# 7.6 Modeling with trigonometric equations  (Page 2/14)

 Page 2 / 14

## Finding the amplitude and period of a function

Find the amplitude and period of the following functions and graph one cycle.

1. $y=2\text{\hspace{0.17em}}\mathrm{sin}\left(\frac{1}{4}x\right)$
2. $y=-3\text{\hspace{0.17em}}\mathrm{sin}\left(2x+\frac{\pi }{2}\right)$
3. $y=\mathrm{cos}\text{\hspace{0.17em}}x+3$

We will solve these problems according to the models.

1. involves sine, so we use the form
$y=A\text{\hspace{0.17em}}\mathrm{sin}\left(Bt+C\right)+D$

We know that is the amplitude, so the amplitude is 2. Period is so the period is

See the graph in [link] .

2. involves sine, so we use the form
$y=A\text{\hspace{0.17em}}\mathrm{sin}\left(Bt-C\right)+D$

Amplitude is so the amplitude is Since is negative, the graph is reflected over the x -axis. Period is so the period is

$\frac{2\pi }{B}=\frac{2\pi }{2}=\pi$

The graph is shifted to the left by units. See [link] .

3. involves cosine, so we use the form
$y=A\text{\hspace{0.17em}}\mathrm{cos}\left(Bt±C\right)+D$

Amplitude is so the amplitude is 1. The period is See [link] . This is the standard cosine function shifted up three units.

What are the amplitude and period of the function

The amplitude is and the period is

## Finding equations and graphing sinusoidal functions

One method of graphing sinusoidal functions is to find five key points. These points will correspond to intervals of equal length representing of the period. The key points will indicate the location of maximum and minimum values. If there is no vertical shift, they will also indicate x -intercepts. For example, suppose we want to graph the function We know that the period is $\text{\hspace{0.17em}}2\pi ,$ so we find the interval between key points as follows.

$\frac{2\pi }{4}=\frac{\pi }{2}$

Starting with we calculate the first y- value, add the length of the interval to 0, and calculate the second y -value. We then add repeatedly until the five key points are determined. The last value should equal the first value, as the calculations cover one full period. Making a table similar to [link] , we can see these key points clearly on the graph shown in [link] .

 $\theta$ $0$ $\frac{\pi }{2}$ $\pi$ $\frac{3\pi }{2}$ $2\pi$ $y=\mathrm{cos}\text{\hspace{0.17em}}\theta$ $1$ $0$ $-1$ $0$ $1$

## Graphing sinusoidal functions using key points

Graph the function using amplitude, period, and key points.

The amplitude is The period is (Recall that we sometimes refer to as One cycle of the graph can be drawn over the interval To find the key points, we divide the period by 4. Make a table similar to [link] , starting with and then adding successively to and calculate See the graph in [link] .

 $x$ $0$ $\frac{1}{2}$ $1$ $\frac{3}{2}$ $2$ $y=-4\text{\hspace{0.17em}}\mathrm{cos}\left(\pi x\right)$ $-4$ $0$ $4$ $0$ $-4$

Graph the function using the amplitude, period, and five key points.

x $3\mathrm{sin}\left(3x\right)$
0 0
$\frac{\pi }{6}$ 3
$\frac{\pi }{3}$ 0
$\frac{\pi }{2}$ $-3$
$\frac{2\pi }{3}$ 0

## Modeling periodic behavior

We will now apply these ideas to problems involving periodic behavior.

## Modeling an equation and sketching a sinusoidal graph to fit criteria

The average monthly temperatures for a small town in Oregon are given in [link] . Find a sinusoidal function of the form $\text{\hspace{0.17em}}y=A\text{\hspace{0.17em}}\mathrm{sin}\left(Bt-C\right)+D\text{\hspace{0.17em}}$ that fits the data (round to the nearest tenth) and sketch the graph.

Month Temperature, ${}^{\text{o}}\text{F}$
January 42.5
February 44.5
March 48.5
April 52.5
May 58
June 63
July 68.5
August 69
September 64.5
October 55.5
November 46.5
December 43.5

Recall that amplitude is found using the formula

Thus, the amplitude is

The data covers a period of 12 months, so $\text{\hspace{0.17em}}\frac{2\pi }{B}=12\text{\hspace{0.17em}}$ which gives $\text{\hspace{0.17em}}B=\frac{2\pi }{12}=\frac{\pi }{6}.$

The vertical shift is found using the following equation.

Thus, the vertical shift is

So far, we have the equation $\text{\hspace{0.17em}}y=13.3\text{\hspace{0.17em}}\mathrm{sin}\left(\frac{\pi }{6}x-C\right)+55.8.$

To find the horizontal shift, we input the $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ values for the first month and solve for $\text{\hspace{0.17em}}C.$

We have the equation $\text{\hspace{0.17em}}y=13.3\text{\hspace{0.17em}}\mathrm{sin}\left(\frac{\pi }{6}x-\frac{2\pi }{3}\right)+55.8.\text{\hspace{0.17em}}$ See the graph in [link] .

#### Questions & Answers

I've run into this: x = r*cos(angle1 + angle2) Which expands to: x = r(cos(angle1)*cos(angle2) - sin(angle1)*sin(angle2)) The r value confuses me here, because distributing it makes: (r*cos(angle2))(cos(angle1) - (r*sin(angle2))(sin(angle1)) How does this make sense? Why does the r distribute once
How can you tell what type of parent function a graph is ?
generally by how the graph looks and understanding what the base parent functions look like and perform on a graph
William
if you have a graphed line, you can have an idea by how the directions of the line turns, i.e. negative, positive, zero
William
y=x will obviously be a straight line with a zero slope
William
y=x^2 will have a parabolic line opening to positive infinity on both sides of the y axis vice versa with y=-x^2 you'll have both ends of the parabolic line pointing downward heading to negative infinity on both sides of the y axis
William
y=x will be a straight line, but it will have a slope of one. Remember, if y=1 then x=1, so for every unit you rise you move over positively one unit. To get a straight line with a slope of 0, set y=1 or any integer.
Aaron
yes, correction on my end, I meant slope of 1 instead of slope of 0
William
what is f(x)=
I don't understand
Joe
Typically a function 'f' will take 'x' as input, and produce 'y' as output. As 'f(x)=y'. According to Google, "The range of a function is the complete set of all possible resulting values of the dependent variable (y, usually), after we have substituted the domain."
Thomas
Sorry, I don't know where the "Â"s came from. They shouldn't be there. Just ignore them. :-)
Thomas
Darius
Thanks.
Thomas
Â
Thomas
It is the Â that should not be there. It doesn't seem to show if encloses in quotation marks. "Â" or 'Â' ... Â
Thomas
Now it shows, go figure?
Thomas
what is this?
i do not understand anything
unknown
lol...it gets better
Darius
I've been struggling so much through all of this. my final is in four weeks 😭
Tiffany
this book is an excellent resource! have you guys ever looked at the online tutoring? there's one that is called "That Tutor Guy" and he goes over a lot of the concepts
Darius
thank you I have heard of him. I should check him out.
Tiffany
is there any question in particular?
Joe
I have always struggled with math. I get lost really easy, if you have any advice for that, it would help tremendously.
Tiffany
Sure, are you in high school or college?
Darius
Hi, apologies for the delayed response. I'm in college.
Tiffany
how to solve polynomial using a calculator
So a horizontal compression by factor of 1/2 is the same as a horizontal stretch by a factor of 2, right?
The center is at (3,4) a focus is at (3,-1), and the lenght of the major axis is 26
The center is at (3,4) a focus is at (3,-1) and the lenght of the major axis is 26 what will be the answer?
Rima
I done know
Joe
What kind of answer is that😑?
Rima
I had just woken up when i got this message
Joe
Can you please help me. Tomorrow is the deadline of my assignment then I don't know how to solve that
Rima
i have a question.
Abdul
how do you find the real and complex roots of a polynomial?
Abdul
@abdul with delta maybe which is b(square)-4ac=result then the 1st root -b-radical delta over 2a and the 2nd root -b+radical delta over 2a. I am not sure if this was your question but check it up
Nare
This is the actual question: Find all roots(real and complex) of the polynomial f(x)=6x^3 + x^2 - 4x + 1
Abdul
@Nare please let me know if you can solve it.
Abdul
I have a question
juweeriya
hello guys I'm new here? will you happy with me
mustapha
The average annual population increase of a pack of wolves is 25.
how do you find the period of a sine graph
Period =2π if there is a coefficient (b), just divide the coefficient by 2π to get the new period
Am
if not then how would I find it from a graph
Imani
by looking at the graph, find the distance between two consecutive maximum points (the highest points of the wave). so if the top of one wave is at point A (1,2) and the next top of the wave is at point B (6,2), then the period is 5, the difference of the x-coordinates.
Am
you could also do it with two consecutive minimum points or x-intercepts
Am
I will try that thank u
Imani
Case of Equilateral Hyperbola
ok
Zander
ok
Shella
f(x)=4x+2, find f(3)
Benetta
f(3)=4(3)+2 f(3)=14
lamoussa
14
Vedant
pre calc teacher: "Plug in Plug in...smell's good" f(x)=14
Devante
8x=40
Chris
Explain why log a x is not defined for a < 0
the sum of any two linear polynomial is what
divide simplify each answer 3/2÷5/4
divide simplify each answer 25/3÷5/12
Momo
how can are find the domain and range of a relations
the range is twice of the natural number which is the domain
Morolake
A cell phone company offers two plans for minutes. Plan A: $15 per month and$2 for every 300 texts. Plan B: $25 per month and$0.50 for every 100 texts. How many texts would you need to send per month for plan B to save you money?
6000
Robert
more than 6000
Robert
For Plan A to reach $27/month to surpass Plan B's$26.50 monthly payment, you'll need 3,000 texts which will cost an additional \$10.00. So, for the amount of texts you need to send would need to range between 1-100 texts for the 100th increment, times that by 3 for the additional amount of texts...
Gilbert
...for one text payment for 300 for Plan A. So, that means Plan A; in my opinion is for people with text messaging abilities that their fingers burn the monitor for the cell phone. While Plan B would be for loners that doesn't need their fingers to due the talking; but those texts mean more then...
Gilbert