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## Algebra 2

### Course: Algebra 2 > Unit 11

Lesson 9: Sinusoidal models- Interpreting trigonometric graphs in context
- Interpreting trigonometric graphs in context
- Trig word problem: modeling daily temperature
- Trig word problem: modeling annual temperature
- Modeling with sinusoidal functions
- Trig word problem: length of day (phase shift)
- Modeling with sinusoidal functions: phase shift
- Trigonometry: FAQ

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# Trig word problem: length of day (phase shift)

CCSS.Math: , , ,

Sal solves a word problem about the annual change in the length of day by modeling it with a sinusoidal function that has a phase shift. Created by Sal Khan.

## Want to join the conversation?

- why is it (t-172) rather than ( t+172)?(11 votes)
- You want the function to be maximized when you plug in 172. In the first case, cos(0) = 1 and the function is maximized. In the second case, you are taking the cosine of 2pi*344/365, which is a little smaller.(11 votes)

- L(55)=−51.875cos(2pi/365(55+11))+727.725

each time i do this i get 675.86 as the answer and not 705.88 which is the answer in the exercise. Any advice on what i am doing wrong? The above formulae is the correct one from the hints.(4 votes)- Everyone's made this mistake including me:
*make sure you're using radian mode on your calculator when it's measuring in radians*. This should be in red.(25 votes)

- So I've done a lot of practice with phase shift but I still have trouble determining the parameters of the equation asin(bx+c)+d when there is a word problem given or when they ask you to graph a trig equation with a phase shift in it. Can someone please help me understand this? :I(13 votes)
- Good question, I have struggled with this concept as well. I will first explain how sketch equation cosine function and then how to sketch involving sin and tan briefly.

Question: How to sketch Cos(2x-pi/3), why is the phase shift not pi/3.

There two transformations going on, the horizontal stretch and the phase shift.

To stretch a function horizontally by factor of n the transformation is just f(x/n).

So let f(x) = cos(x)

=> f(x/(1/2)) = cos(x /(1/2) ) = cos(2x)

So the horizontal stretch is by factor of 1/2.

Since the horizontal stretch is affecting the phase shift pi/3 the actual phase shift is pi/6 to the right as the horizontal sretch is 1/2.

cos(2x-pi/3) = cos(2(x-pi/6))

Let say you now want to sketch cos(-2x+pi/3). Remember that cos theta is even function. A function is even if f(-x) = f(x). Substitute u = 2x-pi/3 =>

cos(-2x+pi/3) = cos(-u) = cos(u) = cos(2x-pi/3)

Similarly you can sketch sin(2x-pi/3) the same way just beware sin is an odd function not an even function. Note a function is odd if f(-x) = -f(x).

If you wanted to sketch an equation tan theta remember that period of tan theta is pi and tan is an odd function.

Basic notes

Note when equation is in the form asin(bx+c) + d the period is just 2pi/b. The formula for the period might be easier to use than thinking of horizontal stretches.

Vertical shift is just d.

The amplitude is the distance from the midline. i.e. amplitude = |a|. The || (absolute value function) makes everything that is negative become positive. As 'a' can be negative we take the absolute value to make sure it is positive. e.g. |-6| = 6, |-pi| = pi, |-2.78| = 2.78. Note that anything that is positive remains positive i.e. |4| = 4, |6| = 6, |pi| = pi. The reason for the absolute value is we are interested in the distance from the midline which is non-negative.

Note it might to help to learn about curve sketching for more detail on things discussed.(3 votes)

- Hey guys i wanna ask a question. In one of the question in "Modeling with sinusoidal functions: phase shift", i found a weird solution (weird because i don't understand).

###the question + solution###

A weight is attached to the end of a spring. Its height after t seconds is given by the equation: -2 sin[(2phi/7) * (t+1)] + 5.

When does the weight first reach its maximum height? Give an exact answer.

The answer says: (2phi/7) * (t+1) = (-phi/2) + (2phi*n)

###end of the question + solution###

can someone explain to me, what does that 2phi*n doing right over there? Thanks(7 votes)- When they ask for a solution, they want a closed phrase describing all possible solutions: that would make any solution theta, theta + 2π * n, (the solvent angle plus any number of full rotations: remember, adding 360 degrees or 2π to an angle doesn't make a new angle. It's just another expression for the reference angle.)(7 votes)

- At2:20, it seems like Sal just decides to use cosine instead of sin, just because it is easier. Is it optional whether to use sine or cosine, or is there a good reason that he used cosine?(7 votes)
- Unless you want a much harder equation and potentially wrong answer, go ahead and use the more convenient function.(5 votes)

- How is the sign of the amplitude determined? My understanding was that a negative dip after the origin would mean a negative amplitude, but that doesn't seem to be the case going by the practice sets.(5 votes)
- Amplitude is simply the distance away from the center line that the tips of the peaks/troughs of the graph are. Because it's a distance, it can't be negative at all. Whether you're looking at a part of the graph where it rises away from the center line or falls below it, the amplitude will be the same.(6 votes)

- could we use a trigonometric function to create a function of the height of a basketball released above the ground as a function of time? while considering that each time it bounces the height get 20% smaller?(5 votes)
- f(t) = (cos(2pi*t)/2+0.5)*0.8^t

I believe this would work.

(cos(2pi*t)/2+0.5) varies between 1 and 0 and 0.8^t decreases by 20%.(5 votes)

- I do not understand how to know if the amplitude is negative or positive? Can someone help me?(2 votes)
- Whether it's a cos or sin function, the first extremum should be positive: if it's not, the function is negated (i.e. -cos or -sin). Don't mistake this for acos or asin.(5 votes)

- Hi, can someone tell me what I am doing wrong? In the practice questions, I get the equations right, but when asked for a specific value at a specific time, my calculator gives me the wrong answer.

The equation could be this: 8.5 cos(2pi/3(t-1.3))-35.5, I plug in the values for when t=2.5, and get -27.0.. but the answer said -42.8(it's correct, I checked with an online graphing calculator).(3 votes)- The questions specify which angle measurement to use: degrees or radians. If you use the wrong one, your answer will be completely off even with the right equation.

Use the right mode on your calculator and you should be fine :-)(6 votes)

- Guys I don’t understand what sal said,I don’t know how to graph and write equations for sinusoidal functions, is different from what we learn before, can anybody help me(2 votes)
- You are modeling behavior using the form "a sin
*or*cos(b(*x_+c)+d." When modeling a real life example of periodic behavior, the greatest challenge is not in finding the amplitude or the period but in determining the type of function and the correct phase shift.*th day of the year or t days after the beginning of the year. Since the earliest value specifically defined with regard to our origin, 0, is the maximum point (longest day of the year) this is a cos function.2:07First, you determine what sort of trig function most easily models this behavior. Remember that time is always the x axis when modeling periodic behavior. Our equation needs to model the function verbally stated: _t

Its horizontal shift is c=-172; remember that shift to the right is negative according to the form since c tells us how much we are going to shift left.

Its vertical shift is the average of the maximum and minimum values. (Or simply the y-coordinate of a point intersecting the mid-line if it is given but in this case it is not).6:22Sal first generates a function*L_ (_u*) to describe the behavior without having to worry about horizontal shift. When dealing with time, it may be simpler to do this and simply rectify your result according to the function actually required after determining everything else (amplitude, period, and mid-line). The advantages are it is far easier to see which function (cos or sin) is more convenient (2:07), and it it might be a better way of determining the c or horizontal shift as well.(7 votes)

## Video transcript

Sal:The longest day of the year in Juneau, Alaska is June 21st.
It's 1,096.5 minutes long. Half a year later when
the days are at their shortest the days are
about 382.5 minutes long. If it's not a leap year
the year is 365 days long and June 21st is the
172nd day of the year. Write a trigonometric function that models the length L of the T-th day of the year. It's going to be L as a function of T, assuming it's not a leap year. I encourage you to pause
this video and try to do this on your own before
I try to work through it. Let me give a go at it. Instead of first starting
at L of T I'm going to actually start with
L as a function of U, where U is another variable
I'll use as an intermediary variable that will help
set it up in a simpler way. Where U is days after June 21st. Let's just think about this a little bit. June 21st. If we're thinking about in
terms of U, U is going to be equal to zero because it's
zero days after June 21st. But if we're thinking
about it in terms of T, June 21st is the 172nd day of the year. 172. What's the relationship between U and T? It's shifted by 172 days. U is going to be equal to T minus 172. Notice when T is 172, U is equal to zero. Let's figure out L of
U first and then later we can just substitute U with T minus 172. First of all what's happening
when U is equal to zero? Let me write all of this down. What is happening when U is equal to zero. U equals zero is June 21st
and that's the maximum point. What trig function hits its maximum point when the input into the
trig function is zero? Well sine of zero is zero
while cosine of zero is one. Cosine hits its maximum point. It seems a little bit easier
to model this with a cosine. It's going to be some amplitude times cosine of, let's say, I'll write some coefficient
C right over here. Actually just let me use a
B since I already used an A. Some coefficient over
here times our U plus some constant that will shift the
entire function up or down. This is the form that our
function of U is going to take. Now we just have to figure out
what each of these parts are. First let's think about what the amplitude and what the mid line is going to be. The mid line is essentially how much we're shifting the function up. Let's get our calculator out. The mid line is going to be half way between these two numbers. We could say 1,096.5 plus 382.5 divided by two gets us to 739.5. That's what C is equal
to. C is equal to 739.5. Now the amplitude is how much
do we vary from that mid line. We could take 1,096 minus this or we could take this minus 382.5. Let's do that. Let's take 1,096.5 minus what we just got, 739.5 and we get 357. This is how
much we vary from that mid line. A is equal to 357. This right over here is equal to 357. What's B going to be equal to? For that I always think,
"What's the behavior of the "function, what's the period
of the function going to be?" Let me make a little table here. Let's put some different inputs from U. When U is zero we're zero
days after June 21st, we're at our maximum point. And we already said that what
we want the cosine function to evaluate to at that
point is essentially we want it to evaluate as 357 times
cosine of zero plus 739.5. Now what's a full period?
A full period is a year. At a year we get to the
same point in the year, which is I guess a little
bit of common sense. You go all the way to 365, when U is 365 we should
have completed a period. We should be back to that maximum point. This should essentially be
357 times cosine of two pi. If we were just thinking in terms of a traditional trig function, if
we just had a theta in here, you complete a period every two pi. This should be equivalent to
what I'm writing out over here. Plus 739.5. One way to think about it is B times 365 should be equal to two pi. Notice this is going to be B times 365, so let's write that down. B times 365. That's the input into the cosine function needs to be equal to two pi. Or B is equal to two pi over 365. B is equal to two pi over 365. We are almost done. We
figured out what A, B and C are and now we just have to
substitute U with T minus 172 to get our function of T. Let's just do that. We get -- we deserve a little
bit of a drum roll now. L of T is equal to A, which is 357 times cosine
of B two pi over 365. Two pi over 365 times not U, but now we're going
to write it in terms of T. We want to think about day of the year, not days after June 21st. So times T minus 172 and then
finally plus our mid line. Plus 739.5. We are done. It seems like a very
complicated expression but if you just break it
down and think about it, make the point that we're
talking about the extreme point, either the minimum or the maximum. Make that when the input into
our function is zero or two pi. Zero is actually the easiest one. Then later you can worry about the shift. Hopefully you found that helpful.