In Boston, the number of hours of daylight D(t) at a particular time of the 1year may be approximated by
sin 210˚ = -
⎡ 2π ⎤ 2
D(t) = 3 sin ⎢ (t - 79) ⎥ + 12 with t in days and t = 0 corresponding to January 1. How many days of
⎣ 365 ⎦ 1
sin 210˚ = -
the year have more than 10.5 hours of daylight? 2
2π 7π
Solution: (t - 79) =
365 6
⎡ 2π ⎤
D(t) = 3 sin ⎢ (t - 79) ⎥ + 12 7π/6
⎣ 365 ⎦ t - 79 =
2π / 365
⎡ π ⎤
10.5 = 3 sin ⎢ (t - 79) ⎥ + 12 365(7)
⎣ 365 ⎦ t - 79 =
12
⎡ 2π ⎤
3 sin ⎢ (t - 79) ⎥ = - 15 t = 213 + 79
⎣ 365 ⎦
⎡ 2π ⎤ t = 292
sin ⎢ (t - 79) ⎥ = - 0.5
⎣ 365 ⎦ 2π 11 π
(t- 79) =
1 365 6
sin θ = -
2 π/6
7π t - 79 =
θ= or 210˚ 2π / 365
6
11(365)
11π t - 79 =
θ= θ 330˚ 12
6
1 t = 414 > 365
sin 210˚ = -
2 t = 414 - 365
1 t = 49
sin 210˚ = -
2
2π 7π There will be at least 10.5 hours of daylight from
(t - 79) = t = 49 to t = 292 days or 292˚ - 49˚ = 243 days of the
365 6 year.
7π/6
t - 79 =
2π / 365
365(7)
t - 79 =
12
t = 213 + 79
t = 292
2π 11 π
(t- 79) =
365 6
π/6
t - 79 =
2π / 365
11(365)
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