finding values of X with unit circle.

O

onehundredeightydegrees

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Hi there,

I have been given a word problem for finding the equation for a sine function of perceivable daylight for a year where i live.
This is the function and it is fine.

[MATH]f\left(x\right)=4\sin\left(\frac{2\pi}{366}\left(x-80\right)\right)+13[/MATH]
For the next question he asks us to use the function to find the month and day when there is 11 hours of daylight.
So i took the inverse of the function and plugged in 11 for x.

[MATH]f^{-1}\left(x\right)=\frac{183\sin^{-1}\left(\frac{x-13}{4}\right)}{\pi}+80[/MATH]
Which gave me 49.5 or the 49th day of the year.

But I can't for the life of me figure out how to get the second x where y= 11.
I am able to find it on desmos, but i can't figure out algebraically.

Thanks in advace.
 
Recall that the inverse sine function has a very limited domain. I would plug in 11 for f(x), and solve that equation for x.
 
There are many angles with a given sine. They are all coterminal with either the inverse sine or its supplement. So try subtracting your inverse sine from pi to get another result.

Incidentally, you are using x for two different things. It would be safer to use two different variables. For instance, [MATH]D = f(x)[/MATH], and [MATH]x = f^{-1}(D)[/MATH], where x is the number of the day of the year (I think) and D is the hours of daylight (I think). You didn't actually define your variables ...
 
Dr.Peterson said:
There are many angles with a given sine. They are all coterminal with either the inverse sine or its supplement. So try subtracting your inverse sine from pi to get another result.

Incidentally, you are using x for two different things. It would be safer to use two different variables. For instance, [MATH]D = f(x)[/MATH], and [MATH]x = f^{-1}(D)[/MATH], where x is the number of the day of the year (I think) and D is the hours of daylight (I think). You didn't actually define your variables ...
Click to expand...

Yes you are correct. My apologies for not adding the variable definitions.
Thank you so much for the response, this is perfect.
 
onehundredeightydegrees said:
[MATH]f\left(x\right)=4\sin\left(\frac{2\pi}{366}\left(x-80\right)\right)+13[/MATH]
Click to expand...
Hello 180º. Your function definition may be simplified (as you did with the inverse function).

\[f(x) = 4 \sin\left(\frac{\pi}{183}(x - 80)\right) + 13\]

finding values of x with unit circle … i can't figure out algebraically.
Click to expand...
Here's the method Romsek suggested, in post #2:

\[11 = 4 \sin\left(\frac{\pi}{183}(x - 80)\right) + 13\]

Isolating the sine term yields:

\[-\frac{1}{2} = \sin\left(\frac{\pi}{183}(x - 80)\right)\]

Now, from the unit circle, we recall the two angles between 0 and 2\(\pi\) whose sine is -1/2. The first is \(\frac{7\pi}{6}\), and the second is \(\frac{11\pi}{6}\). In other words, the sine input must be \(\frac{7\pi}{6}\) or \(\frac{11\pi}{6}\), to get f(x)=11.

\[\frac{\pi}{183}(x - 80) = \frac{7\pi}{6}\]

Solving for x, we get 293.5 (the 293rd day).

\[\frac{\pi}{183}(x - 80) = \frac{11\pi}{6}\]

Solving for x, we get 415.5 (the 415th day). That result lies beyond the first period (it's the third instance when y=11), so subtract one period.

\[415.5 - 366 = 49.5\]

There are 11 hours of daylight on the 49th and 293rd days.

PS: Were you instructed to use 366 days per year? (Using a value like 365.2425 won't change the answer; I'm just curious.)

?
 
Last edited:
Otis said:
Hello 180º. Your function definition may be simplified (as you did with the inverse function).

\[f(x) = 4 \sin\left(\frac{\pi}{183}(x - 80)\right) + 13\]


Here's the method Romsek suggested, in post #2:

\[11 = 4 \sin\left(\frac{\pi}{183}(x - 80)\right) + 13\]

Isolating the sine term yields:

\[-\frac{1}{2} = \sin\left(\frac{\pi}{183}(x - 80)\right)\]

Now, from the unit circle, we recall the two angles between 0 and 2\(\pi\) whose sine is -1/2. The first is \(\frac{7\pi}{6}\), and the second is \(\frac{11\pi}{6}\). In other words, the sine input must be \(\frac{7\pi}{6}\) or \(\frac{11\pi}{6}\), to get f(x)=11.

\[\frac{\pi}{183}(x - 80) = \frac{7\pi}{6}\]

Solving for x, we get 293.5 (the 293rd day).

\[\frac{\pi}{183}(x - 80) = \frac{11\pi}{6}\]

Solving for x, we get 415.5 (the 415th day). That result lies beyond the first period (it's the third instance when y=11), so subtract one period.

\[415.5 - 366 = 49.5\]

There are 11 hours of daylight on the 49th and 293rd days.

PS: Were you instructed to use 366 days per year? (Using a value like 365.2425 won't change the answer; I'm just curious.)

?
Click to expand...
Thanks for the thoughtful reply, this really helps clear up any remaining questions. Yeah, 2020 is a leap year so they want us to calculate 366 days. funny thing is this equation is probably far too simple and is about an hour off the daylight charts for my area. So the extra day probably does not matter. ?
 
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