Mastering Trigonometric Models and Periodic Function Insights

Function rule	What does the function model?
$R (t) = \frac{9 0 0 sin 2 t}{3 2}$	The horizontal distance from a player to a basketball thrown with an initial velocity of $30$ feet per second at an angle with measure $t .$ Negative distance means that the direction of the throw changes.
$s (t) = 3.5 sin \frac{π}{1 8 0} (t + 4) + 15.7$	The average wind speed in a certain city, measured in miles per hour.
$P (t) = 100 - 20 cos \frac{8 π}{3} t$	The blood pressure of a person at rest, measured in millimeters of mercury.
$N (t) = 3.7 sin (\frac{4 t}{5} - 0.7) + 20$	The number of thousands of employees at a sports company.

Function rule

What does the function model?

R (t) = \frac{9 0 0 sin 2 t}{3 2}

The horizontal distance from a player to a basketball thrown with an initial velocity of

30

feet per second at an angle with measure

t .

Negative distance means that the direction of the throw changes.

s (t) = 3.5 sin \frac{π}{1 8 0} (t + 4) + 15.7

The average wind speed in a certain city, measured in miles per hour.

P (t) = 100 - 20 cos \frac{8 π}{3} t

The blood pressure of a person at rest, measured in millimeters of mercury.

N (t) = 3.7 sin (\frac{4 t}{5} - 0.7) + 20

The number of thousands of employees at a sports company.

	Average Monthly Water Temperatures
$t$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$	$10$	$11$	$12$
$D (^{\circ} F)$	$70$	$73$	$75$	$78$	$82$	$85$	$86$	$85$	$84$	$81$	$75$	$73$

Average Monthly Water Temperatures

t

1

2

3

4

5

6

7

8

9

10

11

12

D (^{\circ} F)

70

73

75

78

82

85

86

85

84

81

75

73

Number of Rotations	Time
$3$	$1 minute = 60 seconds$
$1$	$\frac{6 0}{3} = 20$ seconds

Number of Rotations

Time

3

1 minute = 60 seconds

1

\frac{6 0}{3} = 20

seconds

Transformations of $y = cos x$
Vertical Stretch or Shrink	$Vertical stretch, a > 1 y = a cos x$
$Vertical shrink, 0 < a < 1 y = a cos x$
Horizontal Stretch or Shrink	$Horizontal shrink, b > 1 y = cos (b x)$
$Horizontal stretch, 0 < b < 1 y = cos (b x)$

Transformations of

y = cos x

Vertical Stretch or Shrink

Vertical stretch, a > 1 y = a cos x

Vertical shrink, 0 < a < 1 y = a cos x

Horizontal Stretch or Shrink

Horizontal shrink, b > 1 y = cos (b x)

Horizontal stretch, 0 < b < 1 y = cos (b x)

$x$	$y$
$1$	$7.61$
$2$	$7.52$
$3$	$6.86$
$4$	$5.99$
$5$	$5.18$
$6$	$4.67$
$7$	$4.68$
$8$	$5.11$
$9$	$5.67$
$10$	$6.23$
$11$	$6.82$
$12$	$7.41$

x

y

1

7.61

2

7.52

3

6.86

4

5.99

5

5.18

6

4.67

7

4.68

8

5.11

9

5.67

10

6.23

11

6.82

12

7.41

	Average Monthly Precipitation
$t (months)$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$	$10$	$11$	$12$
$P (inches)$	$0.5$	$0.6$	$0.9$	$1.5$	$2.2$	$2.6$	$2.7$	$2.3$	$1.6$	$1.2$	$0.7$	-

Average Monthly Precipitation

t (months)

1

2

3

4

5

6

7

8

9

10

11

12

P (inches)

0.5

0.6

0.9

1.5

2.2

2.6

2.7

2.3

1.6

1.2

0.7

	Average Monthly Water Temperatures
$t$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$	$10$	$11$	$12$
$D (^{\circ} F)$	$70$	$73$	$75$	$78$	$82$	$85$	$86$	$85$	$84$	$81$	$75$	$73$

Average Monthly Water Temperatures

t

1

2

3

4

5

6

7

8

9

10

11

12

D (^{\circ} F)

70

73

75

78

82

85

86

85

84

81

75

73

Consider the function that models the average monthly rainfall for a town in Hawaii.

y = 1.3 sin \frac{π}{6} (x - 9) + 1.6

Here,

x

represents the number of the month such that January

= 1

, February

= 2,

and so on. Rewrite the function so that it involves a cosine function instead of a sine function.

We are given a sine function that models how the average monthly rainfall changes in a town, where x represents the number of the month. y=1.3sin π6(x-9)+1.6 We are asked to rewrite this function such that it involves a cosine instead of the sine. To do so, let's first examine how to rewrite a parent sine function y=sin x as a cosine function. We will plot and compare the graphs of y=sin x and y=cos x.

This graph illustrates that we need to translate the graph of y=cos x by π2 units to the right in order to obtain the graph of y=sin x. sin x = cos ( x-π/2) Now, let's consider the general form of a transformed sine function. y= a sin ( bx- h)+ k When changing sine to cosine in a function, the values of a, b, and k will stay the same. In other words, the amplitude, period, and midline will not change. However, the value of h should be changed to represent the horizontal shift by π2 units to the right. First, let's rewrite the function to identify the current value of h.

Comparing this function to the general equation of a transformed sine function, we can see that h= 3π2. Now, to change sine to cosine, we need to increase the value of h by π2. y=1.3 sin(π/6x-3π/2) + 1.6 [0.1cm] ⇕ [0.1cm] y=1.3 cos (π/6x-3π/2- π/2) +1.6 We can subtract the terms inside the parentheses to simplify the function. y=1.3cos (π/6x-4π/2) +1.6 [0.1cm] ⇕ [0.1cm] y=1.3cos (π/6x-2π) +1.6 Finally, note that since the period of a cosine function is 2π, translating a cosine function horizontally by 2π will result in the same graph. Therefore, the obtained function is equivalent to the following cosine function. y=1.3cosπ/6x +1.6

Depending on the time of day, the depth of the water in bodies of water can vary. The following equation models the water depth

d

in the Bay of Stanley measured in feet.

d = 27 - 18 cos \frac{π}{6} t

Here,

t

is the number of hours from midnight.

What is the graph of this function?

What is the water depth of the Bay of Stanley

4

hours after midnight?

We are given a function that models the water depth d in the Bay of Stanley in feet depending on the numbers of hours from midnight t. d=27-18cos π/6t To graph this function, we will start by comparing it with the general form of a transformed cosine function. General Form y= acos b(x- h)+ k [0.1cm] Given Function d= - 18 cos π/6(t- 0)+ 27 As we can see, a= - 18, b= π6, h= 0, and k= 27. Now, we can use the formulas for the x -intercepts, maximum value, and minimum value of a cosine function of the form y= a cos bx.

	Formula
x-intercepts	(1/42π/b,0), (3/42π/b,0)
Minimum	(0, a), (2π/b, a )
Maximum	(1/2*2π/b, - a)

Our function is in the form y= a cos b(x- h)+ k. The values of h and k represent the vertical and horizontal translations, respectively. Here, it is a translation 27 units up and no translation right or left as h= 0. Therefore, when finding the key points, we will also add 27 to their y -coordinates.

Formula	Substitute	Simplify
(1/42π/b+ h,0+ k), (3/42π/b+ h,0+ k)	(1/42π/π6+ 0,0+ 27), (3/42π/π6+ 0,0+ 27)	(3,27),(9,27)
(0+ h, a+ k), (2π/b+ h, a+ k )	(0+ 0, - 18+ 27), (2π/π6+ 0, - 18+ 27 )	(0,9),(12,9)
(1/2*2π/b+ h,- a+ k)	(1/2*2π/π6+ 0, - ( - 18)+ 27)	(6,45)

Now we know five points that the given function passes through. Let's plot and connect them with a smooth curve.

Since t represents the number of hours since midnight, we will extend the pattern along the x-axis to obtain the graph for all 24 hours in a day.

The obtained graph matches option D.

We are asked to find the water depth of the Bay of Stanley 4 hours after midnight. To do so, let's substitute 4 for t into the given function and solve it for d.

We can conclude that 4 hours after midnight the water will be 9 feet deep.

Mark and his friend Heichi came to a rock climbing center. At one point, Mark was standing $120$ feet from the base of a $90 -$ foot wall, from which Heichi was rappelling down.

Write an equation that models the distance

d

in feet Heichi is from the top of the ceiling as a function of the angle of elevation

θ .

What is the graph of the function from Part A considering the context of the situation?

Graphs of different tangent functions in the first quadrant

Use a graphing calculator to determine the angle of elevation in radians when Heichi has rappelled one-third of the way down the climbing wall. Round the answer to two decimal places.

We are told that Mark was standing 120 feet away from the base of a 90-foot climbing wall while Heichi was rappelling down. Let d represent the distance Heichi is from the top of the ceiling.

To write an equation that models the distance d as a function of the angle of elevation θ, let's take a closer look at the right triangle that Mark, Heichi, and the floor form.

In this triangle, the leg adjacent to ∠ θ is 120 feet long, while the leg opposite ∠ θ is 90-d feet long. To find the relationship between these sides, we can use the tangent ratio. tanθ=opp/adj ⇒ tanθ = 90-d/120 Let's solve the obtained equation for d.

This way we have found an equation that models the distance d in feet Heichi is from the ceiling when rappelling down the climbing wall.

To graph the distance equation that we found in Part A, let's first take a closer look at the equation. We can notice that this tangent function can be graphed by translating the function y =- 120 tan θ by 90 units up. d= 90 - 120tanθ Therefore, let's first graph y=- 120 tan θ and then translate it vertically. When drawing one period of a function of the form y= a tan bθ, we can use the following characteristics.

	Formula
x-intercept	(0,0)
Asymptotes	θ = - π/2\| b\|, θ = π/2\| b\|
Halfway Points	(- π/4 b,- a), (π/4 b, a)

Let's now identify the values of a and b for our function by comparing it with the general form of a tangent function. y&= a tan bθ d&= - 120tan 1 θ As we can see, a= - 120 and b= 1. Now, we can substitute these values into the formulas and determine the characteristics of our function.

Formula	Substitute	Evaluate
Asymptotes θ = - π/2\| b\|, θ = π/2\| b\|	θ = - π/2\| 1\|, θ = π/2\| 1\|	θ = - π/2, θ = π/2
Halfway Points (- π/4 b,- a), (π/4 b, a)	(- π/4( 1),- ( - 120)), (π/4( 1), - 120)	(- π/4,120), (π/4,- 120)

Let's now plot the found points and asymptotes and draw the graph of the function.

Also, note that θ represents an acute angle in a triangle. This means that it can only have values between 0^(∘) and 90^(∘), which corresponds to Quadrant I and IV. Therefore, only the part of the graph in those two quadrants makes sense in the context of the situation.

Lastly, we need to translate the graph of y=- 120 tan θ up by 90 units.

Finally, since d represents distance and cannot be negative, we will only leave the part of the graph in Quadrant I where d is non-negative. We can also express the angle measure in degrees. To do so, recall that π2 = 90^(∘).

This graph matches option D.

We are asked to find the angle of elevation θ when Heichi has rappelled one third of the way by using a graphing calculator. We know that the wall is 90 feet high. Let's multiply this distance by 13! 1/3* 90 feet=30feet Therefore, Heichi has rappelled down 30 feet and he still has 90-30=60 feet to go.

First, we will resize the window by pushing WINDOW and changing the settings. We will use similar dimensions as when graphing the function without a calculator. For that, note that π2 ≈ 1.57. We also need to change the y-scale to 20 so that the ticks do not overlap.

We want to find the angle θ for which the distance from the top is 30 feet. This means finding the point of intersection of the the graphs of y=30 and the distance equation found in Part A. y&=30 y&=90 -120tan θ We will now graph both functions. To draw a graph on a calculator, we first press the Y= button and type in the functions. Then we push GRAPH to draw them. Note that here the argument is x, not θ.

To find the point of intersection of the functions, push 2nd and CALC and choose the fifth option, intersect. Choose the first and second curve and pick a best guess for the point of intersection.

The point of intersection is approximately x=0.46, which means that the the angle when Heichi has rappelled one third of the way is 0.46 radians or 26^(∘).

Applying Trigonometric Functions to Real-World Situations

Catch-Up and Review

Phenomena Modeled by Trigonometric Functions

How to Model Average Monthly Water Temperature?

Using an Electrocardiogram to Calculate a Pulse Rate

Hint

Solution

Periodic Change in the Distance of a Sound Beam

Answer

Hint

Solution

Extra

Sinusoids

Graphing an Earthquake Wave

Hint

Solution

Cosine Model of a Sound Wave

Hint

Solution

Period

Amplitude

How Do Trigonometric Functions Model Chest Compressions?

Hint

Solution

Finding a Curve of Best Fit Using a Calculator

Sinusoidal Regression

Using Sinusoidal Regression to Predict Precipitation

Answer

Hint

Solution

Modeling Periodic Phenomena Using Trigonometric Functions

Answer

Hint

Solution

Scatter Plot

Model

Graph of the Function

Period

Applying Trigonometric Functions to Real-World Situations

Recommended exercises

	12 Theory slides
	8 Exercises - Grade E - A
	Each lesson is meant to take 1-2 classroom sessions