# Thread: Problem with sunlight hours trig equation

1. ## Problem with sunlight hours trig equation

Hi

I have the equation s = -3.2cos(0.01726t + 0.15534) + 12

s is equal to the number of hours of sunlight on a given day
t is equal to the day number of the year

If I am solving for t and s equals 13 hours I have tried to solve the equation like this

13 = -3.2cos(0.01726t + 0.15534) + 12
13 - 12 = -3.2cos0.01726(t + 9)
1/-3.2cos0.01726 = t + 9
-0.31 - 9 = t
- 9.31 = t

As I a have been told you take t and minus it from the mid point date which has the most number of day which is day 174 (as is explained in the question the lowest point in the year for sunlight is the 355th day) so that the two day when it is 13 hours of sunlight is 165th day and 183th day since it is a cos function and 13 hours falls on two different dates

However according to my graph and other online info the days of the year for 13 hours of sunlight should be closer to the 97th day and 251st day of the year.

I have been able to solve for s and the answer always works out correctly so the equation is correct. Why am I having trouble solving for t?

2. Hi

$\displaystyle 1 = -3.2 \cos \left(0.01726(t + 9)\right)$

$\displaystyle \cos \left(0.01726(t + 9)\right) = -\frac{1}{3.2}$

$\displaystyle 0.01726(t + 9) = \arccos \left(-\frac{1}{3.2}\right)$ or $\displaystyle 0.01726(t + 9) = 2\pi-\arccos \left(-\frac{1}{3.2}\right)$