Sinusoidal Curve Fitting and Graphical Analysis

Section 2.3 Sinusoidal Curve Fitting and Graphical Analysis

In this chapter, we learn that sinusoidal patterns exist in various aspects throughout our world. One example is the moon, which undergoes phases oscillating from no illumination, to waxing (increasing illumination), reaching a fully lit moon, and then waning (decreasing illumination) until it completes its cycle with no illumination again. Indigenous cultures across the world have deeply connected with these lunar cycles, shaping cultural practices aligned with the moon’s phases. For example, the Māori of New Zealand and the Hopi Tribe in northeastern Arizona, USA, both time activities such as planting and harvesting specific plants for each moon cycle based on generations of lunar observations.

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In Section 2.1, we graphed sinusoidal functions and determined their values for any given input value. The ability to formulate a sinusoidal equation modeling the moon’s phases allows us to predict the moon’s phase on any date. In this section, we will explore the process of developing sinusoidal functions based on provided information. This will enable us to model real-world phenomena using real data.

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Subsection 2.3.1 Finding Sinusoidal Equations from Characteristics

To begin finding sinusoidal equations of the form

y = A \sin (B (x - C)) + D and y = A \cos (B (x - C)) + D

we will use the characteristics of sine and cosine functions, as described in Remark 2.1.38 and summarized below, to determine the values of

A,

B,

C,

and

D .

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Amplitude and Vertical Stretch/Compression: $| A |$
- $| A |$ is the value of the amplitude.
- If $| A | > 1,$ there is vertical stretching.
- If $0 < | A | < 1,$ there is vertical compression.
Period and Horizontal Stretch/Compression: $| B |$
- The period is $\frac{2 π}{| B |} .$
- If $| B | > 1,$ there is horizontal compression and the period is shortened.
- If $0 < | B | < 1,$ there is horizontal stretching and the period is lengthened.
Phase Shift: $C$
- If $C$ is positive, there is a shift to the right.
- If $C$ is negative, there is a shift to the left.
Vertical Shift: $D$
- If $D$ is positive, there is a shift upward.
- If $D$ is negative, there is a shift downward.
Reflection about the $x$ -axis:
- If $A$ is negative ( $A < 0$ ), there is a reflection about the $x$ -axis.
Reflection about the $y$ -axis:
- If $B$ is negative ( $B < 0$ ), there is a reflection about the $y$ -axis.

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Example 2.3.1. Finding an Equation of a Sine Function Using its Characteristics.

Write an equation of a sine function with an amplitude of 2, period of 8, phase shift of

\frac{π}{3},

and vertical shift of 4.

Solution.

To determine an equation of the form

f (x) = A \sin (B (x - C)) + D

we first examine the information that we are given: amplitude of 2, period of 8, phase shift of

\frac{π}{3},

and vertical shift of 4.

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Since

| A |

represents the value of the amplitude, we get

| A | = 2 .

Thus we either have

A = 2

A = - 2 .

Since we are not given a reflection about the

x

-axis, we can conclude that

A

is not negative, thus

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A = 2 .

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Next, since the period of a sine function is given by

\frac{2 π}{| B |},

we get

\frac{2 π}{| B |} = 8 .

Solving for

| B |,

we get

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| B | = \frac{2 π}{8} = \frac{π}{4} .

Since there is no reflection about the

y

-axis, we have that

B

must be positive:

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B = \frac{π}{4} .

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Finally, since

C

and

D

represent the phase shift and vertical shift, respectively, we get

C = \frac{π}{3}, D = 4 .

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Combining these, the sine function becomes:

f (x) = 2 \sin (\frac{π}{4} (x - \frac{π}{3})) + 4 .

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Subsection 2.3.2 Finding Sinusoidal Equations from Graphs

Sometimes we are not explicitly given the characteristics of the function, but are provided with the graph. Examining a graph can reveal its characteristics, allowing us to find the equation of a function.

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In this next example, we’ll explore how to find an equation of a cosine function based on the graph of a sine function.

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Example 2.3.2. Finding an Equation of a Cosine Function Using the Graph of a Sine Function.

Below is the graph of

y = \sin (x) .

Find an equation of the form

y = A \cos (B (x - C)) + D

that fits the graph.

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Solution.

The cosine function has its maximum when

x = 0,

decreases to its minimum at

x = π,

and then increases before completing one period at

2 π .

This graph starts at 0 when

x = 0,

then increases to its maximum at

x = \frac{π}{2},

and then has the shape of the cosine function starting at

x = 0 .

This represents a phase shift to the right. There no reflections about the

x

-axis or

y

-axis. We will find the equation of a function with characteristics of a phase shift to the right.

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Amplitude ( $A$ ): Since the amplitude of the cosine function will be the same as the amplitude of the sine function, which is 1, we have

| A | = 1 .

Since there is no reflection about the

x

-axis, we choose the positive value to get

A = 1 .

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Vertical Shift (D): The vertical shift of the cosine function will be the same as that of the sine function, which is 0. So, we have

D = 0 .

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Period (B): The period of the cosine function will also be the same as the period of the sine function, which is

2 π .

Since

2 π = \frac{2 π}{| B |},

we have

| B | = 1 .

Since there is no reflection about the

y

-axis, we get

B = 1 .

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Therefore, we have

y = 1 \cos (1 x) + 0 = \cos (x) .

Overlapping the graph of

y = \cos (x)

onto the original graph will help us determine the phase shift.

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Phase Shift (C): From the graph, the cosine function will have a phase shift of

\frac{π}{2}

radians to the right to match the graph of the sine function. This is because the cosine function reaches its maximum value at

x = 0,

while the sine function reaches its maximum at

x = \frac{π}{2} .

So, we have

C = \frac{π}{2} .

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Therefore, the equation of the cosine function that fits the graph of

y = \sin (x)

is:

y = \cos (x - \frac{π}{2})

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Remark 2.3.3.

Since this graph is the graph of

y = \sin (x),

we get

\sin (x) = \cos (x - \frac{π}{2}) .

This solution demonstrates that any sine function can be written as a cosine function with an appropriate phase shift. In this case, the phase shift of

\frac{π}{2}

radians to the right transforms the cosine function into its corresponding sine function.

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Example 2.3.4. Finding an Equation of a Sine Function Using its Graphs.

Given the graph below, find an equation that represents the graph in the following forms:

(a)

f (x) = A \sin (B (x - C)) + D .

Solution.

The sine function starts at the origin, increases to its maximum when

x = \frac{π}{2},

decreases to its minimum at

x = \frac{3 π}{2},

and then increases again before completing one period at

2 π .

In the given graph, the function has a value of 8 when

x = 0,

then increases to its maximum value of 16 at

x = 10,

decreases to its minimum value of 0 at

x = 30,

and then increases again before completing one period at

x = 40 .

This indicates a vertical shift, a vertical stretch, and a horizontal stretch of the period. There are no reflections about the

x

-axis or

y

-axis. Therefore, we need to find the equation of a function with these characteristics."

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Amplitude ( $A$ ): Recall from Definition 2.1.24 that the amplitude is defined as half of the difference between the maximum and minimum values of the function. Here, the maximum value is 16 and the minimum value is 0, so the amplitude is:

| A | = \frac{16 - 0}{2} = 8 .

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Since there are no reflections about the

x

-axis, we use the positive value to get

A = 8 .

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Vertical Shift ( $D$ ): The midline, calculated as

y = \frac{maximum + minimum}{2} = \frac{16 + 0}{2} = 8,

represents the vertical shift of the graph. Therefore,

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D = 8 .

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Period ( $B$ ): The period of a function is the length of one full repetition (or cycle). We can identify the period on the graph by measuring the horizontal distance between corresponding points where the graph completes a cycle. In this case, we’ll use two corresponding peaks at

x = 10

and

x = 50

to obtain:

period = 50 - 10 = 40 .

Note that other points on the graph, such as the minimum values or the points where the graph crosses the midline, could also be used to determine the period.

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Since Definition 2.1.28 defines a period as

\frac{2 π}{| B |},

we have:

40 = \frac{2 π}{| B |} .

Thus,

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| B | = \frac{2 π}{40} = \frac{π}{20} .

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Since there are no reflections about the

y

-axis, we get

B = \frac{π}{20} .

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Phase Shift ( $C$ ): The phase shift of the graph refers to its horizontal translation. The cycle of a sine function typically starts at

x = 0

on the midline, increases to the maximum, decreases, passes the midline to the minimum, and then completes a cycle back at the midline. Since this graph follows these characteristics without any horizontal translation, there is no phase shift. Therefore,

C = 0 .

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Thus, the graph can be described by the following sine function:

f (x) = 8 \sin (\frac{π}{20} x) + 8 .

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(b)

g (x) = A \cos (B (x - C)) + D .

Solution.

Since the cosine function has its maximum value when

x = 0

before decreasing to its minimum and then increasing to complete one period, the graph provides several options for phase shifts. By selecting a peak and determining the direction and value of the phase shift needed for the cosine function to reach that peak, we can align it with the given graph. For example, if we choose the peak at

x = - 30,

the cosine function will shift 30 units to the left. Similarly, for peaks at

x = 10

x = 50,

the cosine function would need to shift 10 or 50 units, respectively, to the right. Since the peak at

x = 10

is the closest to the peak of the original cosine function when

x = 0,

we opt for this phase shift. Additionally, as the graph of the sine function, there is a vertical shift, a vertical stretch, and a horizontal stretch, with no reflections about the

y

-axis or the

x

-axis.

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Amplitude ( $A$ ): The amplitude of a cosine function is the same as that of the corresponding sine function. Thus,

| A | = 8 .

Since there are no reflections about the

x

-axis, we use the positive value to get

A = 8 .

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Vertical Shift ( $D$ ): The vertical shift of a cosine function is the same as that of the corresponding sine function. Thus,

D = 8 .

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Period ( $B$ ): The period of a cosine function is also the same as that of the corresponding sine function. In this case,

| B | = \frac{π}{20} .

Since there are no reflections about the

y

-axis, we determine

B = \frac{π}{20} .

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Before we examine the phase shift, let’s summarize what we found so far:

g (x) = 8 \cos (\frac{π}{20} (x)) + 8

and overlap this graph with the given graph.

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Phase Shift ( $C$ ): From the graph, we see that the starting point of the cosine function’s cycle is at its maximum value, unlike the starting point of a sine function’s cycle, which is at its midline. In this case, the graph has a horizontal shift to the right of 10 units. Thus,

C = 10 .

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We can now describe the graph with the following cosine function:

g (x) = 8 \cos (\frac{π}{20} (x - 10)) + 8 .

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Subsection 2.3.3 Finding Sinusoidal Equations from Data

Example 2.3.5. Modeling the Daylight Hours in Munda.

In Section 2.1 we learned that Earth’s axis is tilted, and as the Earth orbits the sun, this axial tilt causes seasons, which are periodic. Another effect of the axial tilt and orbit is the amount of daylight each part of Earth experiences, which is also periodic.

The total sunlight duration in Munda, on the island of New Georgia in the Solomon Islands in 2025, is plotted in Figure 2.3.6. On the 21st of June 2025 (the 172nd day of the year), which is the shortest day of the year in Munda, the total sunlight duration is 11 hours, 38 minutes, and 17 seconds (approximately 11.64 hours). Conversely, on the 22nd of December 2025 (the 356th day of the year), which is the longest day of the year, the total sunlight duration is 12 hours, 36 minutes, and 44 seconds (approximately 12.61 hours). Find a function of the form

y = A \cos (B (x - C)) + D

to model the total hours of daylight in Munda in 2025, assuming that one period represents one year or 365 days.

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Figure 2.3.6. Hours of daylight in Munda, Solomon Islands. Source: NOAA Solar Calculator.

Solution.

Amplitude ( $A$ ): From the definition of amplitude (Definition 2.1.24),

| A | = \frac{maximum - minimum}{2} = \frac{12.61 - 11.64}{2} = 0.485 .

Assuming no reflections about the

x

-axis, we have:

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A = 0.485 .

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Period ( $B$ ): We assume a period of one year or 365 days. Additionally, we can assume no reflections about the

y

-axis, leading to a positive value for

B .

Thus we have

\frac{2 π}{B} = 365

B = \frac{2 π}{365} \approx 0.0172 .

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Phase Shift ( $C$ ): The characteristic of a cosine function is that at

x = 0,

the function is at its maximum value. However, in this case, the maximum value occurs on Day 356. This represents a phase shift to the right by 356 days, thus:

C = 356 .

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Vertical Shift ( $D$ ): The value of the midline represents the average duration of daylight, which is the vertical shift:

D = \frac{maximum + minimum}{2} = \frac{12.61 + 11.64}{2} = 12.125 .

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Therefore, the equation of our function is

y = 0.485 \cos (0.0172 (x - 356)) + 12.125 .

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Remark 2.3.7.

It’s important to note that the average duration of daylight is not exactly 12 hours but instead approximately 12.125 hours. This discrepancy arises due to atmospheric refractions, which cause the apparent sunrise and sunset to occur slightly before and after, respectively, the sun crosses the horizon—the actual sunrise and sunset times.

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Example 2.3.8. Modeling the Temperature in Christchurch.

Another effect of axial tilt besides daylight hours is temperature, which is also periodic. The average monthly temperature for Christchurch, New Zealand is given in Table 2.3.9 (Source: National Institute of Water and Atmospheric Research (NIWA). Retrieved 18 March 2024). Find a sinusoidal function of the form

y = A \cos (B (x - C)) + D

to model the average monthly temperature of Christchurch.

Table 2.3.9. The average monthly temperature for Christchurch, New Zealand.

Month, $x$	Temperature ( $^{\circ}$ C)
January, $1$	$17.5$
February, $2$	$17.2$
March, $3$	$15.5$
April, $4$	$12.7$
May, $5$	$9.8$
June, $6$	$7.1$
July, $7$	$6.6$
August, $8$	$7.9$
September, $9$	$10.3$
October, $10$	$12.2$
November, $11$	$14.1$
December, $12$	$16.1$

Solution.

We begin by plotting the points in Table 2.3.9.

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Examining the plot of the data points, we see this looks like the graph of a cosine function with no reflections about the

x

-axis or the

y

-axis.

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Amplitude ( $A$ ): The amplitude is

| A | = \frac{maximum - minimum}{2} = \frac{17.5 - 6.6}{2} = 5.45 .

Since there are no reflections about the

x

-axis, we get

A = 5.45

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Period (B): Since the temperatures repeat every 12 months, the period is 12 and so

\frac{2 π}{| B |} = 12 .

Since there are no reflections about the

y

-axis, we keep the positive value to obtain

B = \frac{2 π}{12} = \frac{π}{6} .

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Vertical Shift ( $D$ ): The vertical shift is the average of the maximum and minimum values of the data, which corresponds to the midline of the graph:

D = \frac{maximum + minimum}{2} = \frac{17.5 + 6.6}{2} = 12.05 .

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Phase Shift ( $C$ ): The maximum temperature in the data occurs in January (

x = 1

). However, since the graph of cosine reaches its maximum value at

x = 0,

we have a phase shift of 1 to the right to align the peak with the maximum temperature data point. Thus,

C = 1 .

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Our function now becomes

y = 5.45 \cos (\frac{π}{6} (x - 1)) + 12.05 .

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Finally, we plot our function and the data together.

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Remark 2.3.10. Steps for Deriving Sinusoidal Models from Data.

Graph the data points.
Determine the characteristics of the data, including vertical and horizontal stretching, phase shifts, vertical shifts, and reflections about the $x$ -axis or the $y$ -axis.
Amplitude: Calculate the amplitude using the formula

$| A | = \frac{maximum - minimum}{2} .$

Use the positive value if there is no reflection about the $x$ -axis, and use the negative value if there is a reflection.
Vertical Shift ( $D$ ): The vertical shift is the average of the maximum and minimum values of the data, which corresponds to the midline of the graph. We can calculate the vertical shift using the formula:

$D = \frac{maximum + minimum}{2} .$
Period: Calculate the period from the data, then find

$| B | = \frac{2 π}{period} .$

Use the positive value if there is no reflection about the $y$ -axis, and use the negative value if there is a reflection.
Phase Shift: Determine the phase shift, $C,$ by calculating the distance between points from the data and their corresponding points on the sine or cosine function, such as the maximum, minimum, or values on the midline.

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Subsection 2.3.4 Finding Sinusoidal Equations with Technology

Some graphing utilities, such as the TI-83 calculator or Desmos Graphing Calculator

Desmos Graphing Calculator

, have functions that allow us to find a sinusoidal best-fit function given data. While some devices can find the best-fit for both sine and cosine functions, others calculate the best-fit line only for sine. For example, the TI-83 uses the SinReg function to calculate the sine function.

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Example 2.3.11. Finding the Sinusoidal Equation using Desmos.

Utilize a graphing utility to determine the best-fit cosine function for the data provided in Table 2.3.9.

Solution.

Open a new table in the Desmos Graphing Calculator by either typing “table” in a blank expression line or by clicking the Add Item menu in the upper left corner and selecting Table.
Enter the values from Table 2.3.9 into the table, where $x_{1}$ represents the month and $y_{1}$ represents the temperature.
Use the Zoom Fit icon (a magnifying glass with a + symbol) at the bottom left corner of the table to automatically adjust the graph settings window to best display the data.
In a blank expression line, type “ $y_{1} \sim A \cos (B (x_{1} - C)) + D$ ” to fit a cosine function to the data.
The parameters for the best-fit function will be returned: $A = 5.21412,$ $B = 0.534145,$ $C = 1.19926,$ $D = 12.1509 .$ Thus, the best-fit cosine function is:

$y = 5.21412 \cos (0.534145 (x - 1.19926)) + 12.1509 .$

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Figure 2.3.12 displays an interactive Desmos Graphing Calculator with the completed table and the curve of best fit plotted together.

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Figure 2.3.12. Given data points in a table, Desmos can create a sinusoidal function to model the data. If you are viewing the PDF or a printed copy, you can scan the QR code or follow the “Standalone” link to explore the interactive version online.

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Exercises 2.3.5 Exercises

x

-axis.

Answer.

y = - 1.5 \sin (12 x)

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4.

Amplitude:

2;

Period:

3 π;

Reflection about the

x

-axis.

Answer.

y = - 2 \sin (\frac{2}{3} x)

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5.

Amplitude:

4;

Period:

\frac{2 π}{3};

Vertical Shift

2 .

Answer.

y = 4 \sin (3 x) + 2

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6.

Amplitude:

3;

Period:

\frac{5}{2};

Vertical Shift:

- \frac{2}{3} .

Answer.

y = 3 \sin (\frac{4 π}{5} x) - \frac{2}{3}

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7.

Amplitude:

\frac{1}{2};

Period:

2 π;

Phase Shift:

\frac{π}{4} .

Answer.

y = \frac{1}{2} \sin (x - \frac{π}{4})

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8.

Amplitude:

1;

Period:

\frac{5 π}{3};

Phase Shift:

\frac{π}{6} .

Answer.

y = \sin (\frac{6}{5} (x - \frac{π}{6}))

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9.

Amplitude:

2,

Period:

\frac{3 π}{2};

Phase Shift:

\frac{π}{3};

Vertical Shift:

- 1 .

Answer.

y = 2 \sin (\frac{4}{3} (x - \frac{π}{3})) - 1

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10.

Amplitude:

2,

Period:

20;

Phase Shift:

- \frac{π}{4};

Vertical Shift: 3.

Amplitude: 2; Period:

\frac{7 π}{4};

Reflection in the

x

-axis.

Answer.

y = - 2 \cos (\frac{8}{7} x)

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14.

Amplitude: 1.5; Period:

\frac{4 π}{5};

Reflection in the

x

-axis.

Answer.

y = - 1.5 \cos (\frac{5}{2} x)

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15.

Amplitude: 4; Period:

\frac{5 π}{3};

Vertical Shift: -3.

Answer.

y = 4 \cos (\frac{6}{5} x) - 3

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16.

Amplitude: 3.5; Period:

\frac{4 π}{3};

Vertical Shift: 2.

Answer.

y = 3.5 \cos (\frac{3}{2} x) + 2

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17.

Amplitude: 3; Period:

\frac{3 π}{2};

Phase Shift:

\frac{π}{6} .

Answer.

y = 3 \cos (\frac{4}{3} (x - \frac{π}{6}))

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18.

Amplitude: 2.5; Period:

\frac{π}{3};

Phase Shift:

\frac{π}{4} .

Answer.

y = 2.5 \cos (6 (x - \frac{π}{4}))

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19.

Amplitude: 2; Period:

3 π;

Phase Shift:

- \frac{π}{2};

Vertical Shift: 2.

Answer.

y = 2 \cos (\frac{2}{3} (x + \frac{π}{2})) + 2

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20.

Amplitude: 1.2; Period:

\frac{5 π}{2};

Phase Shift:

- \frac{π}{3};

Vertical Shift: 3.

Answer.

y = 1.2 \cos (\frac{4}{5} (x + \frac{π}{3})) + 3

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Exercise Group.

For each given graph, identify the amplitude, period, phase shift, and vertical shift, and write an equation of the form

y = A \sin (B (x - C)) + D

that represents these characteristics.

21.

Answer.

Amplitude:

A = 4;

Period:

8;

Phase Shift:

C = 0;

Vertical Shift:

D = 0;

y = 4 \sin (\frac{π}{4} x)

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22.

Answer.

Amplitude:

A = 3;

Period:

π;

Phase Shift:

C = 0;

Vertical Shift:

D = 1;

y = 3 \sin (2 x) + 1

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23.

Answer.

Amplitude:

A = 1.5;

Period:

π;

Phase Shift:

C = \frac{π}{2};

Vertical Shift:

D = - 1;

y = 1.5 \sin (2 (x - \frac{π}{2})) - 1

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24.

Answer.

Amplitude:

A = 4;

Period:

12;

Phase Shift:

C = 3;

Vertical Shift:

D = 2;

y = 4 \sin (\frac{π}{6} (x - 3)) + 2

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Exercise Group.

For each given graph, identify the amplitude, period, phase shift, and vertical shift. Write an equation that represents these characteristics of the form

y = A \cos (B (x - C)) + D .

25.

Answer.

Amplitude:

A = 2;

Period:

\frac{2 π}{3};

Phase Shift:

C = 0;

Vertical Shift:

D = 1;

y = 2 \cos (3 x) + 1

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26.

Answer.

Amplitude:

A = 5;

Period:

12;

Phase Shift:

C = 2;

Vertical Shift:

D = 0;

y = 5 \cos (\frac{π}{6} (x - 2))

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27.

Answer.

Amplitude:

A = 2;

Period:

4 π;

Phase Shift:

C = π;

Vertical Shift:

D = - 4;

y = 2 \cos (\frac{1}{2} (x - π)) - 4

🔗

28.

Answer.

Amplitude:

A = 4;

Period:

8 π;

Phase Shift:

C = 2 π;

Vertical Shift:

D = 2;

y = 4 \cos (\frac{1}{4} (x - 2 π)) + 2

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Hours of daylight.

For each of the following questions, the number of daylight hours for a pair of islands in 2025 is given. These islands share the same latitude or are close to it, with one island located north of the equator and the other south of it. Find a sinusoidal function of the form

y = A \cos (B (x - C)) + D

to model the daylight hours for each island. This data is sourced from the NOAA Solar Calculator.

29.

Pohnpei, situated at 6.9° North latitude in the Federated States of Micronesia, experiences its longest day, lasting 12.52 hours, on 21 June 2025 (the 172nd day of the year), and its shortest day, lasting 11.72 hours, on 22 December 2025 (the 356th day of the year).

Nanumanga, located at 6.3° South latitude in Tuvalu, experiences its longest day, lasting 12.49 hours, on 22 December 2025 (the 356th day of the year), and its shortest day, lasting 11.76 hours, on 21 June 2025 (the 172nd day of the year).

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Answer.

Pohnpei:

y = 0.4 \cos (0.0172 (x - 172)) + 12.12;

Nanumanga:

y = 0.365 \cos (0.0172 (x - 356)) + 12.125

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30.

Saipan, positioned at 15.2° North latitude in the Northern Mariana Islands, has its longest day, lasting 13.03 hours, on 21 June 2025 (the 172nd day of the year), and its shortest day, lasting 11.23 hours, on 22 December 2025 (the 356th day of the year).

Espiritu Santo, located at 15.4° South latitude in Vanuatu, experiences its longest day, lasting 13.04 hours, on 22 December 2025 (the 356th day of the year), and its shortest day, lasting 11.21 hours, on 21 June 2025 (the 172nd day of the year).

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Answer.

Saipan:

y = 0.9 \cos (0.0172 (x - 172)) + 12.13;

Espiritu Santo:

y = 0.915 \cos (0.0172 (x - 356)) + 12.125

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31.

Kauaʻi, situated at 22.1° North latitude in Hawaiʻi, has its longest day, lasting 13.48 hours, on 20 June 2025 (the 171st day of the year), and its shortest day, lasting 10.78 hours, on 21 December 2025 (the 355th day of the year).

Mangaia, located at 21.9° South latitude in the Cook Islands, experiences its longest day, lasting 13.47 hours, on 21 December 2025 (the 355th day of the year), and its shortest day, lasting 10.79 hours, on 20 June 2025 (the 171st day of the year).

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It’s noteworthy that Kauaʻi and Mangaia are situated east of the International Date Line, causing them to experience the winter and summer solstice one day earlier than islands located west of the International Date Line.

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Answer.

Kauaʻi:

y = 1.35 \cos (0.0172 (x - 171)) + 12.13;

Mangaia:

y = 1.34 \cos (0.0172 (x - 355)) + 12.13

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32.

What patterns do you observe when comparing the graphs of daylight hours for pairs of islands across different latitudes? Additionally, how does the variation in daylight hours change with increasing latitude, that is, as one moves away from the equator?

Answer.

The islands exhibit mirrored patterns in daylight hours, where one island experiences longer days while the other experiences shorter days. This is due to their opposite positions relative to the equator.

As latitude increases (moving away from the equator), the variation in daylight hours also increases. Islands closer to the equator experience less variation in daylight hours throughout the year, while islands further away from the equator experience more changes in daylight hours between seasons.

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Exercise Group.

The islands of Oʻahu and Rarotonga are located at similar distances from the equator. However, they experience different climates due to their locations relative to the equator. Oʻahu is at 21.26° North, directly mirrored by Rarotonga at 21.26° South. The table below gives the average monthly temperatures (in °C) for each island. Use the data in the table to answer the following questions. Source: http://www.worldclimate.com, retrieved on 18 March, 2024.

Month ( $x$ )	Oʻahu (21.3° N)	Rarotonga (21.2° S)
January (1)	22.7	25.7
February (2)	22.7	26.1
March (3)	23.5	25.8
April (4)	24.3	24.9
May (5)	25.2	23.5
June (6)	26.3	22.3
July (7)	26.9	21.7
August (8)	27.4	21.6
September (9)	27.2	22.1
October (10)	26.4	22.9
November (11)	25.1	23.8
December (12)	23.3	24.8

33.

Determine a sinusoidal function of the form

y = A \sin (B (x - C)) + D

to represent the average monthly temperatures provided in the table for each island.

Answer.

Oʻahu:

y = 2.35 \sin (0.5236 (x - 5)) + 25.05;

Rarotonga:

y = 2.25 \sin (0.5236 (x - 11)) + 23.85

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34.

Utilize a graphing utility to identify the best-fit sinusoidal function of the form

y = A \sin (B (x - C)) + D

for each island.

Answer.

Oʻahu:

y = 2.3727 \sin (0.502443 (x - 4.72583)) + 25.005;

Rarotonga:

y = 2.24202 \sin (0.521233 (x + 1.16396)) + 23.7744

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35.

How do the temperature patterns of Oʻahu, situated in the northern hemisphere, compare with those of Rarotonga, positioned in the southern hemisphere?

Answer.

During Oʻahu’s winter, Rarotonga experiences its summer, and similarly, during Oʻahu’s summer, Rarotonga experiences its winter.

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Exercise Group.

Use the table below, which gives the average monthly temperatures (in Celsius) at various latitudes in the South Pacific, to answer the following questions. Express your answers in the form

y = A \sin (B (x - C)) + D .

Source: Data for Apia, Suva, Nukuʻalofa, and Rapa Nui obtained from http://www.worldclimate.com, retrieved on 18 March 2024; Data for Whangārei and Dunedin obtained from National Institute of Water and Atmospheric Research (NIWA)

National Institute of Water and Atmospheric Research (NIWA)

, retrieved on 18 March 2024.

	Apia	Suva	Nukuʻalofa	Rapa Nui	Whangārei	Dunedin
Month	13.8°S	18.1°S	21.1°S	27.1°S	35.7°S	45.9°S
Jan (1)	27.6	26.7	25.6	23.3	19.9	15.3
Feb (2)	27.6	26.9	26	23.7	20.2	15
Mar (3)	27.8	26.7	25.8	23.1	18.8	13.7
Apr (4)	27.8	26	24.9	21.9	16.6	11.7
May (5)	27.4	24.8	23.1	20.1	14.4	9.3
Jun (6)	27.1	24	22.4	18.9	12.4	7.3
Jul (7)	26.7	23.2	21.3	18	11.6	6.6
Aug (8)	26.5	23.2	21.2	17.9	11.9	7.7
Sep (9)	26.7	23.7	21.7	18.3	13.3	9.5
Oct (10)	27	24.4	22.4	19	14.6	10.9
Nov (11)	27.4	25.2	23.5	20.4	16.4	12.4
Dec (12)	27.4	26.1	24.7	21.8	18.5	13.9

y = 4.35 \sin (0.5236 (x - 11)) + 10.95

Answer.

As latitude increases (moving away from the equator towards the poles), the sinusoidal function’s amplitude increases, signifying greater temperature variability, while its vertical shift decreases, indicating lower average temperatures at higher latitudes.

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