Eureka Math Algebra 2 Module 2 Lesson 10 Answer Key

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Eureka Math Algebra 2 Module 2 Lesson 10 Problem Set Answer Key

Question 1.
Describe the values of x for which each of the following Is true.
a. The cosine function has a relative maximum.
Answer:
The cosine function has a relative maximum at all x = 2nπ, where n is an integer.

b. The sine function has a relative maximum.
Answer:
The sine function has a relative maximum at all x = \(\frac{\pi}{2}\) + 2πn, where n is an integer.

Question 2.
Without using a calculator, rewrite each of the following in order from least to greatest. Use the graph to explain your reasoning.
sin\(\left(\frac{\pi}{2}\right)\) sin\(\left(-\frac{2 \pi}{3}\right)\) sin\(\left(\frac{\pi}{4}\right)\) sin\(\left(-\frac{\pi}{2}\right)\)
Eureka Math Algebra 2 Module 2 Lesson 10 Problem Set Answer Key 1
Answer:
At –\(\frac{\pi}{2}\), the sine function takes on its smallest value, and at \(\frac{\pi}{2}\) the sine function takes on its largest value. Additionally, sin\(\) is negative, and sin\(\frac{\pi}{4}\) is positive. Then, these four values in increasing order are
sin\(\left(-\frac{\pi}{2}\right)\), sin\(\left(-\frac{2 \pi}{3}\right)\), sin\(\left(\frac{\pi}{4}\right)\), sin\(\left(\frac{\pi}{2}\right)\).

Question 3.
Without using a calculator, rewrite each of the following in order from least to greatest. Use the graph to explain your reasoning.
cos\(\left(\frac{\pi}{2}\right)\) cos\(\left(\frac{5 \pi}{4}\right)\) cos\(\left(\frac{\pi}{4}\right)\) cos(5π)
Eureka Math Algebra 2 Module 2 Lesson 10 Problem Set Answer Key 2
Answer:
At 5π, the cosine function takes on its smallest value because 5π = 2π + 2π + π, so cos(5π) = cos(π) = – 1. We can see that cos \(\left(\frac{5 \pi}{4}\right)\) is negative, cos \(\left(\frac{\pi}{2}\right)\) = 0, and cos \(\left(\frac{\pi}{4}\right)\) is positive. Then, these four values in increasing order are
cos(5π), cos \(\left(\frac{5 \pi}{4}\right)\), cos \(\left(\frac{\pi}{2}\right)\), cos \(\left(\frac{\pi}{4}\right)\).

Question 4.
Evaluate each of the following without a calculator using a trigonometric identity when needed.
cos\(\left(\frac{\pi}{6}\right)\)
Answer:
\(\frac{\sqrt{3}}{2}\)

cos\(\left(-\frac{\pi}{6}\right)\)
Answer:
\(\frac{\sqrt{3}}{2}\)

cos\(\left(\frac{7 \pi}{6}\right)\)
Answer:
–\(\frac{\sqrt{3}}{2}\)

cos\(\left(\frac{13 \pi}{6}\right)\)
Answer:
\(\frac{\sqrt{3}}{2}\)

Question 5.
Evaluate each of the following without a calculator using a trigonometric identity when needed.
sin\(\left(\frac{3 \pi}{4}\right)\)
Answer:
\(\frac{\sqrt{2}}{2}\)

sin\(\left(\frac{11 \pi}{4}\right)\)
Answer:
\(\frac{\sqrt{2}}{2}\)

sin\(\left(\frac{7 \pi}{4}\right)\)
Answer:
–\(\frac{\sqrt{2}}{2}\)

sin\(\left(\frac{-5 \pi}{4}\right)\)
Answer:
\(\frac{\sqrt{2}}{2}\)

Question 6.
Use the rotation through θ radians shown to answer each of the following questions.
Eureka Math Algebra 2 Module 2 Lesson 10 Problem Set Answer Key 3
a. Explain why sin(-θ) = -sin(θ) for all real numbers θ.
Answer:
If the initial ray is rotated through -θ radians, It is rotated B radians in the clockwise direction, If the terminal ray of a rotation through θ radians intersects the unit circle at the point (a, b), then the terminal ray of rotation through -θ radians will intersect the unit circle at the point (a, -b). Then, sin(θ) = b and sin(-θ) = -b, so sin(-θ) = -sin(θ).

b. What symmetry does this identity demonstrate about the graph of y = sin(x)?
Answer:
It demonstrates that the graph of y = sin(x) is symmetric with respect to the origin because if the point (x, y) is on the graph, then the point (-x, -y) is also on the graph.

Question 7.
Use the same rotation shown in Problem 6 to answer each of the following questions.
a. Explain why cos(-θ) = cos(θ).
Answer:
If the initial ray is rotated through -θ radians, -t is rotated θ radians in the clockwise direction. If the terminal ray of a rotation through θ radians intersects the unit circle at the point (a, b), then the terminal ray of rotation through -θ radians will intersect the unit circle at the point (a, -b). Then, cos(θ) = a and cos(-θ) = a, so cos(-θ) = cos(θ).

b. What symmetry does this identity demonstrate about the graph of y = cos(θ)?
Answer:
It demonstrates that the graph of y = cos(x) is symmetric with respect to the y-axis because if the point (x, y) is on the graph, then the point (-x, y) is also on the graph.

Question 8.
Find equations of two different functions that can be represented by the graph shown below-one sine and one cosine-using different horizontal transformations.
Eureka Math Algebra 2 Module 2 Lesson 10 Problem Set Answer Key 4
Answer:
y = sin(x – \(\frac{\pi}{2}\)) y = cos(x – π) (other correct answers are possible.)

Question 9.
Find equations of two different functions that can be represented by the graph shown below-one sine and one cosine-using different horizontal translations.
Eureka Math Algebra 2 Module 2 Lesson 10 Problem Set Answer Key 5
Answer:
y = cos(x + \(\frac{\pi}{2}\)) y = sin(x – π) (other correct answers are possible.)

Eureka Math Algebra 2 Module 2 Lesson 10 Exit Ticket Answer Key

Question 1.
Demonstrate how to evaluate cos \(\left(\frac{8 \pi}{3}\right)\) by using a trigonometric identity.
Answer:
cos\(\left(\frac{8 \pi}{3}\right)\) = cos(\(\frac{2 \pi}{3}\) + 2π) = cos\(\left(\frac{2 \pi}{3}\right)\) = –\(\frac{1}{2}\)

Question 2.
Determine if the following statement is true or false, without using a calculator.
Sin\(\left(\frac{8 \pi}{7}\right)\) = Sin\(\left(\frac{\pi}{7}\right)\)
Answer:
False. sin\(\left(\frac{8 \pi}{7}\right)\) = Sin(\(\frac{\pi}{7}\) + π) = -Sin\(\left(\frac{\pi}{7}\right)\) ≠ Sin\(\left(\frac{\pi}{7}\right)\)

Question 3.
If the graph of the cosine function is translated to the right \(\frac{\pi}{2}\) units, the resulting graph is that of the sine function, which leads to the identity: For all x, sin(x) = cos (x – \(\frac{\pi}{2}\)). Write another identity for sin(x) using a different horizontal shift.
Answer:
sin(x) = cos(x + \(\frac{3\pi}{2}\))

Eureka Math Algebra 2 Module 2 Lesson 10 Exploratory Challenge Answer Key

Exploratory Challenge 1
Consider the function f(x) = sin(x) where x is measured in radians.
Graph f(x) = sin(x) on the interval [-π, 5π] by constructing a table of values. Include all intercepts, relative maximum points, and relative minimum points of the graph. Then, use the graph to answer the questions that follow.
Eureka Math Algebra 2 Module 2 Lesson 10 Exploratory Challenge Answer Key 6
Eureka Math Algebra 2 Module 2 Lesson 10 Exploratory Challenge Answer Key 7
a. Using one of your colored pencils, mark the point on the graph at each of the following pairs of x-values.
x = –\(\frac{\pi}{2}\) and x = –\(\frac{\pi}{2}\) + 2π
x = π and x = π + 2π
x = \(\frac{7 \pi}{4}\) and x = \(\frac{7 \pi}{4}\) + 2π
Answer:

b. What can be said about the y-values for each pair of x-values marked on the graph?
Answer:
For each pair of x-values, the y-values are the same.

c. Will this relationship hold for any two x-values that differ by 2π? Explain how you know.
Answer:
Yes. Since the sine function repeats every 2π units, then the relationship described in part (b) will hold for any two x-values that differ by 2π.

d. Based on these results, make a conjecture by filling in the blank below.
For any real number x, sin(x + 2π) = _______ .
Answer:
sin(x + 2π) = sin(x)

e. Test your conjecture by selecting another x-value from the graph and demonstrating that the equation from part (d) holds for that value of x.
Answer:
sin\(\left(\frac{5 \pi}{2}\right)\) = Sin(\(\frac{\pi}{2}\) + 2π) = sin\(\left(\frac{\pi}{2}\right)\) = 1 (Answers will vary.)

f. How does the conjecture in part (d) support the claim that the sine function is a periodic function?
Answer:
The sine function repeats every 2π units because 2π radians is one full turn. Thus, if we rotate the initial ray through x + 2π radians, the terminal ray is in the same position as if we had rotated by x radians.

g. Use this identity to evaluate sin\(\left(\frac{13 \pi}{6}\right)\)
Answer:
sin\(\left(\frac{13 \pi}{6}\right)\) = sin(\(\frac{\pi}{6}\) + 2π) = sin\(\left(\frac{\pi}{6}\right)\) = \(\frac{1}{2}\)

h. Using one of your colored pencils, mark the point on the graph at each of the following pairs of x-values.
x = –\(\frac{\pi}{4}\) and x = –\(\frac{\pi}{4}\) + π
x = 2π and x = 2π + π
x = \(\frac{5 \pi}{2}\) and x = \(\frac{5 \pi}{2}\) + π
Answer:

i. What can be said about the y-values for each pair of x-values marked on the graph?
Answer:
For each pair of x-values, the y-values have the same magnitude but opposite sign.

j. Will this relationship hold for any two x-values that differ by π? Explain how you know.
Answer:
Yes. Since rotating by an additional π radians produces a point in the opposite quadrant with the same reference angle, the sine of the two numbers will have the same magnitude and opposite sign.

k. Based on these results, make a conjecture by filling in the blank below.
For any real number x, sin(x + π) = _________ .
Answer:
sin(x + π) = -sin(x)

i. Test your conjecture by selecting another x-value from the graph and demonstrating that the equation from part (k) holds for that value of x.
Answer:
sin\(\left(\frac{3 \pi}{2}\right)\) = sin(\(\frac{\pi}{2}\) + π) = -sin\(\left(\frac{\pi}{2}\right)\) = -1

m. Is the following statement true or false? Use the conjecture from (k) to explain your answer.
sin\(\left(\frac{4 \pi}{3}\right)\) = -sin\(\left(\frac{\pi}{3}\right)\)
Answer:
This statement is true: sin\(\left(\frac{4 \pi}{3}\right)\) = sin(\(\frac{\pi}{3}\) + π) = -sin\(\left(\frac{\pi}{3}\right)\).

n. Using one of your colored pencils, mark the point on the graph at each of the following pairs of x-values.
x = –\(\frac{3 \pi}{4}\) and x = \(\frac{3 \pi}{4}\)
x = –\(\frac{\pi}{2}\) and x = \(\frac{\pi}{2}\)
Answer:

o. What can be said about the y-values for each pair of x-values marked on the graph?
Answer:
For each pair of x-values, the y-values have the same magnitude but with the opposite sign.

p. Will this relationship hold for any two x-values with the same magnitude but opposite sign? Explain how you know.
Answer:
Yes. If rotation by x radians produces the point (a, b) on the unit circle, then rotation by -x radians will produce a point (a, -b) on the unit circle.

q. Based on these results, make a conjecture by filling in the blank below.
For any real number x, sin(-x) = _________.
Answer:
sin(-x) = – sin(x)

r. Test your conjecture by selecting another x-value from the graph and demonstrating that the equation from part (q) holds for that value of x.
Answer:
For example, sin\(\left(-\frac{3 \pi}{4}\right)\) = \(\frac{\sqrt{2}}{2}\) and sin \(\left(\frac{3 \pi}{4}\right)\) = \(\frac{\sqrt{2}}{2}\), so sin\(\left(-\frac{3 \pi}{4}\right)\) = -sin\(\left(\frac{3 \pi}{4}\right)\).

s. Is the sine function an odd function, even function, or neither? Use the identity from part (q) to explain.
Answer:
The sine function is an odd function because sin(-x) = -sin(x) and because the graph is symmetric with respect to the origin.

t. Describe the x-intercepts of the graph of the sine function.
Answer:
The graph of the sine function has x-intercepts at all x-values such that x = nπ, where n is an integer.

u. Describe the end behavior of the sine function.
Answer:
As x increases to ∞ or as x decreases to -∞, the sine function cycles between the values of -1 and 1.

Exploratory Challenge 2.
Consider the function g(x) = cos(x) where x Is measured in radians.
Graph g(x) = cos(x) on the interval [-π, 5π] by constructing a table of values. Include all intercepts, relative maximum points, and relative minimum points. Then, use the graph to answer the questions that follow.
Eureka Math Algebra 2 Module 2 Lesson 10 Exploratory Challenge Answer Key 8
Eureka Math Algebra 2 Module 2 Lesson 10 Exploratory Challenge Answer Key 9
a. Using one of your colored pencils, mark the point on the graph at each of the following pairs of x-values.
x = –\(\frac{\pi}{2}\) and x = –\(\frac{\pi}{2}\) + 2π
x = π and x = π + 2π
x = \(\frac{7 \pi}{4}\) and x = \(\frac{7 \pi}{4}\) + 2π
Answer:

b. What can be said about the y-values for each pair of x-values marked on the graph?
Answer:
For each pair of x-values, the y-values are the same.

c. Will this relationship hold for any two x-values that differ by 2π? Explain how you know.
Answer:
Yes. Since the sine function repeats every 2π units, then the relationship described in part (b) will hold for any two x-values that differ by 2π.

d. Based on these results, make a conjecture by filling in the blank below.
For any real number x, cos(x + 2π) = _______ .
Answer:
cos(x + 2π) = cos(x)

e. Test your conjecture by selecting another x-value from the graph and demonstrating that the equation from part (d) holds for that value of x.
Answer:
cos (0) = cos(0 + 2π) = 1 (Answers will vary.)

f. How does the conjecture in part (d) support the claim that the cosine function is a periodic function?
Answer:
Like the sine function, the cosine function repeats every 2π units because 2π radians is one full turn. Thus, if we rotate the Initial ray through x + 2π radians, the terminal ray is in the same position as if we had rotated by x radians.

g. Use this identity to evaluate cos\(\left(\frac{9 \pi}{4}\right)\)
Answer:
cos\(\left(\frac{9 \pi}{4}\right)\) = cos(\(\frac{\pi}{4}\) + 2π) = cos\(\left(\frac{\pi}{4}\right)\) = \(\frac{\sqrt{2}}{2}\)

h. Using one of your colored pencils, mark the point on the graph at each of the following pairs of x-values.
x = –\(\frac{\pi}{4}\) and x = –\(\frac{\pi}{4}\) + π
x = 2π and x = 2π + π
x = \(\frac{5 \pi}{2}\) and x = \(\frac{5 \pi}{2}\) + π
Answer:

i. What can be said about the y-values for each pair of x-values marked on the graph?
Answer:
For each pair of x-values, the y-values have the same magnitude but opposite sign.

j. Will this relationship hold for any two x-values that differ by π? Explain how you know.
Answer:
Yes. Since rotating by an additional π radians produces a point in the opposite quadrant with the same reference angle, the sine of the two numbers will have the same magnitude and opposite sign.

k. Based on these results, make a conjecture by filling in the blank below.
For any real number x, cos(x + π) = _________ .
Answer:
cos(x + π) = -cos(x)

i. Test your conjecture by selecting another x-value from the graph and demonstrating that the equation from part (k) holds for that value of x.
Answer:
cos (3π) = cos(2π + π) = -cos(2π) = -1

m. Is the following statement true or false? Use the conjecture from (k) to explain your answer.
cos\(\left(\frac{5 \pi}{3}\right)\) = -cos\(\left(\frac{2 \pi}{3}\right)\)
Answer:
This statement is true: cos\(\left(\frac{5 \pi}{3}\right)\) = cos(\(\frac{2 \pi}{3}\) + π) = -cos\(\left(\frac{2 \pi}{3}\right)\).

n. Using one of your colored pencils, mark the point on the graph at each of the following pairs of x-values.
x = –\(\frac{3 \pi}{4}\) and x = \(\frac{3 \pi}{4}\)
x = -π and x = π
Answer:

o. What can be said about the y-values for each pair of x-values marked on the graph?
Answer:
For each pair of x-values, the y-values have the same magnitude but with the opposite sign.

p. Will this relationship hold for any two x-values with the same magnitude and same sign? Explain how you know.
Answer:
Yes. If rotation by x radians produces the point (a, b) on the unit circle, then rotation by -x radians will produce a point (a, -b) on the unit circle.

q. Based on these results, make a conjecture by filling in the blank below.
For any real number, cos(-x) = _________.
Answer:
cos(-x) = cos(x)

r. Test your conjecture by selecting another x-value from the graph and demonstrating that the identity is true for that value of x.
Answer:
cos(-2π) = cos(2π) = 1

s. Is the cosine function an odd function, even function, or neither? Use the identity from part (n) to explain.
Answer:
The cosine function is an odd function because cosin(-x) = cos(x) and because the graph is symmetric with respect to the origin.

t. Describe the x-intercepts of the graph of the sine function.
Answer:
The graph of the sine function has x-intercepts at all x-values such that x = \(\frac{\pi}{2}\) + nπ, where n is an integer.

u. Describe the end behavior of g(x) = cos(x).
Answer:
As x increases to ∞ or as x decreases to -∞, the cosine function cycles between the values of -1 and 1.

Exploratory Challenge 3
Graph both f(x) sin(x) and g(x) = cos(x) on the graph below. Then, use the graphs to answer the questions that follow.
Eureka Math Algebra 2 Module 2 Lesson 10 Exploratory Challenge Answer Key 10
a. List ways in which the graphs of the sine and cosine functions are alike.
Answer:
Both functions are periodic and have a period of 2π. Both functions have a domain of all real numbers and a range of [-1,1]. Both functions cycle between -1 and 1 as x increases to ∞ or as x decreases to -∞.
Both have similar identities such as sin(x + 2π) = sin(x) and cos(x + 2π) = cos(x).

b. List ways in which the graphs of the sine and cosine functions are different.
Answer:
We stated above that f(x) = sin(x) is an odd function, and g(x) = cos(x) is an even function. Where the sine function is at a maximum or minimum point, the cosine function has an x-intercept and vice versa.

c. What type of transformation would be required to make the graph of the sine function coincide with the graph of the cosine function?
Answer:
A horizontal shift

d. What is the smallest possible horizontal translation required to make the graph of f(x) = sin(x) coincide with the graph of g(x) = cos(x)?
Answer:
A horizontal shift \(\frac{\pi}{2}\) units to the left

e. What is the smallest possible horizontal translation required to make the graph of g(x) = cos(x) coincide with the graph of f(x) = sin(x)?
Answer:
A horizontal shift \(\frac{\pi}{2}\) units to the right

f. Use your answers from parts (d) and (e) to fill in the blank below.
For any real number x, _______ = cos (x – \(\frac{\pi}{2}\)).
For any real number x, ________= sin (x + \(\frac{\pi}{2}\)).
Answer:
sin(x) = cos (x – \(\frac{\pi}{2}\))
cos(x)= sin (x + \(\frac{\pi}{2}\))

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