pre calc help please

Question

Outside temperature over a day can be modeled as a sinusoidal function. Suppose you know the high temperature of 72 degrees occurs at 3 PM and the average temperature for the day is 55 degrees. Find the temperature, to the nearest degree, at 11 AM.

Answer 1

average temp = 55 = D= (max+min)/2

Amplitude = (max-min)/2 = (72-min)/2 = 36 - min/2= 36-38/2 = 36-19 = 17 = A

minimum = 55 - (72-55) = 55-17 = 38

B = 24, 2pi/24 = pi/12

high temp = 72 at 3pm

what is the temp at 11am?? that's 4 hours before high temp. when t= -1

T = temp, t= time counting from 12 noon

C = horizontal shift = 13pi/12

T(t)= Asin(2pit/B + C) + D

72 = 17sin(3pi/12 +C) + 55

3pi/12 + C =pi/2, C = pi/2 - pi/4 = pi/4

= 17sin(pit/12 + pi/4) + 55

T(-1) = 17sin(-pi/12 + pi/4) +55

= 17sin(pi/6) +55

= 17(1/2) +55 = 8.5 +55

= 63.5 degrees at 11am

if you mistakenly used C=0 you'd get 46.5 at 11 am

Answer 2

°Lowest temperature is 2·55 - 72 = 38°. Then temperature T = 17°°sin(π/6t) + 55.

At 11AM time t = -1, because we count time from 12PM. SO, T = 17sin(-π/6) + 55 = - 8.5 + 55 = 46.5°