I need help trying to answer this question please

Question

Outside temperature over a day can be modeled as a sinusoidal function. Suppose you know the temperature varies between 59 and 71 degrees during the day and the average daily temperature first occurs at 8 AM. How many hours after midnight, to two decimal places, does the temperature first reach 61 degrees?

Answer 1

T = 6sin(π/12t - 2π/3) + 65 = 61; sin(πt/12 - 2π/3) = - 2/3; π/12t - 2π/3 = - sin^-1(2/3);

t = 12/π(2π/3 - sin^-1(2/3)) = 5.21 h

So, at 5.21 AM temperature reaches 61 degrees.

Answer 2

Let T be temperature and t be time in hours

T = Asin(B(t+C))+D

D = (51+79)/2 = 65

High = A(1)+D

71= A+65

A = 6

Average temperature when B(8+C) = 0 --> C = -8

B=2π/period = 2π/24 = π/12

T = 6sin((π/12)(t-8))+65

61 = 6sin((π/12)(t-8))+65

sin((π/12)(t-8)) = -4/6 = -2/3

t = sin^-1(-2/3)(12/π)+8

t=(-0.72972)(12/π)+8 = 5.21 hours

Make sure your calculator is set to radians when calculating sin^-1(-2/3).

Answer 3

f(t) = temperature t hours after midnight

f((t) = Asin(Bt + C) + D

D = average temperature = (59 + 71) / 2 = 65

A = amplitude = 71 - 65 = 6

Period = 24 hours = 2π/B. So, B = π/12.

Phase shift = -C/B = 8. So, C = -8B = -2π/3.

So, f(t) = 6sin [ (π/12)t - (2π/3) ] + 65