NYS COMMON CORE MATHEMATICS CURRICULUM M2 Lesson 10
ALGEBRA II
Lesson 10: Basic Trigonometric Identities from Graphs
144
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Lesson 10: Basic Trigonometric Identities from Graphs
Student Outcomes
Students observe identities from graphs of sine and cosine basic trigonometric identities and relate those
identities to periodicity, even and odd properties, intercepts, end behavior, and cosine as a horizontal
translation of sine.
Lesson Notes
Students have previous experience with graphing the sine and cosine functions in degrees and have been introduced to
radian measure in the previous lesson. For the remainder of the module, students will use radians to work with and
graph trigonometric functions. The purpose of this lesson is to increase studentsβ comfort with radians and to formalize
the characteristics of periodicity, end behavior, intercepts, and relative extrema of the sine and cosine functions through
the observation and conjecture of some basic trigonometric identities. As students work through the lesson, make sure
that they grasp the following identities that are valid for all real numbers π₯:
sin(π₯ + 2π) = sin(π₯) sin(π₯ + π) = β sin(π₯) sin(βπ₯) = βsin(π₯)
cos(π₯ + 2π) = cos(π₯) cos(π₯ + π) = β cos(π₯) cos(βπ₯) = cos(π₯)
and that they know how to use these identities to evaluate sine and cosine for a variety of values of π₯. Discovering these
six trigonometric identities through observation is the focus of this lesson. Discovering and proving other identities will
be the focus of Lessons 15, 16, and 17 of this module.
It is important to note that an identity is a statement that two functions are equal on a common domain. Thus, to
specify an identity, students need to specify both an equation and a set of values for the variable for which the
statement is true. That is, the statement sin(βπ₯) = βsin(π₯) itself is not an identity, but the statement
sin(βπ₯) = βsin(π₯) for all real numbers π₯ is an identity.
Materials
Supply students with colored pencils for the activity.
Classwork
Opening (2 minutes)
Students should be able to largely work through these explorations in groups without too much assistance. It may be
necessary to help students get started in constructing their table of values by reminding them of the values where the
sine and cosine functions take on the values 0, 1, and β1. Also, remind students that they have seen these graphs
earlier in the module, but the graphs were constructed using degree measures. Today (and for the remainder of the
module) graphing uses radian measures.
15
NYS COMMON CORE MATHEMATICS CURRICULUM M2 Lesson 10
ALGEBRA II
Lesson 10: Basic Trigonometric Identities from Graphs
145
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Exploratory Challenge 1 (18 minutes)
Allow students to work in groups through this problem. Because students are repeating a similar process to discover
three different identities, assign groups different parts of the Exploratory Challenge, and then have them report their
results to the class. If students are having trouble with the table, help them get started by reminding them that the
maximum and minimum values and π₯-intercepts can be found by evaluating sine at the extreme points on the circle.
Circulate around the room to ensure that students are comfortable working with radian values and that they are using
the graph to discover the identities. Debrief after this activity to ensure that students have discovered the identities
correctly.
Exploratory Challenge 1
Consider the function π(π) = π¬π’π§(π) where π is measured in radians.
Graph π(π) = π¬π’π§(π) on the interval [βπ , ππ ] 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.
π
π(π)
a. Using one of your colored pencils, mark the point on the graph at each of the
following pairs of π-values.
π = βπ π
and π = βπ π
+ ππ
π = π and π = π + ππ
π =ππ π
and π =ππ π
+ ππ
b. What can be said about the π-values for each pair of π-values marked on the
graph?
For each pair of π-values, the π-values are the same.
c. Will this relationship hold for any two π-values that differ by ππ ? Explain how
you know.
Yes. Since the sine function repeats every ππ units, then the relationship
described in part (b) will hold for any two π-values that differ by ππ .
Scaffolding:
This could also be accomplished using technology. Students could observe these identities by using the table or graph feature of a graphing calculator.
Advanced learners could be asked to write similar identities for the cosecant function. For example, since sin(π₯ + π) = βsin(π₯), does it follow that csc(π₯ + π) = βcsc(π₯)? Test it either by using test values or by exploring the graph of cosecant.
MP.8
NYS COMMON CORE MATHEMATICS CURRICULUM M2 Lesson 10
ALGEBRA II
Lesson 10: Basic Trigonometric Identities from Graphs
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d. Based on these results, make a conjecture by filling in the blank below.
For any real number π, π¬π’π§(π + ππ ) = _______________.
π¬π’π§(π + ππ ) = π¬π’π§(π)
e. Test your conjecture by selecting another π-value from the graph and demonstrating that the equation from
part (d) holds for that value of π.
π¬π’π§ (ππ π
) = π¬π’π§ (π π
+ ππ ) = π¬π’π§ (π π
) = π (Answers will vary.)
f. How does the conjecture in part (d) support the claim that the sine function is a periodic function?
The sine function repeats every ππ units because ππ radians is one full turn. Thus, if we rotate the initial ray
through π + ππ radians, the terminal ray is in the same position as if we had rotated by π radians.
g. Use this identity to evaluate π¬π’π§ (πππ
π).
π¬π’π§ (πππ
π) = π¬π’π§ (
π
π+ ππ ) = π¬π’π§ (
π
π) =
π
π
h. Using one of your colored pencils, mark the point on the graph at each of the following pairs of π-values.
π = βπ π
and π = βπ π
+ π
π = ππ and π = ππ + π
π =ππ π
and π =ππ π
+ π
i. What can be said about the π-values for each pair of π-values marked on the graph?
For each pair of π-values, the π-values have the same magnitude but opposite sign.
j. Will this relationship hold for any two π-values that differ by π ? Explain how you know.
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 π, π¬π’π§(π + π ) = _______________.
π¬π’π§(π + π ) = βπ¬π’π§(π)
l. Test your conjecture by selecting another π-value from the graph and demonstrating that the equation from
part (k) holds for that value of π.
π¬π’π§ (ππ
π) = π¬π’π§ (
π
π+ π ) = β π¬π’π§ (
π
π) = βπ
m. Is the following statement true or false? Use the conjecture from (k) to explain your answer.
π¬π’π§ (πππ
) = βπ¬π’π§ (ππ
)
This statement is true: π¬π’π§ (ππ π
) = π¬π’π§ (π π
+ π ) = β π¬π’π§ (π π
).
MP.8
NYS COMMON CORE MATHEMATICS CURRICULUM M2 Lesson 10
ALGEBRA II
Lesson 10: Basic Trigonometric Identities from Graphs
147
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n. Using one of your colored pencils, mark the point on the graph at each of the following pairs of π-values.
π = βππ π
and π =ππ π
π = βπ π
and π =π π
o. What can be said about the π-values for each pair of π-values marked on the graph?
For each pair of π-values, the π-values have the same magnitude but with the opposite sign.
p. Will this relationship hold for any two π-values with the same magnitude but opposite sign? Explain how you
know.
Yes. If rotation by π radians produces the point (π, π) on the unit circle, then rotation by βπ radians will
produce a point (π, βπ) on the unit circle.
q. Based on these results, make a conjecture by filling in the blank below.
For any real number π, π¬π’π§(βπ) = _______________.
π¬π’π§(βπ) = βπ¬π’π§(π)
r. Test your conjecture by selecting another π-value from the graph and demonstrating that the equation from
part (q) holds for that value of π.
For example, π¬π’π§ (βππ π
) = ββππ
and π¬π’π§ (ππ π
) =βππ
, so π¬π’π§ (βππ π
) = βπ¬π’π§ (ππ π
).
s. Is the sine function an odd function, even function, or neither? Use the identity from part (q) to explain.
The sine function is an odd function because π¬π’π§(βπ) = βπ¬π’π§(π) and because the graph is symmetric with
respect to the origin.
t. Describe the π-intercepts of the graph of the sine function.
The graph of the sine function has π-intercepts at all π-values such that π = ππ , where π is an integer.
u. Describe the end behavior of the sine function.
As π increases to β or as π decreases to ββ, the sine function cycles between the values of βπ and π.
During the debriefing, record (or have students record) the key results on the board.
For all π₯: sin(π₯ + 2π) = sin(π₯) sin(π₯ + π) = β sin(π₯) sin(βπ₯) = βsin(π₯)
NYS COMMON CORE MATHEMATICS CURRICULUM M2 Lesson 10
ALGEBRA II
Lesson 10: Basic Trigonometric Identities from Graphs
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Exploratory Challenge 2 (10 minutes)
Allow students to work in groups through this problem. This exploration should go more quickly as it is the same process
that students went through in Exploratory Challenge 1. As suggested above, assign groups different parts of the
Exploratory Challenge, and then have them report their results to the class. Debrief after this activity to ensure that
students have completed the identities correctly.
Exploratory Challenge 2
Consider the function π(π) = ππ¨π¬(π) where π is measured in radians.
Graph π(π) = ππ¨π¬(π) on the interval [βπ , ππ ] 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.
π
π(π)
a. Using one of your colored pencils, mark the point on the graph at each of the following pairs of π-values.
π = βπ π
and π = βπ π
+ ππ
π = π and π = π + ππ
π =ππ π
and π =ππ π
+ ππ
b. What can be said about the π-values for each pair of π-values marked on the graph?
For each pair of π-values, the π-values are the same.
c. Will this relationship hold for any two π-values that differ by ππ ? Explain how you know.
Yes. Since the sine function repeats every ππ units, then the relationship in part (b) will hold for any two
π-values that differ by ππ .
d. Based on these results, make a conjecture by filling in the blank below.
For any real number π, ππ¨π¬(π + ππ ) = _______________.
ππ¨π¬(π + ππ ) = ππ¨π¬(π)
NYS COMMON CORE MATHEMATICS CURRICULUM M2 Lesson 10
ALGEBRA II
Lesson 10: Basic Trigonometric Identities from Graphs
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e. Test your conjecture by selecting another π-value from the graph and demonstrating that the equation from
part (d) holds for that value of π.
ππ¨π¬(π) = ππ¨π¬(π + ππ ) = π (Answers will vary.)
f. How does the conjecture from part (d) support the claim that the cosine function is a periodic function?
Like the sine function, the cosine function repeats every ππ units because ππ radians is one full turn. Thus, if
we rotate the initial ray through π + ππ radians, the terminal ray is in the same position as if we had rotated
by π radians.
g. Use this identity to evaluate ππ¨π¬ (ππ π
).
ππ¨π¬ (ππ
π) = ππ¨π¬ (
π
π+ ππ ) = ππ¨π¬ (
π
π) =
βπ
π
h. Using one of your colored pencils, mark the point on the graph at each of the following pairs of π-values.
π = βπ π
and π = βπ π
+ π
π = ππ and π = ππ + π
π =ππ π
and π =ππ π
+ π
i. What can be said about the π-values for each pair of π-values marked on the graph?
For each pair of π-values, the π-values have the same magnitude but opposite sign.
j. Will this relationship hold for any two π-values that differ by π ? Explain how you know.
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 π, ππ¨π¬ (π + π ) = _______________.
ππ¨π¬(π + π ) = βππ¨π¬(π)
l. Test your conjecture by selecting another π-value from the graph and demonstrating that the equation from
part (k) holds for that value of π.
ππ¨π¬(ππ ) = ππ¨π¬(ππ + π ) = βππ¨π¬(ππ ) = βπ
m. Is the following statement true or false? Use the identity from part (k) to explain your answer.
ππ¨π¬ (ππ π
) = βππ¨π¬ (ππ π
)
This statement is true: ππ¨π¬ (ππ π
) = ππ¨π¬ (ππ π
+ π ) = βππ¨π¬ (ππ π
).
n. Using one of your colored pencils, mark the point on the graph at each of the following pairs of π-values.
π = βππ π
and π =ππ π
π = βπ and π = π
NYS COMMON CORE MATHEMATICS CURRICULUM M2 Lesson 10
ALGEBRA II
Lesson 10: Basic Trigonometric Identities from Graphs
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o. What can be said about the π-values for each pair of π-values marked on the graph?
For each pair of π-values, the π-values have the same magnitude and the same sign.
p. Will this relationship hold for any two π-values with the same magnitude and same sign? Explain how you
know.
Yes. If rotation by π radians produces the point (π, π) on the unit circle, then rotation by βπ radians will
produce a point (π, βπ) on the unit circle.
q. Based on these results, make a conjecture by filling in the blank below.
For any real number , ππ¨π¬(βπ) = _______________.
ππ¨π¬(βπ) = ππ¨π¬(π)
r. Test your conjecture by selecting another π-value from the graph and demonstrating
that the identity is true for that value of π.
ππ¨π¬(βππ ) = ππ¨π¬(ππ ) = π
s. Is the cosine function an odd function, even function, or neither? Use the identity
from part (n) to explain.
The cosine function is an even function because ππ¨π¬(βπ) = ππ¨π¬(π) and because the
graph is symmetric with respect to the π-axis.
t. Describe the π-intercepts of the graph of the cosine function.
The graph of the cosine function has π-intercepts at all π-values such that π =π π
+ ππ , where π is an integer.
u. Describe the end behavior of π(π) = ππ¨π¬(π).
As π increases to β or as π decreases to ββ, the cosine function cycles between the values of βπ and π.
During the debriefing, record (or have students record) the key results on the board.
For all π₯: cos(π₯ + 2π) = cos(π₯) cos(π₯ + π) = β cos(π₯) cos(βπ₯) = cos(π₯)
Exploratory Challenge 3 (8 minutes)
Allow students to work in groups through this problem. To save time, provide students with a paper that already has the
two functions graphed together. Debrief after this exercise to ensure that students have completed the identities
correctly.
Scaffolding:
Advanced learners could be asked to write similar identities for the tangent function. For example, does tan(βπ₯) equal tan(π₯) or tan(βπ₯) or neither? Prove it by applying the properties of sine and cosine.
NYS COMMON CORE MATHEMATICS CURRICULUM M2 Lesson 10
ALGEBRA II
Lesson 10: Basic Trigonometric Identities from Graphs
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Exploratory Challenge 3
Graph both π(π) = π¬π’π§(π) and π(π) = ππ¨π¬(π) on the graph below. Then, use the graphs to answer the questions that
follow.
a. List ways in which the graphs of the sine and cosine functions are alike.
Both functions are periodic and have a period of ππ . Both functions have a domain of all real numbers and a
range of [βπ, π]. Both functions cycle between βπ and π as π increases to β or as π decreases to ββ.
Both have similar identities such as π¬π’π§(π + ππ ) = π¬π’π§(π) and ππ¨π¬(π + ππ ) = ππ¨π¬(π).
b. List ways in which the graphs of the sine and cosine functions are different.
We stated above that π(π) = π¬π’π§(π) is an odd function, and π(π) = ππ¨π¬(π) is an even function. Where the
sine function is at a maximum or minimum point, the cosine function has an π-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?
A horizontal shift
d. What is the smallest possible horizontal translation required to make the graph of π(π) = π¬π’π§(π) coincide
with the graph of π(π) = ππ¨π¬(π)?
A horizontal shift π
π units to the left
e. What is the smallest possible horizontal translation required to make the graph of π(π) = ππ¨π¬(π) coincide
with the graph of π(π) = π¬π’π§(π)?
A horizontal shift π
π units to the right
f. Use your answers from parts (d) and (e) to fill in the blank below.
For any real number π, __________ = ππ¨π¬ (π βπ π
).
For any real number π, ________= π¬π’π§ (π +π π
).
π¬π’π§(π) = ππ¨π¬ (π βπ
π)
ππ¨π¬(π) = π¬π’π§ (π +π
π)
Note during the debriefing that there are many different horizontal shifts that could be used in order to make the sine
function coincide with the cosine function or vice versa. Ask students for other examples.
NYS COMMON CORE MATHEMATICS CURRICULUM M2 Lesson 10
ALGEBRA II
Lesson 10: Basic Trigonometric Identities from Graphs
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Closing (2 minutes)
Ask students to explain to a partner or record an explanation on paper. If time permits, ask them to demonstrate using a
specific example.
Explain how the identities For all real numbers π₯, cos(π₯ + 2π) = cos(π₯) and For all real numbers π₯,
sin(π₯ + 2π) = sin(π₯) support the idea that both sine and cosine are periodic.
These identities both state that if 2π is added to the value of π₯, the value of sine or cosine does not
change. This confirms that the functions are periodic and repeat every 2π units.
How do these identities help us to evaluate sine and cosine for various π₯-values? For example, how can we
use the fact that sin (π4
) =β22
to find sin (5π4
)?
sin (5π4
) = sin (π4
+ π) = β sin (π4
) = ββ22
Exit Ticket (5 minutes)
Lesson Summary
For all real numbers π:
π¬π’π§(π + ππ ) = π¬π’π§(π) ππ¨π¬(π + ππ ) = ππ¨π¬(π)
π¬π’π§(π + π ) = βπ¬π’π§(π) ππ¨π¬(π + π ) = βππ¨π¬(π)
π¬π’π§(βπ) = βπ¬π’π§(π) ππ¨π¬(βπ) = βππ¨π¬(π)
π¬π’π§ (π +π π
) = ππ¨π¬(π) ππ¨π¬ (π βπ π
) = π¬π’π§(π)
MP.2
NYS COMMON CORE MATHEMATICS CURRICULUM M2 Lesson 10
ALGEBRA II
Lesson 10: Basic Trigonometric Identities from Graphs
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Name Date
Lesson 10: Basic Trigonometric Identities from Graphs
Exit Ticket
1. Demonstrate how to evaluate cos (8π3
) by using a trigonometric identity.
2. Determine if the following statement is true or false, without using a calculator.
sin (8π
7) = sin (
π
7)
3. If the graph of the cosine function is translated to the right π
2 units, the resulting graph is that of the sine function,
which leads to the identity: For all π₯, sin (π₯) = cos (π₯ βπ2
). Write another identity for sin(π₯) using a different
horizontal shift.
NYS COMMON CORE MATHEMATICS CURRICULUM M2 Lesson 10
ALGEBRA II
Lesson 10: Basic Trigonometric Identities from Graphs
154
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Exit Ticket Sample Solutions
1. Demonstrate how to evaluate ππ¨π¬ (ππ π
) by using a trigonometric identity.
ππ¨π¬ (ππ
π) = ππ¨π¬ (
ππ
π+ ππ ) = ππ¨π¬ (
ππ
π) = β
π
π
2. Determine if the following statement is trueor false, without using a calculator.
π¬π’π§ (ππ π
) = π¬π’π§ (π π
)
False. π¬π’π§ (ππ π
) = π¬π’π§ (π π
+ π ) = βπ¬π’π§ (π π
) β π¬π’π§ (π π
)
3. If the graph of the cosine function is translated to the right π
π units, the resulting graph is that of the sine function,
which leads to the identity: For all π, π¬π’π§(π) = ππ¨π¬ (π β ππ
). Write another identity for π¬π’π§(π) using a different
horizontal shift.
π¬π’π§(π) = ππ¨π¬ (π +ππ
π)
Problem Set Sample Solutions
1. Describe the values of π for which each of the following is true.
a. The cosine function has a relative maximum.
The cosine function has a relative maximum at all π = ππ π, where π is an integer.
b. The sine function has a relative maximum.
The sine function has a relative maximum at all π =π π
+ ππ π, where π is an integer.
2. Without using a calculator, rewrite each of the following in order from least to greatest. Use the graph to explain
your reasoning.
π¬π’π§ (π π
) π¬π’π§ (βππ π
) π¬π’π§ (π π
) π¬π’π§ (βπ π
)
At βπ π
, the sine function takes on its
smallest value, and at π
π the sine function
takes on its largest value. Additionally,
π¬π’π§ (βππ π
) is negative, and π¬π’π§ (π π
) is
positive. Then, these four values in
increasing order are
π¬π’π§ (βπ π
), π¬π’π§ (βππ π
), π¬π’π§ (π π
), π¬π’π§ (π π
).
NYS COMMON CORE MATHEMATICS CURRICULUM M2 Lesson 10
ALGEBRA II
Lesson 10: Basic Trigonometric Identities from Graphs
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3. Without using a calculator, rewrite each of the following in order from least to greatest. Use the graph to explain
your reasoning.
ππ¨π¬ (π π
) ππ¨π¬ (ππ π
) ππ¨π¬ (π π
) ππ¨π¬(ππ )
At ππ , the cosine function takes on its smallest value because ππ = ππ + ππ + π , so ππ¨π¬(ππ ) = ππ¨π¬(π ) = βπ.
We can see that ππ¨π¬ (ππ π
) is negative, ππ¨π¬ (π π
) = π, and ππ¨π¬ (π π
) is positive. Then, these four values in increasing
order are
ππ¨π¬(ππ ), ππ¨π¬ (ππ π
), ππ¨π¬ (π π
), ππ¨π¬ (π π
).
4. Evaluate each of the following without a calculator using a trigonometric identity when needed.
ππ¨π¬ (π
π) ππ¨π¬ (β
π
π) ππ¨π¬ (
ππ
π) ππ¨π¬ (
πππ
π)
βπ
π
βπ
π β
βπ
π
βπ
π
5. Evaluate each of the following without a calculator using a trigonometric identity when needed.
π¬π’π§ (ππ
π) π¬π’π§ (
πππ
π) π¬π’π§ (
ππ
π) π¬π’π§ (
βππ
π)
βπ
π
βπ
π β
βπ
π
βπ
π
6. Use the rotation through π½ radians shown to answer each of the following
questions.
a. Explain why π¬π’π§(βπ½) = β π¬π’π§(π½) for all real numbers π½.
If the initial ray is rotated through βπ½ radians, it is rotated π½
radians in the clockwise direction. If the terminal ray of a rotation
through π½ radians intersects the unit circle at the point (π, π), then
the terminal ray of rotation through βπ½ radians will intersect the
unit circle at the point (π, βπ). Then, π¬π’π§(π½) = π and
π¬π’π§(βπ½) = βπ, so π¬π’π§(βπ½) = βπ¬π’π§(π½).
b. What symmetry does this identity demonstrate about the graph of
π = π¬π’π§(π)?
It demonstrates that the graph of π = π¬π’π§(π) is symmetric with respect to the origin because if the point
(π, π) is on the graph, then the point (βπ, βπ) is also on the graph.
NYS COMMON CORE MATHEMATICS CURRICULUM M2 Lesson 10
ALGEBRA II
Lesson 10: Basic Trigonometric Identities from Graphs
156
This work is derived from Eureka Math β’ and licensed by Great Minds. Β©2015 Great Minds. eureka-math.org This file derived from ALG II-M2-TE-1.3.0-08.2015
This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
7. Use the same rotation shown in Problem 6 to answer each of the following questions.
a. Explain why ππ¨π¬(βπ½) = ππ¨π¬(π½).
If the initial ray is rotated through βπ½ radians, it is rotated π½ radians in the clockwise direction. If the
terminal ray of a rotation through π½ radians intersects the unit circle at the point (π, π), then the terminal ray
of rotation through βπ½ radians will intersect the unit circle at the point (π, βπ). Then, ππ¨π¬(π½) = π and
ππ¨π¬(βπ½) = π, so ππ¨π¬(βπ½) = ππ¨π¬(π½).
b. What symmetry does this identity demonstrate about the graph of π = ππ¨π¬(π)?
It demonstrates that the graph of π = ππ¨π¬(π) is symmetric with respect to the π-axis because if the point
(π, π) is on the graph, then the point (βπ, π) is also on the graph.
8. Find equations of two different functions that can be represented by the graph shown belowβone sine and one
cosineβusing different horizontal transformations.
π = π¬π’π§ (π βπ π
) π = ππ¨π¬(π β π ) (Other correct answers are possible.)
NYS COMMON CORE MATHEMATICS CURRICULUM M2 Lesson 10
ALGEBRA II
Lesson 10: Basic Trigonometric Identities from Graphs
157
This work is derived from Eureka Math β’ and licensed by Great Minds. Β©2015 Great Minds. eureka-math.org This file derived from ALG II-M2-TE-1.3.0-08.2015
This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
9. Find equations of two different functions that can be represented by the graph shown belowβone sine and one
cosineβusing different horizontal translations.
π = ππ¨π¬ (π +π π
) π = π¬π’π§(π β π ) (Other correct answers are possible.)