NYS COMMON CORE MATHEMATICS CURRICULUM M2 Lesson 7
ALGEBRA II
Lesson 7: Secant and the Co-Functions
99
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Lesson 7: Secant and the Co-Functions
Student Outcomes
Students define the secant function and the co-functions in terms of points on the unit circle. They relate the
names for these functions to the geometric relationships among lines, angles, and right triangles in a unit circle
diagram.
Students use reciprocal relationships to relate the trigonometric functions to each other and use these
relationships to evaluate trigonometric functions at multiples of 30, 45, and 60 degrees.
Lesson Notes
The geometry of the unit circle and its related triangles provide a clue as to how the different reciprocal functions got
their names. This lesson draws out the connections among tangent and secant lines of a circle, angle relationships, and
the trigonometric functions. The names for the various trigonometric functions make more sense to students when
viewed through the lens of geometric figures, providing students with an opportunity to practice MP.7. Students make
sense of the domain and range of these functions and use the definitions to evaluate the trigonometric functions for
rotations that are multiples of 30, 45, and 60 degrees.
The relevant vocabulary upon which this lesson is based appears below.
SECANT FUNCTION (description): The secant function,
sec: {π₯ β β | π₯ β 90 + 180π for all integers π} β β
can be defined as follows: Let π be any real number such that π β 90 + 180π for all integers π. In the
Cartesian plane, rotate the initial ray by π degrees about the origin. Intersect the resulting terminal ray with the
unit circle to get a point (π₯π , π¦π). The value of sec(πΒ°) is 1
π₯π.
COSECANT FUNCTION (description): The cosecant function,
csc: {π₯ β β | π₯ β 180π for all integers π} β β
can be defined as follows: Let π be any real number such that π β 180π for all integers π. In the Cartesian
plane, rotate the initial ray by π degrees about the origin. Intersect the resulting terminal ray with the unit
circle to get a point (π₯π , π¦π). The value of csc(πΒ°) is 1
π¦π.
COTANGENT FUNCTION (description): The cotangent function,
cot: {π₯ β β | π₯ β 180π for all integers π} β β
can be defined as follows: Let π be any real number such that π β 180π for all integers π. In the Cartesian
plane, rotate the initial ray by π degrees about the origin. Intersect the resulting terminal ray with the unit
circle to get a point (π₯π , π¦π). The value of cot(πΒ°) is π₯π
π¦π.
NYS COMMON CORE MATHEMATICS CURRICULUM M2 Lesson 7
ALGEBRA II
Lesson 7: Secant and the Co-Functions
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Classwork
Opening Exercise (5 minutes)
Give students a short time to work independently on this Opening Exercise (about 2 minutes). Lead a whole-class
discussion afterward to reinforce the reasons why the different segments have the given measures. If students have
difficulty naming these segments in terms of trigonometric functions they have already studied, remind them of the
conclusions of the previous few lessons.
Opening Exercise
Find the length of each segment below in terms of the value of a trigonometric function.
πΆπΈ = π·πΈ = πΉπΊ =
Debrief this exercise with a short discussion. For the purposes of this section, limit the values of π to be between 0
and 90. Make sure that each student has labeled the proper line segments on his paper as sin(πΒ°), cos(πΒ°), tan (πΒ°),
and sec(πΒ°) before moving on to Example 1.
Why is ππ = sin(πΒ°)? Why is ππ = cos(πΒ°)?
The values of the sine and cosine functions correspond to the π¦- and π₯-coordinates of a point on the
unit circle where the terminal ray intersects the circle after a rotation of π degrees about the origin.
Why is π π = tan(πΒ°)?
In Lesson 5, we used similar triangles to show that π πΜ Μ Μ Μ had length ππ
ππ=
sin(πΒ°)
cos(πΒ°).
This quotient is tan(πΒ°).
NYS COMMON CORE MATHEMATICS CURRICULUM M2 Lesson 7
ALGEBRA II
Lesson 7: Secant and the Co-Functions
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Scaffolding:
Have students use different-
colored highlighters or pencils
to mark proportional
segments.
Consider writing out the
proportional relationships
using the segment names first.
For example, ππ
ππ=
ππ
ππ.
Provide a numeric example for
students to work first.
For more advanced students,
skip the Opening Exercise and
start by stating that a new
function, sec( πΒ°) = ππ , is
being introduced since ππ Μ Μ Μ Μ is
on the secant line that passes
through the center of the
circle.
Since there are ways to calculate lengths for nearly every line segment in this diagram
using the length of the radius, or the cosine, sine, or tangent functions, it makes sense
to find the length of ππ Μ Μ Μ Μ , the line segment on the terminal ray that intersects the
tangent line.
What do you call a line that intersects a circle at more than one point?
It is called a secant line.
In Lesson 6, we saw that π π = tan(πΒ°), where π πΜ Μ Μ Μ lies on the line tangent to
the unit circle at (1,0), which helped to explain how this trigonometric
function got its name. Letβs introduce a new function sec(πΒ°), the secant
of π, to be the length of ππ Μ Μ Μ Μ since this segment is on the secant line that
contains the terminal ray. Then the secant of π is sec(πΒ°) = ππ .
Example 1 (5 minutes)
Direct students to label the appropriate segments on their papers if they have not
already done so. Reproduce the Opening Exercise diagram on chart paper, and label
the segments on the diagram in a different color than the rest of the diagram so they
are easy to see. Refer to this chart often in the next section of the lesson.
Example 1
Use similar triangles to find the value of π¬ππ(π½Β°) in terms of one other trigonometric
function.
Approach 1:
By similar triangles, πΆπΉ
πΆπΊ=
πΆπ·
πΆπΈ, so
π¬ππ(π½Β°)
π=
π
ππ¨π¬(π½Β°); thus, π¬ππ(π½Β°) =
πππ¨π¬(π½Β°)
.
Approach 2:
By similar triangles πΆπΉ
πΆπΊ=
πΆπ·
πΆπΈ, so
π¬ππ(π½Β°)
π=
πππ§(π½Β°)
π¬π’π§(π½Β°)=
π¬π’π§(π½Β°)/ππ¨π¬(π½Β°)
π¬π’π§(π½Β°)=
π¬π’π§(π½Β°)
ππ¨π¬(π½Β°)β
π
π¬π’π§(π½Β°)=
π
ππ¨π¬(π½Β°);
thus, π¬ππ(π½Β°) =π
ππ¨π¬(π½Β°).
Exercise 1 (5 minutes)
Following the same technique as in the previous lesson with the tangent function, use this working definition of the
secant function to extend it outside of the first quadrant. Present the definition of secant, and then ask students to
answer the following questions with a partner or in writing. Circulate around the classroom to informally assess
understanding and provide guidance. Lead a whole-class discussion based on these questions after giving individuals or
partners a few minutes to record their thoughts. Encourage students to revise what they wrote as the discussion
progresses. Start a bulleted list of the main points of this discussion on the board. Encourage students to draw a picture
representing both the circle description of the secant function and its relationship to cos(πΒ°) to assist them in answering
the question. A series of guided questions follows that help in scaffolding this discussion.
NYS COMMON CORE MATHEMATICS CURRICULUM M2 Lesson 7
ALGEBRA II
Lesson 7: Secant and the Co-Functions
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Exercise 1
The definition of the secant function is offered below. Answer the questions to better understand this definition and the
domain and range of this function. Be prepared to discuss your responses with others in your class.
a. What is the domain of the secant function?
The domain of the secant function is all real numbers π½ such that π½ β ππ + ππππ, for all integers π.
b. The domains of the secant and tangent functions are the same. Why?
Both the tangent and secant functions can be written as rational expressions with ππ¨π¬(π½Β°) in the
denominator. That is, πππ§(π½Β°) =π¬π’π§(π½Β°)
ππ¨π¬(π½Β°), and π¬ππ(π½Β°) =
πππ¨π¬(π½Β°)
. Since the restricted values in their domains
occur when the denominator is zero, it makes sense that they have the same domain.
c. What is the range of the secant function? How is this range related to the range of the cosine function?
Since π¬ππ(π½Β°) =π
ππ¨π¬(π½Β°), and βπ β€ ππ¨π¬(π½Β°) β€ π, we see that either π¬ππ(π½Β°) β₯ π, or π¬ππ(π½Β°) β€ βπ.
Thus, the range of the secant function is (ββ, βπ] βͺ [π, β).
d. Is the secant function a periodic function? If so, what is its period?
Yes, its period is πππ.
Discussion (5 minutes)
Use these questions to scaffold the discussion to debrief the preceding exercise.
What are the values of sec(πΒ°) when the terminal ray is horizontal? When the terminal ray is vertical?
When the terminal ray is horizontal, it will coincide with the positive or negative π₯-axis, and
sec(πΒ°) =11
= 1, or sec(πΒ°) =1
β1= β1.
When the terminal ray is vertical, the π₯-coordinate of point π is zero, so sec(πΒ°) =10 , which is
undefined. If we use the geometric interpretation of sec(πΒ°) as the length of the segment from the
origin to the intersection of the terminal ray and the tangent line to the circle at (1,0), then sec(πΒ°) is
undefined because the secant line containing the terminal ray and the tangent line will be parallel and
will not intersect.
Let π½ be any real number.
In the Cartesian plane, rotate the nonnegative
π-axis by π½ degrees about the origin. Intersect this
new ray with the unit circle to get a point (ππ½, ππ½).
If ππ½ β π, then the value of π¬ππ(π½Β°) is π
ππ½. Otherwise,
π¬ππ(π½Β°) is undefined.
In terms of the cosine function, π¬ππ(π½Β°) =π
ππ¨π¬(π½Β°)
for ππ¨π¬(π½Β°) β π.
MP.3
NYS COMMON CORE MATHEMATICS CURRICULUM M2 Lesson 7
ALGEBRA II
Lesson 7: Secant and the Co-Functions
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Name several values of π for which sec(πΒ°) is undefined. Explain your reasoning.
The secant of π is undefined when cos(πΒ°) = 0 or when the terminal ray intersects the unit circle at
the π¦-axis. Some values of π that meet these conditions include 90Β°, 270Β°, and 450Β°.
What is the domain of the secant function?
Answers will vary but should be equivalent to all real numbers except 90 + 180π, where π is an
integer. Students may also use set-builder notation such as {π β β| cos(πΒ°) β 0}.
The domains of the tangent and secant functions are the same. Why?
Approach 1 (circle approach):
The tangent and secant segments are defined based on their intersection, so at times when
they are parallel, these segments do not exist.
Approach 2 (working definition approach):
Both the tangent and secant functions are defined as rational expressions with cos(π) in the
denominator. That is, tan(πΒ°) =sin(πΒ°)
cos(πΒ°), and sec(πΒ°) =
1cos(πΒ°)
. Since the restricted values in
their domains occur when the denominator is zero, it makes sense that they have the same
domain.
How do the values of the secant and cosine functions vary with each other? As cos(πΒ°) gets larger, what
happens to the value of sec(πΒ°)? As cos(πΒ°) gets smaller but stays positive, what happens to the value of
sec(πΒ°)? What about when cos(πΒ°) < 0?
We know that sec(πΒ°) and cos(πΒ°) are reciprocals of each other, so as the magnitude of one gets
larger, the magnitude of the other gets smaller and vice versa. As cos(πΒ°) increases, sec(πΒ°) gets
closer to 1. As cos(πΒ°) decreases but stays positive, sec(πΒ°) increases without bound. When cos(πΒ°) is
negative, as cos(πΒ°) increases but stays negative, sec(πΒ°) decreases without bound. As cos(πΒ°)
decreases, sec(πΒ°) gets closer to β1.
What is the smallest positive value of sec(πΒ°)? Where does this occur?
The smallest positive value of sec(πΒ°) is 1 and occurs when cos(πΒ°) = 1, that is, when π = 360π,
for π an integer.
What is the largest negative value of sec(πΒ°)? Where does this occur?
The largest negative value of sec(πΒ°) = β1 and occurs when cos(πΒ°) = β1; that is,
when π = 180 + 360π, for π an integer.
What is the range of the secant function?
All real numbers outside of the interval (β1,1), including β1 and 1.
Is the secant function a periodic function? If so, what is its period?
The secant function is periodic, and the period is 360Β°.
Exercise 2 (5 minutes)
These questions get students thinking about the origin of the names of the co-functions. They start with the diagram
from the Opening Exercise and ask students how the diagram below compares to it. Then, the point of Example 2 is to
introduce the sine, secant, and tangent ratios of the complement to π and justify why we name these functions cosine,
cosecant, and cotangent.
NYS COMMON CORE MATHEMATICS CURRICULUM M2 Lesson 7
ALGEBRA II
Lesson 7: Secant and the Co-Functions
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Exercise 2
In the diagram, the horizontal blue line is tangent to the unit circle at (π, π).
a. How does this diagram compare to the one given in the Opening Exercise?
It is basically the same diagram with different angles and lengths marked. We can
consider a rotation through π· degrees measured clockwise from the π-axis instead of
a rotation through π½ degrees measured counterclockwise from the π-axis. It is
essentially a reflection of the diagram from the Opening Exercise across the diagonal
line with equation π = π.
b. What is the relationship between π· and π½?
Angles π»πΆπ· and πΊπΆπ· are complements, so π· + π½ = ππ.
c. Which segment in the figure has length π¬π’π§(π½Β°)? Which segment has length ππ¨π¬(π½Β°)?
πΆπΌ = π¬π’π§(π½Β°), and πΌπ· = ππ¨π¬(π½Β°).
d. Which segment in the figure has length π¬π’π§(π·Β°)? Which segment has length ππ¨π¬(π·Β°)?
πΌπ· = π¬π’π§(π·Β°), and πΆπΌ = ππ¨π¬(π½Β°).
e. How can you write π¬π’π§(π½Β°) and ππ¨π¬(π½Β°) in terms of the trigonometric functions of π·?
π¬π’π§(π½Β°) = ππ¨π¬(π·Β°), and ππ¨π¬(π½Β°) = π¬π’π§(π·Β°).
Briefly review the solutions to these exercises before moving on. Reinforce the result of part (e) that the cosine of an
acute angle is the same as the sine of its complement. Ask students to consider whether the trigonometric ratios of the
complement of an angle might be related to the original angle in a similar fashion. Record student responses to the
exercises and any other predictions or thoughts on another sheet of chart paper.
Scaffolding:
Students struggling to see
the complementary
relationships among the
trigonometric functions
may benefit from drawing
the unit circle on a
transparency or patty
paper and reflecting across
the diagonal line π¦ = π₯ so
that the π₯-axis and π¦-axis
are interchanged.
Students may need to be
shown where sin(πΒ°) is on
the diagram as well.
MP.7
NYS COMMON CORE MATHEMATICS CURRICULUM M2 Lesson 7
ALGEBRA II
Lesson 7: Secant and the Co-Functions
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Example 2 (5 minutes)
Use similar triangles to show that if π + π½ = 90, then sec(π½Β°) =1
sin(πΒ°) and tan(π½Β°) =
1tan(πΒ°)
. Depending on studentsβ
level, provide additional scaffolding or just have groups work independently on this. Circulate around the classroom as
they work.
Example 2
The horizontal blue line is tangent to the circle at (π, π).
a. If two angles are complements with measures π· and π½ as shown in the diagram, use similar triangles to show
that π¬ππ(π·Β°) =π
π¬π’π§(π½Β°).
Consider triangles β³ πΆπ·πΌ and β³ πΆπΉπ». Then πΆπΉ
πΆπ·=
π»πΉ
πΌπ·, so
π¬ππ(π·Β°)
π=
πππ§(π·Β°)
ππ¨π¬(π½Β°). Since ππ¨π¬(π½Β°) = π¬π’π§(π·Β°) and
π¬π’π§(π½Β°) = ππ¨π¬(π·Β°), we can write
π¬ππ(π·Β°) =
(π¬π’π§(π·Β°)
ππ¨π¬(π·Β°))
ππ¨π¬(π½Β°)
=
(ππ¨π¬(π½Β°)
π¬π’π§(π½Β°))
ππ¨π¬(π½Β°)
=π
π¬π’π§(π½Β°).
b. If two angles are complements with measures π· and π½ as shown in the diagram, use similar triangles to show
that πππ§(π·Β°) =π
πππ§(π½Β°).
Using triangles β³ πΆπ·πΌ and β³ πΆπΉπ», we have πΉπ»
πΆπ»=
πΌπ·
πΆπΌ. Then
πππ§(π·Β°)
π=
ππ¨π¬(π½Β°)
π¬π’π§(π½Β°). Since πππ§(π½Β°) =
π¬π’π§(π½Β°)
ππ¨π¬(π½Β°),
then π
πππ§(π½Β°)=
ππ¨π¬(π½Β°)
π¬π’π§(π½Β°); so, πππ§(π·Β°) =
ππππ§(π½Β°)
.
NYS COMMON CORE MATHEMATICS CURRICULUM M2 Lesson 7
ALGEBRA II
Lesson 7: Secant and the Co-Functions
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Discussion (7 minutes)
The reciprocal functions 1
sin(πΒ°) and
1
tan(πΒ°) are called the cosecant and cotangent functions. The cosine function has
already been defined in terms of a point on the unit circle. The same can be done with the cosecant and cotangent
functions. All three functions introduced in this lesson are defined below.
For each of the reciprocal functions, use the facts that sec(πΒ°) =1
cos(πΒ°),
csc(πΒ°) =1
sin(πΒ°), and cot(πΒ°) =
1tan(πΒ°)
to extend the definitions of these functions
beyond the first quadrant and to help with the following ideas. Have students work in
pairs to answer the following questions before sharing their conclusions with the class.
Discussion
Definitions of the cosecant and cotangent functions are offered below. Answer the questions to better understand the
definitions and the domains and ranges of these functions. Be prepared to discuss your responses with others in your
class.
The secant, cosecant, and cotangent functions are often
referred to as reciprocal functions. Why do you think these functions are so named?
These functions are the reciprocals of the three main trigonometric functions:
π¬ππ(π½Β°) =π
ππ¨π¬(π½Β°), ππ¬π(π½Β°) =
ππ¬π’π§(π½Β°)
, and ππ¨π(π½Β°) =π
πππ§(π½Β°).
Why are the domains of these functions restricted?
We restrict the domain to prevent division by zero.
We restrict the domain because the geometric shapes defining the functions must make sense (i.e., based on the
intersection of distinct parallel lines).
The domains of the cosecant and cotangent functions are the same. Why?
Both the cosecant and cotangent functions are equal to rational expressions that have π¬π’π§(π½Β°) in the denominator.
What is the range of the cosecant function? How is this range related to the range of the sine function?
The sine function has range [βπ, π], so the cosecant function has range (ββ, βπ] βͺ [π, β). The two ranges only intersect
at βπ and at π.
Let π½ be any real number such that π½ β ππππ, for all
integers π.
In the Cartesian plane, rotate the initial ray by
π½ degrees about the origin. Intersect the resulting
terminal ray with the unit circle to get a point (ππ½, ππ½).
The value of ππ¬π(π½Β°) is π
ππ½.
The value of ππ¨π(π½Β°) is ππ½
ππ½.
Scaffolding:
To help students keep track of
the reciprocal definitions,
mention that every
trigonometric function has a
co-function.
NYS COMMON CORE MATHEMATICS CURRICULUM M2 Lesson 7
ALGEBRA II
Lesson 7: Secant and the Co-Functions
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What is the range of the cotangent function? How is this range related to the range of the tangent function?
The range of the cotangent function is all real numbers. This is the same as the range of the tangent function.
Why are the secant, cosecant, and cotangent functions called reciprocal functions?
Each one is a reciprocal of cosine, sine, or tangent according to their definitions. For example, csc(πΒ°)
is equal to 1
π¦π, the reciprocal of sin(πΒ°).
Which two of the reciprocal functions share the same domain? Why? What is their domain?
The cosecant and cotangent functions share the same domain. Both functions can be written as a
rational expression with sin(πΒ°) in the denominator, so whenever sin(πΒ°) = 0, they are undefined. The
domain is all real numbers except π such that sin(πΒ°) = 0. More specifically, the domain is {π β β|π β 180π, for all integers π}.
What is the smallest positive value of the cosecant function?
As with the secant function, the smallest positive value of the cosecant function is 1. This is because the
cosecant function is the reciprocal of the sine function, which has a maximum value of 1. Also, the
secant and cosecant functions have values based on the length of line segments from the center of the
circle to the intersection with the tangent line. Thus, when positive, they must always be greater than
or equal to the radius of the circle.
What is the greatest negative value of the cosecant function?
The greatest negative value of the cosecant function is β1.
What is the range of the cosecant function? What other trigonometric function has this range?
The range of the cosecant function is all real numbers except between β1 and 1; that is, the range of
the cosecant function is (ββ, β1] βͺ [1, β), which is the same as the range of the secant function.
Can sec(πΒ°) or csc(πΒ°) be a number between 0 and 1? Can cot(πΒ°)? Explain why or why not.
No, sec(πΒ°) and csc(πΒ°) cannot be between 0 and 1, but cot(πΒ°) can. Both the secant and cosecant
functions are reciprocals of functions that range from β1 to 1, while the cotangent function is the
reciprocal of a function that ranges across all real numbers. Whenever tan(πΒ°) > 1, 0 < cot(πΒ°) < 1.
What is the value of cot (90Β°)?
cot(90Β°) =cos(90Β°)
sin(90Β°)=
01
= 0.
How does the range of the cotangent function compare to the range of the tangent function? Why?
The ranges of the tangent and cotangent functions are the same. When tan(πΒ°) is close to 0, cot(πΒ°) is
far from 0, either positive or negative depending on the sign of tan(πΒ°). When tan(πΒ°) is far from 0,
then cot(πΒ°) is close to 0. Finally, when one function is undefined, the other is 0.
Closing (4 minutes)
The secant and cosecant functions are reciprocals of the cosine and sine functions, respectively. The tangent
and cotangent functions are also reciprocals of each other. It is helpful to draw the circle diagram used to
define the tangent and secant functions to ensure mistakes are not made with the relationships derived from
similar triangles.
NYS COMMON CORE MATHEMATICS CURRICULUM M2 Lesson 7
ALGEBRA II
Lesson 7: Secant and the Co-Functions
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This is a good time to summarize all six trigonometric functions. With a partner, in writing, or as a class, have
students summarize the definitions for sine, cosine, tangent, and the three reciprocal functions along with
their domains. Use this as an opportunity to check for any gaps in understanding. Use the following summary
as a model:
Let π½ be any real number. In the Cartesian plane, rotate the initial ray by π½ degrees about the origin.
Intersect the resulting terminal ray with the unit circle to get a point (ππ½, ππ½). Then:
Function Value For any π½ such that β¦ Formula
Sine ππ½ π½ is a real number
Cosine ππ½ π½ is a real number
Tangent ππ½
ππ½
π½ β ππ + ππππ, for all
integers π πππ§(π½Β°) =
π¬π’π§(π½Β°)
ππ¨π¬(π½Β°)
Secant π
ππ½
π½ β ππ + ππππ, for all
integers π π¬ππ(π½Β°) =
π
ππ¨π¬(π½Β°)
Cosecant π
ππ½
π½ β ππππ, for all integers π ππ¬π(π½Β°) =π
π¬π’π§(π½Β°)
Cotangent ππ½
ππ½
π½ β ππππ, for all integers π ππ¨π(π½Β°) =ππ¨π¬(π½Β°)
π¬π’π§(π½Β°)
Exit Ticket (4 minutes)
NYS COMMON CORE MATHEMATICS CURRICULUM M2 Lesson 7
ALGEBRA II
Lesson 7: Secant and the Co-Functions
109
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Name Date
Lesson 7: Secant and the Co-Functions
Exit Ticket
Consider the following diagram, where segment π΄π΅ is tangent to the circle at π·. Right triangles π΅π΄π, π΅ππ·, ππ΄π·, and
ππ·πΆ are similar. Identify each length π΄π·, ππ΄, ππ΅, and π΅π· as one of the following: tan(πΒ°), cot(πΒ°), sec(πΒ°), and
csc(πΒ°).
NYS COMMON CORE MATHEMATICS CURRICULUM M2 Lesson 7
ALGEBRA II
Lesson 7: Secant and the Co-Functions
110
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Exit Ticket Sample Solutions
Consider the following diagram, where segment π¨π© is tangent to the circle at π«. Right triangles π©π¨πΆ, π©πΆπ«, πΆπ¨π«,
and πΆπ«πͺ are similar. Identify each length π¨π«, πΆπ¨, πΆπ©, and π©π« as one of the following: πππ§(π½Β°), ππ¨π(π½Β°), π¬ππ(π½Β°),
and ππ¬π(π½Β°).
Since β³ πΆπ«πͺ ~ β³ πΆπ«π¨, π¨π«
πΆπ«=
πͺπ«
πΆπͺ so
π¨π«
π=
π¬π’π§(π½Β°)
ππ¨π¬(π½Β°) and thus π¨π« = πππ§(π½Β°).
Since β³ πΆπ«πͺ ~ β³ πΆπ«π¨, πΆπ¨
πΆπ«=
πΆπ«
πΆπͺ so
πΆπ¨
π=
π
ππ¨π¬(π½Β°) and thus πΆπ¨ = π¬ππ(π½Β°).
Since β³ πΆπ«πͺ ~ β³ π©π¨πΆ, πΆπ©
πΆπ¨=
πΆπͺ
πͺπ« so
πΆπ©
π¬ππ(π½Β°)=
ππ¨π¬(π½Β°)
π¬π’π§(π½Β°) and thus πΆπ¨ = π¬ππ(π½Β°)
ππ¨π¬(π½Β°)
π¬π’π§(π½Β°)=
ππ¬π’π§(π½Β°)
= ππ¬π(π½Β°).
Since β³ πΆπ«πͺ ~ β³ π©πΆπ«, π©π«
πΆπ«=
πΆπͺ
πͺπ« so
π©π«
π=
ππ¨π¬(π½Β°)
π¬π’π§(π½Β°) and thus π©π« = ππ¨π(π½Β°).
NYS COMMON CORE MATHEMATICS CURRICULUM M2 Lesson 7
ALGEBRA II
Lesson 7: Secant and the Co-Functions
111
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Problem Set Sample Solutions
1. Use the reciprocal interpretations of π¬ππ(π½Β°),
ππ¬π(π½Β°), and ππ¨π(π½Β°) and the unit circle
provided to complete the table.
π½, in degrees π¬ππ(π½Β°) ππ¬π(π½Β°) ππ¨π(π½Β°)
π π Undefined Undefined
ππ πβπ
π π βπ
ππ βπ βπ π
ππ π πβπ
π
βπ
π
ππ Undefined π π
πππ βπ πβπ
π β
βπ
π
πππ βπ Undefined Undefined
πππ ββπ ββπ π
πππ βπ βπβπ
π
βπ
π
πππ Undefined βπ π
πππ βπ ββπ βπ
πππ πβπ
π βπ ββπ
NYS COMMON CORE MATHEMATICS CURRICULUM M2 Lesson 7
ALGEBRA II
Lesson 7: Secant and the Co-Functions
112
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2. Find the following values from the information given.
a. π¬ππ(π½Β°); ππ¨π¬(π½Β°) = π. π
π¬ππ(π½Β°) =
π
π. π=
π
π/ππ=
ππ
π
b. ππ¬π(π½Β°); π¬π’π§(π½Β°) = βπ. ππ
ππ¬π(π½Β°) =
π
βπ. ππ=
π
βπ/ππ= βππ
c. ππ¨π(π½Β°); πππ§(π½Β°) = ππππ
ππ¨π(π½Β°) =
π
πππ§(π½Β°)=
π
ππππ
d. π¬ππ(π½Β°); ππ¨π¬(π½Β°) = βπ. π
π¬ππ(π½Β°) =
π
βπ. π=
π
βπ/ππ= β
ππ
π
e. ππ¬π(π½Β°); π¬π’π§(π½Β°) = π
ππ¬π(π½Β°) =
ππ¬π’π§(π½Β°)
=ππ
, but this is undefined.
f. ππ¨π(π½Β°); πππ§(π½Β°) = βπ. ππππ
ππ¨π(π½Β°) =
π
βπ. ππππ=
π
βπ/πππππ= β
πππππ
π= βππππ
3. Choose three π½ values from the table in Problem π for which π¬ππ(π½Β°), ππ¬π(π½Β°), and πππ§(π½Β°) are defined and not
zero. Show that for these values of π½, π¬ππ(π½Β°)
ππ¬π(π½Β°)= πππ§(π½Β°).
For π½ = ππΒ°, πππ§(π½Β°) =βππ
, and π¬ππ(π½Β°)
ππ¬π(π½Β°)=
πβπ/π
π=
βπ
π. Thus, for π½ = ππ,
π¬ππ(π½Β°)
ππ¬π(π½Β°)= πππ§(π½Β°).
For π½ = ππΒ°, πππ§(π½Β°) = π, and π¬ππ(π½Β°)
ππ¬π(π½Β°)=
βπβπ
= π. Thus, for π½ = ππ, π¬ππ(π½Β°)
ππ¬π(π½Β°)= πππ§(π½Β°).
For π½ = ππΒ°, πππ§(π½Β°) = βπ, and π¬ππ(π½Β°)
ππ¬π(π½Β°)=
π
πβπ
π
=π
βπ= βπ. Thus, for π½ = ππ,
π¬ππ(π½Β°)
ππ¬π(π½Β°)= πππ§(π½Β°).
4. Find the value of π¬ππ(π½Β°)ππ¨π¬(π½Β°) for the following values of π½.
a. π½ = πππ
We know that ππ¨π¬(πππΒ°) = βππ
, so π¬ππ(πππΒ°) = βπ, and then π¬ππ(πππΒ°)ππ¨π¬(πππΒ°) = π.
b. π½ = πππ
We know that ππ¨π¬(πππΒ°) = ββππ
, so sππ(πππΒ°) = ββπ, and then π¬ππ(πππΒ°)ππ¨π¬(πππΒ°) = π.
c. π½ = πππ
We know that ππ¨π¬(πππΒ°) =βππ
, so π¬ππ(πππΒ°) =π
βπ, and then π¬ππ(πππΒ°)ππ¨π¬(πππΒ°) = π.
d. Explain the reasons for the pattern you see in your responses to parts (a)β(c).
If ππ¨π¬(π½Β°) β π, then π¬ππ(π½Β°) =π
ππ¨π¬(π½Β°), so we know that π¬ππ(π½Β°) ππ¨π¬(π½Β°) =
πππ¨π¬(π½Β°)
β ππ¨π¬(π½Β°) = π.
NYS COMMON CORE MATHEMATICS CURRICULUM M2 Lesson 7
ALGEBRA II
Lesson 7: Secant and the Co-Functions
113
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5. Draw a diagram representing the two values of π½ between π and πππ so that π¬π’π§(π½Β°) = ββππ
. Find the values of
πππ§(π½Β°), π¬ππ(π½Β°), and ππ¬π(π½Β°) for each value of π½.
πππ§(πππΒ°) = βπ, π¬ππ(πππΒ°) = βπ, and ππ¬π(πππΒ°) = βπβπ
π.
πππ§(πππΒ°) = ββπ, π¬ππ(πππΒ°) = π, and ππ¬π(πππΒ°) = βπβπ
π.
6. Find the value of (π¬ππ(π½Β°))π
β (πππ§(π½Β°))π when π½ = πππ.
(π¬ππ(πππΒ°))π
β (πππ§(πππΒ°))π
= (βπ)π
β ππ = π
7. Find the value of (ππ¬π(π½Β°))π
β (ππ¨π(π½Β°))π when π½ = πππ.
(ππ¬π(π½Β°))π
β (ππ¨π(π½Β°))π
= (βπ)π β (ββπ)π
= π β π = π
Extension:
8. Using the formulas π¬ππ(π½Β°) =π
ππ¨π¬(π½Β°), ππ¬π(π½Β°) =
ππ¬π’π§(π½Β°)
, and ππ¨π(π½Β°) =π
πππ§(π½Β°), show that
π¬ππ(π½Β°)
ππ¬π(π½Β°)= πππ§(π½Β°),
where these functions are defined and not zero.
π¬ππ(π½Β°)
ππ¬π(π½Β°)=
πππ¨π¬(π½Β°)
ππ¬π’π§(π½Β°)
=π
ππ¨π¬(π½Β°)β π¬π’π§(π½Β°) =
π¬π’π§(π½Β°)
ππ¨π¬(π½Β°)= πππ§(π½Β°)
9. Tara showed that π¬ππ(π½Β°)
ππ¬π(π½Β°)= πππ§(π½Β°), for values of π½ for which the functions are defined and ππ¬π(π½Β°) β π, and
then concluded that π¬ππ(π½Β°) = π¬π’π§(π½Β°) and ππ¬π(π½Β°) = ππ¨π¬(π½Β°). Explain what is wrong with her reasoning.
Just because π¨
π©=
πͺ
π«, this does not mean that π¨ = πͺ and π© = π«. It only means that they are equivalent fractions.
For instance, π
π=
π
ππ. Substituting any value of π½ verifies that Tara is incorrect. Also, since the only values of π½ in
the range of both secant and sine are β π and π, we would have either π¬ππ(π½Β°) = π¬π’π§(π½Β°) = βπ
or π¬ππ(π½Β°) = π¬π’π§(π½Β°) = π. Since the value of the sine function is either β π ππ π, the value of the cosine function
would be zero. Since π¬ππ(π½Β°) =π
ππ¨π¬(π½Β°), we would have ππ¨π¬(π½Β°) = π or ππ¨π¬(π½Β°) = βπ. Since ππ¨π¬(π½Β°) cannot be
simultaneously zero and nonzero, it is impossible to have π¬ππ(π½Β°) = π¬π’π§(π½Β°) and ππ¬π(π½Β°) = ππ¨π¬(π½Β°).
NYS COMMON CORE MATHEMATICS CURRICULUM M2 Lesson 7
ALGEBRA II
Lesson 7: Secant and the Co-Functions
114
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10. From Lesson 6, Ren remembered that the tangent function is odd, meaning that βπππ§(π½Β°) = πππ§(βπ½Β°) for all π½ in
the domain of the tangent function. He concluded because of the relationship between the secant function,
cosecant function, and tangent function developed in Problem 9, it is impossible for both the secant and the
cosecant functions to be odd. Explain why he is correct.
If we assume that both the secant and cosecant functions are odd, then the tangent function could not be odd. That
is, we would get
πππ§(βπ½Β°) =π¬ππ(βπ½Β°)
ππ¬π(βπ½Β°)=
βπ¬ππ(π½Β°)
βππ¬π(π½Β°)=
π¬ππ(π½Β°)
ππ¬π(π½Β°)= πππ§(π½Β°),
but that would contradict βπππ§(π½Β°) = πππ§(π½Β°). Thus, it is impossible for both the secant and cosecant functions to
be odd.