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Eureka Math Grade 3 Module 7 Answer Key | Engage NY Math 3rd Grade Module 7 Answer Key

Eureka Math Grade 3 Module 7 Answer Key provided lays a deeper understanding of concepts. Practice using the Engage NY Math 3rd Grade Module 7 Solutions and develop the Knowledge that builds upon itself throughout the learning process. Avail the Problem-Solving Methods used in Eureka Math Grade 3 Answer Key Module 7 and solve related problems in no time. Detailed Solutions provided in the 3rd Grade Engage NY Module 7 Answer Key make it easy for you to retain the concepts for a long time.

Download the Eureka Math 3rd Grade Module 7 Geometry and Measurement for free of cost prepared by subject experts adhering to the latest common core curriculum. Solving the Engage NY Math Third Grade Module 7 Answers helps you to be focused no matter where you are. Look no further and begin your preparation right away taking the help of these online resources.

EngageNY Math Grade 3 Module 7 Answer Key | Eureka Math 3rd Grade Module 7 Answer Key

Engage NY Math Grade 3 module 7 Geometry and Measurement Answer Key provides a flexible, knowledge-building curriculum along with any time learning environment. Third Grade Eureka Math Module 7 Answer Key has questions from Lessons, Extra Practice, Mid Module, End Module Assessments, Review Tests, Assessments, etc. You can access the Topicwise Engage NY Eureka Math 3rd Grade Module 7 Solutions via the direct links available. Just tap on them and learn the underlying concepts within them efficiently and score better grades in exams.

Eureka Math Grade 3 Module 7 Geometry and Measurement Word Problems

Eureka Math Grade 3 Module 7 Topic A Solving Word Problems

• Eureka Math Grade 3 Module 7 Lesson 1 Answer Key
• Eureka Math Grade 3 Module 7 Lesson 2 Answer Key
• Eureka Math Grade 3 Module 7 Lesson 3 Answer Key

Eureka Math 3rd Grade Module 7 Topic B Attributes of Two-Dimensional Figures

• Eureka Math Grade 3 Module 7 Lesson 4 Answer Key
• Eureka Math Grade 3 Module 7 Lesson 5 Answer Key
• Eureka Math Grade 3 Module 7 Lesson 6 Answer Key
• Eureka Math Grade 3 Module 7 Lesson 7 Answer Key
• Eureka Math Grade 3 Module 7 Lesson 8 Answer Key
• Eureka Math Grade 3 Module 7 Lesson 9 Answer Key

Engage NY Math 3rd Grade Module 7 Topic C Problem Solving with Perimeter

• Eureka Math Grade 3 Module 7 Lesson 10 Answer Key
• Eureka Math Grade 3 Module 7 Lesson 11 Answer Key
• Eureka Math Grade 3 Module 7 Lesson 12 Answer Key
• Eureka Math Grade 3 Module 7 Lesson 13 Answer Key
• Eureka Math Grade 3 Module 7 Lesson 14 Answer Key
• Eureka Math Grade 3 Module 7 Lesson 15 Answer Key
• Eureka Math Grade 3 Module 7 Lesson 16 Answer Key
• Eureka Math Grade 3 Module 7 Lesson 17 Answer Key

Eureka Math Grade 3 Module 7 Mid Module Assessment Answer Key

EngageNY Math Grade 3 Module 7 Topic D Recording Perimeter and Area Data on Line Plots

• Eureka Math Grade 3 Module 7 Lesson 18 Answer Key
• Eureka Math Grade 3 Module 7 Lesson 19 Answer Key
• Eureka Math Grade 3 Module 7 Lesson 20 Answer Key
• Eureka Math Grade 3 Module 7 Lesson 21 Answer Key
• Eureka Math Grade 3 Module 7 Lesson 22 Answer Key

3rd Grade Eureka Math Module 7 Topic E Problem Solving with Perimeter and Area

• Eureka Math Grade 3 Module 7 Lesson 23 Answer Key
• Eureka Math Grade 3 Module 7 Lesson 24 Answer Key
• Eureka Math Grade 3 Module 7 Lesson 25 Answer Key
• Eureka Math Grade 3 Module 7 Lesson 26 Answer Key
• Eureka Math Grade 3 Module 7 Lesson 27 Answer Key
• Eureka Math Grade 3 Module 7 Lesson 28 Answer Key
• Eureka Math Grade 3 Module 7 Lesson 29 Answer Key
• Eureka Math Grade 3 Module 7 Lesson 30 Answer Key

Eureka Math Grade 3 Module 7 End of Module Assessment Answer Key

Engage NY Grade 3 Module 7 Topic F Year in Review

• Eureka Math Grade 3 Module 7 Lesson 31 Answer Key
• Eureka Math Grade 3 Module 7 Lesson 32 Answer Key
• Eureka Math Grade 3 Module 7 Lesson 33 Answer Key
• Eureka Math Grade 3 Module 7 Lesson 34 Answer Key

Wrapping Up

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Learn, Practice, Succeed

Learn, Practice, and Succeed from   Eureka Math™   offer teachers multiple ways to differentiate instruction, provide extra practice, and assess student learning. These versatile companions to   A Story of Units®   (Grades K–5) guide teachers in response to intervention (RTI), provide extra practice, and inform instruction.

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The Learn book serves as a student’s in-class companion where they show their thinking, share what they know, and watch their knowledge build every day!

Application Problems:  Problem solving in a real-world context is a daily part of   Eureka Math , building student confidence and perseverance as students apply their knowledge in new and varied ways.

Problem Sets :  A carefully sequenced Problem Set provides an in-class opportunity for independent work, with multiple entry points for differentiation.

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Eureka Math   fluency activities provide differentiated practice through a variety of formats—some are conducted orally, some use manipulatives, others use a personal whiteboard, or a handout and paper-and-pencil format.

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Answer Key 3.7

• [latex]2x+5=25[/latex]
• [latex]4x+12=36[/latex]
• [latex]3x-8=22[/latex]
• [latex]6x-8=22[/latex]
• [latex]x-8=\dfrac{x}{2}[/latex]
• [latex]x-4=\dfrac{x}{2}[/latex]
• [latex]x+x+1+x+2=21[/latex]
• [latex]x+x+2-(x+4)=5[/latex]
• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrrrrr} 5&+&3x&=&17& \\ -5&&&&-5& \\ \hline &&3x&=&12& \\ \\ &&x&=&\dfrac{12}{3}&\text{or } 4 \end{array}[/latex]
• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrrrrr} 3x&-&5&=&10& \\ &+&5&&+5& \\ \hline &&3x&=&15& \\ \\ &&x&=&\dfrac{15}{3}&\text{or } 5 \end{array}[/latex]
• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrrrrrr} 60&+&9x&=&10x&-&2 \\ +2&-&9x&&-9x&+&2 \\ \hline &&62&=&x&& \end{array}[/latex]
• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrrrrrr} 7x&-&11&=&6x&+&5 \\ -6x&+&11&&-6x&+&11 \\ \hline &&x&=&16&& \end{array}[/latex]
• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrrrrrrrrrl} x&+&x&+&1&+&x&+&2&=&108 \\ &&&&&&3x&+&3&=&108 \\ &&&&&&&-&3&&-3 \\ \hline &&&&&&&&3x&=&105 \\ \\ &&&&&&&&x&=&\dfrac{105}{3}\text{ or }35 \\ \\ &&&&&&&&&\therefore &35, 36, 37 \end{array}[/latex]
• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrrrrrrrrrl} x&+&x&+&1&+&x&+&2&=&-126 \\ &&&&&&3x&+&3&=&-126 \\ &&&&&&&-&3&&-\phantom{00}3 \\ \hline &&&&&&&&3x&=&-129 \\ \\ &&&&&&&&x&=&\dfrac{-129}{3}\text{ or }-43 \\ \\ &&&&&&&&&\therefore &-43, -42, -41 \end{array}[/latex]
• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrrrrrrrrrl} x&+&2(x&+&1)&+&3(x&+&2)&=&-76 \\ x&+&2x&+&2&+&3x&+&6&=&-76 \\ &&&&&&6x&+&8&=&-76 \\ &&&&&&&-&8&&-\phantom{0}8 \\ \hline &&&&&&&&6x&=&-84 \\ \\ &&&&&&&&x&=&\dfrac{-84}{6}\text{ or }-14 \\ \\ &&&&&&&&&\therefore &-14, -13, -12 \end{array}[/latex]
• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrrrrrrrrrl} x&+&2(x&+&2)&+&3(x&+&4)&=&\phantom{0}70 \\ x&+&2x&+&4&+&3x&+&12&=&\phantom{0}70 \\ &&&&&&6x&+&16&=&\phantom{0}70 \\ &&&&&&&-&16&&-16 \\ \hline &&&&&&&&6x&=&54 \\ \\ &&&&&&&&x&=&\dfrac{54}{6}\text{ or }9 \\ \\ &&&&&&&&&\therefore &9,11,13 \end{array}[/latex]

Intermediate Algebra Copyright © 2020 by Terrance Berg is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.

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7.1 Solving Trigonometric Equations with Identities

sin 2 θ − 1 tan θ sin θ − tan θ = ( sin θ + 1 ) ( sin θ − 1 ) tan θ ( sin θ − 1 ) = sin θ + 1 tan θ sin 2 θ − 1 tan θ sin θ − tan θ = ( sin θ + 1 ) ( sin θ − 1 ) tan θ ( sin θ − 1 ) = sin θ + 1 tan θ

This is a difference of squares formula: 25 − 9 sin 2 θ = ( 5 − 3 sin θ ) ( 5 + 3 sin θ ) . 25 − 9 sin 2 θ = ( 5 − 3 sin θ ) ( 5 + 3 sin θ ) .

7.2 Sum and Difference Identities

2 + 6 4 2 + 6 4

2 − 6 4 2 − 6 4

1 − 3 1 + 3 1 − 3 1 + 3

cos ( 5 π 14 ) cos ( 5 π 14 )

7.3 Double-Angle, Half-Angle, and Reduction Formulas

cos ( 2 α ) = 7 32 cos ( 2 α ) = 7 32

cos 4 θ − sin 4 θ = ( cos 2 θ + sin 2 θ ) ( cos 2 θ − sin 2 θ ) = cos ( 2 θ ) cos 4 θ − sin 4 θ = ( cos 2 θ + sin 2 θ ) ( cos 2 θ − sin 2 θ ) = cos ( 2 θ )

cos ( 2 θ ) cos θ = ( cos 2 θ − sin 2 θ ) cos θ = cos 3 θ − cos θ sin 2 θ cos ( 2 θ ) cos θ = ( cos 2 θ − sin 2 θ ) cos θ = cos 3 θ − cos θ sin 2 θ

10 cos 4 x = 10 cos 4 x = 10 ( cos 2 x ) 2              = 10 [ 1 + cos ( 2 x ) 2 ] 2 Substitute reduction formula for cos 2 x .              = 10 4 [ 1 + 2 cos ( 2 x ) + cos 2 ( 2 x ) ]              = 10 4 + 10 2 cos ( 2 x ) + 10 4 ( 1 + cos 2 ( 2 x ) 2 ) Substitute reduction formula for cos 2 x .              = 10 4 + 10 2 cos ( 2 x ) + 10 8 + 10 8 cos ( 4 x )              = 30 8 + 5 cos ( 2 x ) + 10 8 cos ( 4 x )              = 15 4 + 5 cos ( 2 x ) + 5 4 cos ( 4 x ) 10 cos 4 x = 10 cos 4 x = 10 ( cos 2 x ) 2              = 10 [ 1 + cos ( 2 x ) 2 ] 2 Substitute reduction formula for cos 2 x .              = 10 4 [ 1 + 2 cos ( 2 x ) + cos 2 ( 2 x ) ]              = 10 4 + 10 2 cos ( 2 x ) + 10 4 ( 1 + cos 2 ( 2 x ) 2 ) Substitute reduction formula for cos 2 x .              = 10 4 + 10 2 cos ( 2 x ) + 10 8 + 10 8 cos ( 4 x )              = 30 8 + 5 cos ( 2 x ) + 10 8 cos ( 4 x )              = 15 4 + 5 cos ( 2 x ) + 5 4 cos ( 4 x )

− 2 5 − 2 5

7.4 Sum-to-Product and Product-to-Sum Formulas

1 2 ( cos 6 θ + cos 2 θ ) 1 2 ( cos 6 θ + cos 2 θ )

1 2 ( sin 2 x + sin 2 y ) 1 2 ( sin 2 x + sin 2 y )

− 2 − 3 4 − 2 − 3 4

2 sin ( 2 θ ) cos ( θ ) 2 sin ( 2 θ ) cos ( θ )

tan θ cot θ − cos 2 θ = ( sin θ cos θ ) ( cos θ sin θ ) − cos 2 θ = 1 − cos 2 θ = sin 2 θ tan θ cot θ − cos 2 θ = ( sin θ cos θ ) ( cos θ sin θ ) − cos 2 θ = 1 − cos 2 θ = sin 2 θ

7.5 Solving Trigonometric Equations

x = 7 π 6 , 11 π 6 x = 7 π 6 , 11 π 6

π 3 ± π k π 3 ± π k

θ ≈ 1.7722 ± 2 π k θ ≈ 1.7722 ± 2 π k and θ ≈ 4.5110 ± 2 π k θ ≈ 4.5110 ± 2 π k

cos θ = − 1 , θ = π cos θ = − 1 , θ = π

π 2 , 2 π 3 , 4 π 3 , 3 π 2 π 2 , 2 π 3 , 4 π 3 , 3 π 2

7.6 Modeling with Trigonometric Functions

The amplitude is 3 , 3 , and the period is 2 3 . 2 3 .

y = 8 sin ( π 12 t ) + 32 y = 8 sin ( π 12 t ) + 32 The temperature reaches freezing at noon and at midnight.

initial displacement =6, damping constant = -6, frequency = 2 π 2 π

y = 10 e − 0.5 t cos ( π t ) y = 10 e − 0.5 t cos ( π t )

y = 5 cos ( 6 π t ) y = 5 cos ( 6 π t )

7.1 Section Exercises

All three functions, F F , G G , and H , H , are even.

This is because F ( − x ) = sin ( − x ) sin ( − x ) = ( − sin x ) ( − sin x ) = sin 2 x = F ( x ) , G ( − x ) = cos ( − x ) cos ( − x ) = cos x cos x = cos 2 x = G ( x ) F ( − x ) = sin ( − x ) sin ( − x ) = ( − sin x ) ( − sin x ) = sin 2 x = F ( x ) , G ( − x ) = cos ( − x ) cos ( − x ) = cos x cos x = cos 2 x = G ( x ) and H ( − x ) = tan ( − x ) tan ( − x ) = ( − tan x ) ( − tan x ) = tan 2 x = H ( x ) . H ( − x ) = tan ( − x ) tan ( − x ) = ( − tan x ) ( − tan x ) = tan 2 x = H ( x ) .

When cos t = 0 , cos t = 0 , then sec t = 1 0 , sec t = 1 0 , which is undefined.

sin x sin x

sec x sec x

csc t csc t

sec 2 x sec 2 x

sin 2 x + 1 sin 2 x + 1

1 sin x 1 sin x

1 cot x 1 cot x

tan x tan x

− 4 sec x tan x − 4 sec x tan x

± 1 cot 2 x + 1 ± 1 cot 2 x + 1

± 1 − sin 2 x sin x ± 1 − sin 2 x sin x

Answers will vary. Sample proof:

cos x − cos 3 x = cos x ( 1 − cos 2 x ) cos x − cos 3 x = cos x ( 1 − cos 2 x ) = cos x sin 2 x = cos x sin 2 x

Answers will vary. Sample proof: 1 + sin 2 x cos 2 x = 1 cos 2 x + sin 2 x cos 2 x = sec 2 x + tan 2 x = tan 2 x + 1 + tan 2 x = 1 + 2 tan 2 x 1 + sin 2 x cos 2 x = 1 cos 2 x + sin 2 x cos 2 x = sec 2 x + tan 2 x = tan 2 x + 1 + tan 2 x = 1 + 2 tan 2 x

Answers will vary. Sample proof: cos 2 x − tan 2 x = 1 − sin 2 x − ( sec 2 x − 1 ) = 1 − sin 2 x − sec 2 x + 1 = 2 − sin 2 x − sec 2 x cos 2 x − tan 2 x = 1 − sin 2 x − ( sec 2 x − 1 ) = 1 − sin 2 x − sec 2 x + 1 = 2 − sin 2 x − sec 2 x

Proved with negative and Pythagorean Identities

True 3 sin 2 θ + 4 cos 2 θ = 3 sin 2 θ + 3 cos 2 θ + cos 2 θ = 3 ( sin 2 θ + cos 2 θ ) + cos 2 θ = 3 + cos 2 θ 3 sin 2 θ + 4 cos 2 θ = 3 sin 2 θ + 3 cos 2 θ + cos 2 θ = 3 ( sin 2 θ + cos 2 θ ) + cos 2 θ = 3 + cos 2 θ

7.2 Section Exercises

The cofunction identities apply to complementary angles. Viewing the two acute angles of a right triangle, if one of those angles measures x , x , the second angle measures π 2 − x . π 2 − x . Then sin x = cos ( π 2 − x ) . sin x = cos ( π 2 − x ) . The same holds for the other cofunction identities. The key is that the angles are complementary.

sin ( − x ) = − sin x , sin ( − x ) = − sin x , so sin x sin x is odd. cos ( − x ) = cos ( 0 − x ) = cos x , cos ( − x ) = cos ( 0 − x ) = cos x , so cos x cos x is even.

6 − 2 4 6 − 2 4

− 2 − 3 − 2 − 3

− 2 2 sin x − 2 2 cos x − 2 2 sin x − 2 2 cos x

− 1 2 cos x − 3 2 sin x − 1 2 cos x − 3 2 sin x

csc θ csc θ

cot x cot x

tan ( x 10 ) tan ( x 10 )

sin ( a − b ) = ( 4 5 ) ( 1 3 ) − ( 3 5 ) ( 2 2 3 ) = 4 − 6 2 15 sin ( a − b ) = ( 4 5 ) ( 1 3 ) − ( 3 5 ) ( 2 2 3 ) = 4 − 6 2 15 cos ( a + b ) = ( 3 5 ) ( 1 3 ) − ( 4 5 ) ( 2 2 3 ) = 3 − 8 2 15 cos ( a + b ) = ( 3 5 ) ( 1 3 ) − ( 4 5 ) ( 2 2 3 ) = 3 − 8 2 15

cot ( π 6 − x ) cot ( π 6 − x )

cot ( π 4 + x ) cot ( π 4 + x )

sin x 2 + cos x 2 sin x 2 + cos x 2

They are the same.

They are the different, try g ( x ) = sin ( 9 x ) − cos ( 3 x ) sin ( 6 x ) . g ( x ) = sin ( 9 x ) − cos ( 3 x ) sin ( 6 x ) .

They are the different, try g ( θ ) = 2 tan θ 1 − tan 2 θ . g ( θ ) = 2 tan θ 1 − tan 2 θ .

They are different, try g ( x ) = tan x − tan ( 2 x ) 1 + tan x tan ( 2 x ) . g ( x ) = tan x − tan ( 2 x ) 1 + tan x tan ( 2 x ) .

− 3 − 1 2 2 ,  or  − 0.2588 − 3 − 1 2 2 ,  or  − 0.2588

1 + 3 2 2 , 1 + 3 2 2 , or 0.9659

tan ( x + π 4 ) = tan x + tan ( π 4 ) 1 − tan x tan ( π 4 ) = tan x + 1 1 − tan x ( 1 ) = tan x + 1 1 − tan x tan ( x + π 4 ) = tan x + tan ( π 4 ) 1 − tan x tan ( π 4 ) = tan x + 1 1 − tan x ( 1 ) = tan x + 1 1 − tan x

cos ( a + b ) cos a cos b = cos a cos b cos a cos b − sin a sin b cos a cos b = 1 − tan a tan b cos ( a + b ) cos a cos b = cos a cos b cos a cos b − sin a sin b cos a cos b = 1 − tan a tan b

cos ( x + h ) − cos x h = cos x cosh − sin x sinh − cos x h = cos x ( cosh − 1 ) − sin x sinh h = cos x cos h − 1 h − sin x sin h h cos ( x + h ) − cos x h = cos x cosh − sin x sinh − cos x h = cos x ( cosh − 1 ) − sin x sinh h = cos x cos h − 1 h − sin x sin h h

True. Note that sin ( α + β ) = sin ( π − γ ) sin ( α + β ) = sin ( π − γ ) and expand the right hand side.

7.3 Section Exercises

Use the Pythagorean identities and isolate the squared term.

1 − cos x sin x , sin x 1 + cos x , 1 − cos x sin x , sin x 1 + cos x , multiplying the top and bottom by 1 − cos x 1 − cos x and 1 + cos x , 1 + cos x , respectively.

a) 3 7 32 3 7 32 b) 31 32 31 32 c) 3 7 31 3 7 31

a) 3 2 3 2 b) − 1 2 − 1 2 c) − 3 − 3

cos θ = − 2 5 5 , sin θ = 5 5 , tan θ = − 1 2 , csc θ = 5 , sec θ = − 5 2 , cot θ = − 2 cos θ = − 2 5 5 , sin θ = 5 5 , tan θ = − 1 2 , csc θ = 5 , sec θ = − 5 2 , cot θ = − 2

2 sin ( π 2 ) 2 sin ( π 2 )

2 − 2 2 2 − 2 2

2 − 3 2 2 − 3 2

2 + 3 2 + 3

− 1 − 2 − 1 − 2

a) 3 13 13 3 13 13 b) − 2 13 13 − 2 13 13 c) − 3 2 − 3 2

a) 10 4 10 4 b) 6 4 6 4 c) 15 3 15 3

120 169 , – 119 169 , – 120 119 120 169 , – 119 169 , – 120 119

2 13 13 , 3 13 13 , 2 3 2 13 13 , 3 13 13 , 2 3

cos ( 74 ∘ ) cos ( 74 ∘ )

cos ( 18 x ) cos ( 18 x )

3 sin ( 10 x ) 3 sin ( 10 x )

− 2 sin ( − x ) cos ( − x ) = − 2 ( − sin ( x ) cos ( x ) ) = sin ( 2 x ) − 2 sin ( − x ) cos ( − x ) = − 2 ( − sin ( x ) cos ( x ) ) = sin ( 2 x )

sin ( 2 θ ) 1 + cos ( 2 θ ) tan 2 θ = 2 sin ( θ ) cos ( θ ) 1 + cos 2 θ − sin 2 θ tan 2 θ = 2 sin ( θ ) cos ( θ ) 2 cos 2 θ tan 2 θ = sin ( θ ) cos θ tan 2 θ = tan θ tan 2 θ = tan 3 θ sin ( 2 θ ) 1 + cos ( 2 θ ) tan 2 θ = 2 sin ( θ ) cos ( θ ) 1 + cos 2 θ − sin 2 θ tan 2 θ = 2 sin ( θ ) cos ( θ ) 2 cos 2 θ tan 2 θ = sin ( θ ) cos θ tan 2 θ = tan θ tan 2 θ = tan 3 θ

1 + cos ( 12 x ) 2 1 + cos ( 12 x ) 2

3 + cos ( 12 x ) − 4 cos ( 6 x ) 8 3 + cos ( 12 x ) − 4 cos ( 6 x ) 8

2 + cos ( 2 x ) − 2 cos ( 4 x ) − cos ( 6 x ) 32 2 + cos ( 2 x ) − 2 cos ( 4 x ) − cos ( 6 x ) 32

3 + cos ( 4 x ) − 4 cos ( 2 x ) 3 + cos ( 4 x ) + 4 cos ( 2 x ) 3 + cos ( 4 x ) − 4 cos ( 2 x ) 3 + cos ( 4 x ) + 4 cos ( 2 x )

1 − cos ( 4 x ) 8 1 − cos ( 4 x ) 8

3 + cos ( 4 x ) − 4 cos ( 2 x ) 4 ( cos ( 2 x ) + 1 ) 3 + cos ( 4 x ) − 4 cos ( 2 x ) 4 ( cos ( 2 x ) + 1 )

( 1 + cos ( 4 x ) ) sin x 2 ( 1 + cos ( 4 x ) ) sin x 2

4 sin x cos x ( cos 2 x − sin 2 x ) 4 sin x cos x ( cos 2 x − sin 2 x )

2 tan x 1 + tan 2 x = 2 sin x cos x 1 + sin 2 x cos 2 x = 2 sin x cos x cos 2 x + sin 2 x cos 2 x = 2 tan x 1 + tan 2 x = 2 sin x cos x 1 + sin 2 x cos 2 x = 2 sin x cos x cos 2 x + sin 2 x cos 2 x = 2 sin x cos x . cos 2 x 1 = 2 sin x cos x = sin ( 2 x ) 2 sin x cos x . cos 2 x 1 = 2 sin x cos x = sin ( 2 x )

2 sin x cos x 2 cos 2 x − 1 = sin ( 2 x ) cos ( 2 x ) = tan ( 2 x ) 2 sin x cos x 2 cos 2 x − 1 = sin ( 2 x ) cos ( 2 x ) = tan ( 2 x )

sin ( x + 2 x ) = sin x cos ( 2 x ) + sin ( 2 x ) cos x = sin x ( cos 2 x − sin 2 x ) + 2 sin x cos x cos x = sin x cos 2 x − sin 3 x + 2 sin x cos 2 x = 3 sin x cos 2 x − sin 3 x sin ( x + 2 x ) = sin x cos ( 2 x ) + sin ( 2 x ) cos x = sin x ( cos 2 x − sin 2 x ) + 2 sin x cos x cos x = sin x cos 2 x − sin 3 x + 2 sin x cos 2 x = 3 sin x cos 2 x − sin 3 x

1 + cos ( 2 t ) sin ( 2 t ) − cos t = 1 + 2 cos 2 t − 1 2 sin t cos t − cos t = 2 cos 2 t cos t ( 2 sin t − 1 ) = 2 cos t 2 sin t − 1 1 + cos ( 2 t ) sin ( 2 t ) − cos t = 1 + 2 cos 2 t − 1 2 sin t cos t − cos t = 2 cos 2 t cos t ( 2 sin t − 1 ) = 2 cos t 2 sin t − 1

( cos 2 ( 4 x ) − sin 2 ( 4 x ) − sin ( 8 x ) ) ( cos 2 ( 4 x ) − sin 2 ( 4 x ) + sin ( 8 x ) ) = = ( cos ( 8 x ) − sin ( 8 x ) ) ( cos ( 8 x ) + sin ( 8 x ) ) = cos 2 ( 8 x ) − sin 2 ( 8 x ) = cos ( 16 x ) ( cos 2 ( 4 x ) − sin 2 ( 4 x ) − sin ( 8 x ) ) ( cos 2 ( 4 x ) − sin 2 ( 4 x ) + sin ( 8 x ) ) = = ( cos ( 8 x ) − sin ( 8 x ) ) ( cos ( 8 x ) + sin ( 8 x ) ) = cos 2 ( 8 x ) − sin 2 ( 8 x ) = cos ( 16 x )

7.4 Section Exercises

Substitute α α into cosine and β β into sine and evaluate.

Answers will vary. There are some equations that involve a sum of two trig expressions where when converted to a product are easier to solve. For example: sin ( 3 x ) + sin x cos x = 1. sin ( 3 x ) + sin x cos x = 1. When converting the numerator to a product the equation becomes: 2 sin ( 2 x ) cos x cos x = 1 2 sin ( 2 x ) cos x cos x = 1

8 ( cos ( 5 x ) − cos ( 27 x ) ) 8 ( cos ( 5 x ) − cos ( 27 x ) )

sin ( 2 x ) + sin ( 8 x ) sin ( 2 x ) + sin ( 8 x )

1 2 ( cos ( 6 x ) − cos ( 4 x ) ) 1 2 ( cos ( 6 x ) − cos ( 4 x ) )

2 cos ( 5 t ) cos t 2 cos ( 5 t ) cos t

2 cos ( 7 x ) 2 cos ( 7 x )

2 cos ( 6 x ) cos ( 3 x ) 2 cos ( 6 x ) cos ( 3 x )

1 4 ( 1 + 3 ) 1 4 ( 1 + 3 )

1 4 ( 3 − 2 ) 1 4 ( 3 − 2 )

1 4 ( 3 − 1 ) 1 4 ( 3 − 1 )

cos ( 80° ) − cos ( 120° ) cos ( 80° ) − cos ( 120° )

1 2 ( sin ( 221° ) + sin ( 205° ) ) 1 2 ( sin ( 221° ) + sin ( 205° ) )

2 cos ( 31° ) 2 cos ( 31° )

2 cos ( 66.5 ° ) sin ( 34.5 ° ) 2 cos ( 66.5 ° ) sin ( 34.5 ° )

2 sin ( −1.5° ) cos ( 0.5° ) 2 sin ( −1.5° ) cos ( 0.5° )

2 sin ( 7 x ) − 2 sin x = 2 sin ( 4 x + 3 x ) − 2 sin ( 4 x − 3 x ) = 2 ( sin ( 4 x ) cos ( 3 x ) + sin ( 3 x ) cos ( 4 x ) ) − 2 ( sin ( 4 x ) cos ( 3 x ) − sin ( 3 x ) cos ( 4 x ) ) = 2 sin ( 4 x ) cos ( 3 x ) + 2 sin ( 3 x ) cos ( 4 x ) ) − 2 sin ( 4 x ) cos ( 3 x ) + 2 sin ( 3 x ) cos ( 4 x ) ) = 4 sin ( 3 x ) cos ( 4 x ) 2 sin ( 7 x ) − 2 sin x = 2 sin ( 4 x + 3 x ) − 2 sin ( 4 x − 3 x ) = 2 ( sin ( 4 x ) cos ( 3 x ) + sin ( 3 x ) cos ( 4 x ) ) − 2 ( sin ( 4 x ) cos ( 3 x ) − sin ( 3 x ) cos ( 4 x ) ) = 2 sin ( 4 x ) cos ( 3 x ) + 2 sin ( 3 x ) cos ( 4 x ) ) − 2 sin ( 4 x ) cos ( 3 x ) + 2 sin ( 3 x ) cos ( 4 x ) ) = 4 sin ( 3 x ) cos ( 4 x )

sin x + sin ( 3 x ) = 2 sin ( 4 x 2 ) cos ( − 2 x 2 ) = sin x + sin ( 3 x ) = 2 sin ( 4 x 2 ) cos ( − 2 x 2 ) = 2 sin ( 2 x ) cos x = 2 ( 2 sin x cos x ) cos x = 2 sin ( 2 x ) cos x = 2 ( 2 sin x cos x ) cos x = 4 sin x cos 2 x 4 sin x cos 2 x

2 tan x cos ( 3 x ) = 2 sin x cos ( 3 x ) cos x = 2 ( .5 ( sin ( 4 x ) − sin ( 2 x ) ) ) cos x 2 tan x cos ( 3 x ) = 2 sin x cos ( 3 x ) cos x = 2 ( .5 ( sin ( 4 x ) − sin ( 2 x ) ) ) cos x = 1 cos x ( sin ( 4 x ) − sin ( 2 x ) ) = sec x ( sin ( 4 x ) − sin ( 2 x ) ) = 1 cos x ( sin ( 4 x ) − sin ( 2 x ) ) = sec x ( sin ( 4 x ) − sin ( 2 x ) )

2 cos ( 35 ∘ ) cos ( 23 ∘ ) ,  1 .5081 2 cos ( 35 ∘ ) cos ( 23 ∘ ) ,  1 .5081

− 2 sin ( 33 ∘ ) sin ( 11 ∘ ) , − 0.2078 − 2 sin ( 33 ∘ ) sin ( 11 ∘ ) , − 0.2078

1 2 ( cos ( 99 ∘ ) − cos ( 71 ∘ ) ) , − 0.2410 1 2 ( cos ( 99 ∘ ) − cos ( 71 ∘ ) ) , − 0.2410

It is and identity.

It is not an identity, but 2 cos 3 x 2 cos 3 x is.

tan ( 3 t ) tan ( 3 t )

2 cos ( 2 x ) 2 cos ( 2 x )

− sin ( 14 x ) − sin ( 14 x )

Start with cos x + cos y . cos x + cos y . Make a substitution and let x = α + β x = α + β and let y = α − β , y = α − β , so cos x + cos y cos x + cos y becomes cos ( α + β ) + cos ( α − β ) = cos α cos β − sin α sin β + cos α cos β + sin α sin β = 2 cos α cos β cos ( α + β ) + cos ( α − β ) = cos α cos β − sin α sin β + cos α cos β + sin α sin β = 2 cos α cos β

Since x = α + β x = α + β and y = α − β , y = α − β , we can solve for α α and β β in terms of x and y and substitute in for 2 cos α cos β 2 cos α cos β and get 2 cos ( x + y 2 ) cos ( x − y 2 ) . 2 cos ( x + y 2 ) cos ( x − y 2 ) .

cos ( 3 x ) + cos x cos ( 3 x ) − cos x = 2 cos ( 2 x ) cos x − 2 sin ( 2 x ) sin x = − cot ( 2 x ) cot x cos ( 3 x ) + cos x cos ( 3 x ) − cos x = 2 cos ( 2 x ) cos x − 2 sin ( 2 x ) sin x = − cot ( 2 x ) cot x

cos ( 2 y ) − cos ( 4 y ) sin ( 2 y ) + sin ( 4 y ) = − 2 sin ( 3 y ) sin ( − y ) 2 sin ( 3 y ) cos y = 2 sin ( 3 y ) sin ( y ) 2 sin ( 3 y ) cos y = tan y cos ( 2 y ) − cos ( 4 y ) sin ( 2 y ) + sin ( 4 y ) = − 2 sin ( 3 y ) sin ( − y ) 2 sin ( 3 y ) cos y = 2 sin ( 3 y ) sin ( y ) 2 sin ( 3 y ) cos y = tan y

cos x − cos ( 3 x ) = − 2 sin ( 2 x ) sin ( − x ) = 2 ( 2 sin x cos x ) sin x = 4 sin 2 x cos x cos x − cos ( 3 x ) = − 2 sin ( 2 x ) sin ( − x ) = 2 ( 2 sin x cos x ) sin x = 4 sin 2 x cos x

tan ( π 4 − t ) = tan ( π 4 ) − tan t 1 + tan ( π 4 ) tan ( t ) = 1 − tan t 1 + tan t tan ( π 4 − t ) = tan ( π 4 ) − tan t 1 + tan ( π 4 ) tan ( t ) = 1 − tan t 1 + tan t

7.5 Section Exercises

There will not always be solutions to trigonometric function equations. For a basic example, cos ( x ) = −5. cos ( x ) = −5.

If the sine or cosine function has a coefficient of one, isolate the term on one side of the equals sign. If the number it is set equal to has an absolute value less than or equal to one, the equation has solutions, otherwise it does not. If the sine or cosine does not have a coefficient equal to one, still isolate the term but then divide both sides of the equation by the leading coefficient. Then, if the number it is set equal to has an absolute value greater than one, the equation has no solution.

π 3 , 2 π 3 π 3 , 2 π 3

3 π 4 , 5 π 4 3 π 4 , 5 π 4

π 4 , 5 π 4 π 4 , 5 π 4

π 4 , 3 π 4 , 5 π 4 , 7 π 4 π 4 , 3 π 4 , 5 π 4 , 7 π 4

π 4 , 7 π 4 π 4 , 7 π 4

7 π 6 , 11 π 6 7 π 6 , 11 π 6

π 18 π 18 , 5 π 18 5 π 18 , 13 π 18 13 π 18 , 17 π 18 17 π 18 , 25 π 18 25 π 18 , 29 π 18 29 π 18

3 π 12 , 5 π 12 , 11 π 12 , 13 π 12 , 19 π 12 , 21 π 12 3 π 12 , 5 π 12 , 11 π 12 , 13 π 12 , 19 π 12 , 21 π 12

1 6 , 5 6 , 13 6 , 17 6 , 25 6 , 29 6 , 37 6 1 6 , 5 6 , 13 6 , 17 6 , 25 6 , 29 6 , 37 6

0 , π 3 , π , 5 π 3 0 , π 3 , π , 5 π 3

π 3 , π , 5 π 3 π 3 , π , 5 π 3

π 3 , 3 π 2 , 5 π 3 π 3 , 3 π 2 , 5 π 3

0 , π 0 , π

π − sin − 1 ( − 1 4 ) , 7 π 6 , 11 π 6 , 2 π + sin − 1 ( − 1 4 ) π − sin − 1 ( − 1 4 ) , 7 π 6 , 11 π 6 , 2 π + sin − 1 ( − 1 4 )

1 3 ( sin − 1 ( 9 10 ) ) , π 3 − 1 3 ( sin − 1 ( 9 10 ) ) , 2 π 3 + 1 3 ( sin − 1 ( 9 10 ) ) , π − 1 3 ( sin − 1 ( 9 10 ) ) , 4 π 3 + 1 3 ( sin − 1 ( 9 10 ) ) , 5 π 3 − 1 3 ( sin − 1 ( 9 10 ) ) 1 3 ( sin − 1 ( 9 10 ) ) , π 3 − 1 3 ( sin − 1 ( 9 10 ) ) , 2 π 3 + 1 3 ( sin − 1 ( 9 10 ) ) , π − 1 3 ( sin − 1 ( 9 10 ) ) , 4 π 3 + 1 3 ( sin − 1 ( 9 10 ) ) , 5 π 3 − 1 3 ( sin − 1 ( 9 10 ) )

θ = sin - 1 2 3 , π - sin - 1 2 3 , π + sin - 1 2 3 , 2 π - sin - 1 2 3 θ = sin - 1 2 3 , π - sin - 1 2 3 , π + sin - 1 2 3 , 2 π - sin - 1 2 3

3 π 2 , π 6 , 5 π 6 3 π 2 , π 6 , 5 π 6

0 , π 3 , π , 4 π 3 0 , π 3 , π , 4 π 3

There are no solutions.

cos − 1 ( 1 3 ( 1 − 7 ) ) , 2 π − cos − 1 ( 1 3 ( 1 − 7 ) ) cos − 1 ( 1 3 ( 1 − 7 ) ) , 2 π − cos − 1 ( 1 3 ( 1 − 7 ) )

tan − 1 ( 1 2 ( 29 − 5 ) ) , π + tan − 1 ( 1 2 ( − 29 − 5 ) ) , π + tan − 1 ( 1 2 ( 29 − 5 ) ) , 2 π + tan − 1 ( 1 2 ( − 29 − 5 ) ) tan − 1 ( 1 2 ( 29 − 5 ) ) , π + tan − 1 ( 1 2 ( − 29 − 5 ) ) , π + tan − 1 ( 1 2 ( 29 − 5 ) ) , 2 π + tan − 1 ( 1 2 ( − 29 − 5 ) )

0 , 2 π 3 , 4 π 3 0 , 2 π 3 , 4 π 3

sin − 1 ( 3 5 ) , π 2 , π − sin − 1 ( 3 5 ) , 3 π 2 sin − 1 ( 3 5 ) , π 2 , π − sin − 1 ( 3 5 ) , 3 π 2

cos − 1 ( − 1 4 ) cos − 1 ( − 1 4 ) , 2 π − cos − 1 ( − 1 4 ) 2 π − cos − 1 ( − 1 4 )

π 3 , cos − 1 ( − 3 4 ) , 2 π − cos − 1 ( − 3 4 ) , 5 π 3 π 3 , cos − 1 ( − 3 4 ) , 2 π − cos − 1 ( − 3 4 ) , 5 π 3

cos − 1 ( 3 4 ) , cos − 1 ( − 2 3 ) , 2 π − cos − 1 ( − 2 3 ) cos − 1 ( 3 4 ) , cos − 1 ( − 2 3 ) , 2 π − cos − 1 ( − 2 3 ) , 2 π − cos − 1 ( 3 4 ) 2 π − cos − 1 ( 3 4 )

0 , π 2 , π , 3 π 2 0 , π 2 , π , 3 π 2

π 3 , cos −1 ( − 1 4 ) , 2 π − cos −1 ( − 1 4 ) , 5 π 3 π 3 , cos −1 ( − 1 4 ) , 2 π − cos −1 ( − 1 4 ) , 5 π 3

π + tan −1 ( −2 ) π + tan −1 ( −2 ) , π + tan −1 ( − 3 2 ) , 2 π + tan −1 ( −2 ) , 2 π + tan −1 ( − 3 2 ) π + tan −1 ( − 3 2 ) , 2 π + tan −1 ( −2 ) , 2 π + tan −1 ( − 3 2 )

2 π k + 0.2734 , 2 π k + 2.8682 2 π k + 0.2734 , 2 π k + 2.8682

π k − 0.3277 π k − 0.3277

0.6694 , 1.8287 , 3.8110 , 4.9703 0.6694 , 1.8287 , 3.8110 , 4.9703

1.0472 , 3.1416 , 5.2360 1.0472 , 3.1416 , 5.2360

0.5326 , 1.7648 , 3.6742 , 4.9064 0.5326 , 1.7648 , 3.6742 , 4.9064

sin − 1 ( 1 4 ) , π − sin − 1 ( 1 4 ) , 3 π 2 sin − 1 ( 1 4 ) , π − sin − 1 ( 1 4 ) , 3 π 2

π 2 , 3 π 2 π 2 , 3 π 2

7.2 ∘ 7.2 ∘

5.7 ∘ 5.7 ∘

82.4 ∘ 82.4 ∘

31.0 ∘ 31.0 ∘

88.7 ∘ 88.7 ∘

59.0 ∘ 59.0 ∘

36.9 ∘ 36.9 ∘

7.6 Section Exercises

Physical behavior should be periodic, or cyclical.

Since cumulative rainfall is always increasing, a sinusoidal function would not be ideal here.

y = − 3 cos ( π 6 x ) − 1 y = − 3 cos ( π 6 x ) − 1

5 sin ( 2 x ) + 2 5 sin ( 2 x ) + 2

y = 4 - 6 cos ( x π 2 ) y = 4 - 6 cos ( x π 2 )

y = tan ( x π 8 ) y = tan ( x π 8 )

tan ( x π 12 ) tan ( x π 12 )

From June 15 through November 16

From day 31 through day 58

Floods: April 16 to July 15. Drought: October 16 to January 15.

Amplitude: 8, period: 1 3 , 1 3 , frequency: 3 Hz

Amplitude: 4, period: 4 , 4 , frequency: 1 4 1 4 Hz

P ( t ) = − 19 cos ( π 6 t ) + 800 + 160 12 t P ( t ) = − 19 cos ( π 6 t ) + 800 + 40 3 t P ( t ) = − 19 cos ( π 6 t ) + 800 + 160 12 t P ( t ) = − 19 cos ( π 6 t ) + 800 + 40 3 t

P ( t ) = − 33 cos ( π 6 t ) + 900 + ( 1.07 ) t P ( t ) = − 33 cos ( π 6 t ) + 900 + ( 1.07 ) t

D ( t ) = 10 ( 0.85 ) t cos ( 36 π t ) D ( t ) = 10 ( 0.85 ) t cos ( 36 π t )

D ( t ) = 17 ( 0.9145 ) t cos ( 28 π t ) D ( t ) = 17 ( 0.9145 ) t cos ( 28 π t )

15.4 seconds

Spring 2 comes to rest first after 7.3 seconds.

234.3 miles, at 72.2°

y = 6 ( 4 ) x + 5 sin ( π 2 x ) y = 6 ( 4 ) x + 5 sin ( π 2 x )

y = 4 ( – 2 ) x + 8 sin ( π 2 x ) y = 4 ( – 2 ) x + 8 sin ( π 2 x )

y = 3 ( 2 ) x cos ( π 2 x ) + 1 y = 3 ( 2 ) x cos ( π 2 x ) + 1

Review Exercises

sin − 1 ( 3 3 ) , π − sin − 1 ( 3 3 ) , π + sin − 1 ( 3 3 ) , 2 π − sin − 1 ( 3 3 ) sin − 1 ( 3 3 ) , π − sin − 1 ( 3 3 ) , π + sin − 1 ( 3 3 ) , 2 π − sin − 1 ( 3 3 )

sin − 1 ( 1 4 ) , π − sin − 1 ( 1 4 ) sin − 1 ( 1 4 ) , π − sin − 1 ( 1 4 )

cos ( 4 x ) − cos ( 3 x ) cos x = cos ( 2 x + 2 x ) − cos ( x + 2 x ) cos x                                    = cos ( 2 x ) cos ( 2 x ) − sin ( 2 x ) sin ( 2 x ) − cos x cos ( 2 x ) cos x + sin x sin ( 2 x ) cos x                                    = ( cos 2 x − sin 2 x ) 2 − 4 cos 2 x sin 2 x − cos 2 x ( cos 2 x − sin 2 x ) + sin x ( 2 ) sin x cos x cos x                                    = ( cos 2 x − sin 2 x ) 2 − 4 cos 2 x sin 2 x − cos 2 x ( cos 2 x − sin 2 x ) + 2 sin 2 x cos 2 x                                    = cos 4 x − 2 cos 2 x sin 2 x + sin 4 x − 4 cos 2 x sin 2 x − cos 4 x + cos 2 x sin 2 x + 2 sin 2 x cos 2 x                                    = sin 4 x − 4 cos 2 x sin 2 x + cos 2 x sin 2 x                                    = sin 2 x ( sin 2 x + cos 2 x ) − 4 cos 2 x sin 2 x                                    = sin 2 x − 4 cos 2 x sin 2 x cos ( 4 x ) − cos ( 3 x ) cos x = cos ( 2 x + 2 x ) − cos ( x + 2 x ) cos x                                    = cos ( 2 x ) cos ( 2 x ) − sin ( 2 x ) sin ( 2 x ) − cos x cos ( 2 x ) cos x + sin x sin ( 2 x ) cos x                                    = ( cos 2 x − sin 2 x ) 2 − 4 cos 2 x sin 2 x − cos 2 x ( cos 2 x − sin 2 x ) + sin x ( 2 ) sin x cos x cos x                                    = ( cos 2 x − sin 2 x ) 2 − 4 cos 2 x sin 2 x − cos 2 x ( cos 2 x − sin 2 x ) + 2 sin 2 x cos 2 x                                    = cos 4 x − 2 cos 2 x sin 2 x + sin 4 x − 4 cos 2 x sin 2 x − cos 4 x + cos 2 x sin 2 x + 2 sin 2 x cos 2 x                                    = sin 4 x − 4 cos 2 x sin 2 x + cos 2 x sin 2 x                                    = sin 2 x ( sin 2 x + cos 2 x ) − 4 cos 2 x sin 2 x                                    = sin 2 x − 4 cos 2 x sin 2 x

tan ( 5 8 x ) tan ( 5 8 x )

− 24 25 , − 7 25 , 24 7 − 24 25 , − 7 25 , 24 7

2 ( 2 + 2 ) 2 ( 2 + 2 )

2 10 , 7 2 10 , 1 7 , 3 5 , 4 5 , 3 4 2 10 , 7 2 10 , 1 7 , 3 5 , 4 5 , 3 4

cot x cos ( 2 x ) = cot x ( 1 − 2 sin 2 x )                      = cot x − cos x sin x ( 2 ) sin 2 x                      = − 2 sin x cos x + cot x                      = − sin ( 2 x ) + cot x cot x cos ( 2 x ) = cot x ( 1 − 2 sin 2 x )                      = cot x − cos x sin x ( 2 ) sin 2 x                      = − 2 sin x cos x + cot x                      = − sin ( 2 x ) + cot x

10 sin x − 5 sin ( 3 x ) + sin ( 5 x ) 8 ( cos ( 2 x ) + 1 ) 10 sin x − 5 sin ( 3 x ) + sin ( 5 x ) 8 ( cos ( 2 x ) + 1 )

− 2 2 − 2 2

1 2 ( sin ( 6 x ) + sin ( 12 x ) ) 1 2 ( sin ( 6 x ) + sin ( 12 x ) )

2 sin ( 13 2 x ) cos ( 9 2 x ) 2 sin ( 13 2 x ) cos ( 9 2 x )

3 π 4 , 7 π 4 3 π 4 , 7 π 4

0 , π 6 , 5 π 6 , π 0 , π 6 , 5 π 6 , π

3 π 2 3 π 2

No solution

0.2527 , 2.8889 , 4.7124 0.2527 , 2.8889 , 4.7124

1.3694 1.3694 , 1.9106 1.9106 , 4.3726 4.3726 , 4.9137 4.9137

3 sin ( x π 2 ) − 2 3 sin ( x π 2 ) − 2

71.6 ∘ 71.6 ∘

P ( t ) = 950 − 450 sin ( π 6 t ) P ( t ) = 950 − 450 sin ( π 6 t )

Amplitude: 3, period: 2, frequency: 1 2 1 2 Hz

C ( t ) = 20 sin ( 2 π t ) + 100 ( 1.4427 ) t C ( t ) = 20 sin ( 2 π t ) + 100 ( 1.4427 ) t

Practice Test

π 2 , 3π 2 π 2 , 3π 2

2 cos ( 3 x ) cos ( 5 x ) 2 cos ( 3 x ) cos ( 5 x )

x = cos –1 ( 1 5 ) x = cos –1 ( 1 5 )

3 5 , − 4 5 , − 3 4 3 5 , − 4 5 , − 3 4

tan 3 x – tan x sec 2 x = tan x ( tan 2 x – sec 2 x ) = tan x ( tan 2 x – ( 1 + tan 2 x ) ) = tan x ( tan 2 x – 1 – tan 2 x ) = – tan x = tan ( – x ) = tan – x ) tan 3 x – tan x sec 2 x = tan x ( tan 2 x – sec 2 x ) = tan x ( tan 2 x – ( 1 + tan 2 x ) ) = tan x ( tan 2 x – 1 – tan 2 x ) = – tan x = tan ( – x ) = tan – x )

sin ( 2 x ) sin x – cos ( 2 x ) cos x = 2 sin x cos x sin x – 2 cos 2 x – 1 cos x = 2 cos x – 2 cos x + 1 cos x = 1 cos x = sec x = sec x sin ( 2 x ) sin x – cos ( 2 x ) cos x = 2 sin x cos x sin x – 2 cos 2 x – 1 cos x = 2 cos x – 2 cos x + 1 cos x = 1 cos x = sec x = sec x

Amplitude: 1 4 1 4 , period 1 60 1 60 , frequency: 60 Hz

Amplitude: 8 8 , fast period: 1 500 1 500 , fast frequency: 500 Hz, slow period: 1 10 1 10 , slow frequency: 10 Hz

D ( t ) = 20 ( 0.9086 ) t cos ( 4 π t ) D ( t ) = 20 ( 0.9086 ) t cos ( 4 π t ) , 31 seconds

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Access for free at https://openstax.org/books/precalculus/pages/1-introduction-to-functions

• Authors: Jay Abramson
• Publisher/website: OpenStax
• Book title: Precalculus
• Publication date: Oct 23, 2014
• Location: Houston, Texas
• Book URL: https://openstax.org/books/precalculus/pages/1-introduction-to-functions
• Section URL: https://openstax.org/books/precalculus/pages/chapter-7

© Dec 8, 2021 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.