The daily number of hours of daylight can be given by h=(41/4)+(5/3sin(2pi x/365), where x is the number of days after March 21(disregarding leap year).

Question

The daily number of hours of daylight can be given by h=(41/4)+(5/3sin(2pi x/365), where x is the number of days after March 21(disregarding leap year). On what day or days will there be approximately 10 hours of daylight?

Answer 1

Let's start by re-writing your equation with 10 in place of h:

[1] 10 = 41/4 + 5/3[sin(2·pi·x/365)]

Now subtract 41/4 from each side:

[2] -1/4 = 5/3[sin(2·pi·x/365)]

Next, multiply both sides by 3/5:

[3] -3/20 = sin(2·pi·x/365)

In order to continue isolating the 'x' we need to eliminate the sine function. We can do that by taking the arcsine of each side. It's important to remember that sine and cosine equations typically have multiple solutions and arcsine can only determine one of them. More on that later.

[4] arcsin(-3/20) = 2·pi·x/365

[5] arcsin(-3/20) is approximately -0.15 radians, so:

[6] -0.15 ≈ 2·pi·x/365

Now multiply both sides by 365/(2·pi):

[7] -0.15·365/(2·pi) ≈ -8.7 ≈ x

So we have 10 hours of daylight about 9 days before March 21: March 12.

It would seem logical that if we have 10 hours of daylight 9 days before the vernal equinox, we should also have 10 hours of daylight approximately 9 days after the autumnal equinox. Let's see if we can show that mathematically.

Remember the earlier comment about arcsine. When you take the arcsine of a value between -1 and 1, the result is an angle between -pi/2 and pi/2 (the right-hand side of the Unit Circle). There is also an angle on the left side of the Unit Circle that will have the same sine value. In our case, arcsine gave us an angle in the 4th quadrant that is roughly .15 radians below the horizontal. The angle in the 3rd quadrant that is .15 radians below the horizontal would have the same sine value, so our other angle is pi+.15.

Let's start again at step [6] with our new angle:

[6A] pi + .15 ≈ 2·pi·x/365

Again, multiply both sides by 365/(2·pi):

[7A] 365(pi + .15)/(2·pi) ≈ 191.2 ≈ x

So we have a second solution 191 days after March 21 which would be Sept. 28th!