Calculating Daylight Hours Using Trigonometry

[TonyC]

The number of hours of daylight in a town on the west coast of North America can be modeled by:

h=3.75sin[2pi/365(d-79)] + 12

Where h is the number of hours of daylight in a day and d is the day of the year, with d=1 representing January 1 (assume 28 days in Feb). What is the total accumlated number of hours of daylight by a town between Mar 29 and June 29.

I worked the problem and came up with 1355.6 hours.

Am I good to go?

 

[LENIN]

You proberlly ment the formula more like this: h=3.75sin[(2pi/365)(d-79)] + 12. In the previous form it would be undefined for the 79th day and as far as I know all days have a real number of hours. As for your resoult I can't even imagine a way to get more than 15.75 hours. You probelly just made a mistake when tiping it in your calculator. I would sugest taht you do it in steps and not all at once .

 

[Fermat]

I don't know how you got your result but I wrote a program to work it out!

Taking March 29 as day 88 and June 29 as day 180, I got the following,

day 88 to day 180 inclusive: total hrs = 1369.77
day 89 to day 180 inclusive: total hrs = 1357.18

 

[Tide]

You can write a general formula by summing the sequence of sines. Use Euler's formula for the trig functions and sum it as geometric series.

 

[AKG]

The hours of sunlight from day n to day N inclusive is:

12(N - n + 1) + 3.75\sum _{d = n} ^{d = N}\sin \left (\frac{2\pi (d - 79)}{365}\right ) = 12(N - n + 1) + 3.75\sum _{d = n - 79} ^{N - 79}\sin \left (\frac{2\pi d}{365}\right )

 

[Tide]

This may help:

\sum_{n = a}^{b} \sin nx = \frac { \cos x(a-1/2) - \cos x(b+1/2)} {2 \sin x/2}

(Edited: YIKES! Sorry, akg! I went dyslexic when I typed it and interchanged the x with the a and b!)

 

[AKG]

Wait a minute, what if a = b = 1?

sin(x) = ^{[cos(x - 1/2) - cos(x + 1/2)]}/_2sin(x/2)

= ^{[cos(x)cos(1/2) + sin(x)sin(1/2) - cos(x)cos(1/2) + sin(x)sin(1/2)]}/_2sin(x/2)

= ^{sin(x)sin(1/2)}/_sin(x/2)​

sin(x/2) = sin(1/2)

That doesn't seem right. Where did you get that formula from?

 

[Fermat]

It seems right now.

I used tide's sum of sines formula in akg's expression for hours of sunlight and got the same results as my computer.