SOLUTION: the sun rises at 09:17 on dec. 21 and at 04:35 on June 22> there is no daylight savings time, the time the sun rises on any other date can be predicted from a sinusoidal grpah with

the sun rises at 09:17 on dec. 21 and at 04:35 on June 22> there is no daylight
savings time, the time the sun rises on any other date can be predicted from a
sinusoidal graph with period of 365 days.

Write a sinusoidal equation that relates the time the sun rises to the day of
the year

==================================================================

This is obviously referring to some location on earth nearer the North pole
than the continental US.  Does that perhaps happen in northern Alaska?  

Anyway,

Let S(t) represent the time of dawn on day number t.

To find the number of day of the year that June 22 is, we add up the numbers
of days of the months up through May

Jan has 31
Feb has 28
Mar has 31
Apr has 30
May has 31
----------
       151

and then add another 22 to 151 and get that June 22 is the 173rd of the 
year.

Since December has 31 days, December 21 is 10 days before the end of the
year, so we subtract 10 from 365 and get that December 21 is the 355th 
day of the year. 

4:35 AM is 4 + 35/60 or 55/12 hours. 

So when t = 173, S(t) = 55/12 

9:17 AM is 9 + 17/60 or 557/60 hours.

So when t = 355, S(t) = 557/60

Use the equation 

S(t) = A sin(Bt + C) + D

The maximum value of the sine function is 1 which occurs when the angle
is p/2, so when t = 355, Bt + C = p/2

B·(355) + C = p/2
355B + C = p/2

The minimum value of the sine function is -1 which occurs when the angle
is 3p/2, so when t = 173, Bt + C = 3p/2

B·(173) + C = p/2
173B + C = p/2

So we solve the system of equations
(Best to use the MATRIX operations on a graphical calculator) 

355B + C =  p/2
173B + C = 3p/2

and get

B = -.0172514981
C = 7.698628151

It is convenient to use a graphical caluculator and store
those two values respectively as B and C.

so    

Now when t = 355, S(t) = 557/60

So we plug that into

S(t) = A sin(Bt + C) + D

and get

S(355) = 557/60 = A sin(355B + C) + D

Also when t = 173, S(t) = 55/12

So we plug that into

S(t) = 55/12 = A sin(173B + C) + D

Now we have this system of two equations in two unknowns:

A sin(355B + C) + D = 557/60 
A sin(173B + C) + D = 55/12

Using the matrix utility of a graphing calculator with
the stored values of B and C, and we get

A = 2.35, D = 6.9333333333

So the equation is

S(t) = A sin(Bt + C) + D

with A = 2.35, B = -.0172614981, C = 7.698628151, D = 6.933333333

Edwin
AnlytcPhil@aol.com