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Texas Go Math Grade 5 Lesson 8.3 Answer Key Multiplication and Division Equations

Refer to our Texas Go Math Grade 5 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 5 Lesson 8.3 Answer Key Multiplication and Division Equations.

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Jake is buying a movie ticket and a box of popcorn for $12. The ticket costs 2 times as much as the popcorn. How much does the popcorn cost? How much does the ticket cost? The cost of the tickets is 2 × p, where p is the price of the popcorn.

• How does knowing that a ticket costs 2 times as much as the popcorn help you choose an operation to write an expression? ________________________ The expression is 2 x p + p = $12

Use a related equation.

The popcorn costs p is $4 . Label the strip for the popcorn. It represents 1 unit. The cost of the ticket is 2 x p times as much. Label the strips for the ticket. The total cost is $12 . 3 units are shown in the model.

To find a missing factor, We have to write a division equation that is related to the multiplication equation. Then divide. 12 ÷ 3 = p 4 = p So, 1 bag of popcorn costs  $4. The cost 1 tickets is 2 times the cost of 1 bag of popcorn. 2 x $4 = $8 So, 1 ticket costs $8.

Use a related equation. Katrina buys a package of trading cards. She divides the cards into 4 equal piles. Each pile contains 12 cards. The equation that describes the number of cards in the package is c ÷ 4 = 12. How many cards were in the package when Katrina bought it?

SOLVE Write a related multiplication equation. Multiplication equation: _____________ c = ____________ So, the package Katrina bought had _________ trading cards in it. Answer: c ÷ 4 = 12 Multiplication equation: c = 12 x 4 c = 48 So, the package Katrina bought had 48 trading cards in it. Explanation: Katrina buys a package of trading cards. She divides the cards into 4 equal piles. Each pile contains 12 cards. The equation that describes the number of cards in the package is c ÷ 4 = 12. The multiplication equation is c = 12 x 4 the product is 48. So the value of c is 48. So, the package Katrina bought had 48 trading cards in it.

Math Talk Mathematical Processes

Explain how the strip diagrams for multiplication and division are related. Answer:

Share and Show

Use the strip diagrams to write an equation. Then solve.

Problem Solving

Practice: Copy and Solve Use a strip diagram or a related equation to solve. Check your solution.

Question 3. 128 = 8 × d Answer: 128 = 8 × d Related equation: 128 ÷  8 = d 16 = d Explanation: The given equation is 128 = 8 x d. We have to find out the d value. So, the related equation for the given equation is 128 ÷  8 = d. Perform division operation on 128 ÷  8 the quotient is 16. The value of d is 16.

Question 4. r ÷ 9 = 17 Answer: r ÷ 9 = 17 Related equation: r = 17 x 9 r = 153 Explanation: The given equation is r ÷ 9 = 17. We have to find out the r value. So, the related equation for the given equation is r = 17 x 9. Perform multiplication operation with 17 and 9 the product is 153. The value of r is 153.

Question 5. 6m = 78 Answer: 6m = 78 Related equation: m = 78 ÷ 6 m = 13 Explanation: The given equation is 6m = 78. We have to find out the m value. So, the related equation for the given equation is 78 ÷  6 = m. Perform division operation on 78 ÷  6 the quotient is 13. The value of m is 13.

Go Math 5th Grade Multiplication Equation Lesson 8.3 Question 6. 7 = b ÷ 17 Answer: 7 = b ÷ 17 Related equation: 7 x 17 = b 119 = b Explanation: The given equation is 7 = b ÷ 17. We have to find out the b value. So, the related equation for the given equation is b = 7 x 17. Perform multiplication operation with 17 and 7 the product is 119. The value of b is 119.

Question 7. How can you justify that j = 348 is the solution to j ÷ 12 = 29? Explain. Answer: The equation is j ÷ 12 = 29 we know that j = 348 Related equation j = 29 x 12 j = 348 The value j = 348 is the solution to j ÷ 12 = 29. Explanation: The given equation is j ÷ 12 = 29. Here j is equal to 348. So, the related equation for the given equation is j = 29 x 12 Perform multiplication operation with 29 and 12 the product is 348. Hence the value j is equal to 348 is the solution to j ÷ 12 = 29.

Question 8. Multi-Step On Friday, the snack bar made $992 selling large buckets of popcorn. How many large buckets of popcorn did the snack bar sell on Friday? In your equation1 let p represent the number of large buckets of popcorn sold. Answer:

Question 10. H.O.T. Multi-Step Asher wants to buy a video game console. In order to save the money needed to buy it in 5 months, he divides the cost by 5. He finds he needs to save $37 a month. What is the total cost, c, of the game console? Answer: Equation: c ÷ 5 = $37 Related equation: c = $37 x 5 c = $185 The total cost of the video game console is $185. Explanation: Asher wants to buy a video game console. In order to save the money needed to buy it in 5 months, he divides the cost by 5. He finds he needs to save $37 a month. The equation is c ÷ 5 = $37. The related equation for the given equation is c = $37 x 5. Perform multiplication operation with 37 and 5 the product is $185. The total cost of the video game console is $185.

Go Math Grade 5 Lesson 8.3 Answer Key Question 11. Representations Mariah bought a bag of buttons. She divides the buttons equally among 15 containers. Each container has 10 buttons. How many buttons does Mariah have? In your equation, let b represent the number of buttons Mariah has. Answer: Equation: b ÷ 15 = 10 Related equation: b = 10 x 15 b = 150 Here b represents the number of buttons does Mariah have. Mariah have 150 buttons. Explanation: Mariah bought a bag of buttons. She divides the buttons equally among 15 containers. Each container has 10 buttons. The equation is b ÷ 15 = 10. The related equation for the given equation is b = 10 x 15. Perform multiplication operation with 10 and 15 the product is 150. Here b represents the number of buttons does Mariah have. Mariah have 150 buttons.

Daily Assessment Task

Fill in the bubble completely to show your answer.

Texas Test Prep

Texas Go Math Grade 5 Lesson 8.3 Homework and Practice Answer Key

Use a strip diagram or a related equation to solve.

Check your solution.

Question 1. c ÷ 5 = 13 Answer: c ÷ 5 = 13 Related equation: c = 13 x 5 c = 65 Explanation: The given equation is c ÷ 5 = 13. We have to find out the c value. So, the related equation for the given equation is c = 13 x 5. Perform multiplication operation with 13 and 5 the product is 65. The value of c is 65.

Question 2. 112 = 7 × b Answer: 112 = 7 × b Related equation: 112 ÷ 7 = b 16 = b Explanation: The given equation is 112 = 7 x b. We have to find out the b value. So, the related equation for the given equation is 112 ÷ 7 = b . Perform division operation 112 by 7 the result is 16. The value of b is 16.

Texas Go Math Practice and Homework Lesson 8.3 Answer Key Question 3. 4p = 68 Answer: 4p = 68 Related equation: p = 68 ÷ 4 p = 17 Explanation: The given equation is 4p = 68. We have to find out the p value. So, the related equation for the given equation is 68 ÷ 4 =p . Perform division operation on 68 by 4 the result is 17. The value of p is 17.

Question 4. 9 = d ÷ 21 Answer: 9 = d ÷ 21 Related equation: 9 x 21 = d 189 = d Explanation: The given equation is 9 = d ÷ 21. We have to find out the d value. So, the related equation for the given equation is d = 9 x 21. Perform multiplication operation with 9 and 21 the product is 189. The value of d is 189.

Question 5. 105 = 3 × a Answer: 105 = 3 x a Related equation: 105 ÷ 3 = a 35 = a Explanation: The given equation is 105 = 3 x a. We have to find out the a value. So, the related equation for the given equation is 105 ÷ 3 =a . Perform division operation on 105 by 3 the result is 35. The value of a is 35.

Question 6. 9g = 99 Answer: 9g = 99 Related equation: g = 99 ÷ 9  g = 11 Explanation: The given equation is 9g = 99. We have to find out the g value. So, the related equation for the given equation is g = 99 ÷ 9. Perform division operation on 99 by 9 the result is 11. The value of g is 11.

Question 7. m ÷ 10 = 16 Answer: m ÷ 10 = 16 Related equation: m = 16 x 10  m = 160 Explanation: The given equation is m ÷ 10 = 16. We have to find out the m value. So, the related equation for the given equation is m = 16 x 10. Perform multiplication operations with 16 and 10 the product is 160. The value of m is 160.

Go Math Lesson 8.3 5th Grade Division Answer Key Question 8. 22 = n ÷ 7 Answer: 22 = n ÷ 7 Related equation: 22 x 7 = n 154 = n Explanation: The given equation is 22 = n ÷ 7. We have to find out the n value. So, the related equation for the given equation is n = 22 x 7. Perform multiplication operation with 22 and 7 the product is 154. The value of n is 154.

Question 9. 92 = f × 4 Answer: 92 = f x 4 Related equation: 92 ÷ 4 = f 23 = f Explanation: The given equation is 92 = f x 4. We have to find out the f value. So, the related equation for the given equation is f = 92 ÷ 4. Perform division operation on 92 by 4 the result is 23. The value of f is 23.

Question 10. h ÷ 9 = 14 Answer: h ÷ 9 = 14 Related equation: h = 14 x 9 h = 126 Explanation: The given equation is h ÷ 9 = 14. We have to find out the h value. So, the related equation for the given equation is h = 14 x 9. Perform multiplication operation with 14 and 9 the product is 126. The value of h is 126.

Question 11. 13 = j ÷ 8 Answer: 13 = j ÷ 8 Related equation: 13 x 8 = j 104 = j Explanation: The given equation is 13 = j ÷ 8. We have to find out the j value. So, the related equation for the given equation is 13 x 8 = j. Perform multiplication operation with 13 and 8 the product is 104. The value of j is 104.

Question 12. 165 = 11r Answer: 165 = 11r Related equation: 165 ÷ 11 = r 15 = r Explanation: The given equation is 165 = 11r. We have to find out the r value. So, the related equation for the given equation is r = 165 ÷ 11 . Perform division operation on 165 by 11 the result is 15. The value of r is 15.

Question 13. Sierra arranges fabric squares to sew together to make a blanket for her baby sister. She divides the fabric squares equally among 12 rows. Each row has 8 squares. How many fabric squares does Sierra have? In your equation, let f represent the number of fabric squares Sierra has. Answer: f ÷ 12 = 8 Related equation: f = 8 x 12 f = 96 Here f represent the number of fabric squares does Sierra have. Sierra have 96 fabric squares. Explanation: Sierra arranges fabric squares to sew together to make a blanket for her baby sister. She divides the fabric squares equally among 12 rows. Each row has 8 squares. The equation is f ÷ 12 = 8. The related equation is f = 8 x 12. Multiply 8 with 12 the product is 96. Here f represent the number of fabric squares. Sierra have 96 fabric squares.

Go Math Grade 5 Lesson 8.3 Division Strip Diagrams Question 14. A school district purchased 90 new computers. The computers were divided equally among 6 classrooms. How many new computers are in each classroom? In your equation, let c represent the number of computers in each classroom. Answer: 90 ÷ 6 = c 15 = c Here c represent the number of computers in each classroom. In each classroom there are 15 computers. Explanation: A school district purchased 90 new computers. The computers were divided equally among 6 classrooms. The equation is 90 ÷ 6 = c. Divide 90 by 6 the result is 15. Here c represent the number of computers in each classroom. In each classroom there are 15 computers.

Question 15. Write a real-world problem for the equation 4 × y = 60. Then solve. Answer:

Lesson Check

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8.3 Add and Subtract Rational Expressions with a Common Denominator

Learning objectives.

By the end of this section, you will be able to:

• Add rational expressions with a common denominator
• Subtract rational expressions with a common denominator
• Add and subtract rational expressions whose denominators are opposites

Be Prepared 8.9

Before you get started, take this readiness quiz.

If you miss a problem, go back to the section listed and review the material.

Add: y 3 + 9 3 . y 3 + 9 3 . If you missed this problem, review Example 1.77 .

Be Prepared 8.10

Subtract: 10 x − 2 x . 10 x − 2 x . If you missed this problem, review Example 1.79 .

Be Prepared 8.11

Factor completely: 8 n 5 − 20 n 3 . 8 n 5 − 20 n 3 . If you missed this problem, review Example 7.59 .

Be Prepared 8.12

Factor completely: 45 a 3 − 5 a b 2 . 45 a 3 − 5 a b 2 . If you missed this problem, review Example 7.62 .

Add Rational Expressions with a Common Denominator

What is the first step you take when you add numerical fractions? You check if they have a common denominator. If they do, you add the numerators and place the sum over the common denominator. If they do not have a common denominator, you find one before you add.

It is the same with rational expressions. To add rational expressions, they must have a common denominator. When the denominators are the same, you add the numerators and place the sum over the common denominator.

Rational Expression Addition

If p , q , and r p , q , and r are polynomials where r ≠ 0 r ≠ 0 , then

To add rational expressions with a common denominator, add the numerators and place the sum over the common denominator.

We will add two numerical fractions first, to remind us of how this is done.

Example 8.30

Add: 5 18 + 7 18 . 5 18 + 7 18 .

Try It 8.59

Add: 7 16 + 5 16 . 7 16 + 5 16 .

Try It 8.60

Add: 3 10 + 1 10 . 3 10 + 1 10 .

Remember, we do not allow values that would make the denominator zero. What value of y y should be excluded in the next example?

Example 8.31

Add: 3 y 4 y − 3 + 7 4 y − 3 . 3 y 4 y − 3 + 7 4 y − 3 .

The numerator and denominator cannot be factored. The fraction is simplified.

Try It 8.61

Add: 5 x 2 x + 3 + 2 2 x + 3 . 5 x 2 x + 3 + 2 2 x + 3 .

Try It 8.62

Add: x x − 2 + 1 x − 2 . x x − 2 + 1 x − 2 .

Example 8.32

Add: 7 x + 12 x + 3 + x 2 x + 3 . 7 x + 12 x + 3 + x 2 x + 3 .

Try It 8.63

Add: 9 x + 14 x + 7 + x 2 x + 7 . 9 x + 14 x + 7 + x 2 x + 7 .

Try It 8.64

Add: x 2 + 8 x x + 5 + 15 x + 5 . x 2 + 8 x x + 5 + 15 x + 5 .

Subtract Rational Expressions with a Common Denominator

To subtract rational expressions, they must also have a common denominator. When the denominators are the same, you subtract the numerators and place the difference over the common denominator.

Rational Expression Subtraction

To subtract rational expressions, subtract the numerators and place the difference over the common denominator.

We always simplify rational expressions. Be sure to factor, if possible, after you subtract the numerators so you can identify any common factors.

Example 8.33

Subtract: n 2 n − 10 − 100 n − 10 . n 2 n − 10 − 100 n − 10 .

Try It 8.65

Subtract: x 2 x + 3 − 9 x + 3 . x 2 x + 3 − 9 x + 3 .

Try It 8.66

Subtract: 4 x 2 2 x − 5 − 25 2 x − 5 . 4 x 2 2 x − 5 − 25 2 x − 5 .

Be careful of the signs when you subtract a binomial!

Example 8.34

Subtract: y 2 y − 6 − 2 y + 24 y − 6 . y 2 y − 6 − 2 y + 24 y − 6 .

Try It 8.67

Subtract: n 2 n − 4 − n + 12 n − 4 . n 2 n − 4 − n + 12 n − 4 .

Try It 8.68

Subtract: y 2 y − 1 − 9 y − 8 y − 1 . y 2 y − 1 − 9 y − 8 y − 1 .

Example 8.35

Subtract: 5 x 2 − 7 x + 3 x 2 − 3 x – 18 − 4 x 2 + x − 9 x 2 − 3 x – 18 . 5 x 2 − 7 x + 3 x 2 − 3 x – 18 − 4 x 2 + x − 9 x 2 − 3 x – 18 .

Try It 8.69

Subtract: 4 x 2 − 11 x + 8 x 2 − 3 x + 2 − 3 x 2 + x − 3 x 2 − 3 x + 2 . 4 x 2 − 11 x + 8 x 2 − 3 x + 2 − 3 x 2 + x − 3 x 2 − 3 x + 2 .

Try It 8.70

Subtract: 6 x 2 − x + 20 x 2 − 81 − 5 x 2 + 11 x − 7 x 2 − 81 . 6 x 2 − x + 20 x 2 − 81 − 5 x 2 + 11 x − 7 x 2 − 81 .

Add and Subtract Rational Expressions whose Denominators are Opposites

When the denominators of two rational expressions are opposites, it is easy to get a common denominator. We just have to multiply one of the fractions by −1 −1 −1 −1 .

Let’s see how this works.

Example 8.36

Add: 4 u − 1 3 u − 1 + u 1 − 3 u . 4 u − 1 3 u − 1 + u 1 − 3 u .

Try It 8.71

Add: 8 x − 15 2 x − 5 + 2 x 5 − 2 x . 8 x − 15 2 x − 5 + 2 x 5 − 2 x .

Try It 8.72

Add: 6 y 2 + 7 y − 10 4 y − 7 + 2 y 2 + 2 y + 11 7 − 4 y . 6 y 2 + 7 y − 10 4 y − 7 + 2 y 2 + 2 y + 11 7 − 4 y .

Example 8.37

Subtract: m 2 − 6 m m 2 − 1 − 3 m + 2 1 − m 2 . m 2 − 6 m m 2 − 1 − 3 m + 2 1 − m 2 .

Try It 8.73

Subtract: y 2 − 5 y y 2 − 4 − 6 y − 6 4 − y 2 . y 2 − 5 y y 2 − 4 − 6 y − 6 4 − y 2 .

Try It 8.74

Subtract: 2 n 2 + 8 n − 1 n 2 − 1 − n 2 − 7 n − 1 1 − n 2 . 2 n 2 + 8 n − 1 n 2 − 1 − n 2 − 7 n − 1 1 − n 2 .

Section 8.3 Exercises

Practice makes perfect.

In the following exercises, add.

2 15 + 7 15 2 15 + 7 15

4 21 + 3 21 4 21 + 3 21

7 24 + 11 24 7 24 + 11 24

7 36 + 13 36 7 36 + 13 36

3 a a − b + 1 a − b 3 a a − b + 1 a − b

3 c 4 c − 5 + 5 4 c − 5 3 c 4 c − 5 + 5 4 c − 5

d d + 8 + 5 d + 8 d d + 8 + 5 d + 8

7 m 2 m + n + 4 2 m + n 7 m 2 m + n + 4 2 m + n

p 2 + 10 p p + 2 + 16 p + 2 p 2 + 10 p p + 2 + 16 p + 2

q 2 + 12 q q + 3 + 27 q + 3 q 2 + 12 q q + 3 + 27 q + 3

2 r 2 2 r − 1 + 15 r − 8 2 r − 1 2 r 2 2 r − 1 + 15 r − 8 2 r − 1

3 s 2 3 s − 2 + 13 s − 10 3 s − 2 3 s 2 3 s − 2 + 13 s − 10 3 s − 2

8 t 2 t + 4 + 32 t t + 4 8 t 2 t + 4 + 32 t t + 4

6 v 2 v + 5 + 30 v v + 5 6 v 2 v + 5 + 30 v v + 5

2 w 2 w 2 − 16 + 8 w w 2 − 16 2 w 2 w 2 − 16 + 8 w w 2 − 16

7 x 2 x 2 − 9 + 21 x x 2 − 9 7 x 2 x 2 − 9 + 21 x x 2 − 9

In the following exercises, subtract.

y 2 y + 8 − 64 y + 8 y 2 y + 8 − 64 y + 8

z 2 z + 2 − 4 z + 2 z 2 z + 2 − 4 z + 2

9 a 2 3 a − 7 − 49 3 a − 7 9 a 2 3 a − 7 − 49 3 a − 7

25 b 2 5 b − 6 − 36 5 b − 6 25 b 2 5 b − 6 − 36 5 b − 6

c 2 c − 8 − 6 c + 16 c − 8 c 2 c − 8 − 6 c + 16 c − 8

d 2 d − 9 − 6 d + 27 d − 9 d 2 d − 9 − 6 d + 27 d − 9

3 m 2 6 m − 30 − 21 m − 30 6 m − 30 3 m 2 6 m − 30 − 21 m − 30 6 m − 30

2 n 2 4 n − 32 − 18 n − 16 4 n − 32 2 n 2 4 n − 32 − 18 n − 16 4 n − 32

6 p 2 + 3 p + 4 p 2 + 4 p − 5 − 5 p 2 + p + 7 p 2 + 4 p − 5 6 p 2 + 3 p + 4 p 2 + 4 p − 5 − 5 p 2 + p + 7 p 2 + 4 p − 5

5 q 2 + 3 q − 9 q 2 + 6 q + 8 − 4 q 2 + 9 q + 7 q 2 + 6 q + 8 5 q 2 + 3 q − 9 q 2 + 6 q + 8 − 4 q 2 + 9 q + 7 q 2 + 6 q + 8

5 r 2 + 7 r − 33 r 2 − 49 − 4 r 2 + 5 r + 30 r 2 − 49 5 r 2 + 7 r − 33 r 2 − 49 − 4 r 2 + 5 r + 30 r 2 − 49

7 t 2 − t − 4 t 2 − 25 − 6 t 2 + 12 t − 44 t 2 − 25 7 t 2 − t − 4 t 2 − 25 − 6 t 2 + 12 t − 44 t 2 − 25

10 v 2 v − 1 + 2 v + 4 1 − 2 v 10 v 2 v − 1 + 2 v + 4 1 − 2 v

20 w 5 w − 2 + 5 w + 6 2 − 5 w 20 w 5 w − 2 + 5 w + 6 2 − 5 w

10 x 2 + 16 x − 7 8 x − 3 + 2 x 2 + 3 x − 1 3 − 8 x 10 x 2 + 16 x − 7 8 x − 3 + 2 x 2 + 3 x − 1 3 − 8 x

6 y 2 + 2 y − 11 3 y − 7 + 3 y 2 − 3 y + 17 7 − 3 y 6 y 2 + 2 y − 11 3 y − 7 + 3 y 2 − 3 y + 17 7 − 3 y

z 2 + 6 z z 2 − 25 − 3 z + 20 25 − z 2 z 2 + 6 z z 2 − 25 − 3 z + 20 25 − z 2

a 2 + 3 a a 2 − 9 − 3 a − 27 9 − a 2 a 2 + 3 a a 2 − 9 − 3 a − 27 9 − a 2

2 b 2 + 30 b − 13 b 2 − 49 − 2 b 2 − 5 b − 8 49 − b 2 2 b 2 + 30 b − 13 b 2 − 49 − 2 b 2 − 5 b − 8 49 − b 2

c 2 + 5 c − 10 c 2 − 16 − c 2 − 8 c − 10 16 − c 2 c 2 + 5 c − 10 c 2 − 16 − c 2 − 8 c − 10 16 − c 2

Everyday Math

Sarah ran 8 miles and then biked 24 miles. Her biking speed is 4 mph faster than her running speed. If r r represents Sarah’s speed when she ran, then her running time is modeled by the expression 8 r 8 r and her biking time is modeled by the expression 24 r + 4 . 24 r + 4 . Add the rational expressions 8 r + 24 r + 4 8 r + 24 r + 4 to get an expression for the total amount of time Sarah ran and biked.

If Pete can paint a wall in p p hours, then in one hour he can paint 1 p 1 p of the wall. It would take Penelope 3 hours longer than Pete to paint the wall, so in one hour she can paint 1 p + 3 1 p + 3 of the wall. Add the rational expressions 1 p + 1 p + 3 1 p + 1 p + 3 to get an expression for the part of the wall Pete and Penelope would paint in one hour if they worked together.

Writing Exercises

Donald thinks that 3 x + 4 x 3 x + 4 x is 7 2 x . 7 2 x . Is Donald correct? Explain.

Explain how you find the Least Common Denominator of x 2 + 5 x + 4 x 2 + 5 x + 4 and x 2 − 16 . x 2 − 16 .

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?

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Lesson 8.1.1, lesson 8.1.2, lesson 8.1.3, lesson 8.1.4, lesson 8.1.5, lesson 8.2.1, lesson 8.2.2, lesson 8.2.3, lesson 8.2.4, lesson 8.2.5.

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1) Explain the basis for the cofunction identities and when they apply.

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

2) Is there only one way to evaluate \(\cos \left (\dfrac{5\pi }{4} \right )\) ?   Explain how to set up the solution in two different ways, and then compute to make sure they give the same answer.

3) Explain to someone who has forgotten the even-odd properties of sinusoidal functions how the addition and subtraction formulas can determine this characteristic for \(f(x)=\sin (x)\) and \(g(x)=\cos (x)\) .   (Hint: \(0-x=-x\))

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

For exercises 4-9, find the exact value.

4) \(\cos \left (\dfrac{7\pi }{12} \right)\)

5) \(\cos \left (\dfrac{\pi }{12} \right)\)

\(\dfrac{\sqrt{2}+\sqrt{6}}{4}\)

6) \(\sin \left (\dfrac{5\pi }{12} \right)\)

7) \(\sin \left (\dfrac{11\pi }{12} \right)\)

\(\dfrac{\sqrt{6}-\sqrt{2}}{4}\)

8) \(\tan \left (-\dfrac{\pi }{12} \right)\)

9) \(\tan \left (\dfrac{19\pi }{12} \right)\)

\(-2-\sqrt{3}\)

For exercises 10-13, rewrite in terms of \(\sin x\) and \(\cos x\)

10) \(\sin \left (x+\dfrac{11\pi }{6} \right)\)

11) \(\sin \left (x-\dfrac{3\pi }{4} \right)\)

\(-\dfrac{\sqrt{2}}{2}\sin x-\dfrac{\sqrt{2}}{2}\cos x\)

12) \(\cos \left (x-\dfrac{5\pi }{6} \right)\)

13) \(\cos \left (x+\dfrac{2\pi }{3} \right)\)

\(-\dfrac{1}{2}\cos x-\dfrac{\sqrt{3}}{2}\sin x\)

For exercises 14-19, simplify the given expression.

14) \(\csc \left (\dfrac{\pi }{2}-t \right)\)

15) \(\sec \left (\dfrac{\pi }{2}-\theta \right)\)

\(\csc \theta\)

16) \(\cot \left (\dfrac{\pi }{2}-x \right)\)

17) \(\tan \left (\dfrac{\pi }{2}-x \right)\)

18) \(\sin(2x)\cos(5x)-\sin(5x)\cos(2x)\)

19) \(\dfrac{\tan \left (\dfrac{3}{2}x \right)-\tan \left (\dfrac{7}{5}x \right)}{1+\tan \left (\dfrac{3}{2}x \right)\tan \left (\dfrac{7}{5}x \right)}\)

\(\tan \left (\dfrac{x}{10} \right)\)

For exercises 20-21, find the requested information.

20) Given that \(\sin a=\dfrac{2}{3}\) and \(\cos b=-\dfrac{1}{4}\) , with \(a\) and \(b\) both in the interval \(\left [ \dfrac{\pi }{2}, \pi \right )\) , find \(\sin (a+b)\) and \(\cos (a-b)\).

21) Given that \(\sin a=\dfrac{4}{5}\) and \(\cos b=\dfrac{1}{3}\) , with \(a\) and \(b\) both in the interval \(\left [ 0, \dfrac{\pi }{2} \right )\) , find \(\sin (a-b)\) and \(\cos (a+b)\).

\(\sin (a-b)=\left ( \dfrac{4}{5} \right )\left ( \dfrac{1}{3} \right )-\left ( \dfrac{3}{5} \right )\left ( \dfrac{2\sqrt{2}}{3} \right )=\dfrac{4-6\sqrt{2}}{15}\)

\(\cos (a+b)=\left ( \dfrac{3}{5} \right )\left ( \dfrac{1}{3} \right )-\left ( \dfrac{4}{5} \right )\left ( \dfrac{2\sqrt{2}}{3} \right )=\dfrac{3-8\sqrt{2}}{15}\)

For exercises 22-24, find the exact value of each expression.

22) \(\sin \left ( \cos^{-1}\left ( 0 \right )- \cos^{-1}\left ( \dfrac{1}{2} \right )\right )\)

23) \(\cos \left ( \cos^{-1}\left ( \dfrac{\sqrt{2}}{2} \right )+ \sin^{-1}\left ( \dfrac{\sqrt{3}}{2} \right )\right )\)

\(\dfrac{\sqrt{2}-\sqrt{6}}{4}\)

24) \(\tan \left ( \sin^{-1}\left ( \dfrac{1}{2} \right )- \cos^{-1}\left ( \dfrac{1}{2} \right )\right )\)

For exercises 25-32, simplify the expression, and then graph both expressions as functions to verify the graphs are identical.

25) \(\cos \left ( \dfrac{\pi }{2}-x \right )\)

26) \(\sin (\pi -x)\)

27) \(\tan \left ( \dfrac{\pi }{3}+x \right )\)

\(\cot \left ( \dfrac{\pi }{6}-x \right )\)

28) \(\sin \left ( \dfrac{\pi }{3}+x \right )\)

29) \(\tan \left ( \dfrac{\pi }{4}-x \right )\)

\(\cot \left ( \dfrac{\pi }{4}+x \right )\)

30) \(\cos \left ( \dfrac{7\pi }{6}+x \right )\)

31) \(\sin \left ( \dfrac{\pi }{4}+x \right )\)

\(\dfrac{\sin x}{\sqrt{2}}+\dfrac{\cos x}{\sqrt{2}}\)

32) \(\cos \left ( \dfrac{5\pi }{4}+x \right )\)

For exercises 33-41, use a graph to determine whether the functions are the same or different. If they are the same, show why. If they are different, replace the second function with one that is identical to the first. (Hint: think \(2x=x+x\))

33) \(f(x)=\sin(4x)-\sin(3x)\cos x, g(x)=\sin x \cos(3x)\)

They are the same.

34) \(f(x)=\cos(4x)+\sin x \sin(3x), g(x)=-\cos x \cos(3x)\)

35) \(f(x)=\sin(3x)\cos(6x), g(x)=-\sin(3x)\cos(6x)\)

They are different, try \(g(x)=\sin(9x)-\cos(3x)\sin(6x)\)

36) \(f(x)=\sin(4x), g(x)=\sin(5x)\cos x-\cos(5x)\sin x\)

37) \(f(x)=\sin(2x), g(x)=2 \sin x \cos x\)

38) \(f(\theta )=\cos(2\theta ), g(\theta )=\cos^2\theta -\sin^2\theta\)

39) \(f(\theta )=\tan(2\theta ), g(\theta )=\dfrac{\tan \theta }{1+\tan^2\theta }\)

They are different, try \(g(\theta )=\dfrac{2\tan \theta }{1-\tan^2\theta }\)

40) \(f(x)=\sin(3x)\sin x, g(x)=\sin^2(2x)\cos^2x-\cos^2(2x)\sin2x\)

41) \(f(x)=\tan(-x), g(x)=\dfrac{\tan x-\tan(2x)}{1-\tan x \tan(2x)}\)

They are different, try \(g(x)=\dfrac{\tan x-\tan(2x)}{1+\tan x \tan(2x)}\)

For the exercises 42-46, find the exact value algebraically, and then confirm the answer with a calculator to the fourth decimal point.

42) \(\sin (75^{\circ})\)

43) \(\sin (195^{\circ})\)

\(-\dfrac{\sqrt{3}-1}{2\sqrt{2}}\), or \(-0.2588\)

44) \(\cos (165^{\circ})\)

45) \(\cos (345^{\circ})\)

\(\dfrac{1+\sqrt{3}}{2\sqrt{2}}\), or \(-0.9659\)

46) \(\tan (-15^{\circ})\)

For the exercises 47-51, prove the identities provided.

47) \(\tan \left ( x+\dfrac{\pi }{4} \right )=\dfrac{\tan x+1}{1-\tan x}\)

\(\begin{align*} \tan \left ( x+\dfrac{\pi }{4} \right ) &= \\ \dfrac{\tan x + \tan\left (\tfrac{\pi}{4} \right )}{1-\tan x \tan\left (\tfrac{\pi}{4} \right )} &= \\ \dfrac{\tan x+1}{1-\tan x(1)} &= \dfrac{\tan x+1}{1-\tan x} \end{align*}\)

48) \(\dfrac{\tan (a+b)}{\tan (a-b)}=\dfrac{\sin a \cos a + \sin b \cos b}{\sin a \cos a - \sin b \cos b}\)

49) \(\dfrac{\cos (a+b)}{\cos a \cos b}=1-\tan a \tan b\)

\(\begin{align*} \dfrac{\cos (a+b)}{\cos a \cos b} &= \\ \dfrac{\cos a \cos b}{\cos a \cos b}- \dfrac{\sin a \sin b}{\cos a \cos b} &= 1-\tan a \tan b \end{align*}\)

50) \(\cos(x+y)\cos(x-y)=\cos^2x-\sin^2y\)

51) \(\dfrac{\cos(x+h)-\cos(x)}{h}=\cos x\dfrac{\cos h-1}{h}-\sin x \dfrac{\sin h}{h}\)

\(\begin{align*} \dfrac{\cos(x+h)-\cos(x)}{h} &= \\ \dfrac{\cos x\cosh - \sin x\sinh -\cos x}{h} &= \\ \dfrac{\cos x(\cosh-1) - \sin x(\sinh-1)}{h} &= \cos x\dfrac{\cos h-1}{h}-\sin x \dfrac{\sin h}{h} \end{align*}\)

For the exercises 52-, prove or disprove the statements.

52) \(\tan (u+v)=\dfrac{\tan u+\tan v}{1-\tan u \tan v}\)

53) \(\tan (u-v)=\dfrac{\tan u-\tan v}{1+\tan u \tan v}\)

54) \(\dfrac{\tan (x+y)}{1+\tan x \tan x}=\dfrac{\tan x + \tan y}{1-\tan^2 x \tan^2 y}\)

55) If \(\alpha ,\beta\) , and \(\gamma\) are angles in the same triangle, then prove or disprove  \(\sin(α+β)=\sin γ\).

True. Note that \(\sin (\alpha +\beta )=\sin (\pi -\gamma )\) and expand the right hand side.

56) If \(\alpha ,\beta\) , and \(\gamma\) are angles in the same triangle, then prove or disprove \(\tan \alpha +\tan \beta +\tan \gamma =\tan \alpha \tan \beta \tan \gamma\).