Calculus
posted by Sara .
I'm having a little trouble with this problem...it would be great if you could point out where I'm going wrong.
Hours of Daylight as a Function of Latitude.
Let S(x) be the number of sunlight hours on a cloudless June 21st, as a function of latitude, x, measured in degrees.
a.) What is S(0)?
(b) Let x0 be the latitude of the Arctic Circle. In the northern hemisphere, S (x)
is given, for some constants a and b, by the formula:
S (x) = bracket (piecewise function)
a + b arcsin*(tan x/tan x0)
for 0 ≤ x < x0
24 for x0 ≤ x ≤ 90.
Find a and b so that S (x) is continuous.
(c) Calculate S (x) for Tucson, Arizona (x = 32◦ 13′ ) and Walla Walla, Washington (46◦ 4′ ).
(d) Graph S (x), for 0 ≤ x ≤ 90.
(e) Does S (x) appear to be differentiable?
My answers:
a.) @ 0 degrees latitude, you're at the equator, so S(0) = 12 hours daylight
b.) Since S(0)=12,
12 = a + b arcsin * (tan0/tan x0)
12 = a + b arcsin 0
12 = a + b0
12 = a,
and for b, because the function must be continuous,
24 = 12 + b arcsin* (tan 66/tan 66)
12 = b arcsin* 1
7.639 (rounded) = b.
But now when I try to find the answer to part C,
latitude of Tucson= 32 degrees
S(x)= 12 + 7.639 arcsin* (tan 32/tan 66)
...I get a nonreal answer for the arcsin of (tan 32/tan 66).
Can anyone please tell me where I'm going wrong? Thanks!

"I get a nonreal answer for the arcsin of (tan 32/tan 66)"
"arcsin of (tan 32/tan 66)" exists and equals 16.15338661° as long as the angles 32 and 66 are in degrees. I guess that your calculator has been set to radians or grad. Please check.
Post if you have other queries.
I am in the process of going through your other answers, so unable to comment for now. 
Oh, I am sorry! I had my calculator on radian mode instead of degree mode. Silly me. Thanks so much for the help!
I think I have managed to do the remainder of the problem, except for part E. How would I be able to tell if that function were differentiable or not? 
How would you draw the graph of this problem?