, 3) lies in Quadrant I, so a line through that point forms a 45° angle with the positive x-axis Thus, θ = 45° + n ⋅ 360°, n any integer 61 (–5, 5) lies in Quadrant II, so a line through that point forms a 90° + 45° = 135° angle with the positive x-axis Thus, θ = 135° + n ⋅ 360°, n any integer 62 (–2, –2) lies in Quadrant III, so a line through that point forms a 180° + 45° = 225° angle with the positive x-axis Thus, θ = 225° + n ⋅ 360°, n any integer 63 (1, 0) lies on the positive x-axis, so a line through that point forms a 0° angle with the positive x-axis Thus, θ = 0° + n ⋅ 360°, n any integer 64 (0, 2) lies on the positive y-axis, so a line through that point forms a 90° angle with the positive x-axis Thus, θ = 90° + n ⋅ 360°, n any integer 65 (–3, 0) lies on the negative x-axis, so a line through that point forms a 180° angle with the positive x-axis Thus, θ = 180° + n ⋅ 360°, n any integer 66 (0, –4) lies on the negative y-axis, so a line through that point forms a 270° angle with the positive x-axis Thus, θ = 270° + n ⋅ 360°, n any integer 1.1 B Exercises: Applying the Concepts 67 60 sec ⋅ 20 = 1200 sec 360° = 2160° rev = rev ⋅ rev Now use a proportion x° 10 sec 10 ⋅ 2160 = ⇒x= = 18° 1200 sec 2160° 1200 20 = 360° = 180, 000° rev Now use a proportion x° sec 180, 000 = ⇒x= = 3000° 60 sec 180, 000° 60 68 500 rev = 500 rev ⋅ 69 There are 50 minutes or of an hour from 4:20 pm to 5:10 pm The minute hand travels 360° in one hour, so it travels 56 ⋅ 360° = 300° in the given time period 60 (4, –4) lies in Quadrant IV, so a line through that point forms a 270° + 45° = 315° angle with the positive x-axis Thus, θ = 315° + n ⋅ 360°, n any integer Copyright © 2011 Pearson Education Inc Publishing as Addison-Wesley Chapter Trigonometric Functions 70 There are hours 30 minutes from 3:15 pm to 7:45 pm The hour hand travels 360° in 12 4.5 ⋅ 360° = 135° in the hours, so it travels 12 given time period Since α and γ are vertical angles, they are equal and γ = 102° Since β and θ are vertical angles, they are equal and θ = 78° Use the figure below for exercises 75 and 76 71 The sum of the angles is 180°, so we have ( x + 30) + 75 + ( x + 15) = 180 ⇒ x + 120 = 180 ⇒ x = 60 ⇒ x = 30 72 The sum of the angles is 180°, so we have (3x + 15) + 40 + (2 x + 20) = 180 ⇒ x + 75 = 180 ⇒ x = 105 ⇒ x = 21 The marked angles are (3x + 15) = 3(21) + 15 = 78°, and (2x + 20) = 2(21) + 20 = 62° Use the figure below for exercises 73 and 74 73 α = 3β ; α + β = 180° ⇒ 3β + β = 180° ⇒ β = 180° ⇒ β = 45° α + 45° = 180° ⇒ α = 135° Since α and γ are vertical angles, they are 75 The angle vertical with α is the interior angle on the same side of the transversal as β , so the angles are supplementary α = (5 x + 70) °, β = (4 x + 29) ° α + β = 180° ⇒ (5 x + 70) + (4 x + 29) = 180° ⇒ x + 99 = 180° ⇒ x = 81° ⇒ x = 9° α = (5 ⋅ + 70) ° = 115° β = ( ⋅ + 29) ° = 65° 76 The angle vertical with α is the interior angle on the same side of the transversal as β , so the angles are supplementary α = (7 x + 94) °, β = (5 x − 10) ° α + β = 180° ⇒ (7 x + 94) + (5 x − 10) = 180° ⇒ 12 x + 84 = 180° ⇒ 12 x = 96° ⇒ x = 8° α = (7 ⋅ + 94) ° = 150° β = (5 ⋅ − 10) ° = 30° 77 F equal and γ = 135° Since β and θ are vertical angles, they are equal and θ = 45° 74 α = ( x + 2) °; β = (3x + 3) ° α + β = 180° ⇒ (4 x + 2) + (3x + 3) = 180° ⇒ m∠BOF = α and m∠COF = β , so m∠BOC = α + β = 40° + 30° = 70° x + = 180° ⇒ x = 175° ⇒ x = 25° α = ( ⋅ 25 + 2) ° = 102° β = (3 ⋅ 25 + 3) ° = 78° Copyright © 2011 Pearson Education Inc Publishing as Addison-Wesley Section 1.1 Angles ∠ABQ + ∠BQD + ∠CDQ = (∠ABP + ∠PBQ ) + ∠BQD + (∠CDP + ∠PDQ ) = (∠PBQ + ∠BQD + ∠PDQ ) + ( ∠ABP + ∠CDP ) Recall that the sum of the interior angles of a quadrilateral is 360°, so ∠PBQ + ∠BQD + ∠PDQ + ∠BPD = 360° ⇒ ∠PBQ + ∠BQD + ∠PDQ + 80° = 360° ⇒ ∠PBQ + ∠BQD + ∠PDQ = 280° From part (a), we know that ∠ABP + ∠CDP = 80°, so (∠PBQ + ∠BQD + ∠PDQ ) + (∠ABP + ∠CDP ) = 280° + 80° = 360° 78 α = m∠BCD ⇒ m∠BCD = 60° ⇒ m∠DCE = 30° Angles CEF and DCE are interior angles on the same side of the transversal, so they are supplementary m∠CEF + 30° = 180° ⇒ m∠CEF = 150° 1.1 C Exercises: Beyond the Basics 79 81 The sum of the measures of angles AOC, COE, and BOE is 180°, so we have x + 90 + x = 180 ⇒ x = 90 ⇒ x = 30 3 Thus, m∠AOC = ⋅ 30° = 20° and m∠BOE = ⋅ 30° = 70° Angles AOC and DOB are vertical angles, so their measures are equal m∠DOB = y = 20° Angles DOB and DOA are supplements, so m∠DOA + m∠DOB = 180° ⇒ z + 20° = 180° ⇒ z = 160° 80 a From exercise 77, we know that m∠BPD = m∠ABP + m∠CDP = 45° + 35° = 80° b ∠ABQ = ∠ABP + ∠PBQ ∠CDQ = ∠CDP + ∠PDQ Adding the two equations and ∠BQD , we have MP bisects angle AMN, MQ bisects angle BMN, NQ bisects angle DNM, and NP bisects angle CNM Therefore, ∠AMP = ∠NMP, ∠BMQ = ∠NMQ, ∠MNQ = ∠DNQ, and ∠MNP = ∠CNP Since the sum of the interior angles on the same side of a transversal is 180°, we have ∠BMN + ∠DNM = 180° ⇒ (∠BMQ + ∠NMQ ) + ( ∠DNQ + ∠MNQ ) = 180° ⇒ (∠NMQ + ∠NMQ ) + ( ∠MNQ + ∠MNQ ) = 180° ⇒ (∠NMQ + ∠MNQ ) = 180° ⇒ (1) ∠NMQ + ∠MNQ = 90° Similarly, we can show that ∠MNP + ∠PNM = 90° (2) In triangle MQN, we have ∠NMQ + ∠MNQ + ∠Q = 180° ⇒ 90° + ∠Q = 180° ⇒ ∠Q = 90° In triangle MPN, we have ∠MNP + ∠PNM + ∠P = 180° ⇒ 90° + ∠P = 180° ⇒ ∠P = 90° (continued on next page) Copyright © 2011 Pearson Education Inc Publishing as Addison-Wesley Chapter Trigonometric Functions (continued from page 7) Now we must show that angles PMQ and PNQ are also right angles We know that angles AMN and DNM are equal since they are alternate interior angles Each angle is bisected, so we can conclude that angle PMN = angle MNQ Similarly, we can deduce that angle QMN = angle MNP Adding equations (1) and (2) then substituting, we have ∠NMQ + ∠MNQ + ∠MNP + ∠PNM = 180° ⇒ ∠ + ∠ NMQ PNM ( ) + (∠MNQ + ∠MNP ) = 180° ⇒ 2∠NMQ + 2∠MNP = 180° ⇒ ∠NMQ + ∠MNP = ∠NMQ + ∠NMP = 90° ⇒ ∠PMQ = 90° Similarly, we can show that angle PNQ = 90° Thus, MPNQ is a rectangle 1.1 Critical Thinking 84 The maximum number of points of intersection is You may want to investigate the formula for determining the maximum number of points of n (n − 1) intersection of n distinct lines, 82 1.2 Triangles 1.2 Practice Problems From exercise 81, we know that MPNQ is a rectangle Since MN ⊥ AB, angles AMN and BMN each measure 90° MP bisects angle AMN, so angle PMN = 45° MQ bisects angle BMN, so angle QMN = 45° Thus QMN ≅ PMN by AAS, and therefore, MP = MQ From geometry, we know that a rectangle with two consecutive equal sides is a square, so MPNQ is a square △ △ (33 − x ) + (11 + x) + 110 = 180 ⇒ 44 + x = 70 ⇒ x = 26 ⇒ x = 13 33 − 13 = 20; 11 + (13) = 50 The three angles are 20°, 50°, and 110° 83 Since AB CD, ∠ABC = ∠BCD ⇒ α + β = 70° Since CD EF , ∠CEF + ∠ECD = 180° ⇒ 160° + β = 180° ⇒ β = 20° α + 20 = 70° ⇒ α = 50° Therefore, α − β = 50° − 20° = 30° (3 + x )2 + 10 = (5 + x )2 x + x + + 100 = x + 10 x + 25 x + 109 = 10 x + 25 84 = x ⇒ 21 = x The sides of the triangle are 10, + 21 = 24, and + 21 = 26 Copyright © 2011 Pearson Education Inc Publishing as Addison-Wesley Section 1.2 Triangles Both triangles contain A, so triangles ADE and ABC are similar because they have equal corresponding angles b To find x, we use the proportion AE AD = ⇒ = ⇒ + x 12 AC AB 108 = (9 + x ) ⇒ 108 = 72 + x ⇒ 36 = =x 36 = x ⇒ To find y, we use the proportion AD DE y = ⇒ = ⇒ 128 = 12 y ⇒ AB BC 12 16 y = 10 3420 3420 = ⇒ 2 x = 1710 ≈ 2418 ft x = 3420 ⇒ x = 1.2 A Exercises: Basic Skills and Concepts The sum of the measures of the three angle of a triangle is 180° AC is the leg opposite the 60° angle, so 6 6=a 3⇒a= = = ≈ 3.5 ft 3 c = 2a = ≈ 6.9 ft In an isosceles triangle, the two angles opposite equal sides are equal or congruent In a 30°-60°-90° triangle, the hypotenuse is two time the length of the shortest side and the side opposite the 60° angle is time the length of the shortest side In similar triangles, the lengths of the corresponding sides are proportional True y = = x 12 = ⇒ x = 18 ⇒ x = x y = ⇒ 24 = y ⇒ = y 12 a Because BC is parallel to DE, AB and AC are transversals Thus, ∠ADE = ∠ABC and ∠AED = ∠ACB False The corresponding sides of two similar triangles are proportional while the corresponding sides of two congruent triangles are equal Note that congruent triangles are also similar A + B + C = 180° ⇒ 50° + 72° + C = 180° ⇒ 122° + C = 180° ⇒ C = 58° A + B + C = 180° ⇒ A + 64° + 48° = 180° ⇒ A + 112° = 180° ⇒ A = 68° A + B + C = 180° ⇒ 48°15′ + B + 98° = 180° ⇒ 146°15′ + B = 180° ⇒ B = 33°45′ 10 A + B + C = 180° ⇒ 34° + 67°45′ + C = 180° ⇒ 101°45′ + C = 180° ⇒ C = 78°15′ 11 A + B + C = 180° ⇒ 46.72° + B + 65° = 180° ⇒ 111.72° + B = 180° ⇒ B = 68.28° Copyright © 2011 Pearson Education Inc Publishing as Addison-Wesley 10 12 13 14 Chapter Trigonometric Functions A + B + C = 180° ⇒ 69° + 54.67° + C = 180° ⇒ 123.67° + C = 180° ⇒ C = 56.33° 23 (5 + x ) + ( + x ) = (5 + x ) ⇒ 25 + 10 x + x + 36 + 36 x + x = 25 + 40 x + 16 x ⇒ 10 x + 46 x + 61 = 16 x + 40 x + 25 ⇒ −6 x + x + 36 = ⇒ x − x − = ⇒ ( x − 3)( x + 2) = ⇒ x = 3, − If x = –2, then b = + 3(–2) = 0, which is not possible, so reject x = –2 If x = 3, then a = + = 8, b = + 3(3) = 15, and c = + 4(3) = 17 A + B + C = 180° ⇒ 60° + (57 − x ) ° + (25 + 3x ) ° = 180° ⇒ 142° + x = 180° ⇒ x = 38° ⇒ x = 19° B = 57 − 19 = 38° C = 25 + ⋅ 19 = 82° A + B + C = 180° ⇒ (50 + x ) ° + 50° + (80 − x ) ° = 180° ⇒ 180° + x = 180° ⇒ x = 0° ⇒ x = 0° A = 50°; C = 80° 15 A + B + C = 180° ⇒ x + x + 3x = 180° ⇒ x = 180° ⇒ x = 30° A = 30°; B = ⋅ 30° = 60°; C = ⋅ 30° = 90° a + b2 = c2 ⇒ 24 a + b2 = c2 ⇒ (2 x − 1) + (10 + x )2 = (6 + x ) ⇒ x − x + + 100 + 120 x + 36 x = 36 + 84 x + 49 x ⇒ 40 x + 116 x + 101 = 49 x + 84 x + 36 ⇒ −9 x + 32 x + 65 = ⇒ x − 32 x − 65 = ⇒ 13 ( x − 5)(9 x + 13) = ⇒ x = 5, − 13 If x = − , then a and c are negative, so reject this solution If x = 5, then a = 2(5) – = 9, b = 10 + 6(5) = 40, and c = + 7(5) = 41 16 A + B + C = 180° ⇒ x + 3x + x = 180° ⇒ x = 180° ⇒ x = 20° A = ⋅ 20° = 40°; B = ⋅ 20° = 60° C = ⋅ 20° = 800° 17 a + b = c ⇒ + 12 = c ⇒ 25 + 144 = c ⇒ 169 = c ⇒ c = 13 18 a + b = c ⇒ + 24 = c ⇒ 49 + 576 = c ⇒ 625 = c ⇒ c = 25 19 a + b = c ⇒ + b = 132 ⇒ 25 + b = 169 ⇒ b = 144 ⇒ b = 12 20 a + b = c ⇒ 20 + b = 29 ⇒ 400 + b = 841 ⇒ b = 441 ⇒ b = 21 25 Since each leg has length 4, the hypotenuse has length a + b = c ⇒ (8 − x ) + (9 + x ) = 132 ⇒ 26 Since each leg has length 5, the hypotenuse has length 21 2 64 − 16 x + x + 81 + 18 x + x = 169 ⇒ x + x + 145 = 169 ⇒ x + x − 24 = ⇒ x + x − 12 = ⇒ ( x + 4)( x − 3) = ⇒ x = −4, If x = –4, then a = – (–4) = 12 and b = + (–4) = If x = 3, then a = – = and b = + =12 22 a + b2 = c2 ⇒ (17 − x )2 + 24 = (15 + x) ⇒ 289 − 68 x + x + 576 = 225 + 60 x + x ⇒ x − 68 x + 865 = x + 60 x + 225 ⇒ 640 = 128 x ⇒ = x a = 17 − ⋅ = 7; c = 15 + ⋅ = 25 In exercises 25–32, recall that in a 45°-45°-90° triangle, both legs are equal and the length of the hypotenuse is times the length of a leg 27 Since each leg has length has length the hypotenuse , the hypotenuse 28 Since each leg has length has length , 29 Since the hypotenuse has length 2, each leg has length 2 = 30 Since the hypotenuse has length 2, each leg has length 2 = Copyright © 2011 Pearson Education Inc Publishing as Addison-Wesley Section 1.5 Fundamental Trigonometric Identities 19 26 Since θ is in quadrant II, cot θ is negative 1 = = sec α 17 17 sin α sin α tan α = ⇒ = ⇒ cos α 4 17 cos α = csc θ − = cot θ ⇒ 32 − = = cot θ ⇒ cot θ = − = −2 27 Since θ is in quadrant III, sin θ , cos θ , sec θ , and csc θ are negative, and tan θ and cot θ are positive csc θ = =− sin θ 17 17 sin α = = = 17 17 1 = = csc α 34 34 cos α cos α cot α = ⇒ = ⇒ sin α 3 34 20 sin α = cos α = 5 34 = 34 34 21 Since θ is in quadrant III, cos θ is negative ⎛ 12 ⎞ sin θ + cos θ = ⇒ ⎜ − ⎟ + cos θ = ⇒ ⎝ 13 ⎠ 144 25 + cos θ = ⇒ cos θ = ⇒ 169 169 25 cos θ = − =− 169 13 22 Since θ is in quadrant II, sin θ is positive ⎛ ⎞ sin θ + cos θ = ⇒ sin θ + ⎜ − =1⇒ ⎝ 13 ⎟⎠ ⇒ sin θ + = ⇒ sin θ = 13 13 3 13 sin θ = = = 13 13 13 23 Since θ is in quadrant III, sec θ is negative + tan θ = sec θ ⇒ + 32 = 10 = sec θ ⇒ sec θ = − 10 24 Since θ is in quadrant IV, tan θ is negative ⎛5⎞ sec θ − = tan θ ⇒ ⎜ ⎟ − = = tan θ ⇒ ⎝4⎠ 16 tan θ = − =− 16 25 Since θ is in quadrant III, csc θ is negative ⎛1⎞ + cot θ = csc θ ⇒ + ⎜ ⎟ = = csc θ ⇒ ⎝2⎠ csc θ = − 5 =− 33 ⎛ 3⎞ sin θ + cos θ = ⇒ ⎜ − ⎟ + cos θ = ⇒ ⎝ 5⎠ 16 + cos θ = ⇒ cos θ = ⇒ 25 25 16 cos θ = − = − ; sec θ = =− 25 cos θ sin θ −3 tan θ = = = ; cot θ = = cos θ −4 tan θ 28 Since θ is in quadrant II, cos θ , tan θ , cot θ , and sec θ are negative, and sin θ and csc θ are positive 13 =− sec θ = cos θ 12 ⎛ 12 ⎞ sin θ + cos θ = ⇒ sin θ + ⎜ − ⎟ = ⇒ ⎝ 13 ⎠ 144 25 = ⇒ sin θ = ⇒ sin θ + 169 169 25 13 = ; csc θ = = sin θ = 169 13 sin θ sin θ 13 = =− tan θ = cos θ −12 13 12 12 =− cot θ = tan θ 29 Since θ is in quadrant I, all the trigonometric functions of θ are positive 1 cot θ = = tan θ + tan θ = sec θ ⇒ sec θ = + 2 = ⇒ sec θ = 5; cos θ = 1 = = sec θ 5 ⎛1⎞ + cot θ = csc θ ⇒ + ⎜ ⎟ = = csc θ ⇒ ⎝2⎠ csc θ = 5 2 ; sin θ = = = = csc θ 5 Copyright © 2011 Pearson Education Inc Publishing as Addison-Wesley 34 Chapter Trigonometric Functions 30 Since θ is in quadrant II, cos θ , tan θ , cot θ , and sec θ are negative, and sin θ and csc θ are positive 1 cos θ = =− sec θ + cot θ = csc θ ⇒ + (−2) = = csc θ ⇒ csc θ = 5; sin θ = 33 ⎛ 1⎞ sin θ + cos θ = ⇒ sin θ + ⎜ − ⎟ = ⇒ ⎝ 3⎠ sin θ + = ⇒ sin θ = ⇒ 9 2 sin θ = = 3 csc θ = = = sin θ 2 tan θ = sin θ 2 = = −2 cos θ −1 cot θ = 1 =− =− tan θ 2 31 Since θ is in quadrant III, sin θ , cos θ , sec θ , and csc θ are negative, and tan θ and cot θ are positive sin θ = =− csc θ ⎛ 2⎞ sin θ + cos θ = ⇒ ⎜ − ⎟ + cos θ = ⇒ ⎝ 3⎠ + cos θ = ⇒ cos θ = ⇒ 9 5 =− cos θ = − 3 =− =− sec θ = cos θ 5 −2 sin θ 2 = = = tan θ = cos θ − 5 cot θ = = tan θ 32 Since θ is in quadrant II, cos θ , tan θ , cot θ , and sec θ are negative, and sin θ and csc θ are positive 1 tan θ = = − ; + tan θ = sec θ ⇒ cot θ 2 5 ⎛ 1⎞ sec θ = + ⎜ − ⎟ = ⇒ sec θ = − ⎝ 2⎠ cos θ = 2 =− =− sec θ 5 1 = = csc θ 5 cos θ < and sin θ > ⇒ θ is in quadrant II Thus, cos θ , tan θ , cot θ , and sec θ are negative, and sin θ and csc θ are positive 5 = − ; csc θ = = sec θ = cos θ sin θ sin θ 45 = =− tan θ = cos θ −3 3 =− cot θ = tan θ 34 cos θ > and sin θ < ⇒ θ is in quadrant IV Thus, sin θ , tan θ , cot θ , and csc θ are negative, and cos θ and sec θ are positive 5 = ; csc θ = =− sec θ = cos θ sin θ sin θ −4 tan θ = = =− cos θ 35 3 cot θ = =− tan θ 35 sin θ > and cos θ > ⇒ θ is in quadrant I Thus, all trigonometric functions of θ are positive = = csc θ = sin θ 3 = = sec θ = cos θ sin θ 3 = = = tan θ = cos θ 2 = = cot θ = tan θ 36 cos θ < and sin θ > ⇒ θ is in quadrant II Thus, cos θ , tan θ , cot θ , and sec θ are negative, and sin θ and csc θ are positive = = csc θ = sin θ 3 =− =− cos θ sin θ 3 = =− =− tan θ = cos θ − 2 =− =− cot θ = tan θ sec θ = Copyright © 2011 Pearson Education Inc Publishing as Addison-Wesley Section 1.5 Fundamental Trigonometric Identities 37 tan α > and sec θ < ⇒ θ is in quadrant III Thus, sin α , cos α , sec α , and csc θ are negative, and tan α and cot α are positive 41 2 =− =− sec α 5 sin α sin α ⇒ = ⇒ sin α = − tan α = cos α −2 5 5 =− = − 5; cot α = =2 csc α = sin α tan α cos α = 38 tan α > and sec α > ⇒ θ is in quadrant I Thus, all trigonometric functions of α are positive 144 144 12 ⇒ cos θ = − =− 169 169 13 13 =− sec θ = cos θ 12 cos θ = 42 Thus, sin β , tan β , cot β , and csc β are negative, and cos β and sec β are positive 1 = sec α =− = −2 tan β = cot β cos β = ⎛1⎞ sin β + cos β = ⇒ sin β + ⎜ ⎟ = ⇒ ⎝3⎠ 8 2 sin β = ⇒ sin β = − =− 9 3 csc α = =− =− sin α 2 2 cot θ < and sin θ < ⇒ θ is in quadrant IV Thus, sin θ , tan θ , cot θ , and csc θ are negative, and cos θ and sec θ are positive 13 = − ; csc θ = =− tan θ = cot θ 12 sin θ ⎛ 5⎞ sin θ + cos θ = ⇒ ⎜ − ⎟ + cos θ = ⇒ ⎝ 13 ⎠ 39 sec β > and cotβ > ⇒ θ is in quadrant I 40 sec β > and cotβ < ⇒ θ is in quadrant IV cot θ > and sin θ < ⇒ θ is in quadrant III Thus, sin θ , cos θ , sec θ , and csc θ are negative, and tan θ and cot θ are positive 13 = ; csc θ = =− tan θ = cot θ 12 sin θ ⎛ 5⎞ sin θ + cos θ = ⇒ ⎜ − ⎟ + cos θ = ⇒ ⎝ 13 ⎠ 2 = = cos α = sec α 5 sin α sin α ⇒ = ⇒ sin α = tan α = cos α 2 5 5 = = 5; cot α = =2 csc α = sin α tan α Thus, all trigonometric functions of β are positive 1 = ; tan β = = =2 cos β = sec α cot β 2 ⎛1⎞ sin β + cos β = ⇒ sin β + ⎜ ⎟ = ⇒ ⎝3⎠ 8 2 sin β = ⇒ sin β = = 9 3 = = csc α = sin α 2 35 144 144 12 ⇒ cos θ = = 169 169 13 13 = sec θ = cos θ 12 cos θ = 43 cos θ < and sin θ > ⇒ θ is in quadrant II Thus, cos θ , tan θ , cot θ , and sec θ are negative, and sin θ and csc θ are positive = csc θ = sin θ ⎛3⎞ sin θ + cos θ = ⇒ ⎜ ⎟ + cos θ = ⇒ ⎝4⎠ 7 ⇒ cos θ = − =− 16 16 4 =− sec θ = − 7 sin θ 34 3 = =− =− tan θ = cos θ − 7 cos θ = cot θ = cos θ − 7 = =− sin θ 34 Copyright © 2011 Pearson Education Inc Publishing as Addison-Wesley 36 44 Chapter Trigonometric Functions cos θ > and tanθ < ⇒ θ is in quadrant IV Thus, sin θ , tan θ , cot θ , and csc θ are negative, and cos θ and sec θ are positive = sec θ = cos θ 47 sec θ > and sin θ < ⇒ θ is in quadrant IV Thus, sin θ , tan θ , cot θ , and csc θ are negative, and cos θ and sec θ are positive = cos θ = sec θ 2 ⎛ 4⎞ sin θ + cos θ = ⇒ sin θ + ⎜ ⎟ = ⇒ ⎝5⎠ ⎛2⎞ sin θ + cos θ = ⇒ sin θ + ⎜ ⎟ = ⇒ ⎝5⎠ 9 ⇒ sin θ = − =− 25 25 5 =− csc θ = sin θ sin θ −3 = =− tan θ = cos θ 45 cos θ 45 = =− cot θ = sin θ −3 sin θ = 21 21 21 ⇒ sin θ = − =− 25 25 5 21 =− =− csc θ = sin θ 21 21 sin θ − 21 21 = =− tan θ = cos θ 25 sin θ = 45 tan θ > and sin θ < ⇒ θ is in quadrant III Thus, sin θ , cos θ , sec θ , and csc θ are negative, and tan θ and cot θ are positive 1 = cot θ = tan θ tan θ + = sec θ ⇒ 2 + = = sec θ ⇒ cot θ = 48 ⎛1⎞ sin θ + cos θ = ⇒ ⎜ ⎟ + cos θ = ⇒ ⎝2⎠ 3 ⇒ cos θ = − =− 4 2 =− =− sec θ = cos θ 3 sin θ 12 = =− =− tan θ = cos θ − 3 cos θ = ⎛1⎞ cot θ + = csc θ ⇒ ⎜ ⎟ + = = csc θ ⇒ ⎝2⎠ 5 =− 2 =− =− sin θ = csc θ 5 csc θ = − 46 cot θ < and sec θ > ⇒ θ is in quadrant IV Thus, sin θ , tan θ , cot θ , and csc θ are negative, and cos θ and sec θ are positive 1 =− tan θ = cot θ 2 ⎛ 1⎞ tan θ + = sec θ ⇒ ⎜ − ⎟ + = = sec θ ⇒ ⎝ 2⎠ sec θ = 5 2 = = = ; cos θ = sec θ 5 cot θ + = csc θ ⇒ (−2) + = = csc θ ⇒ 1 =− =− csc θ = − 5; sin θ = csc θ 5 csc θ > and tan θ < ⇒ θ is in quadrant II Thus, cos θ , tan θ , cot θ , and sec θ are negative, and sin θ and csc θ are positive 1 = sin θ = csc θ 1 =− =− sec θ = − 5; cos θ = sec θ 5 cos θ 25 2 21 = =− =− sin θ − 21 21 21 cot θ = 49 cos θ − = =− sin θ 12 (1 − sin θ )(1 + sin θ ) + sin θ = − sin θ + sin θ = 50 (cos θ + 1)(cos θ − 1) + sin θ ( ) = cos θ − + sin θ = sin θ + cos θ − = 1−1 = 51 (1 + tan θ )(1 − tan θ ) + sec θ ( ) = − tan θ + sec θ = + sec θ − tan θ = tan θ − tan θ = 2 Copyright © 2011 Pearson Education Inc Publishing as Addison-Wesley Section 1.5 Fundamental Trigonometric Identities 52 (sec θ − 1)(sec θ + 1) − tan θ ( ) 60 = sec θ − − tan θ = sec θ − tan θ − = 1−1 = 53 (sec θ + tan θ )(sec θ − tan θ ) = sec θ − tan θ = 54 (csc θ + cot θ )(csc θ − cot θ ) = csc θ − cot θ = 55 56 sec θ − − sec θ sec θ − (sec θ − 2)(sec θ + 2) = − sec θ sec θ − = sec θ + − sec θ = − csc θ + csc θ + csc θ (3 + csc θ )(3 − csc θ ) = + csc θ + csc θ = − csc θ + csc θ = 57 sin θ cos θ ( tan θ + cot θ ) ⎛ sin θ cos θ ⎞ = sin θ cos θ ⎜ + ⎝ cos θ sin θ ⎟⎠ ⎛ sin θ ⎞ ⎛ cos θ ⎞ = sin θ cos θ ⎜ + sin θ cos θ ⎜ ⎝ cos θ ⎟⎠ ⎝ sin θ ⎟⎠ = sin θ + cos θ = 58 59 sec θ csc θ (sin θ + cos θ ) sec θ + csc θ ⎞ ⎛ + sec θ csc θ ⎜ ⎝ csc θ sec θ ⎟⎠ = sec θ + csc θ ⎛ ⎞ ⎛ ⎞ + sec θ csc θ ⎜ sec θ csc θ ⎜ ⎝ csc θ ⎟⎠ ⎝ sec θ ⎟⎠ = sec θ + csc θ sec θ + csc θ = =1 sec θ + csc θ 1 + − 2sec θ sec θ − tan θ sec θ + tan θ (sec θ + tan θ ) + (sec θ − tan θ ) = − sec θ (sec θ + tan θ )(sec θ − tan θ ) 2sec θ = − 2sec θ sec θ − tan θ 2sec θ = − 2sec θ = 61 62 37 1 − − cot θ csc θ − cot θ csc θ + cot θ (csc θ + cot θ ) − (csc θ − cot θ ) = − cot θ (csc θ + cot θ )(csc θ − cot θ ) cot θ = − cot θ csc θ − cot θ cot θ = − cot θ = tan θ − tan θ − ( tan θ − 3)( tan θ + 1) = tan θ + tan θ + = tan θ − tan θ + sec θ − sec θ − + sec θ − = sec θ − sec θ − sec θ + sec θ − = sec θ − sec θ + 2)(sec θ − 1) ( = sec θ − = sec θ + 63 If (sin θ + cos θ ) = sin θ + cos θ is an identity, then it must be true for all values of θ When θ = 30°, we have (sin 30° + cos 30°) ⎛1 3⎞ =⎜ + ⎟ ⎝2 ⎠ ⎛ 3⎞ 3 2+ = + 2⎜ ⎟ + = 1+ = 4 ⎝ ⎠ while sin θ + cos θ = Therefore, the equation is not an identity 64 If (1 − cos θ ) = − cos θ is an identity, then it must be true for all values of θ When θ = 30°, we have ⎛ (1 − cos 30°)2 = ⎜1 − ⎝ 3⎞ ≈ 0.0179 while ⎟⎠ ⎛ 3⎞ 1 − cos θ = − ⎜ ⎟ = Therefore, the ⎝ ⎠ equation is not an identity 65 If sin (θ + 45°) = sin θ + sin 45° is an identity, then it must be true for all values of θ When θ = 45°, we have sin ( 45° + 45°) = sin 90° = while sin 45° + sin 45° = Therefore, the equation is not an identity Copyright © 2011 Pearson Education Inc Publishing as Addison-Wesley 38 Chapter Trigonometric Functions 66 If cos (θ + 45°) = cos θ + cos 45° is an identity, then it must be true for all values of θ When θ = 45°, we have cos (45° + 45°) = cos 90° = 72 If csc θ + = csc θ + is an identity, then it must be true for all values of θ When θ = 45°, we have while cos 45° + cos 45° = Therefore, the equation is not an identity 67 If sin 2θ = sin θ is an identity, then it must be true for all values of θ When θ = 45°, we have sin ( ⋅ 45°) = sin 90° = while csc 45° + = while csc 45° + = + Therefore, the equation is not an identity 1.5 B Exercises: Applying the Concepts 73 x= 20 = 20 csc θ sin θ 74 x= 60 = 60 cot θ tan θ 75 A = nr 69 If tan θ − = sec θ is an identity, then it must be true for all values of θ When 76 ⎛ 180° ⎞ A = 32 tan ⎜ = 36 tan 45° = 36 sq ft ⎝ ⎟⎠ θ = 45°, we have tan (45°) − = while 77 sin 45° = Therefore, the equation is not an identity 68 If cos 2θ = cos θ is an identity, then it must be true for all values of θ When θ = 45°, we have cos (2 ⋅ 45°) = cos 90° = while cos 45° = Therefore, the equation is not an identity sec 45° = Therefore, the equation is not an identity ( ) (1 + m1m2 ) sin θ (1 + m1m2 ) sin θ (1 + m1m2 ) sin θ θ = 45°, we have sec 45° − = while = cos θ ( m1 − m2 ) = tan θ = θ = 90°, we have cot (90°) + = while 71 If sec θ − = sec θ − is an identity, then it must be true for all values of θ When = m1 cos θ − m2 cos θ cos θ (m1 − m2 ) cos θ cos θ (1 + m1m2 ) tan θ = m1 − m2 70 If cot θ + = sec θ is an identity, then it must be true for all values of θ When sec 90° is undefined Therefore, the equation is not an identity (180n ° ) = nr tan ⎛ 180° ⎞ ⎜⎝ ⎟ ° n ⎠ cos (180 n ) sin 78 m1 = 4, m2 = m1 − m2 + m1m2 4− 17 m1 − m2 = = =1 + m1m2 + 17 5 tan θ = () tan θ = ⇒ θ = 45° or θ = 135° sec 45° − = − Therefore, the equation is not an identity Copyright © 2011 Pearson Education Inc Publishing as Addison-Wesley Section 1.5 Fundamental Trigonometric Identities 1.5 C Beyond the Basics 79 (sin α cos β + cos α sin β ) + (cos α cos β − sin α sin β ) = sin α cos β + sin α cos β cos α sin β + cos α sin β + cos α cos β − cos α cos β sin α sin β + sin α sin β 2 2 = sin α cos β + cos α sin β + cos α cos β + sin α sin β ( ) ( ) 2 β ) + cos α (sin β + cos β ) = sin α + cos α = = sin α cos β + sin α sin β + cos α sin β + cos α cos β ( = sin α cos β + sin 80 2 (sin α cos β − cos α sin β )2 + (cos α cos β + sin α sin β )2 = sin α cos β − sin α cos β cos α sin β + cos α sin β + cos α cos β + cos α cos β sin α sin β + sin α sin β = sin α cos β + cos α sin β + cos α cos β + sin α sin β ( ) ( ) 2 = sin α (cos β + sin β ) + cos α (sin β + cos β ) = sin α + cos α = (sin α + cos α ) − (sin α + cos α ) ⎡ ⎤ = ⎢(sin α + cos α ) − sin α cos α ⎥ − ⎡(sin α + cos α )(sin α − sin α cos α + cos α )⎤ ⎣ ⎦ ⎣ ⎦ 2 2 = (1 − sin α cos α ) − (sin α − sin α cos α + cos α ) = sin α cos β + sin α sin β + cos α sin β + cos α cos β 81 2 = − sin α cos α − sin α + sin α cos α − cos α = − sin α − sin α cos α − cos α ( = − (sin α + cos α ) = − = = − sin α + sin α cos α + cos α ) 82 sin α + cos α + 3cos α − 3cos α ( )( ) = sin α + cos α sin α − sin α cos α + cos α + 3cos α − 3cos α = sin α − sin α cos α + cos α + cos α − 3cos α 2 4 = sin α − sin α cos α − cos α + 3cos α ( )( ) = sin α − cos α sin α + cos α + 3cos α = sin α − cos α + 3cos α = sin α + cos α = (1 − cos θ ) + (1 + cos θ ) cos θ − cos θ + cos θ + − cot θ = − + cos θ − cos θ − cos θ sin θ 83 = = = − cos θ + cos θ + + cos θ + cos θ − cos θ + cos θ sin θ − cos θ sin θ − = cos θ sin θ ( = − cos θ sin θ − cos θ sin θ + cos θ − cos θ sin θ ) = sin θ = sin θ Copyright © 2011 Pearson Education Inc Publishing as Addison-Wesley 39 40 84 Chapter Trigonometric Functions cos θ + sin θ cos θ − sin θ + cos θ + sin θ cos θ − sin θ (cos θ = ( ( + sin θ ) cos θ − cos θ sin θ + sin θ cos θ + sin θ ) ( ) + (cosθ − sin θ ) (cos θ + cos θ sin θ + sin θ ) = cos θ − cos θ sin θ + sin θ + cos θ + cos θ sin θ + sin θ 2 ( 2 ) ) cos θ − sin θ = cos θ + sin θ = cos θ + sin θ = sec θ + tan θ 85 + tan θ = cos θ 1+ + sin θ + sin θ + sin θ cos θ = + sin θ = = =1 3sin θ cos θ + 3sin θ − sin θ + 3sin θ + sin θ cos θ csc α + sec α 86 87 csc α − sec α ( − + tan α 1+ sin α 2 2 sin α cos α − cos α = cos α + sin α − cos α + sin α = = 1 − tan α sin α cos α − sin α cos α − sin α − 1− 2 sin α cos α cos α cos α − cos α ( + ) ( + sin α − sin α ) ) ( = cos α − 24 cos α + 16 cos α + sin α − 24 sin α + 16 sin α ) = cos α − 24 cos α + 16 cos α + sin α − 24 sin α + 16 sin α ( ) = cos α + sin α − 24 cos α − 24 sin α + 16 cos α + 16 sin α ( ) ( ) ( ) = − 24 cos α − 24 sin α + 16 (cos α + sin α )(cos α − cos α sin α + sin α ) = cos α + sin α − 24 cos α + sin α + 16 cos α + sin α = − 24 cos α − 24 sin α + 16 cos α − 16 cos α sin α + 16 sin α ( = − cos α − 16 cos α sin α − 8sin α = − cos α + cos α sin α + sin α ( = − cos α + sin α 88 (sin θ ) ) = 9−8 =1 + csc θ ) + (cos θ + sec θ ) − tan θ − cot θ 2 = sin θ + sin θ csc θ + csc θ + cos θ + cos θ sec θ + sec θ − tan θ − cot θ = sin θ + + csc θ + cos θ + + sec θ − tan θ − cot θ ( ) ( ) ( ) = sin θ + cos θ + sec θ − tan θ + csc θ − cot θ + = + + + = 89 Start by squaring both sides of tan θ + cot θ = 90 Start by squaring both sides of sec θ + cos θ = ( tan θ + cot θ )2 = 2 ⇒ (sec θ + cos θ )2 = 2 ⇒ tan θ + tan θ cot θ + cot θ = ⇒ tan θ + + cot θ = ⇒ tan θ + cot θ = sec θ + sec θ cos θ + cos θ = ⇒ sec θ + + cos θ = ⇒ sec θ + cos θ = Copyright © 2011 Pearson Education Inc Publishing as Addison-Wesley Chapter Review Exercises 91 Start by dividing the numerator and denominator by cos θ 5sin θ 3cos θ − 5sin θ − 3cos θ cos cos θ θ = cos θ sin θ + cos θ sin θ + cos θ cos θ tan θ − − = = = tan θ + + 92 Start by dividing the numerator and denominator by sin θ sin θ cos θ − sin θ − cos θ sin θ sin θ = − cot θ = cos θ + cot θ sin θ + cos θ sin θ + sin θ sin θ cot θ = ⇒ cot θ = − 1 − cot θ 14 = = = + cot θ + 7 93 csc θ − sec θ − = sin θ csc θ + sec θ + sin θ cos θ − = cos θ + 2 cos θ cos θ sin θ ( = cos θ − − cos θ − cos θ 95 a The equation is an identity for ≤ θ ≤ 90° because cos θ ≥ in the given interval b The equation is not an identity for ≤ θ ≤ 180° because the left side is negative for 90° < θ ≤ 180° while the right side is positive 96 False We can show that cos (α − β ) = cos α − cos β is not an identity by letting α = 60° and β = 30° Then cos (α − β ) = cos (60° − 30°) = cos 30° = while cos α − cos β = cos 60° − cos 30° = − 2 97 No, sin θ , cos θ , and tan θ cannot all be negative for an angle θ Note that if two of the sin θ are negative, then functions in tan θ = cos θ the third has to be positive ) 180° – 123.4° = 56.6° 15 30 ⎞ ⎛ 64°15′ 30′′ = ⎜ 64 + + ⎟ ° ≈ 64.26° ⎝ 60 3600 ⎠ 34.742° = 34° + 0.742(60′ ) = 34° + 44.52′ = 34° + 44′ + 0.52(60′′ ) ≈ 34°44′ 31′′ a Quadrant III sin θ = + sin θ + sin θ Now divide the numerator and denominator by sin θ sin θ 1 sin θ = = 2 2 sin θ + csc θ + + sin θ sin θ sin θ 1 = = = 11 = 2 + cot θ + + + 11 ( ) ( , 90° – 67.8° = 22.2° = cos θ − 2 = −1 = −1 sec θ + tan θ 2 = −1 = −1 = −1 = 87 4 1+ 94 1.5 Critical Thinking Chapter Review Exercises sin θ = cos θ − sin θ 41 ) Copyright © 2011 Pearson Education Inc Publishing as Addison-Wesley Chapter Trigonometric Functions 42 b Quadrant III the length of a leg Therefore, the length of the other leg is cm, and the length of the hypotenuse is cm c Quadrant III 10 In a 30°-60°-90° triangle, if the length of the shorter leg (the leg opposite the 30° angle) is x, then the length of the longer leg (the angle opposite the 60° angle) is x 3, and the length of the hypotenuse is 2x The shortest side is cm, so the other leg has length cm, and the hypotenuse has length 12 cm A + B + C = 180° ⇒ (2 x + 10) ° + (3x − 20) ° + (3x + 30) ° = 180° ⇒ x + 20° = 180° ⇒ x = 160° ⇒ x = 20° The measures of the angles are A = ⋅ 20 + 10 = 50°, B = ⋅ 20 − 20 = 40°, and C = ⋅ 20 + 30 = 90° 11 True d Quadrant IV 12 False A scalene triangle is a triangle whose sides have different lengths A scalene triangle may be acute, right, or obtuse 13 a 580° = 220° + 360°, so 550° is coterminal with 220° 14 b 1460° = 20° + ⋅ 360°, so 1460° is coterminal with 20° c −675° = 45° − ⋅ 360°, so –675° is coterminal with 45° d −1345° = 95° − ⋅ 360°, so –1345° is coterminal with 95° a (–2, 0) lies on the negative x-axis, so a line through that point forms a 180° angle with the positive x-axis Thus, θ = 180° + n ⋅ 360°, n any integer b (3, –3) lies in Quadrant IV, so a line through that point forms a 270° + 45° = 315° angle with the positive x-axis Thus, θ = 315° + n ⋅ 360°, n any integer 15 x = −3, y = ⇒ r = (−3) + ⇒ r = 4 sin θ = cos θ = − tan θ = − 5 5 csc θ = sec θ = − cot θ = − 4 x = −5, y = −12 ⇒ r = ( −5) + ( −12) ⇒ r = 13 12 12 sin θ = − cos θ = − tan θ = 13 13 13 13 csc θ = − sec θ = − cot θ = 12 12 2 x = 2, y = −3 ⇒ r = (2) + ( −3) ⇒ r = 13 3 13 =− 13 13 2 13 = cos θ = 13 13 tan θ = − sin θ = − In a 45°-45°-90° triangle, both legs are equal and the length of the hypotenuse is times Copyright © 2011 Pearson Education Inc Publishing as Addison-Wesley csc θ = − 13 13 2 cot θ = − sec θ = Chapter Review Exercises 16 x = 3, y = ⇒ r = (3) + (5) ⇒ r = 34 2 5 34 =− 34 34 3 34 cos θ = = 34 34 tan θ = sin θ = 17 18 19 20 csc θ = 34 sec θ = 34 cot θ = 26 sin θ < and cot θ < ⇒ θ is in quadrant IV 27 , θ in Quadrant II ⇒ x < 0, y > 12 x = −5, y = 12 ⇒ r = (−5) + 12 ⇒ r = 13 cot θ = − 12 13 13 csc θ = 12 sin θ = x = 2, y = ⇒ r = ( 2) + (0) ⇒ r = 2 sin θ = = csc θ = , undefined 2 cos θ = = sec θ = = 2 tan θ = = cot θ = , undefined 43 13 13 sec θ = − cos θ = − 12 5 cot θ = − 12 tan θ = − x = 0, y = ⇒ r = (0) + (3) ⇒ r = 3 sin θ = = csc θ = = 3 cos θ = = sec θ = , undefined 3 tan θ = , undefined cot θ = = 0 28 , θ in Quadrant IV ⇒ x > 0, y < 12 x = 5, y = −12 ⇒ r = + ( −12) ⇒ r = 13 cot θ = − 12 13 13 csc θ = − 12 sin θ = − 13 13 sec θ = cos θ = 12 5 cot θ = − 12 tan θ = − x = −4, y = ⇒ r = ( −4) + (0) ⇒ r = 4 sin θ = − = csc θ = − , undefined 4 cos θ = − = −1 sec θ = − = −1 4 −4 =0 tan θ = cot θ = , undefined −4 2 x = 0, y = −5 ⇒ r = (0) + (−5) ⇒ r = 5 sin θ = − = −1 csc θ = − = −1 5 cos θ = = sec θ = , undefined −5 =0 tan θ = , undefined cot θ = −5 2 21 tan θ < and sin θ > ⇒ θ is in quadrant II 22 cot θ > and csc θ < ⇒ θ is in quadrant III 23 cot θ > and sec θ < ⇒ θ is in quadrant III 24 sec θ < and csc θ > ⇒ θ is in quadrant II 25 sec θ > and tan θ < ⇒ θ is in quadrant IV and cos θ > ⇒ θ is in quadrant I Thus, x > and y > 29 sin θ = y = 3, r = ⇒ = x + 32 ⇒ x = 5 csc θ = sin θ = 5 sec θ = cos θ = 4 cot θ = tan θ = 13 and sin θ < ⇒ θ is in 12 quadrant IV Thus, x > and y < 30 sec θ = x = 12, r = 13 ⇒ 132 = 12 + y ⇒ y = −5 12 cos θ = 13 13 13 13 csc θ = − sec θ = 12 sin θ = − 12 12 cot θ = − tan θ = − 31 θ = 260° ⇒ θ ′ = 260° − 180° = 80° 32 530° = 170° + 360°, so the 530° angle is coterminal with 170° Since 170° lies in quadrant II, the reference angle is determined by 180° – 170° = 10° 33 –275° + 360° = 85° Since 85° lies in quadrant I, the reference angle is 85° 34 –1315° + 4(360°) = 125° Since 125° lies in quadrant II, the reference angle is determined by 180° – 125° = 55° Copyright © 2011 Pearson Education Inc Publishing as Addison-Wesley 44 Chapter Trigonometric Functions 35 390° = 360° + 30°, so 390° is coterminal with 30°, which is the reference angle In quadrant I, all trigonometric functions of θ are positive sin 390° = sin 30° = cos 390° = cos 30° = tan 390° = tan 30° = csc 390° = csc 30° = 38 1020° = 300° + ⋅ 360°, so 1020° is coterminal with 300° 300° lies in quadrant IV, so the reference angle is 360° – 300° = 60° In quadrant IV, sin θ , tan θ , cot θ , and csc θ are negative, and cos θ and sec θ are positive sin1020° = − sin 60° = − tan1020° = − tan 60° = − cos1020° = cos 60° = 3 cot 390° = cot 30° = sec 390° = sec 30° = csc1020° = − csc 60° = − sec1020° = sec 60° = 36 –390° is coterminal with –390° + 2(360°) = 330° 330° lies in quadrant IV, so the reference angle is 360° – 330° = 30° In quadrant IV, sin θ , tan θ , cot θ , and csc θ are negative, and cos θ and sec θ are positive sin ( −390°) = − sin 30° = − cos (−390°) = cos 30° = tan (−390°) = − tan 30° = − csc (−390°) = − csc 30° = −2 3 cot ( −390°) = − cot 30° = − cot1020° = − cot 60° = − sin ( −495°) = − sin 45° = − 2 cos (−495°) = − cos 45° = − 2 tan ( −495°) = tan 45° = 3 3 12 > and cos θ = > ⇒ θ is in 13 13 quadrant I Thus, all trigonometric functions of θ are positive 13 csc θ = = sin θ 13 sec θ = = cos θ 12 sin θ 13 tan θ = = = cos θ 12 13 12 12 cot θ = = tan θ 39 sin θ = sec ( −390°) = sec 30° = 37 –495° is coterminal with –495° + 2(360°) = 225° 225° lies in quadrant III, so the reference angle is 225° – 180° = 45° In quadrant III, sin θ , cos θ , sec θ , and csc θ are negative, and tan θ and cot θ are positive 40 5 > and sec θ = > ⇒ θ is in quadrant I Thus, all trigonometric functions of θ are positive sin θ = = csc θ cos θ = = sec θ sin θ tan θ = = = cos θ 4 cot θ = = tan θ csc θ = csc (−495°) = − csc 45° = − sec ( −495°) = − sec 45° = − cot ( −495°) = cot 45° = Copyright © 2011 Pearson Education Inc Publishing as Addison-Wesley Chapter Review Exercises and θ in quadrant II Thus, cos θ , tan θ , cot θ , and sec θ are negative, and sin θ and csc θ are positive csc θ = =2 sin θ 17 ⎛1⎞ cot θ + = csc θ ⇒ ⎜ ⎟ + = = csc θ ⇒ ⎝4⎠ 16 41 sin θ = 17 17 =− 16 4 17 sin θ = =− =− csc θ 17 17 csc θ = − ⎛1⎞ sin θ + cos θ = ⇒ ⎜ ⎟ + cos θ = ⇒ ⎝2⎠ 44 3 ⇒ cos θ = − =− 4 2 sec θ = =− =− cos θ 3 sin θ 12 tan θ = = =− =− cos θ − 3 cos θ = cot θ = 42 8 2 ⇒ cos θ = − =− 9 3 sec θ = =− =− cos θ 2 sin θ 13 tan θ = = =− =− cos θ −2 2 cos θ = cos θ − = =− sin θ 12 and θ in quadrant IV Thus, sin θ , tan θ , cot θ , and csc θ are negative, and cos θ and sec θ are positive =2 sec θ = cos θ ⎛1⎞ sin θ + cos θ = ⇒ sin θ + ⎜ ⎟ = ⇒ ⎝ 2⎠ 3 ⇒ sin θ = − =− 4 2 csc θ = =− =− sin θ 3 sin θ − tan θ = = =− cos θ 12 sin θ = 43 cos θ 12 = =− =− sin θ − 3 tan θ = and θ in quadrant III Thus, sin θ , cos θ , sec θ , and csc θ are negative, and tan θ and cot θ are positive 1 cot θ = = tan θ tan θ + = sec θ ⇒ + = 17 = sec θ ⇒ sec θ = − 17; cos θ = csc θ = and θ in quadrant II Thus, cos θ , tan θ , cot θ , and sec θ are negative, and sin θ and csc θ are positive 1 sin θ = = csc θ ⎛1⎞ sin θ + cos θ = ⇒ ⎜ ⎟ + cos θ = ⇒ ⎝3⎠ cos θ = cot θ = 45 cot θ = 45 cos θ −2 = = −2 sin θ 13 cot θ = −2 and θ in quadrant IV Thus, sin θ , tan θ , cot θ , and csc θ are negative, and cos θ and sec θ are positive 1 tan θ = =− cot θ 2 ⎛ 1⎞ sec θ = + tan θ ⇒ sec θ = + ⎜ − ⎟ ⇒ ⎝ 2⎠ 5 ⇒ sec θ = = 4 2 cos θ = = = sec θ 5 sec θ = csc θ = + cot θ ⇒ csc θ = + ( −2) ⇒ csc θ = ⇒ csc θ = − sin θ = 1 =− =− csc θ 5 1 17 =− =− sec θ 17 17 Copyright © 2011 Pearson Education Inc Publishing as Addison-Wesley Chapter Trigonometric Functions 46 46 and θ in quadrant II Thus, cos θ , tan θ , cot θ , and sec θ are negative, and sin θ and csc θ are positive =− cot θ = tan θ tan θ = − 0° + n ⋅ 360°, 90° + n ⋅ 360° 180° + n ⋅ 360°, 270° + n ⋅ 360° ⎛ 7⎞ sec θ = + tan θ ⇒ sec θ = + ⎜ − ⎟ ⇒ ⎝ 4⎠ 65 65 65 ⇒ sec θ = − =− 16 16 4 65 =− =− cos θ = sec θ 65 65 sec θ = ∠1 + ∠ + ∠3 = 260° Angles and are vertical angles, so they are equal Angles and are supplements, so ∠ = 180° − ∠1 Now substitute ∠1 + 180° − ∠1 + ∠1 = 260° ⇒ ∠1 + 180° = 260° ⇒ ∠1 = 80° ∠ = 100°, ∠3 = 80°, ∠4 = 100° ⎛ 4⎞ csc θ = + cot θ ⇒ csc θ = + ⎜ − ⎟ ⇒ ⎝ 7⎠ 65 65 ⇒ csc θ = 49 7 65 = = sin θ = csc θ 65 65 csc θ = 47 (1 − cos θ )(1 + cos θ ) − sin θ = − cos θ − sin θ ( 3x + 90° = 180° ⇒ x = 90° ⇒ x = 30° The angle measures are 30° + 15° = 45°, 30° + 30° = 60°, and 30° + 45° = 75° ) = − cos θ + sin θ = − = 48 (csc θ − 1)(csc θ + 1) − cot θ = csc θ − − cot θ ( ) = csc θ − + cot θ = csc θ − csc θ = 49 50 cot θ + cot θ − − cot θ cot θ − (cot θ + 2)(cot θ − 1) = − cot θ cot θ − = cot θ + − cot θ = sin θ sin θ − − tan θ − sin θ + sin θ sin θ (1 + sin θ ) − sin θ (1 − sin θ ) = − tan θ − sin θ sin θ + sin θ − sin θ + sin θ = − tan θ − sin θ sin θ = − tan θ = tan θ − tan θ = cos θ Chapter Test 180° − 61°31′ = 179°60′ − 61°31′ = 118°29′ ( x + 15) ° + ( x + 30) ° + ( x + 45) ° = 180° ⇒ x + 3x + x = 180° ⇒ 10 x = 180° ⇒ x = 18° The measures of the angles are 2(18°) = 36°, 3(18°) = 54°, and 5(18°) = 90° In an isosceles right triangle, both legs are equal and the length of the hypotenuse is times the length of a leg Therefore, the length 20 = 10 ≈ 14.1 cm of legs = True Two congruent triangles are similar because the corresponding angles are equal and the lengths of the corresponding sides are proportional AD represents the length of the shadow AD = ⇒ 12 AD = AD + 13.5 ⇒ AD + 4.5 12 AD = 13.5 ⇒ AD = 1.5 The shadow is 1.5 feet long Copyright © 2011 Pearson Education Inc Publishing as Addison-Wesley Chapter Test x = 2, y = −1 ⇒ r = (2) + ( −1) ⇒ r = 2 16 θ is in quadrant IV, so sec θ is positive ⎛ 5⎞ sec θ = + tan θ ⇒ sec θ = + ⎜ − ⎟ ⇒ ⎝ 12 ⎠ 169 169 13 ⇒ sec θ = = sec θ = 144 144 12 y =− sin θ = = − r 5 10 x = −2, y = −3 ⇒ r = ( −2) + ( −3) ⇒ 2 r = 13 csc θ = 17 If + tan θ = sec θ is an identity, then it must be true for all values of θ When θ = 45°, we r 13 =− y have tan (45°) + = while sec 45° = Therefore, the equation is not an identity 11 sin θ > and sec θ < ⇒ θ lies in quadrant II 12 cot θ < and csc < ⇒ θ lies in quadrant IV 13 845° = (360°) + 125° 18 If csc θ = + cot θ is an identity, then it must be true for all values of θ When θ = 225°, we have csc 225° = − while 125° lies in quadrant II, so the reference angle is 180° – 125° = 55° 14 47 −640° + (360°) = 80° Since 80° lies in quadrant I, it is the reference angle 15 θ is in quadrant II, so cos θ is negative ⎛4⎞ sin θ + cos θ = ⇒ ⎜ ⎟ + cos θ = ⇒ ⎝7⎠ 33 33 33 cos θ = ⇒ cos θ = − =− 49 49 + cot (225°) = Therefore, the equation is not an identity 19 sin θ cot θ − sec θ cos θ ⎛ cos θ ⎞ ⎛ ⎞ = sin θ ⎜ − cos θ ⎝ sin θ ⎟⎠ ⎜⎝ cos θ ⎟⎠ = cos θ − cos θ = 20 2sin θ + sin θ − − sin θ 2sin θ − (2sin θ − 1)(sin θ + 1) = − sin θ sin θ − = sin θ + − sin θ = Copyright © 2011 Pearson Education Inc Publishing as Addison-Wesley ... ADE by SAS Since ∠B = ∠B, △ ABC ∼△DBF by SAS Since ∠C = ∠C , △ ABC ∼△EFC by SAS Since each side in triangle DEF is the By observation, we can conclude that the number of triangles formed by the... triangle ABC, ABC ∼ FDE by SSS △ △ Use the figure below for exercises 75 and 76 △ △ 80 75 We are given that ABC ∼ XYZ Therefore, BC AC AB = = and ∠C = ∠Z YZ XZ XY ADC ∼ XWZ by AA, so AD AC AD BC... given by 60° + n ⋅ 360° or 240° + n ⋅ 360° 1.4 A Exercises: Basic Skills and Concepts The reference angle θ ′ for a nonquadrantal angle θ in standard position is the acute angle formed by the