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Eureka Math™ Homework Helper

2015–2016

Algebra IIModule 2

Lessons 1–17

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2015-16

M2 ALGEBRA II

Lesson 1: Ferris Wheels—Tracking the Height of a Passenger Car


Suppose that a Ferris wheel is 30 feet in diameter and rotates counterclockwise. When a passenger car is at the bottom of the wheel, it is located 4 feet above the ground.

a. Sketch a graph of a function that represents the height of a passenger car that starts at the 9 o’clock position on the wheel for one turn.

b. The sketch you created in part (a) represents a graph of a function. What is the domain of the function? What is the range?

Domain: [𝟎𝟎,𝟑𝟑𝟑𝟑𝟎𝟎] Range: [𝟒𝟒,𝟑𝟑𝟒𝟒]

c. Describe how the graph of the function in part (a) would change if you sketched the graph for three turns.

The graph would repeat itself twice on the intervals [𝟑𝟑𝟑𝟑𝟎𝟎,𝟕𝟕𝟕𝟕𝟎𝟎] and [𝟕𝟕𝟕𝟕𝟎𝟎,𝟏𝟏𝟎𝟎𝟏𝟏𝟎𝟎].

d. Describe how the function in part (a) would change if the diameter of the wheel was 60 feet and the car was 3 feet above the ground at its lowest point.

The range would change to [𝟑𝟑,𝟑𝟑𝟑𝟑]. The graph’s minimum value would be 𝟑𝟑 instead of 𝟒𝟒, and the maximum value would be 𝟑𝟑𝟑𝟑 instead of 𝟒𝟒𝟕𝟕.

I can plot the height of the Ferris wheel after every 1

4 rotation and

then connect the points with a smooth curve to represent its height for one complete turn.

I know the height of the car will repeat the pattern for the second and third rotations.

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© 2015 Great Minds eureka-math.orgALG II-M2-HWH-1.3.0-09.2015

Homework Helper A Story of Functions

2015-16

M2 ALGEBRA II


e. Describe how the original graph would change if the Ferris wheel traveled clockwise.

The graph would increase to a height of 𝟑𝟑𝟒𝟒 feet after 𝟏𝟏𝟒𝟒 of a turn and then decrease to a height of

𝟒𝟒 feet after 𝟑𝟑𝟒𝟒 of a turn. While the domain and range of the function would not change, the actual

correspondence between rotations and height would change since the direction of rotation is different.

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M2

Lesson 2: The Height and Co-Height Functions of a Ferris Wheel

ALGEBRA II


The Millennium Wheel in London, with an overall height of 443 feet, was the tallest Ferris wheel in the world at the time of its construction in 1999. The diameter of the wheel is 394 feet.

a. Create a diagram that shows the position of a passenger car on the Millennium Wheel as it rotates counterclockwise at 45-degree intervals.

b. On the same set of axes, sketch graphs of the height and co-height functions for a passenger car starting at the 3 o’clock position on the Millennium Wheel and completing one turn.

The blue graph represents the height, and the green graph represents the co-height.

I graphed the position of the passenger car where (0, 0) represents the position of the center of the wheel.

The height graph (in blue) represents the height relative to the center of the wheel. The maximum height is 197 feet, and the minimum height is −197 feet. The co-height graph (in green) represents the horizontal distance from the vertical line through the center of the wheel.

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M2


ALGEBRA II

c. How can you use the graphs to determine the radius of the wheel?

The radius is the distance from the center of the wheel to any point on the graph. This is the vertical distance from the horizontal axis to a maximum or minimum point on the graph. For the Millennium Wheel, the maximum point has coordinates (𝟎𝟎,𝟏𝟏𝟏𝟏𝟏𝟏). Therefore, the radius is 𝟏𝟏𝟏𝟏𝟏𝟏 feet.

d. At which amounts of rotation are the values of the height and co-height functions equal?

The graphs of the height and co-height functions intersect at 𝒙𝒙 = 𝟒𝟒𝟒𝟒 and 𝒙𝒙 = 𝟐𝟐𝟐𝟐𝟒𝟒, which means the height and co-height are equal when the car has rotated counterclockwise by 𝟒𝟒𝟒𝟒° or 𝟐𝟐𝟐𝟐𝟒𝟒°.

e. Suppose operators ran the Millennium Wheel in reverse one day. On the same set of axes, sketch graphs of the height and co-height functions for a passenger car starting at the 3 o’clock position on the Millennium Wheel and completing two turns clockwise.

Because the wheel is running backward, the height goes down initially before beginning to go up. This makes the new height graph appear to be upside down from our original graph. The co-height function is not changed because the horizontal distance from the center of the wheel changes in the same way whether the car goes up or down.

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M2 ALGEBRA II

Lesson 3: The Motion of the Moon, Sun, and Stars—Motivating Mathematics

Lesson 3: The Motion of the Moon, Sun, and Stars—Motivating

Mathematics

1. An Indian astronomer noted that the angle of his line of sight to Jupiter measured 60°. The average distance from Earth to Jupiter is 489 million miles. Use Aryabhata’s jya table to estimate the apparent height of Jupiter. Round your answer to the nearest million miles.

𝜽𝜽, in degrees 48

34 52

12 56

14 60 63

34 67

12

𝐣𝐣𝐣𝐣𝐣𝐣(𝜽𝜽°) 2585 2728 2859 2978 3084 3177

By the table, 𝐣𝐣𝐣𝐣𝐣𝐣(𝟔𝟔𝟔𝟔°) = 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐. Since 𝐣𝐣𝐣𝐣𝐣𝐣(𝟔𝟔𝟔𝟔°) ∙ 𝟒𝟒𝟐𝟐𝟐𝟐𝟑𝟑𝟒𝟒𝟑𝟑𝟐𝟐

= 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐 ∙ 𝟒𝟒𝟐𝟐𝟐𝟐𝟑𝟑𝟒𝟒𝟑𝟑𝟐𝟐

≈ 𝟒𝟒𝟐𝟐𝟒𝟒, the apparent height is 𝟒𝟒𝟐𝟐𝟒𝟒 million miles.

I know that jya(𝜃𝜃°) provides the apparent height of the sun for a rotation of 𝜃𝜃° from the eastern horizon. Since the calculation is based on the distance from Earth to the Sun (approximately 3438 million miles), I need to multiply jya(60°) by the ratio of the distance from Earth to Jupiter to the distance from Earth to the Sun to calculate the apparent height of Jupiter.

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M2 ALGEBRA II

Lesson 3: The Motion of the Moon, Sun, and Stars—Motivating Mathematics

2. While Venus does not orbit Earth, it does pass across the Earth’s sky as it orbits the Sun. Apply Aryabhata’s method to estimate the height of Venus, supposing Venus completes one full rotation around Earth’s night sky roughly every 584 days. The average distance from Venus to the Earth is 162 million miles. Suppose that Venus is just visible on the eastern horizon on January 1, so its apparent height at that time would be 0 miles above the horizon. Use Aryabhata’s jya table to find the apparent height of Venus above the observer at the times listed in the table below. Round to the nearest million miles.

Time (days since January 1)

Angle of Elevation 𝜽𝜽, in degrees

𝐣𝐣𝐣𝐣𝐣𝐣(𝜽𝜽°) Height Calculation

𝐣𝐣𝐣𝐣𝐣𝐣(𝜽𝜽°) ∙𝟏𝟏𝟔𝟔𝟐𝟐𝟑𝟑𝟒𝟒𝟑𝟑𝟐𝟐

Apparent Height,

in millions of miles

0 𝟔𝟔 𝟔𝟔 𝐣𝐣𝐣𝐣𝐣𝐣(𝟔𝟔°) ∙𝟏𝟏𝟔𝟔𝟐𝟐𝟑𝟑𝟒𝟒𝟑𝟑𝟐𝟐 = 𝟔𝟔 𝟔𝟔

73 𝟒𝟒𝟒𝟒 𝟐𝟐𝟒𝟒𝟑𝟑𝟏𝟏 𝐣𝐣𝐣𝐣𝐣𝐣(𝟒𝟒𝟒𝟒°) ∙𝟏𝟏𝟔𝟔𝟐𝟐𝟑𝟑𝟒𝟒𝟑𝟑𝟐𝟐 ≈ 𝟏𝟏𝟏𝟏𝟒𝟒 𝟏𝟏𝟏𝟏𝟒𝟒

146 𝟐𝟐𝟔𝟔 𝟑𝟑𝟒𝟒𝟑𝟑𝟐𝟐 𝐣𝐣𝐣𝐣𝐣𝐣(𝟐𝟐𝟔𝟔°) ∙𝟏𝟏𝟔𝟔𝟐𝟐𝟑𝟑𝟒𝟒𝟑𝟑𝟐𝟐 ≈ 𝟏𝟏𝟔𝟔𝟐𝟐 𝟏𝟏𝟔𝟔𝟐𝟐

219 𝟏𝟏𝟑𝟑𝟒𝟒 𝟐𝟐𝟒𝟒𝟑𝟑𝟏𝟏 𝐣𝐣𝐣𝐣𝐣𝐣(𝟏𝟏𝟑𝟑𝟒𝟒°) ∙𝟏𝟏𝟔𝟔𝟐𝟐𝟑𝟑𝟒𝟒𝟑𝟑𝟐𝟐 ≈ 𝟏𝟏𝟏𝟏𝟒𝟒 𝟏𝟏𝟏𝟏𝟒𝟒

292 𝟏𝟏𝟐𝟐𝟔𝟔 𝟔𝟔 𝐣𝐣𝐣𝐣𝐣𝐣(𝟏𝟏𝟐𝟐𝟔𝟔°) ∙𝟏𝟏𝟔𝟔𝟐𝟐𝟑𝟑𝟒𝟒𝟑𝟑𝟐𝟐 = 𝟔𝟔 𝟔𝟔

I notice the time increases by 73 days for each successive row. Since we assume Venus returns to its initial position after 584 days, and 73

584= 1

8, each

73-day interval corresponds to 18 of a rotation.

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M2 ALGEBRA II

Lesson 4: From Circle-ometry to Trigonometry


1. Fill in the chart. Write in the reference angles and the values of the sine and cosine functions for the indicated values of 𝜃𝜃.

Amount of Rotation, 𝜽𝜽, in degrees

Measure of Reference Angle,

in degrees 𝐜𝐜𝐨𝐨𝐨𝐨(𝜽𝜽°)

𝐨𝐨𝐬𝐬𝐬𝐬(𝛉𝛉°)

150 𝟑𝟑𝟑𝟑 −√𝟑𝟑𝟐𝟐

𝟏𝟏𝟐𝟐

240 𝟔𝟔𝟑𝟑 −𝟏𝟏𝟐𝟐

−√𝟑𝟑𝟐𝟐

315 𝟒𝟒𝟒𝟒 √𝟐𝟐𝟐𝟐

−√𝟐𝟐𝟐𝟐

I can use a special right triangle to find the magnitude of sin(𝜃𝜃°) and cos(𝜃𝜃°) for each reference angle. I can then determine the quadrant in which the terminal ray lies to help me determine the sign of the functions for the given rotational values of 𝜃𝜃.

I know the reference angle is the acute angle formed by the terminal ray and the 𝑥𝑥-axis.

I know the reference angle for 150° has measure 180° − 150°, which is 30°.



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M2 ALGEBRA II


2. Using geometry, Jake correctly calculated that cos(22.5°) = 12�2 + √2. Based on this information, fill in

the chart.

For 𝜽𝜽 = 𝟐𝟐𝟐𝟐.𝟒𝟒:

𝒙𝒙𝟐𝟐 + 𝒚𝒚𝟐𝟐 = 𝟏𝟏 𝒚𝒚𝟐𝟐 = 𝟏𝟏 − 𝒙𝒙𝟐𝟐

= 𝟏𝟏 − �𝟏𝟏𝟐𝟐�𝟐𝟐 + √𝟐𝟐 �

𝟐𝟐

= 𝟏𝟏 −𝟏𝟏𝟒𝟒 �𝟐𝟐 + √𝟐𝟐�

=𝟒𝟒 − 𝟐𝟐 − √𝟐𝟐

𝟒𝟒

=𝟐𝟐 − √𝟐𝟐𝟒𝟒

𝒚𝒚 = ±�𝟐𝟐 − √𝟐𝟐𝟒𝟒

= ±�𝟐𝟐 − √𝟐𝟐

𝟐𝟐

= ±𝟏𝟏𝟐𝟐�𝟐𝟐 − √𝟐𝟐

Because the terminal ray is in the first quadrant, 𝐨𝐨𝐬𝐬𝐬𝐬(𝟐𝟐𝟐𝟐.𝟒𝟒°) > 𝟑𝟑. Thus, 𝐨𝐨𝐬𝐬𝐬𝐬(𝟐𝟐𝟐𝟐.𝟒𝟒°) = 𝟏𝟏

𝟐𝟐�𝟐𝟐 − √𝟐𝟐.

For 𝜽𝜽 = 𝟐𝟐𝟑𝟑𝟐𝟐.𝟒𝟒:

𝒙𝒙𝟐𝟐 + 𝒚𝒚𝟐𝟐 = 𝟏𝟏 𝒚𝒚𝟐𝟐 = 𝟏𝟏 − 𝒙𝒙𝟐𝟐

= 𝟏𝟏 − �−𝟏𝟏𝟐𝟐�𝟐𝟐 + √𝟐𝟐 �

𝟐𝟐

= 𝟏𝟏 −𝟏𝟏𝟒𝟒 �𝟐𝟐 + √𝟐𝟐�

=𝟒𝟒 − 𝟐𝟐 − √𝟐𝟐

𝟒𝟒

=𝟐𝟐 − √𝟐𝟐𝟒𝟒

𝒚𝒚 = ±�𝟐𝟐 − √𝟐𝟐𝟒𝟒

= ±�𝟐𝟐 − √𝟐𝟐

𝟐𝟐

= ±𝟏𝟏𝟐𝟐�𝟐𝟐 − √𝟐𝟐

Because the terminal ray is in the third quadrant, 𝐨𝐨𝐬𝐬𝐬𝐬(𝟐𝟐𝟑𝟑𝟐𝟐.𝟒𝟒°) < 𝟑𝟑. Thus, 𝐨𝐨𝐬𝐬𝐬𝐬(𝟐𝟐𝟑𝟑𝟐𝟐.𝟒𝟒°) = −𝟏𝟏

𝟐𝟐�𝟐𝟐 − √𝟐𝟐.

Amount of Rotation,

𝜽𝜽, in degrees 𝐜𝐜𝐨𝐨𝐨𝐨(𝜽𝜽°) 𝐨𝐨𝐬𝐬𝐬𝐬(𝜽𝜽°)

22.5 𝟏𝟏𝟐𝟐�𝟐𝟐 + √𝟐𝟐

𝟏𝟏𝟐𝟐�𝟐𝟐 − √𝟐𝟐

202.5 −𝟏𝟏𝟐𝟐�𝟐𝟐 + √𝟐𝟐 −

𝟏𝟏𝟐𝟐�𝟐𝟐 − √𝟐𝟐

In this right triangle, 𝑥𝑥 = cos(22.5°) and 𝑦𝑦 = sin(22.5°).

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M2 ALGEBRA II


3. The vertices of △ 𝐴𝐴𝐴𝐴𝐴𝐴 have coordinates 𝐴𝐴(0,0), 𝐴𝐴(12,0), and 𝐴𝐴(12,5). a. Argue that △ 𝐴𝐴𝐴𝐴𝐴𝐴 is a right triangle.

It is apparent that 𝑨𝑨𝑨𝑨�� is horizontal and 𝑨𝑨𝑩𝑩�� is vertical, so △ 𝑨𝑨𝑨𝑨𝑩𝑩 is a right triangle.

b. What are the coordinates where the hypotenuse of △ 𝐴𝐴𝐴𝐴𝐴𝐴 intersects the unit circle with equation 𝑥𝑥2 + 𝑦𝑦2 = 1?

Using similar triangles, the coordinates of the point where the hypotenuse of △ 𝑨𝑨𝑨𝑨𝑩𝑩 intersects the

unit circle are �𝟏𝟏𝟐𝟐𝟏𝟏𝟑𝟑

, 𝟒𝟒𝟏𝟏𝟑𝟑�.

c. Let 𝜃𝜃 denote the number of degrees of rotation from 𝐴𝐴𝐴𝐴��⃗ to 𝐴𝐴𝐴𝐴��⃗ . Calculate sin(𝜃𝜃°) and cos(𝜃𝜃°).

By the answer to part (b), 𝐨𝐨𝐬𝐬𝐬𝐬(𝜽𝜽°) = 𝟒𝟒𝟏𝟏𝟑𝟑

, and 𝐜𝐜𝐨𝐨𝐨𝐨(𝜽𝜽°) = 𝟏𝟏𝟐𝟐𝟏𝟏𝟑𝟑

.

I know the coordinates where the hypotenuse of △ 𝐴𝐴𝐴𝐴𝐴𝐴 intersects the unit circle can be used to form right triangle △ 𝐴𝐴𝐴𝐴𝐴𝐴, which is similar to △ 𝐴𝐴𝐴𝐴𝐴𝐴 with hypotenuse 1.

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M2 ALGEBRA II

Lesson 5: Extending the Domain of Sine and Cosine to All Real Numbers

Lesson 5: Extending the Domain of Sine and Cosine to All Real

Numbers

1. Fill in the chart. Write in the measures of the reference angles and the values of the sine and cosine functions for the indicated values of 𝜃𝜃.

Number of Degrees of Rotation,

𝜽𝜽

Quadrant

Measure of Reference Angle, in degrees

𝐜𝐜𝐜𝐜𝐜𝐜(𝜽𝜽°) 𝐜𝐜𝐬𝐬𝐬𝐬(𝜽𝜽°)

375 I 15 12�2 + √3

𝟏𝟏𝟐𝟐�𝟐𝟐 − √𝟑𝟑

525 II 𝟏𝟏𝟏𝟏 −𝟏𝟏𝟐𝟐�𝟐𝟐 + √𝟑𝟑

𝟏𝟏𝟐𝟐�𝟐𝟐 − √𝟑𝟑

600 III 𝟔𝟔𝟔𝟔 −𝟏𝟏𝟐𝟐

−√𝟑𝟑𝟐𝟐

720 None 𝟔𝟔 𝟏𝟏 𝟔𝟔

930 III 𝟑𝟑𝟔𝟔 −√𝟑𝟑𝟐𝟐

−𝟏𝟏𝟐𝟐

1395 IV 𝟒𝟒𝟏𝟏 √𝟐𝟐𝟐𝟐

−√𝟐𝟐𝟐𝟐

1710 None 𝟗𝟗𝟔𝟔 𝟔𝟔 −𝟏𝟏

I know that a counterclockwise rotation by 𝜃𝜃° produces the same terminal ray as a rotation of (𝜃𝜃 + 360)°. I can then calculate the measure of the reference angle and the outputs for cos(𝜃𝜃°) and sin(𝜃𝜃°) using the methods from Lesson 4.

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M2 ALGEBRA II

Lesson 5: Extending the Domain of Sine and Cosine to All Real Numbers

2. Suppose 𝜃𝜃 represents a number of degrees of rotation and that sin(𝜃𝜃°) = −√22

. List the first six possible positive values that 𝜃𝜃 can take and three possible negative values.

Positive values of 𝜽𝜽: 𝟐𝟐𝟐𝟐𝟏𝟏,𝟑𝟑𝟏𝟏𝟏𝟏,𝟏𝟏𝟓𝟓𝟏𝟏,𝟔𝟔𝟔𝟔𝟏𝟏,𝟗𝟗𝟒𝟒𝟏𝟏,𝟏𝟏𝟔𝟔𝟑𝟑𝟏𝟏

Negative values of 𝜽𝜽: −𝟒𝟒𝟏𝟏,−𝟏𝟏𝟑𝟑𝟏𝟏,−𝟒𝟒𝟔𝟔𝟏𝟏

3. Suppose 𝜃𝜃 represents a number of degrees of rotation. Is it possible that cos(𝜃𝜃°) = 𝟐𝟐𝟑𝟑 and sin(𝜃𝜃°) = 𝟏𝟏

𝟑𝟑?

No. If 𝐜𝐜𝐜𝐜𝐜𝐜(𝜽𝜽°) = 𝟐𝟐𝟑𝟑 and 𝐜𝐜𝐬𝐬𝐬𝐬(𝜽𝜽°) = 𝟏𝟏

𝟑𝟑, then 𝒙𝒙 = 𝟐𝟐

𝟑𝟑 and 𝒚𝒚 = 𝟏𝟏

𝟑𝟑 are

coordinates of a point on the unit circle. However,

�𝟐𝟐𝟑𝟑�𝟐𝟐

+ �𝟏𝟏𝟑𝟑�𝟐𝟐≠ 𝟏𝟏, so this point doesn’t lie on the unit circle.

I know sin(𝜃𝜃°) = √22

for a counterclockwise rotation by 45° and that sin(𝜃𝜃°) is negative in Quadrants III and IV.

I know that, for any point (𝑥𝑥,𝑦𝑦) on the unit circle, 𝑥𝑥2 + 𝑦𝑦2 = 1.

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M2 ALGEBRA II

Lesson 6: Why Call It Tangent?


1. Label the missing side lengths in the given triangle, and use the triangle to find the specified values of the tangent function.

a. tan(30°)

𝐭𝐭𝐭𝐭𝐭𝐭(𝟑𝟑𝟑𝟑°) =𝐬𝐬𝐬𝐬𝐭𝐭(𝟑𝟑𝟑𝟑°)𝐜𝐜𝐜𝐜𝐬𝐬(𝟑𝟑𝟑𝟑°) =

�𝟏𝟏𝟐𝟐�

�√𝟑𝟑𝟐𝟐 �=

𝟏𝟏√𝟑𝟑

=√𝟑𝟑𝟑𝟑

b. tan(60°)

𝐭𝐭𝐭𝐭𝐭𝐭(𝟔𝟔𝟑𝟑°) =𝐬𝐬𝐬𝐬𝐭𝐭(𝟔𝟔𝟑𝟑°) 𝐜𝐜𝐜𝐜𝐬𝐬(𝟔𝟔𝟑𝟑°)

=�√𝟑𝟑𝟐𝟐 �

�𝟏𝟏𝟐𝟐�= √𝟑𝟑

I remember that the side lengths of a 30°-60°-90° right

triangle are 12

, √𝟑𝟑𝟐𝟐

, and 1.

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2. Let 𝜃𝜃 be any real number. In the Cartesian plane, rotate the initial ray by 𝜃𝜃 degrees about the origin. Intersect the resulting terminal ray with the unit circle to get point 𝑃𝑃(𝑥𝑥𝜃𝜃,𝑦𝑦𝜃𝜃). a. Complete the table by finding the slope of the line through the origin and the point 𝑃𝑃.

𝜽𝜽, in degrees Coordinates of point 𝑷𝑷 Slope

0 (𝟏𝟏,𝟑𝟑) 𝟑𝟑

30 �√𝟑𝟑𝟐𝟐

,𝟏𝟏𝟐𝟐�

√𝟑𝟑𝟑𝟑

60 � 𝟏𝟏𝟐𝟐

,√𝟑𝟑𝟐𝟐� √𝟑𝟑

90 (𝟑𝟑,𝟏𝟏) Undefined

240 �− 𝟏𝟏𝟐𝟐

,−√𝟑𝟑𝟐𝟐� √𝟑𝟑

330 �√𝟑𝟑𝟐𝟐

,− 𝟏𝟏𝟐𝟐� −

√𝟑𝟑𝟑𝟑

b. Explain how these slopes are related to the tangent function.

The slope of the line through the origin and 𝑷𝑷(𝒙𝒙𝜽𝜽,𝒚𝒚𝜽𝜽) is 𝐭𝐭𝐭𝐭𝐭𝐭(𝜽𝜽°).

3. Consider the following diagram of a circle of radius 𝑟𝑟 centered at the origin. The line ℓ is tangent to the circle at 𝑆𝑆(𝑟𝑟, 0), so ℓ is perpendicular to the 𝑥𝑥-axis. For the given values of 𝑟𝑟 and 𝜃𝜃, find 𝑡𝑡.

I can use the reference angles for these values of 𝜃𝜃 to help me determine the coordinates of point 𝑃𝑃.

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a. 𝜃𝜃 = 45, 𝑟𝑟 = 10

𝐭𝐭𝐭𝐭𝐭𝐭(𝟒𝟒𝟒𝟒°) =𝒕𝒕𝒓𝒓

=𝒕𝒕𝟏𝟏𝟑𝟑

𝒕𝒕 = 𝟏𝟏𝟑𝟑 ⋅ 𝐭𝐭𝐭𝐭𝐭𝐭(𝟒𝟒𝟒𝟒°) = 𝟏𝟏𝟑𝟑(𝟏𝟏) = 𝟏𝟏𝟑𝟑

b. 𝜃𝜃 = 90, 𝑟𝑟 = 2

Lines 𝑶𝑶𝑶𝑶 and 𝓵𝓵 are distinct parallel lines when 𝜽𝜽 = 𝟗𝟗𝟑𝟑. Thus, they will never intersect, and the line segment defined by their intersection does not exist, which means 𝒕𝒕 is undefined.

4. Using what you know of the tangent function, show that tan(𝜃𝜃°) = tan((180 + 𝜃𝜃)°) for 𝜃𝜃 ≠ 90 + 180𝑘𝑘, for all integers 𝑘𝑘. A counterclockwise rotation 𝜽𝜽 of the positive 𝒙𝒙-axis produces a terminal ray that intersects the unit circle at (𝒙𝒙,𝒚𝒚), and 𝐭𝐭𝐭𝐭𝐭𝐭(𝜽𝜽°) = 𝒚𝒚

𝒙𝒙. A rotation by (𝟏𝟏𝟏𝟏𝟑𝟑 + 𝜽𝜽)° produces a terminal ray that intersects the

unit circle at (−𝒙𝒙,− 𝒚𝒚), and 𝐭𝐭𝐭𝐭𝐭𝐭�(𝟏𝟏𝟏𝟏𝟑𝟑+ 𝜽𝜽)°� = −𝒚𝒚−𝒙𝒙

= 𝒚𝒚𝒙𝒙, which is equivalent to 𝐭𝐭𝐭𝐭𝐭𝐭(𝜽𝜽°).

I know a counterclockwise rotation of the point (𝑥𝑥,𝑦𝑦) by 180° about the origin results in an image point (−𝑥𝑥,−𝑦𝑦).

I can use an isosceles right triangle model to help me calculate tan(45°).

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Lesson 7: Secant and the Co-Functions


1. Use the reciprocal interpretations of sec(𝜃𝜃°), csc(𝜃𝜃°), and cot(𝜃𝜃°) and the unit circle provided to complete the table.

𝜽𝜽, in degrees 𝐬𝐬𝐬𝐬𝐬𝐬(𝜽𝜽°) 𝐬𝐬𝐬𝐬𝐬𝐬(𝜽𝜽°) 𝐬𝐬𝐜𝐜𝐜𝐜(𝜽𝜽°)

135 −√𝟐𝟐 √𝟐𝟐 −𝟏𝟏

300 𝟐𝟐 −𝟐𝟐√𝟑𝟑𝟑𝟑

−√𝟑𝟑𝟑𝟑

360 𝟏𝟏 Undefined Undefined

2. Find the following values from the information given. a. csc(𝜃𝜃°); sin(𝜃𝜃°) = −0.2

𝐬𝐬𝐬𝐬𝐬𝐬(𝜽𝜽°) =𝟏𝟏

𝐬𝐬𝐬𝐬𝐬𝐬(𝜽𝜽°)=

𝟏𝟏−𝟎𝟎.𝟐𝟐

=𝟏𝟏

−𝟏𝟏/𝟓𝟓= −𝟓𝟓

I know that sec(𝜃𝜃°) = 1cos(𝜃𝜃°),

csc(𝜃𝜃°) = 1sin(𝜃𝜃°), and cot(𝜃𝜃°) = 1

tan(𝜃𝜃°).

I know that for a point (𝑥𝑥,𝑦𝑦) on the unit circle, 𝑥𝑥 = cos(𝜃𝜃°), 𝑦𝑦 = sin(𝜃𝜃°) , and 𝑦𝑦

𝑥𝑥= tan(𝜃𝜃°).

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b. sec(𝜃𝜃°); cos(𝜃𝜃°) = 0

𝐬𝐬𝐬𝐬𝐬𝐬(𝜽𝜽°) = 𝟏𝟏𝐬𝐬𝐜𝐜𝐬𝐬(𝜽𝜽°)

= 𝟏𝟏𝟎𝟎, which is undefined.

c. tan(𝜃𝜃°) cot(𝜃𝜃°); 𝜃𝜃 = 300

𝐜𝐜𝐭𝐭𝐬𝐬(𝟑𝟑𝟎𝟎𝟎𝟎°) 𝐬𝐬𝐜𝐜𝐜𝐜(𝟑𝟑𝟎𝟎𝟎𝟎°) = −√𝟑𝟑�−𝟏𝟏√𝟑𝟑

� = 𝟏𝟏

3. From Lesson 6, Ren remembered that the tangent function is odd, meaning that for all 𝜃𝜃 in the domain of the tangent function −tan(𝜃𝜃°) = tan(−𝜃𝜃°). He formed a conjecture that the cotangent function is also odd, meaning that −cot(𝜃𝜃°) = cot(−𝜃𝜃°). Explain why he is correct.

Since 𝐬𝐬𝐜𝐜𝐜𝐜(𝜽𝜽°) = 𝟏𝟏𝐜𝐜𝐭𝐭𝐬𝐬(𝜽𝜽°), it follows that 𝐬𝐬𝐜𝐜𝐜𝐜(−𝜽𝜽°) = 𝟏𝟏

𝐜𝐜𝐭𝐭𝐬𝐬(−𝜽𝜽°) = 𝟏𝟏−𝐜𝐜𝐭𝐭𝐬𝐬(𝜽𝜽°) = −� 𝟏𝟏

𝐜𝐜𝐭𝐭𝐬𝐬(𝜽𝜽°)� = −𝐬𝐬𝐜𝐜𝐜𝐜(𝜽𝜽°).

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Lesson 8: Graphing the Sine and Cosine Functions


1. Graph the sine function on the interval [0, 360] showing all key points of the graph (horizontal and vertical intercepts and maximum and minimum points). Then, use the graph to answer each of the following questions.

a. On the interval [0, 360], what is the range of the function? Why?

The range is [−𝟏𝟏,𝟏𝟏]. On the unit circle, the coordinates of the point at the bottom are (𝟎𝟎,−𝟏𝟏), and the coordinates of the point at the top are (𝟎𝟎,𝟏𝟏), which means all the 𝒚𝒚-coordinates of points on the unit circle lie between −𝟏𝟏 and 𝟏𝟏.

b. Arrange the following values in order from smallest to largest by using their location on the graph. sin(100°) sin(15°) sin(285°) sin(340°)

𝐬𝐬𝐬𝐬𝐬𝐬(𝟐𝟐𝟐𝟐𝟐𝟐°), 𝐬𝐬𝐬𝐬𝐬𝐬(𝟑𝟑𝟑𝟑𝟎𝟎°), 𝐬𝐬𝐬𝐬𝐬𝐬(𝟏𝟏𝟐𝟐°), 𝐬𝐬𝐬𝐬𝐬𝐬(𝟏𝟏𝟎𝟎𝟎𝟎°)

I can see that the graph of the sine function is positive on the interval (0, 180). The value of sin(15°) is close to that of sin(0°),

which is 0 while the value of sin(100°) is close to that of sin(90°), which is 1.

I can see that the graph of the sine function is negative on the interval (180, 360), and the value of sin(285°) is close to that of sin(270°), whose value is −1, while the value of sin(340°) is close to that of sin(360°), whose value is 0.

I know that for a counterclockwise rotation by 𝜃𝜃° on the unit circle, the 𝑦𝑦-coordinate of the point of intersection of the terminal ray and the unit circle has the value sin(𝜃𝜃°).

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c. On the interval (0, 90), is the sine function increasing or decreasing? Based on that, name another interval not included in (0, 90) where the sine function must have the same behavior.

The sine function is increasing on (𝟎𝟎,𝟗𝟗𝟎𝟎). The sine function is also increasing on (𝟑𝟑𝟑𝟑𝟎𝟎,𝟑𝟑𝟐𝟐𝟎𝟎).

2. Graph the cosine function on the interval [0, 360] showing all key points of the graph (horizontal and vertical intercepts and maximum and minimum points). Then, use the graph to answer each of the following questions.

a. If we graphed the cosine function on the interval [−720, 720], what would it look like? Why?

It would repeat the same pattern on intervals of length 𝟑𝟑𝟑𝟑𝟎𝟎 in either direction. So from 𝟑𝟑𝟑𝟑𝟎𝟎 to 𝟕𝟕𝟐𝟐𝟎𝟎, the graph would look exactly like it does from 𝟎𝟎 to 𝟑𝟑𝟑𝟑𝟎𝟎, as it also would from −𝟑𝟑𝟑𝟑𝟎𝟎 to 𝟎𝟎 and from −𝟕𝟕𝟐𝟐𝟎𝟎 to −𝟑𝟑𝟑𝟑𝟎𝟎.

b. How is the graph of the cosine function similar to that of the sine function?

The graphs both have a cycle length of 𝟑𝟑𝟑𝟑𝟎𝟎 and oscillate between −𝟏𝟏 and 𝟏𝟏.

c. How is the graph of the cosine function different from that of the sine function?

The graphs have different 𝒚𝒚-intercepts. The cosine function has a maximum value at (𝟎𝟎,𝟏𝟏), decreases from 𝟏𝟏 to −𝟏𝟏 on the interval [𝟎𝟎,𝟏𝟏𝟐𝟐𝟎𝟎] until it reaches its minimum value at (𝟏𝟏𝟐𝟐𝟎𝟎,−𝟏𝟏), and then increases from −𝟏𝟏 to 𝟏𝟏 on the interval [𝟏𝟏𝟐𝟐𝟎𝟎,𝟑𝟑𝟑𝟑𝟎𝟎]. The sine function intersects the 𝒚𝒚-axis at (𝟎𝟎,𝟎𝟎) and increases to its maximum value at (𝟗𝟗𝟎𝟎,𝟏𝟏). It then decreases from 𝟏𝟏 to −𝟏𝟏 on the interval [𝟗𝟗𝟎𝟎,𝟐𝟐𝟕𝟕𝟎𝟎] until it reaches its minimum value at (𝟐𝟐𝟕𝟕𝟎𝟎,−𝟏𝟏) and then increases from −𝟏𝟏 to 𝟎𝟎 on the interval [𝟐𝟐𝟕𝟕𝟎𝟎,𝟑𝟑𝟑𝟑𝟎𝟎].

I know the graph will repeat the same pattern on intervals of length 360 in either direction, so if the graph increases on the interval (0, 90), it will increase on the interval (0 + 360, 90 + 360).

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Lesson 9: Awkward! Who Chose the Number 360, Anyway?

𝐵𝐵 𝐴𝐴

𝐶𝐶

𝐵𝐵 𝐴𝐴

𝐶𝐶


1. Read the radian protractor to measure the amount of rotation in radians of ray 𝐵𝐵𝐴𝐴 to ray 𝐵𝐵𝐶𝐶 in the indicated direction. Record the measurement to the nearest 0.1 radian. Use negative measures to indicate clockwise rotation.

𝟐𝟐.𝟕𝟕 radians −𝟏𝟏 radian

2. Complete the table below, converting from degrees to radians or from radians to degrees. Where appropriate, give your answers in the form of a fraction of 𝜋𝜋.

Radians Degrees 𝝅𝝅𝟏𝟏𝟐𝟐

𝟏𝟏𝟏𝟏°

𝟕𝟕𝝅𝝅𝟔𝟔 210°

−𝟏𝟏𝝅𝝅𝟏𝟏𝟏𝟏

−𝟏𝟏𝟓𝟓°

𝟏𝟏𝟓𝟓𝝅𝝅 1800°

I can count the tenths markings on the protractor to determine the amount of clockwise rotation.

I can use the relationship 360° = 1 turn = 2𝜋𝜋 rad to complete the table.

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3. Use the unit circle diagram provided and your knowledge of the six trigonometric functions to complete the table below. Give your answers in exact form, as either rational numbers or radical expressions. Select values of 𝜃𝜃 so that 0 ≤ 𝜃𝜃 < 2𝜋𝜋.

𝜽𝜽 𝐜𝐜𝐜𝐜𝐜𝐜(𝜽𝜽) 𝐜𝐜𝐬𝐬𝐬𝐬(𝜽𝜽) 𝐭𝐭𝐭𝐭𝐬𝐬(𝜽𝜽) 𝐜𝐜𝐜𝐜𝐭𝐭(𝜽𝜽) 𝐜𝐜𝐬𝐬𝐜𝐜(𝜽𝜽) 𝐜𝐜𝐜𝐜𝐜𝐜(𝜽𝜽)

𝜋𝜋6

√𝟑𝟑𝟐𝟐

𝟏𝟏𝟐𝟐

√𝟑𝟑𝟑𝟑

√𝟑𝟑 𝟐𝟐√𝟑𝟑𝟑𝟑

𝟐𝟐

5𝜋𝜋4

−√𝟐𝟐𝟐𝟐

−√𝟐𝟐𝟐𝟐

𝟏𝟏 𝟏𝟏 −√𝟐𝟐 −√𝟐𝟐

3𝜋𝜋2

𝟓𝟓 −𝟏𝟏 undefined 𝟓𝟓 undefined −𝟏𝟏

−10𝜋𝜋

3 −

𝟏𝟏𝟐𝟐

√𝟑𝟑𝟐𝟐

−√𝟑𝟑 −√𝟑𝟑𝟑𝟑

−𝟐𝟐 𝟐𝟐√𝟑𝟑𝟑𝟑

𝝅𝝅𝟐𝟐

0 1 undefined 𝟓𝟓 undefined 𝟏𝟏

𝟑𝟑𝝅𝝅𝟒𝟒

−√22

√22

−𝟏𝟏 −𝟏𝟏 −√𝟐𝟐 √𝟐𝟐

I know for a counterclockwise rotation 𝜃𝜃, the point (𝑥𝑥,𝑦𝑦) on the unit circle corresponds to the values of cos(𝜃𝜃) and sin(𝜃𝜃), respectively. This means the

point �− √22

, √22� corresponds to a rotation of 3𝜋𝜋

4 radians counterclockwise.

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4. How many radians does the minute hand of a clock rotate through over 5 minutes? How many degrees?

In 𝟏𝟏 minutes, the minute hand makes 𝟏𝟏𝟏𝟏𝟐𝟐

of a rotation, so it rotates through 𝟏𝟏𝟏𝟏𝟐𝟐

(𝟐𝟐𝝅𝝅) = 𝝅𝝅𝟔𝟔 radians. This is

equivalent to 𝟑𝟑𝟓𝟓°.

5. What is the radian measure of an angle formed by the minute and hour hands of a clock when the clock

reads 2:30? What is the degree measure?

At 2:30, the hour hand is halfway between the 𝟐𝟐 and the 𝟑𝟑, and the minute hand is on the 𝟔𝟔. A hand on the clock rotates through 𝟏𝟏

𝟏𝟏𝟐𝟐 of a rotation as it moves from one number to the next, which is a rotation

of 𝝅𝝅𝟔𝟔 radians.

Since there are 𝟑𝟑 𝟏𝟏𝟐𝟐 of these increments between the two hands of the clock at 2:30, the angle formed by

the two clock hands is 𝟕𝟕𝟐𝟐�𝝅𝝅𝟔𝟔� = 𝟕𝟕𝝅𝝅

𝟏𝟏𝟐𝟐 radians. In degrees, this is 𝟏𝟏𝟓𝟓𝟏𝟏°.

6. What is the radian measure of an angle subtended by an arc of a circle with radius 6 cm if the intercepted arc has length 12 cm? How many degrees?

The intercepted arc is the length of 𝟐𝟐 radii, so the angle subtended by that arc measures 𝟐𝟐 radians. This

is equivalent to 𝟐𝟐�𝟏𝟏𝟏𝟏𝟓𝟓°𝝅𝝅� = 𝟑𝟑𝟔𝟔𝟓𝟓°

𝝅𝝅≈ 𝟏𝟏𝟏𝟏𝟒𝟒.𝟔𝟔°.

I can use the rotational value from Problem 4.

I know a radian is the measure of the central angle of a sector of a circle with arc length of one radius length.

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Lesson 10: Basic Trigonometric Identities from Graphs

I know that sin(−𝜋𝜋) = − sin(𝜋𝜋) = 0.

I know that sin �9𝜋𝜋4� = sin �2𝜋𝜋 + 𝜋𝜋

4� = sin �𝜋𝜋

4� = √2

2.


1. Use the graphs above to answer the following questions: a. Describe the values of 𝑥𝑥 for which the

cosine function has a relative minimum.

The cosine function has a relative minimum at all 𝒙𝒙 = 𝝅𝝅 + 𝟐𝟐𝝅𝝅𝟐𝟐, where 𝟐𝟐 is an integer.

b. Describe the values of 𝑥𝑥 for which the sine function has a value of 0.

The sine function has a value of 𝟎𝟎 at 𝒙𝒙 = 𝝅𝝅𝟐𝟐, where 𝟐𝟐 is an integer.

c. Rewrite each of the following function values in order from least to greatest.

sin �3𝜋𝜋2� sin �5𝜋𝜋

6� sin �9𝜋𝜋

4� sin(−𝜋𝜋)

𝐬𝐬𝐬𝐬𝐬𝐬 �𝟑𝟑𝝅𝝅𝟐𝟐�, 𝐬𝐬𝐬𝐬𝐬𝐬(−𝝅𝝅), 𝐬𝐬𝐬𝐬𝐬𝐬 �𝟓𝟓𝝅𝝅

𝟔𝟔�, 𝐬𝐬𝐬𝐬𝐬𝐬 �𝟗𝟗𝝅𝝅

𝟒𝟒�

I can see from the graph that the sine function has a value of 0 at 𝑥𝑥 = 0,𝜋𝜋, and 2𝜋𝜋.

The graph of the cosine function has a relative minimum at 𝑥𝑥 = π. Since the graph repeats the cycle every 2𝜋𝜋 radians, I know there will be a relative minimum for all 𝑥𝑥-values that are a distance from 𝜋𝜋 that is an integer multiple of 2𝜋𝜋.

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I can use the identity cos(−𝑥𝑥) = cos(𝑥𝑥) to

find the value of cos �− 𝜋𝜋3�.

I can use the identity cos(𝑥𝑥 + 𝜋𝜋) = −cos(𝑥𝑥) to find the value of cos �4𝜋𝜋

3�.

I can use the identity cos(𝑥𝑥 + 2𝜋𝜋𝜋𝜋) = cos(𝑥𝑥)

to find the value of cos �13𝜋𝜋3�.

2. Evaluate each of the following without a calculator, using a trigonometric identity when needed.

cos �𝜋𝜋3� cos �−𝜋𝜋

3� cos �4𝜋𝜋

3� cos �13𝜋𝜋

3�

𝐜𝐜𝐜𝐜𝐬𝐬 �𝝅𝝅𝟑𝟑� =

𝟏𝟏𝟐𝟐

𝐜𝐜𝐜𝐜𝐬𝐬 �−𝝅𝝅𝟑𝟑� = 𝐜𝐜𝐜𝐜𝐬𝐬 �

𝝅𝝅𝟑𝟑� =

𝟏𝟏𝟐𝟐

𝐜𝐜𝐜𝐜𝐬𝐬 �𝟒𝟒𝝅𝝅𝟑𝟑� = 𝐜𝐜𝐜𝐜𝐬𝐬 �𝝅𝝅 +

𝝅𝝅𝟑𝟑� = −𝐜𝐜𝐜𝐜𝐬𝐬 �

𝝅𝝅𝟑𝟑� = −

𝟏𝟏𝟐𝟐

𝐜𝐜𝐜𝐜𝐬𝐬 �𝟏𝟏𝟑𝟑𝝅𝝅𝟑𝟑� = 𝐜𝐜𝐜𝐜𝐬𝐬 �𝟒𝟒𝝅𝝅 +

𝝅𝝅𝟑𝟑� = 𝐜𝐜𝐜𝐜𝐬𝐬 �

𝝅𝝅𝟑𝟑� =

𝟏𝟏𝟐𝟐

3. Find two different equations that represent each of the graphs shown below—one sine and one cosine—using different horizontal transformations. You must find equations that are different from those shown.

𝒚𝒚 = 𝐬𝐬𝐬𝐬𝐬𝐬(𝒙𝒙 + 𝟐𝟐𝝅𝝅) 𝒚𝒚 = 𝐬𝐬𝐬𝐬𝐬𝐬 �𝒙𝒙 + 𝝅𝝅𝟐𝟐�

𝒚𝒚 = 𝐜𝐜𝐜𝐜𝐬𝐬 �𝒙𝒙 − 𝝅𝝅𝟐𝟐� 𝒚𝒚 = 𝐜𝐜𝐜𝐜𝐬𝐬(𝒙𝒙 − 𝟐𝟐𝝅𝝅)

Note: There are several valid equations that could represent each graph.

If the graph of 𝑦𝑦 = sin(𝑥𝑥) is translated 𝜋𝜋

2

units to the left, I know the resulting image will be the cosine graph.

If the graph of 𝑦𝑦 = cos(𝑥𝑥) is translated 𝜋𝜋

2 units to the

right, I know the resulting image will be the sine graph.

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Lesson 11: Transforming the Graph of the Sine Function


1. For each function, indicate the amplitude, frequency, period, phase shift, horizontal and vertical translations, and equation of the midline. Without a calculator, graph the function together with a graph of the sine function 𝑓𝑓(𝑥𝑥) = sin(𝑥𝑥) on the same axes. Graph at least one full period of each function.

a. 𝑔𝑔(𝑥𝑥) = 2 sin �𝑥𝑥 + 𝜋𝜋4�

The amplitude is 𝟐𝟐, the frequency is 𝟏𝟏𝟐𝟐𝝅𝝅

, the period is 𝟐𝟐𝝅𝝅, and the phase shift is −𝝅𝝅𝟒𝟒

. The graph is

translated horizontally 𝝅𝝅𝟒𝟒

units to the left; there is no vertical translation, and the equation of the midline is 𝒚𝒚 = 𝟎𝟎.

b. 𝑔𝑔(𝑥𝑥) = sin(3𝑥𝑥 − 𝜋𝜋) − 1

𝒈𝒈(𝒙𝒙) = 𝐬𝐬𝐬𝐬𝐬𝐬 �𝟑𝟑 �𝒙𝒙 − 𝝅𝝅𝟑𝟑�� − 𝟏𝟏

The amplitude is 𝟏𝟏, the frequency is 𝟑𝟑𝟐𝟐𝝅𝝅

, the period is 𝟐𝟐𝝅𝝅𝟑𝟑

, and the phase shift is 𝝅𝝅𝟑𝟑. The graph is

translated horizontally 𝝅𝝅𝟑𝟑 units to the right and vertically 𝟏𝟏 unit down, and the equation of the

midline is 𝒚𝒚 = −𝟏𝟏.

I know that the graph of the function 𝑓𝑓(𝑥𝑥) = 𝐴𝐴 sin�𝜔𝜔(𝑥𝑥 − ℎ)� + 𝑘𝑘 has an amplitude |𝐴𝐴| and phase shift ℎ. Since the phase shift is negative, the graph is translated horizontally to the left.

I know that if I factor 3 from the expression (3𝑥𝑥 − 𝜋𝜋), I can write the equation of the function in the form 𝑓𝑓(𝑥𝑥) = 𝐴𝐴 sin�𝜔𝜔(𝑥𝑥 − ℎ)� + 𝑘𝑘, which can help me determine the period and phase shift for the graph.

I know that the graph of the function 𝑓𝑓(𝑥𝑥) = 𝐴𝐴 sin�𝜔𝜔(𝑥𝑥 − ℎ)� + 𝑘𝑘 has a midline whose equation is 𝑦𝑦 = 𝑘𝑘, a period of 2𝜋𝜋

|𝜔𝜔|, and a frequency |𝜔𝜔|

2𝜋𝜋.

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2. For each function, indicate the amplitude, frequency, period, phase shift, horizontal and vertical translations, and equation of the midline. Without a calculator, graph the function together with a graph of the cosine function 𝑓𝑓(𝑥𝑥) = cos(𝑥𝑥) on the same axes. Graph at least one full period of each function.

a. 𝑔𝑔(𝑥𝑥) = −2 cos �𝑥𝑥2�

The amplitude is 𝟐𝟐, the frequency is 𝟏𝟏𝟒𝟒𝝅𝝅

, the period is 𝟒𝟒𝝅𝝅, and the phase shift is 𝟎𝟎. There are no horizontal or vertical translations, and the equation for the midline is 𝒚𝒚 = 𝟎𝟎. The graph of 𝒈𝒈 is a vertical scaling of the graph of the cosine function by a factor of 𝟐𝟐 and a horizontal scaling of the graph of the cosine function by a factor of 𝟐𝟐. It is also reflected over the 𝒙𝒙-axis.

b. 𝑔𝑔(𝑥𝑥) = 12

cos(−𝑥𝑥) + 2

𝒈𝒈(𝒙𝒙) =𝟏𝟏𝟐𝟐

𝐜𝐜𝐜𝐜𝐬𝐬(−𝒙𝒙) + 𝟐𝟐

=𝟏𝟏𝟐𝟐

𝐜𝐜𝐜𝐜𝐬𝐬(𝒙𝒙) + 𝟐𝟐

The amplitude is 𝟏𝟏𝟐𝟐, the frequency is 𝟏𝟏

𝟐𝟐𝝅𝝅, the period is 𝟐𝟐𝝅𝝅, and the phase shift is 𝟎𝟎. There is no

horizontal translation, the graph is translated vertically 𝟐𝟐 units up, and the equation for the midline is 𝒚𝒚 = 𝟐𝟐. The graph of 𝒈𝒈 is a vertical scaling of the graph of the cosine function by a factor of 𝟏𝟏

𝟐𝟐.

I know that the graph of 𝑔𝑔 is reflected over the 𝑥𝑥-axis because the value of 𝐴𝐴 in the equation 𝑓𝑓(𝑥𝑥) = 𝐴𝐴 cos�𝜔𝜔(𝑥𝑥 − ℎ)� + 𝑘𝑘 is negative.

I know that cos(−𝑥𝑥) = cos(𝑥𝑥), because the graph of 𝑓𝑓 is symmetric about the 𝑦𝑦-axis.

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3. Find the 𝑥𝑥-intercepts of the graph of the function 𝑓𝑓(𝑥𝑥) = 𝐴𝐴 cos�𝜔𝜔(𝑥𝑥 − ℎ)� in terms of the period 𝑃𝑃, where 𝜔𝜔 > 0.

The 𝒙𝒙-intercepts occur when 𝑨𝑨 𝐜𝐜𝐜𝐜𝐬𝐬�𝝎𝝎(𝒙𝒙 − 𝒉𝒉)� = 𝟎𝟎. This happens when 𝝎𝝎(𝒙𝒙 − 𝒉𝒉) = 𝝅𝝅𝟐𝟐

+ 𝒏𝒏𝝅𝝅 for

integers 𝒏𝒏.

So, 𝒙𝒙 − 𝒉𝒉 = 𝝅𝝅𝟐𝟐𝝎𝝎

+ 𝒏𝒏𝝅𝝅𝝎𝝎

, and then 𝒙𝒙 = 𝝅𝝅𝟐𝟐𝝎𝝎

+ 𝒏𝒏𝝅𝝅𝝎𝝎

+ 𝒉𝒉.

Since 𝑷𝑷 = 𝟐𝟐𝝅𝝅𝝎𝝎

, it follows that 𝒙𝒙 = 𝑷𝑷𝟒𝟒

+ 𝒏𝒏𝑷𝑷𝟐𝟐

+ 𝒉𝒉. Thus, the graph of 𝒇𝒇(𝒙𝒙) = 𝑨𝑨 𝐜𝐜𝐜𝐜𝐬𝐬(𝝎𝝎(𝒙𝒙 − 𝒉𝒉)) has

𝒙𝒙-intercepts at 𝒙𝒙 = 𝑷𝑷𝟒𝟒

+ 𝒏𝒏𝑷𝑷𝟐𝟐

+ 𝒉𝒉 for integer values of 𝒏𝒏.

I can rewrite 𝜋𝜋2𝜔𝜔

+ 𝑛𝑛𝜋𝜋𝜔𝜔

as 14�2𝜋𝜋𝜔𝜔� + 𝑛𝑛

2�2𝜋𝜋𝜔𝜔�,

which will help me write the value of 𝑥𝑥 in terms of 𝑃𝑃.

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Lesson 12: Ferris Wheels—Using Trigonometric Functions to Model Cyclical Behavior

Lesson 12: Ferris Wheels—Using Trigonometric Functions to

Model Cyclical Behavior

1. A nationwide fitness center uses large fans to help minimize the costs of air-conditioning their gyms. The wall-mounted fans have diameters of approximately 10 feet and make a complete counterclockwise rotation every half-second. The center of the fan is 7 feet from the floor. Assume there is a chalk mark on the outer edge of one of the blades. a. Explain why a sinusoidal function could be used to model the vertical distance from the chalk mark

to the floor over time.

Since the fan blade is moving around a circle and the sine function measures vertical displacement of an object moving around a circle, we can use a sine function to track the position of the chalk mark above the floor.

b. Sketch a graph of the vertical position of the chalk mark on the fan as a function of time. Assume 𝑡𝑡 = 0 corresponds to a time when the blade with the chalk mark is at the top of the fan, on the vertical line through the center of the fan. What are the amplitude, period, and midline of the graph? What is the phase shift?

The amplitude of the graph is 𝟓𝟓, the period is 𝟏𝟏𝟐𝟐 , and the midline is the

line 𝒚𝒚 = 𝟕𝟕. Since the chalk mark begins at the highest point, the graph is translated by 𝟏𝟏

𝟖𝟖 unit to the left; thus the phase shift is – 𝟏𝟏

𝟖𝟖.

Since the radius of the fan is 5 feet, and the center of the fan is 7 feet above the floor, the maximum vertical distance from the chalk mark on the fan to the floor is 12 feet.

The chalk mark begins with a vertical position aligned with the center and then moves downward as the fan rotates counterclockwise. This motion can be modeled by the sine function when it is translated horizontally to the left.

I remember that a translation to the left corresponds to a negative phase shift.

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c. Model the vertical position of the chalk marking, 𝑉𝑉, as a function of time, 𝑡𝑡, since the chalk mark was at its highest point.

𝑽𝑽(𝒕𝒕) = 𝟓𝟓 𝐬𝐬𝐬𝐬𝐬𝐬 �𝟒𝟒𝟒𝟒�𝒕𝒕 + 𝟏𝟏𝟖𝟖�� + 𝟕𝟕

2. The London Eye, a Ferris wheel in London, England, opened in 1999. The 135-meter tall wheel has a diameter of 120 meters. A ride on one of its 32 passenger cars lasts 30 minutes, the time it takes the wheel to complete one full rotation. Riders board the passenger cars at the bottom of the wheel. Assume that once the wheel is in motion, it maintains a constant speed for the 30-minute ride and rotates in a counterclockwise direction. a. Sketch a graph of the height of a passenger car on the London Eye as a function of the time since the

ride began.

b. Write a sinusoidal function 𝐻𝐻 that represents the height of a passenger car 𝑡𝑡 minutes after the ride begins.

𝑯𝑯(𝒕𝒕) = 𝟔𝟔𝟔𝟔 𝐬𝐬𝐬𝐬𝐬𝐬�𝟒𝟒𝟏𝟏𝟓𝟓

(𝒕𝒕 − 𝟕𝟕.𝟓𝟓)�+ 𝟕𝟕𝟓𝟓

c. Explain how the parameters of your sinusoidal function relate to the situation.

The amplitude, 𝟔𝟔𝟔𝟔, is the radius of the wheel. The period is 𝟑𝟑𝟔𝟔 minutes, so 𝝎𝝎 = 𝟐𝟐𝟒𝟒𝟑𝟑𝟔𝟔

. The midline is 𝒚𝒚 = 𝟕𝟕𝟓𝟓, which represents how high the center of the wheel is above the ground in meters. To make the graph have a minimal point at 𝒕𝒕 = 𝟔𝟔, the phase shift can be 𝟕𝟕.𝟓𝟓, which corresponds to one-fourth of the period.

I know that 𝜔𝜔 = 2𝜋𝜋𝑃𝑃

and that 𝑃𝑃 = 12.

I can find the equation of the midline by taking the average of the highest and lowest values on the graph.

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d. If you were on this ride, how high would you be above the ground after 5 minutes?

𝑯𝑯(𝒕𝒕) = 𝟔𝟔𝟔𝟔 𝐬𝐬𝐬𝐬𝐬𝐬 � 𝟒𝟒𝟏𝟏𝟓𝟓

(𝟓𝟓 − 𝟕𝟕.𝟓𝟓)� + 𝟕𝟕𝟓𝟓 = 𝟔𝟔𝟔𝟔 𝐬𝐬𝐬𝐬𝐬𝐬 �−𝟒𝟒𝟔𝟔� + 𝟕𝟕𝟓𝟓 = −𝟑𝟑𝟔𝟔 + 𝟕𝟕𝟓𝟓 = 𝟒𝟒𝟓𝟓

I would be 𝟒𝟒𝟓𝟓 meters above the ground after 𝟓𝟓 minutes.

e. Identify two different times when the height of the Ferris wheel is 100 meters above the ground. (Hint: Write an equation, and solve it graphically.)

𝟏𝟏𝟔𝟔𝟔𝟔 = 𝟔𝟔𝟔𝟔 𝐬𝐬𝐬𝐬𝐬𝐬 � 𝟒𝟒𝟏𝟏𝟓𝟓

(𝒕𝒕 − 𝟕𝟕.𝟓𝟓)� + 𝟕𝟕𝟓𝟓

𝒕𝒕 ≈ 𝟗𝟗.𝟓𝟓𝟓𝟓, 𝟏𝟏𝟗𝟗.𝟒𝟒𝟓𝟓

The height of the rider will be 𝟏𝟏𝟔𝟔𝟔𝟔 meters above the ground approximately 𝟗𝟗.𝟓𝟓𝟓𝟓 minutes and again at 𝟏𝟏𝟗𝟗.𝟒𝟒𝟓𝟓 minutes after the rider gets on the Ferris wheel.

I know sin �−𝜋𝜋6� = − sin �𝜋𝜋

6� = −1

2.

I can graph the functions 𝑓𝑓(𝑡𝑡) = 60 sin � 𝜋𝜋15

(𝑡𝑡 − 7.5)� +

75 and 𝑔𝑔(𝑡𝑡) = 100 and find the first two positive intersection points.

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Lesson 13: Tides, Sound Waves, and Stock Markets


1. Find equations of both a sine function and a cosine function that could each represent the graph given below.

𝒇𝒇(𝒙𝒙) = 𝟒𝟒 𝐬𝐬𝐬𝐬𝐬𝐬�𝝅𝝅𝟒𝟒

(𝒙𝒙 − 𝟓𝟓)�

𝒈𝒈(𝒙𝒙) = 𝟒𝟒 𝐜𝐜𝐜𝐜𝐬𝐬�𝝅𝝅𝟒𝟒

(𝒙𝒙 + 𝟏𝟏)�

2. A paddle boarder resting in the ocean notices that she is being moved up and down by the waves as they pass beneath her board. The graph shows the vertical position of the paddle boarder in meters relative to sea level as a function of time in minutes.

Find the equation of a trigonometric function that fits this graph.

𝑯𝑯(𝒕𝒕) = 𝟎𝟎.𝟔𝟔 𝐜𝐜𝐜𝐜𝐬𝐬 �𝟒𝟒𝝅𝝅𝟓𝟓𝒕𝒕�

Because the graph has successive relative maxima at 𝑥𝑥 = −1 and at 𝑥𝑥 = 7, I know the period is 8 units and 𝜔𝜔 = 2𝜋𝜋

8= 𝜋𝜋

4.

I can see that there is no vertical or horizontal displacement of the graph relative to the graph of the cosine function. I also notice that the graph completes two cycles in 5 minutes, so 𝜔𝜔 = 4𝜋𝜋

5.

Since the graph intersects the 𝑦𝑦-axis at a relative maximum value, cosine would be the most natural function to use to model the vertical motion of the paddle boarder.

I know I can also write this function using a phase shift of 7, which would result in the equivalent equation 𝑔𝑔(𝑥𝑥) = 4 cos �𝜋𝜋

4(𝑥𝑥 − 7)�.

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3. The following table represents the average monthly minimum temperature, in degrees Fahrenheit, in Death Valley, California, for each of the twelve months in a year. We wish to model these data with a trigonometric function.

Month Jan. Feb. Mar. Apr. May June July Aug. Sept. Oct. Nov. Dec. °F 37.4 42.8 51.8 62.6 71.6 78.8 87.8 84.2 75.2 62.6 46.4 39.2

a. What appears to be the amplitude of the data?

𝟖𝟖𝟖𝟖.𝟖𝟖 − 𝟑𝟑𝟖𝟖.𝟒𝟒𝟐𝟐

= 𝟐𝟐𝟓𝟓.𝟐𝟐

b. What appears to be the midline of the data (equation of the line through the middle of the graph)?

𝟖𝟖𝟖𝟖.𝟖𝟖 + 𝟑𝟑𝟖𝟖.𝟒𝟒𝟐𝟐

= 𝟔𝟔𝟐𝟐.𝟔𝟔

𝒚𝒚 = 𝟔𝟔𝟐𝟐.𝟔𝟔

c. What is a reasonable approximation for the horizontal shift?

It appears that the peak of the function should be in July. If we use the cosine function to model the data and define month 𝟏𝟏 as January, then 𝟖𝟖 is a reasonable value for 𝒉𝒉. (Other values of 𝒉𝒉 near 𝟖𝟖 would also be reasonable.)

d. Write an equation for a function that fits this set of data.

Since 𝒉𝒉 = 𝟖𝟖, 𝒌𝒌 = 𝟔𝟔𝟐𝟐.𝟔𝟔, 𝑨𝑨 = 𝟐𝟐𝟓𝟓.𝟐𝟐, and 𝝎𝝎 = 𝟐𝟐𝝅𝝅𝟏𝟏𝟐𝟐

= 𝝅𝝅𝟔𝟔

, the function

𝒇𝒇(𝒙𝒙) = 𝟐𝟐𝟓𝟓.𝟐𝟐 𝐜𝐜𝐜𝐜𝐬𝐬 �𝝅𝝅𝟔𝟔

(𝒙𝒙 − 𝟖𝟖)� + 𝟔𝟔𝟐𝟐.𝟔𝟔 fits the given data.

I can calculate 𝜔𝜔 using the cycle length of 12 months.

I know one way to calculate the amplitude is to find half the distance between the greatest and least values.

I know that I can find the midline value by calculating the mean of the greatest and least temperatures.

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e. Sketch a graph of your function from part (d), together with the original data.

I can see from the graph that the function

𝑓𝑓(𝑥𝑥) = 25.2 cos �𝜋𝜋6

(𝑥𝑥 − 7)� + 62.6 is a

reasonable approximation to the data.

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M2 ALGEBRA II

Lesson 14: Graphing the Tangent Function


In each set of axes below, the dashed graph represents the function 𝑓𝑓(𝑥𝑥) = tan(𝑥𝑥). Use what you know about function transformations to sketch a graph of 𝑦𝑦 = 𝑔𝑔(𝑥𝑥) for each specified function 𝑔𝑔 on the interval (0, 2𝜋𝜋).

a. 𝑔𝑔(𝑥𝑥) = 3 tan(𝑥𝑥)

b. 𝑔𝑔(𝑥𝑥) = −13

tan(𝑥𝑥)

I know that the parameter 𝐴𝐴 in the graph of 𝑔𝑔(𝑥𝑥) = 𝐴𝐴 tan(𝑥𝑥) vertically stretches the graph of 𝑓𝑓(𝑥𝑥) = tan(𝑥𝑥) by the factor 𝐴𝐴.

I know that when the parameter 𝐴𝐴 is negative, the graph of 𝑔𝑔(𝑥𝑥) = 𝐴𝐴 tan(𝑥𝑥) is the reflection of 𝑓𝑓(𝑥𝑥) = |𝐴𝐴|tan(𝑥𝑥) over the 𝑥𝑥-axis.

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c. 𝑔𝑔(𝑥𝑥) = tan �𝑥𝑥 − 𝜋𝜋4�

d. 𝑔𝑔(𝑥𝑥) = tan �𝑥𝑥 − 𝜋𝜋6� + 2

e. 𝑔𝑔(𝑥𝑥) = tan(2𝑥𝑥) + 1

I know that the parameter ℎ in 𝑔𝑔(𝑥𝑥) = tan(𝑥𝑥 − ℎ) + 𝑘𝑘 indicates a horizontal translation of the graph of 𝑓𝑓(𝑥𝑥) = tan(𝑥𝑥), and the parameter 𝑘𝑘 indicates a vertical translation.

I know that the graph of 𝑔𝑔(𝑥𝑥) = tan (𝑥𝑥 − ℎ) is the graph of 𝑓𝑓(𝑥𝑥) = tan(𝑥𝑥) shifted horizontally to the right by ℎ units.

I know that the parameter 𝜔𝜔 in 𝑔𝑔(𝑥𝑥) = tan(𝜔𝜔𝑥𝑥) + 1 indicates that the graph of 𝑦𝑦 = 𝑔𝑔(𝑥𝑥) is a horizontal scaling of the graph of 𝑓𝑓(𝑥𝑥) = tan(𝑥𝑥) + 1 by the factor 𝜔𝜔.

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f. 𝑔𝑔(𝑥𝑥) = tan �−2 �𝑥𝑥 + 𝜋𝜋4��

I know that because −2 is negative, the graph of

𝑔𝑔(𝑥𝑥) = tan �−2 �𝑥𝑥 + 𝜋𝜋4�� is the

graph of 𝑦𝑦 = tan �2 �𝑥𝑥 + 𝜋𝜋4��

reflected over the 𝑦𝑦-axis.

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2015-16

M2 ALGEBRA II

Lesson 15: What Is a Trigonometric Identity?


1. Determine whether the statement is a trigonometric identity by graphing the functions on each side of the equation

sin(−𝑥𝑥) = − sin(𝑥𝑥), where the functions on both sides are defined.

This statement is an identity that is defined for all real numbers 𝒙𝒙, as shown by the identical graphs above.

2. Determine the domain of the following trigonometric identities:

a. cot(𝑥𝑥) = 1tan(𝑥𝑥), where the functions on both sides are defined.

This identity is defined only for 𝒙𝒙 ≠ 𝒌𝒌𝒌𝒌, for all integers 𝒌𝒌.

b. csc(𝑥𝑥) = 1sin(𝑥𝑥), where the functions on both sides are defined.

This identity is defined only for 𝒙𝒙 ≠ 𝒌𝒌𝒌𝒌, for all integers 𝒌𝒌.

3. Rewrite sin(𝑥𝑥)cos2(𝑥𝑥) + sin3(𝑥𝑥) as an expression containing a single term.

𝐬𝐬𝐬𝐬𝐬𝐬(𝒙𝒙)𝐜𝐜𝐜𝐜𝐬𝐬𝟐𝟐(𝒙𝒙) + 𝐬𝐬𝐬𝐬𝐬𝐬𝟑𝟑(𝒙𝒙) = 𝐬𝐬𝐬𝐬𝐬𝐬(𝒙𝒙) �𝐬𝐬𝐬𝐬𝐬𝐬𝟐𝟐(𝒙𝒙) + 𝐜𝐜𝐜𝐜𝐬𝐬𝟐𝟐(𝒙𝒙)� = 𝐬𝐬𝐬𝐬𝐬𝐬(𝒙𝒙) (𝟏𝟏) = 𝐬𝐬𝐬𝐬𝐬𝐬(𝒙𝒙)

I know the cotangent function is not defined where tan(𝑥𝑥) = 0, which occurs at all integer multiples of 𝜋𝜋.

I can use the Pythagorean identity sin2(𝜃𝜃) + cos2(𝜃𝜃) = 1 to rewrite this expression.

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4. Use the Pythagorean identity sin2(𝜃𝜃) + cos2(𝜃𝜃) = 1, where 𝜃𝜃 is any real number, to find the following:

tan(𝜃𝜃), given sin(𝜃𝜃) = 45, for 𝜋𝜋

2< 𝜃𝜃 < 𝜋𝜋

From the Pythagorean identity, 𝐜𝐜𝐜𝐜𝐬𝐬 𝟐𝟐(𝜽𝜽) = 𝟏𝟏 − �𝟒𝟒𝟓𝟓�𝟐𝟐

. So, 𝐜𝐜𝐜𝐜𝐬𝐬𝟐𝟐(𝜽𝜽) = 𝟏𝟏 − 𝟏𝟏𝟏𝟏𝟐𝟐𝟓𝟓

= 𝟗𝟗𝟐𝟐𝟓𝟓

, and

𝐜𝐜𝐜𝐜𝐬𝐬(𝜽𝜽) = 𝟑𝟑𝟓𝟓, or 𝐜𝐜𝐜𝐜𝐬𝐬(𝜽𝜽) = −𝟑𝟑

𝟓𝟓. Because 𝐜𝐜𝐜𝐜𝐬𝐬(𝜽𝜽) is negative for 𝒌𝒌

𝟐𝟐< 𝜽𝜽 < 𝒌𝒌, 𝐭𝐭𝐭𝐭𝐬𝐬(𝜽𝜽) =

𝟒𝟒𝟓𝟓

−𝟑𝟑𝟓𝟓= −𝟒𝟒

𝟑𝟑.

I know that tan(𝜃𝜃) = sin(𝜃𝜃)cos(𝜃𝜃)

.

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2015-16

M2 ALGEBRA II

Lesson 16: Proving Trigonometric Identities


1. Use tan(𝑥𝑥) = sin(𝑥𝑥)cos(𝑥𝑥) and identities involving the sine and cosine functions to establish the following

identity for the tangent function. Identify the values of 𝑥𝑥 where the equation is an identity. tan(𝑥𝑥 − 𝜋𝜋) = tan(𝑥𝑥)

Let 𝒙𝒙 be a real number so that 𝒙𝒙 ≠ 𝝅𝝅𝟐𝟐

+ 𝝅𝝅𝝅𝝅, for any integer 𝝅𝝅. Then,

𝐭𝐭𝐭𝐭𝐭𝐭(𝒙𝒙 − 𝝅𝝅) =𝐬𝐬𝐬𝐬𝐭𝐭(𝒙𝒙 − 𝝅𝝅)𝐜𝐜𝐜𝐜𝐬𝐬(𝒙𝒙 − 𝝅𝝅) =

𝐬𝐬𝐬𝐬𝐭𝐭(𝒙𝒙 + 𝝅𝝅)𝐜𝐜𝐜𝐜𝐬𝐬(𝒙𝒙 + 𝝅𝝅) =

−𝐬𝐬𝐬𝐬𝐭𝐭(𝒙𝒙)−𝐜𝐜𝐜𝐜𝐬𝐬(𝒙𝒙) =

𝐬𝐬𝐬𝐬𝐭𝐭(𝒙𝒙)𝐜𝐜𝐜𝐜𝐬𝐬(𝒙𝒙) = 𝐭𝐭𝐭𝐭𝐭𝐭(𝒙𝒙).

2. Rewrite each of the following expressions as a single term. Identify the values of 𝑥𝑥 for which the original expression and your expression are equal: a. tan(𝑥𝑥)csc(𝑥𝑥)cos(𝑥𝑥)

𝐭𝐭𝐭𝐭𝐭𝐭(𝒙𝒙)𝐜𝐜𝐬𝐬𝐜𝐜(𝒙𝒙)𝐜𝐜𝐜𝐜𝐬𝐬(𝒙𝒙) =𝐬𝐬𝐬𝐬𝐭𝐭(𝒙𝒙)𝐜𝐜𝐜𝐜𝐬𝐬(𝒙𝒙) ∙

𝟏𝟏𝐬𝐬𝐬𝐬𝐭𝐭(𝒙𝒙) ∙

𝐜𝐜𝐜𝐜𝐬𝐬(𝒙𝒙)𝟏𝟏

= 𝟏𝟏

The expressions are equal when 𝒙𝒙 ≠ 𝝅𝝅𝟐𝟐𝝅𝝅, for all integers 𝝅𝝅.

b. csc2(𝑥𝑥)− cot2(𝑥𝑥)

𝐜𝐜𝐬𝐬𝐜𝐜𝟐𝟐(𝒙𝒙)− 𝐜𝐜𝐜𝐜𝐭𝐭2(𝒙𝒙) =𝟏𝟏

𝐬𝐬𝐬𝐬𝐭𝐭𝟐𝟐(𝒙𝒙) −𝐜𝐜𝐜𝐜𝐬𝐬𝟐𝟐(𝒙𝒙)𝐬𝐬𝐬𝐬𝐭𝐭𝟐𝟐(𝒙𝒙) =

𝟏𝟏 − 𝐜𝐜𝐜𝐜𝐬𝐬𝟐𝟐(𝒙𝒙)𝐬𝐬𝐬𝐬𝐭𝐭𝟐𝟐(𝒙𝒙) =

𝐬𝐬𝐬𝐬𝐭𝐭𝟐𝟐(𝒙𝒙)𝐬𝐬𝐬𝐬𝐭𝐭𝟐𝟐(𝒙𝒙) = 𝟏𝟏

The expressions are equal where 𝒙𝒙 ≠ 𝝅𝝅𝝅𝝅, for all integers 𝝅𝝅.

I can use the identity sin(𝑥𝑥) = sin(𝑥𝑥 + 2𝜋𝜋) to rewrite sin(𝑥𝑥 − 𝜋𝜋) = sin�(𝑥𝑥 − 𝜋𝜋) + 2𝜋𝜋� = sin(𝑥𝑥 + 𝜋𝜋).

I know either sin(𝑥𝑥) or cos(𝑥𝑥) has a value of 0 for every integer multiple of 𝜋𝜋

2,

so the expression tan(𝑥𝑥)csc(𝑥𝑥)cos(𝑥𝑥) will be undefined at these values.

Representing each expression in terms of the sine and cosine functions can help me rewrite the expression as a single term.

I can use the Pythagorean identity sin2(𝑥𝑥) + cos2(𝑥𝑥) = 1 to rewrite 1 − cos2(𝑥𝑥) as sin2(𝑥𝑥).

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c. Prove that for any two real numbers 𝑎𝑎 and 𝑏𝑏, sin2(𝑎𝑎) sin2(𝑏𝑏) + cos2(𝑎𝑎) sin2(𝑏𝑏) + sin2(𝑎𝑎) cos2(𝑏𝑏) + cos2(𝑎𝑎) cos2(𝑏𝑏) = 1.

Proof: Let 𝒂𝒂 and 𝒃𝒃 be any real numbers. Then,

�𝐬𝐬𝐬𝐬𝐭𝐭𝟐𝟐(𝒂𝒂) 𝐬𝐬𝐬𝐬𝐭𝐭𝟐𝟐(𝒃𝒃) + 𝐜𝐜𝐜𝐜𝐬𝐬𝟐𝟐(𝒂𝒂) 𝐬𝐬𝐬𝐬𝐭𝐭𝟐𝟐(𝒃𝒃)� + �𝐬𝐬𝐬𝐬𝐭𝐭𝟐𝟐(𝒂𝒂) 𝐜𝐜𝐜𝐜𝐬𝐬𝟐𝟐(𝒃𝒃) + 𝐜𝐜𝐜𝐜𝐬𝐬𝟐𝟐(𝒂𝒂) 𝐜𝐜𝐜𝐜𝐬𝐬𝟐𝟐(𝒃𝒃)�

= 𝐬𝐬𝐬𝐬𝐭𝐭𝟐𝟐(𝒃𝒃) �𝐬𝐬𝐬𝐬𝐭𝐭𝟐𝟐(𝒂𝒂) + 𝐜𝐜𝐜𝐜𝐬𝐬𝟐𝟐(𝒂𝒂)� + 𝐜𝐜𝐜𝐜𝐬𝐬𝟐𝟐(𝒃𝒃) �𝐬𝐬𝐬𝐬𝐭𝐭𝟐𝟐(𝒂𝒂) + 𝐜𝐜𝐜𝐜𝐬𝐬𝟐𝟐(𝒂𝒂)�

= �𝐬𝐬𝐬𝐬𝐭𝐭𝟐𝟐(𝒃𝒃) + 𝐜𝐜𝐜𝐜𝐬𝐬𝟐𝟐(𝒃𝒃)��𝐬𝐬𝐬𝐬𝐭𝐭𝟐𝟐(𝒂𝒂) + 𝐜𝐜𝐜𝐜𝐬𝐬𝟐𝟐(𝒂𝒂)�

= 𝟏𝟏.

Thus, for any two real numbers 𝒂𝒂 and 𝒃𝒃,

𝐬𝐬𝐬𝐬𝐭𝐭𝟐𝟐(𝒂𝒂) 𝐬𝐬𝐬𝐬𝐭𝐭𝟐𝟐(𝒃𝒃) + 𝐜𝐜𝐜𝐜𝐬𝐬𝟐𝟐(𝒂𝒂) 𝐬𝐬𝐬𝐬𝐭𝐭𝟐𝟐(𝒃𝒃) + 𝐬𝐬𝐬𝐬𝐭𝐭𝟐𝟐(𝒂𝒂) 𝐜𝐜𝐜𝐜𝐬𝐬𝟐𝟐(𝒃𝒃) + 𝐜𝐜𝐜𝐜𝐬𝐬𝟐𝟐(𝒂𝒂) 𝐜𝐜𝐜𝐜𝐬𝐬𝟐𝟐(𝒃𝒃) = 𝟏𝟏.

3. Prove that the following statements are identities for all values of 𝜃𝜃 for which both sides are defined, and describe that set. a. �1 − sin(𝜃𝜃)��tan(𝜃𝜃) + sec(𝜃𝜃)� = cos(𝜃𝜃)

Proof: Let 𝜽𝜽 be a real number so that 𝜽𝜽 ≠ 𝝅𝝅𝟐𝟐𝝅𝝅, for all integers 𝝅𝝅. Then,

�𝟏𝟏 − 𝐬𝐬𝐬𝐬𝐭𝐭(𝜽𝜽)��𝐭𝐭𝐭𝐭𝐭𝐭(𝜽𝜽) + 𝐬𝐬𝐬𝐬𝐜𝐜(𝜽𝜽)� = �𝟏𝟏 − 𝐬𝐬𝐬𝐬𝐭𝐭(𝜽𝜽)� �𝐬𝐬𝐬𝐬𝐭𝐭 (𝜽𝜽)𝐜𝐜𝐜𝐜𝐬𝐬 (𝜽𝜽) + 𝟏𝟏

𝐜𝐜𝐜𝐜𝐬𝐬(𝜽𝜽)�

= �𝟏𝟏 − 𝐬𝐬𝐬𝐬𝐭𝐭(𝜽𝜽)� �𝐬𝐬𝐬𝐬𝐭𝐭(𝜽𝜽) + 𝟏𝟏𝐜𝐜𝐜𝐜𝐬𝐬(𝜽𝜽) �

=�𝟏𝟏 − 𝐬𝐬𝐬𝐬𝐭𝐭(𝜽𝜽)��𝟏𝟏 + 𝐬𝐬𝐬𝐬𝐭𝐭(𝜽𝜽)�

𝐜𝐜𝐜𝐜𝐬𝐬(𝜽𝜽)

=𝟏𝟏 − 𝐬𝐬𝐬𝐬𝐭𝐭𝟐𝟐(𝜽𝜽)𝐜𝐜𝐜𝐜𝐬𝐬(𝜽𝜽)

=𝐜𝐜𝐜𝐜𝐬𝐬𝟐𝟐(𝜽𝜽)𝐜𝐜𝐜𝐜𝐬𝐬(𝜽𝜽)

= 𝐜𝐜𝐜𝐜𝐬𝐬(𝜽𝜽).

Thus, �𝟏𝟏 − 𝐬𝐬𝐬𝐬𝐭𝐭(𝜽𝜽)��𝐭𝐭𝐭𝐭𝐭𝐭(𝜽𝜽) + 𝐬𝐬𝐬𝐬𝐜𝐜(𝜽𝜽)� = 𝐜𝐜𝐜𝐜𝐬𝐬(𝜽𝜽), for all real numbers 𝜽𝜽 so that 𝜽𝜽 ≠ 𝝅𝝅𝟐𝟐𝝅𝝅, for all

integers 𝝅𝝅.

I can use the Factoring by Grouping method to rewrite the expression.

I recognize the numerator as the factored form of a difference of squares: (𝑎𝑎 + 𝑏𝑏)(𝑎𝑎 − 𝑏𝑏) = 𝑎𝑎2 − 𝑏𝑏2.

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I recognize the expression −1 + cos2(𝑥𝑥) as −(1 − cos2(𝑥𝑥)), which I can rewrite as −sin2(𝑥𝑥). Then I can write the numerator as a single term.

b. 5−csc2(𝑥𝑥)+cot2(𝑥𝑥)sec2(𝑥𝑥) = 4 cos2(𝑥𝑥)

Proof: Let 𝒙𝒙 be a real number so that 𝒙𝒙 ≠ 𝝅𝝅𝟐𝟐𝝅𝝅, for all integers 𝝅𝝅. Then,

𝟓𝟓 − 𝐜𝐜𝐬𝐬𝐜𝐜𝟐𝟐(𝒙𝒙) + 𝐜𝐜𝐜𝐜𝐭𝐭𝟐𝟐(𝒙𝒙)𝐬𝐬𝐬𝐬𝐜𝐜𝟐𝟐(𝒙𝒙) =

�𝟓𝟓 − 𝟏𝟏𝐬𝐬𝐬𝐬𝐭𝐭𝟐𝟐(𝒙𝒙) + 𝐜𝐜𝐜𝐜𝐬𝐬𝟐𝟐(𝒙𝒙)

𝐬𝐬𝐬𝐬𝐭𝐭𝟐𝟐(𝒙𝒙)�

� 𝟏𝟏𝐜𝐜𝐜𝐜𝐬𝐬𝟐𝟐(𝒙𝒙)�

= �𝟓𝟓 𝐬𝐬𝐬𝐬𝐭𝐭

𝟐𝟐(𝒙𝒙) − 𝟏𝟏 + 𝐜𝐜𝐜𝐜𝐬𝐬𝟐𝟐(𝒙𝒙)𝐬𝐬𝐬𝐬𝐭𝐭𝟐𝟐(𝒙𝒙) �


=�𝟓𝟓 𝐬𝐬𝐬𝐬𝐭𝐭

𝟐𝟐(𝒙𝒙) − (𝟏𝟏 − 𝐜𝐜𝐜𝐜𝐬𝐬𝟐𝟐(𝒙𝒙))𝐬𝐬𝐬𝐬𝐭𝐭𝟐𝟐(𝒙𝒙) �


= �𝟓𝟓 𝐬𝐬𝐬𝐬𝐭𝐭

𝟐𝟐(𝒙𝒙) − 𝐬𝐬𝐬𝐬𝐭𝐭𝟐𝟐(𝒙𝒙)𝐬𝐬𝐬𝐬𝐭𝐭𝟐𝟐(𝒙𝒙) �


=�𝟒𝟒 𝐬𝐬𝐬𝐬𝐭𝐭

𝟐𝟐(𝒙𝒙)𝐬𝐬𝐬𝐬𝐭𝐭𝟐𝟐(𝒙𝒙) �


=𝟒𝟒


= 𝟒𝟒 𝐜𝐜𝐜𝐜𝐬𝐬𝟐𝟐(𝒙𝒙).

Therefore 𝟓𝟓−𝐜𝐜𝐬𝐬𝐜𝐜𝟐𝟐(𝒙𝒙)+𝐜𝐜𝐜𝐜𝐭𝐭𝟐𝟐(𝒙𝒙)𝐬𝐬𝐬𝐬𝐜𝐜𝟐𝟐(𝒙𝒙) = 𝟒𝟒 𝐜𝐜𝐜𝐜𝐬𝐬𝟐𝟐(𝒙𝒙), where 𝒙𝒙 ≠ 𝝅𝝅

𝟐𝟐𝝅𝝅, for any integer 𝝅𝝅.

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2015-16

M2 ALGEBRA II

Lesson 17: Trigonometric Identity Proofs

I can apply the formulas for the sine of a sum and the sine of a difference: sin(𝛼𝛼 + 𝛽𝛽) = sin(𝛼𝛼)cos(𝛽𝛽) + cos(𝛼𝛼)sin(𝛽𝛽) sin(𝛼𝛼 − 𝛽𝛽) = sin(𝛼𝛼)cos(𝛽𝛽)− cos(𝛼𝛼)sin(𝛽𝛽).


1. Derive a formula for cot(2𝑢𝑢) for 𝑢𝑢 ≠ 𝑘𝑘𝑘𝑘2

, for all integers 𝑘𝑘.

Proof: Let 𝒖𝒖 be any real number so that 𝒖𝒖 ≠ 𝒌𝒌𝒌𝒌𝟐𝟐

, for all integers 𝒌𝒌.

Because 𝐭𝐭𝐭𝐭𝐭𝐭(𝜶𝜶 + 𝜷𝜷) = 𝐭𝐭𝐭𝐭𝐭𝐭(𝜶𝜶)+𝐭𝐭𝐭𝐭𝐭𝐭(𝜷𝜷)𝟏𝟏−𝐭𝐭𝐭𝐭𝐭𝐭(𝜶𝜶) 𝐭𝐭𝐭𝐭𝐭𝐭(𝜷𝜷), it follows that 𝐜𝐜𝐜𝐜𝐭𝐭(𝜶𝜶 + 𝜷𝜷) = 𝟏𝟏−𝐭𝐭𝐭𝐭𝐭𝐭(𝜶𝜶) 𝐭𝐭𝐭𝐭𝐭𝐭(𝜷𝜷)

𝐭𝐭𝐭𝐭𝐭𝐭(𝜶𝜶)+𝐭𝐭𝐭𝐭𝐭𝐭(𝜷𝜷) .

Replacing 𝜶𝜶 and 𝜷𝜷 both by 𝒖𝒖 in the formula for 𝐜𝐜𝐜𝐜𝐭𝐭(𝜶𝜶 + 𝜷𝜷) gives 𝐜𝐜𝐜𝐜𝐭𝐭(𝟐𝟐𝒖𝒖) = 𝟏𝟏−𝐭𝐭𝐭𝐭𝐭𝐭(𝒖𝒖) 𝐭𝐭𝐭𝐭𝐭𝐭(𝒖𝒖)𝐭𝐭𝐭𝐭𝐭𝐭(𝒖𝒖)+𝐭𝐭𝐭𝐭𝐭𝐭(𝒖𝒖) , which is

equivalent to 𝐜𝐜𝐜𝐜𝐭𝐭(𝟐𝟐𝒖𝒖) = 𝟏𝟏−𝐭𝐭𝐭𝐭𝐭𝐭𝟐𝟐(𝒖𝒖)𝟐𝟐 𝐭𝐭𝐭𝐭𝐭𝐭(𝒖𝒖) for 𝒖𝒖 ≠ 𝒌𝒌𝒌𝒌

𝟐𝟐, for all integers 𝒌𝒌.

2. Prove that cos(2𝑢𝑢) = 1 − 2sin2(𝑢𝑢) for any real number 𝑢𝑢.

Proof: Let 𝒖𝒖 be any real number. From Exercise 5, 𝐜𝐜𝐜𝐜𝐜𝐜(𝟐𝟐𝒖𝒖) = 𝐜𝐜𝐜𝐜𝐜𝐜𝟐𝟐(𝒖𝒖) − 𝐜𝐜𝐬𝐬𝐭𝐭𝟐𝟐(𝒖𝒖), for any real number 𝒖𝒖. Using the Pythagorean identity, 𝐜𝐜𝐜𝐜𝐜𝐜𝟐𝟐(𝒖𝒖) = 𝟏𝟏 − 𝐜𝐜𝐬𝐬𝐭𝐭𝟐𝟐(𝒖𝒖). By substitution,

𝐜𝐜𝐜𝐜𝐜𝐜(𝟐𝟐𝒖𝒖) = 𝟏𝟏 − 𝐜𝐜𝐬𝐬𝐭𝐭𝟐𝟐(𝒖𝒖) − 𝐜𝐜𝐬𝐬𝐭𝐭𝟐𝟐(𝒖𝒖) = 𝟏𝟏 − 𝟐𝟐 𝐜𝐜𝐬𝐬𝐭𝐭𝟐𝟐(𝒖𝒖).

Thus, 𝐜𝐜𝐜𝐜𝐜𝐜(𝟐𝟐𝒖𝒖) = 𝟏𝟏 − 𝟐𝟐 𝐜𝐜𝐬𝐬𝐭𝐭𝟐𝟐(𝒖𝒖), for any real number 𝒖𝒖.

3. Write as a single term: sin �𝜃𝜃 − 𝜋𝜋3� + sin �𝜃𝜃 + 𝜋𝜋

3�.

𝐜𝐜𝐬𝐬𝐭𝐭 �𝜽𝜽 − 𝒌𝒌𝟑𝟑�+ 𝐜𝐜𝐬𝐬𝐭𝐭 �𝜽𝜽 + 𝒌𝒌

𝟑𝟑� = 𝐜𝐜𝐬𝐬𝐭𝐭(𝜽𝜽) 𝐜𝐜𝐜𝐜𝐜𝐜 �𝒌𝒌

𝟑𝟑� − 𝐜𝐜𝐬𝐬𝐭𝐭 �𝒌𝒌

𝟑𝟑� 𝐜𝐜𝐜𝐜𝐜𝐜(𝜽𝜽) + 𝐜𝐜𝐬𝐬𝐭𝐭(𝜽𝜽) 𝐜𝐜𝐜𝐜𝐜𝐜 �𝒌𝒌

𝟑𝟑� + 𝐜𝐜𝐬𝐬𝐭𝐭 �𝒌𝒌

𝟑𝟑� 𝐜𝐜𝐜𝐜𝐜𝐜(𝜽𝜽)

=𝟏𝟏𝟐𝟐𝐜𝐜𝐬𝐬𝐭𝐭(𝜽𝜽) −

√𝟑𝟑𝟐𝟐𝐜𝐜𝐜𝐜𝐜𝐜(𝜽𝜽) +

𝟏𝟏𝟐𝟐𝐜𝐜𝐬𝐬𝐭𝐭(𝜽𝜽) +

√𝟑𝟑𝟐𝟐𝐜𝐜𝐜𝐜𝐜𝐜(𝜽𝜽)

= 𝐜𝐜𝐬𝐬𝐭𝐭(𝜽𝜽).

I can use the formula for tan(α + β) that we derived in Exercise 3 and apply the identity cot(𝜃𝜃) = 1

tan(𝜃𝜃) for all defined values of 𝜃𝜃.

Using this form of the identity allows me to write the expression cos(2𝑢𝑢) in terms of sin(𝑢𝑢).

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2015-16

M2 ALGEBRA II


I can apply the formula for the cosine of a difference: cos(𝛼𝛼 − 𝛽𝛽) = sin(𝛼𝛼)sin(𝛽𝛽) + cos(𝛼𝛼)cos(𝛽𝛽).

4. Use the formula for the cosine of differences to verify the Pythagorean identity: sin2(𝑥𝑥) + cos2(𝑥𝑥) = 1, for every real number 𝑥𝑥. Begin with the formula 𝐜𝐜𝐜𝐜𝐜𝐜(𝜶𝜶 − 𝜷𝜷) = 𝐜𝐜𝐜𝐜𝐜𝐜(𝜶𝜶) 𝐜𝐜𝐜𝐜𝐜𝐜(𝜷𝜷) + 𝐜𝐜𝐬𝐬𝐭𝐭(𝜶𝜶) 𝐜𝐜𝐬𝐬𝐭𝐭(𝜷𝜷), and let 𝜶𝜶 = 𝒙𝒙 and 𝜷𝜷 = 𝒙𝒙. Then,

𝐜𝐜𝐬𝐬𝐭𝐭𝟐𝟐(𝒙𝒙) + 𝐜𝐜𝐜𝐜𝐜𝐜𝟐𝟐(𝒙𝒙) = 𝐜𝐜𝐬𝐬𝐭𝐭(𝒙𝒙) 𝐜𝐜𝐬𝐬𝐭𝐭(𝒙𝒙) + 𝐜𝐜𝐜𝐜𝐜𝐜(𝒙𝒙) 𝐜𝐜𝐜𝐜𝐜𝐜(𝒙𝒙) = 𝐜𝐜𝐜𝐜𝐜𝐜(𝒙𝒙 − 𝒙𝒙) = 𝐜𝐜𝐜𝐜𝐜𝐜(𝟎𝟎) = 𝟏𝟏.

Thus, 𝐜𝐜𝐬𝐬𝐭𝐭𝟐𝟐(𝒙𝒙) + 𝐜𝐜𝐜𝐜𝐜𝐜𝟐𝟐(𝒙𝒙) = 𝟏𝟏, for every real number 𝒙𝒙.

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