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### Course: Algebra (all content) > Unit 14

Lesson 15: Solving sinusoidal models# Trig word problem: solving for temperature

Sal solves a word problem about the annual change in temperature by solving a sinusoidal equation. Created by Sal Khan.

## Want to join the conversation?

- At0:00, Sal says"In the last video,

we were able to model ……". Anyone can teach me How to find that video?(65 votes) - I was doing the exercise after this video and I need some help:

Antonio's toy boat is bobbing in the water under a dock. The vertical distance**H**(in**cm**) between the dock and the top of the boat's mast**t**seconds after its first peak is modeled by the following function. Here,**t**is entered in radians.

***H(t)* = 5cos(2π/3 * t ) - 35.5****How long does it take the toy boat to bob down from its peak to a height of −35 cm?****Round your final answer to the nearest tenth of a second.*

Here's the answer to solve the equation:`H(t) = 5cos(2π/3 * t) - 35.5`

has a period of`2π/2π/3 = 3 seconds`

.

We want to find the first solution to the equation`H(t) = -35`

within the period`0 < t < 3`

.`5cos(2π/3 * t) - 35.5 = H(t)`

5cos(2π/3 * t) - 35.5 = -35

5cos(2π/3 * t) = 0.5

cos(2π/3 * t) = 0.5/5

Rounding the several decimal places (in order to be precise), we get *cos^-1(0.5/5)* ≈ 1.4706`2π/3 * t = 1.4706+2πn`

t ≈ 0.7+3n

Now, let's use the identity and**cos(θ) = cos(2π−θ)* to find the second expression for all possible values of *t**.`2π/3 * t = (2π − 1.4706) + 2πn`

t = 3/2π * (2π - 1.4706 + 2πn)

t ≈ 2.7+3n

Therefore, it takes about 0.7 seconds for the boat to bob down to -35cm.

*********************************************************************************************************

My questions are:

1. Where did the**n**come from in the**2π*?**?

2. Why is *2π* added in after getting *1.4706

I hope my question get responded as soon as possible. I'm really struggling to understand this.

Thanks.(16 votes)- The n is a variable without an answer in this question. 2pi is the same as 2*180 or 360 because pi=180 when using degrees and radians. The sin, cos or tan of any angle will be the same when you add 360 degrees or 2pi. So they added the n so that you know that it could be 1.4706 or 1.4706 times 2pi or 1.4706 times 2*2pi, etc. All of those values will be the same. (It would also be the same if you added 360 or 2*360 or 3*360, etc)

Does that help?(17 votes)

- So when he does the cos^-1(-0.2) he's getting 1.77~. Whenever I do it though, I keep getting 101.~. I've checked it on 3 different calculators and it all says the same thing. How is he getting that?(8 votes)
- The formula is specific to radian measure, so that's how Sal is doing his calculations. 1.77 radians is the same as 101 degrees.(14 votes)

- Towards the end of the video, I input the same equation you display on screen, but i keep getting 2586.339714 as my answer. I looked below and saw someone in a similar situation, so I tried inputting 2 * pi instead. I ended up with the same answer. This also happens in the practice problems that I try and I'm not sure what I'm doing wrong. The steps I took are the same as in the video since I was following along.(5 votes)
- I was having the same issue! Instead of doing Ans*365/2pi, you have to do Ans*365/(2pi). for some reason the calculators I've used do 4.511*365/2 then multiplies it by pi which gets you 2586 days rather than 262. I'm not sure how Sal's calculator does it correctly but that's the only way I've found to make it work. 7 years late but hopefully this can help somebody else!(9 votes)

- This video says the last video.

Why can't I see where the last video is?(7 votes)- I know you posted this one month ago, but this link is for anyone who also wants part one of this video

https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:trig/x2ec2f6f830c9fb89:sinusoidal-models/v/modeling-temperature-fluxtuations(6 votes)

- You start the video by saying "In the last video we..." It would be helpful if there were a link to that last video. I couldn't find it.(7 votes)
- The video before this one is called "Trig word problem: modeling annual temperature". Here is the link: https://www.khanacademy.org/math/trigonometry/trig-function-graphs/constructing-sinusoids/v/modeling-temperature-fluxtuations(4 votes)

- In some of the Trig lessons I see sin(θ)=sin(π−θ) and in others I see sin(θ)=sin(-π−θ). How do I know which property to use?(4 votes)
- Just use whatever property need for situation.

A couple of important identities that you may not have learned yet are:

-sin(theta)=sin(-theta)

-tan(theta) = tan(-theta)

cos(theta) = cos(-theta)

The functions tan and sin are considered odd because -f(x) = f(-x). And Cos is even since f(x)=f(-x). So in this case sin(pi-theta)= - sin(theta-pi) since sine is odd. So base on this we can tell the sine function shifted pi to the right as each value needs have pi added to it retain the same value as sin theta. After this graph flip across the x-axis due to minus sign in front sin(theta-pi). So you can visually see that sin theta = sin(pi-theta).

Now on to the second identity all you just doing shifting the function pi to the left, and since sin has period 2pi what this means is shifting function by the same amount.

So you will need to use the identities depending on the context as there is no difference between them.(5 votes)

- I didn't understand the last step where Sal said that the value of 2pi/365*d=4.511.

Can someone please explain?(4 votes)- (1) We are trying to find where cos(2pi/365*d) = -0.2, but there are two values of (2pi/365*d) in the year that satisfy this.

(2) The two values of x that satisfy cos(x) = -0.2 are x = 1.772 and x = 4.511.

(3) By substituting 2pi/365*d for x, we know that either 2pi/365*d=1.772 or 2pi/365*d=4.511.

(4) From Sal's unit circle work, we know we want the one in quadrant III.

(5) 4.511 falls in quadrant III, so the equation we want is 2pi/365*d=4.511.

Hopefully this helps.(6 votes)

- I am have some trouble with these practice problems... the video is nothing like the practice problems after it. At8:00there is usually at 2pi * n at the end with the practice problems but not this video. so like t = 365/2pi * (0.9273 + 2pi* n) Anyone now how to help me solve this?

At2:30, how does Sal calculate the fraction 1.5/1.7 in his head so fast? I need to develop number sense like that!(5 votes)- It's actually 1.5/7.5, however both fractions can be "simplified" by multiplying the numerator and denominator by 10. Now you would be left with 15/75, and 15 can go into 75 five times, so 1/5. Of course, 1.5 also goes into 7.5 five times, but the decimals often confuse people!

The same logic could be applied to 1.5/1.7. Multiply by 10 to get 15/17. Same number but would be slightly easier to work with on a calculator. Nobody would be able to calculate this number in their head that quickly (except for maybe a rough approximation).(2 votes)

- I'm still a little perplexed as to how to know which trigonometry identity to use for each problem. Someone down below also was wondering the same question, but theirs has not been answered yet. So, I was curious if anyone has an answer to this? I understand that you can draw the unit circle to pinpoint the sine or cosine, but the main issue comes that there are simply many identities for sine and cosine, itself, so how do you distinguish between them based on the problem? This was also a similar issue that I encountered in the last practice of solving sinusoidal equations. Also, about the last practice of solving sinusoidal equations, how were you supposed to figure out the interval the answer was supposed to be in between?(4 votes)

## Video transcript

Voiceover:In the last video, we were able to model the average high temperature in Santiago, Chile as a function of days as we go through the year, where the days were the days after January 7th, so this right over here. Day zero is actually January 7th. But we weren't done. They want us to figure out how many days after January 7th is the first spring day, when the temperature reaches 20 degrees celsius And I told you, be careful. Pay attention to this whole notion of the first spring day and the reason is because there's actually two days where the temperature reaches 20 degrees celsius, So let's say this is 20 degrees celsius right over here, notice you have this day right over here and then you have this
day right over there and which one is in the spring or the first spring day? Well, if this is - we're in summer right over here We're in the southern hemisphere, so our summer is going to be when it is the winter in the northern hemisphere. Summer, what season comes after summer? Well, this is going to be the fall, this is going to be the winter, Now this is going to be the spring. Of course, you go back to summer. So, we want this value, not this value. This one will be the day in fall when the average high temperature is 20 degrees celsius, This is the first day of spring where the average high temperature is 20 degrees celsius. The first spring day. I guess it's not necessarily
the first spring day, it's the first spring day where the temperature
reaches 20 degrees celsius. So, this is happening in the spring, so this is the value we want. So, let's just think about that. As we try to manipulate this a little bit. So, we want to get to 20 degrees celsius, so we could write 20 is equal to 7.5 times cosine of 2 pi over 365 times the days, plus 21.5 Now, we could subtract
21.5 from both sides and we get -1.5 is equal to- and I'll just copy and paste all of this- is going to be equal to that. So, copy and paste. It's going to be equal
to that right over there. Now I could divide both sides by 7.5. Notice, I'm trying to solve for cosine, and eventually solve for D, but we're going to take a pause, once we have this in terms of cosine, we have to be careful here. So we're going to divide both sides by 7.5 And we're going to get, let's see... Actually, I don't even need
a calculator for this.. 1.5 divided by 7 - this is one fifth. 5 times 15, or 5 times 1.5 is 7.5 so this is -1/5 or I could write it as -0.2 is equal to cosine of 2 pi over 365
times days after January 7th. Now, this is where we
have to be very careful instead of just blindly applying the inverse cosine function, we have to make sure which angle we are actually getting. That we're getting the
right angle over here. 'Cause remember, we want the argument to the cosine, that doesn't give us this point, that gives us this point right here. Or that corresponds to this
point right over there. So let's draw a unit circle, just to make sure we know what's going on. And I actually do this all the time, especially if I'm trying to apply the inverse trigonometric functions in a kind of applied context, where I just can't blindly plug it in to my calculator. So, let me draw a unit
circle right over here. X-axis, y-axis... Circle of radius 1, centered at the origin You get the picture we've done this many times before. And January 7th, that corresponds to this point right over here. That's that point right over here, that we are in the summer. Then, as the days go by, our argument to the cosine increases, the angle increases, and this right over here will be the fall. So, this point right over here, so we're at the fall right over here. And then we move on to the winter right over there. And then finally, we go to the spring. This is the spring right over there. And we want the angle that gets us -0.2 So, this is -1. -0.2 is a fifth of the way, so it's -0.2 and notice, there's two angles that get us there. There's this angle, there's that angle right over there. And then there's also, let me draw a little dotted line here, there's that angle. But then there's also this angle. which is going even further. Or another way you could think about it, you could go backwards
to get to that angle and then if you wanted
to go all the way around, to the next spring, you could add 2 pi to it. So which one do we want? Of course, we want the one in the spring. But if we just blindly
apply the inverse cosine, of -0.2, that's going to give us this one. And we can verify that. So, let's see... Inverse cosine of -0.2 is 1.77. Remember, this is zero, this is pi right over, so this is 3.14159, and this is 1.77, so notice it's a little
bit more than half of pi which is exactly what it gave- it gave this angle right over here. So this is approximately 1.77 radiants is this angle, is that
angle right over there. But we don't want that one, we want this one, so how do we figure that out? Well, we could view it as, we could go all the way around to 2 pi, and then subtract 1.77, so we could say 2 pi minus 1.77 roughly to get this angle, so let's do that. Let's take 2 times pi, and then subtract this number and I'm going to do second answer, so answer is just the
previous answer that I got so that way I have good precision here. So I get 4.511, and we can kind of make
sure that is the right thing because that's going to be between pi and 2 pi. 2 pi is 2 times 3.14159,
so it's going to be 6.28 something. This is 3.14159, this is the right angle. Now, we're not done yet. That's just the angle, that's the argument that we need to give in here to get to that point, but what's the days going to be. Well 2 pi over 365 days is going to be equal to this number 4.511, so let me write that down. So, this right over here is going to be approximately equal to 4.511. So, we scroll down a little bit We can say 2 pi over 365 times the days after January 7th is approximately going
to be equal to 4.511 to solve for days, we can just multiply both sides by the reciprocal of the coefficient, so we're going to multiply both sides by 365 over 2 pi. So that's going to cancel, and now we can use our
calculator for this. So, we've got much better precision there. So let's take our previous answer, times 365, we deserve a drumroll here... Divided by 2 pi, and we get if we round to the nearest day 262 days after January 7th. So 262 days. and we are done.