Precalc/trig help

Question

The temperature during the day can be modeled by a sinusoid. Answer the following question given that the low temperature of 29 degrees occurs at 4 AM and the high temperature for the day is 53 degrees.

Find the temperature, to the nearest degree, at 8 AM.

Answer 1

29 at 4am high of 53 would be at 4pm given it's a sinosoidal curve

highs and lows are at half the period. Period = 24 hours

8am the temperature was 1/3 on its way to half the period

about 51.392 = 51 degrees at 8am, rounded off to nearest degree

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

A = amplitude = 12

average of max & min = (53+29)/2/ = 82/2 = 41 = midline

53-41 = 12 = amplitude

f(t) = 12sin(B(t+C)) + 41 D= midline = 41,

C=phase shift =4 = 12-8

pi/B = period, B=pi/12

t = 0 at 4am, t=12 at 4pm, t=4 at 8am

f(4) = 12 sin((pi/12)(4 + 4)) + 41

= 12sin(8pi/12)+ 41

=12sin2pi/3 +41

=12sqr3/2 +41

=6sqr3 +41

=10.392 +41

= 51.392

= 51 degrees rounded off to nearest degree