Starting point for trig function

Question

For the trigonometric function describing the average daily temperature in Toronto in the year 2016

 

The coldest day occured on February 1^st with a temperature of -11 degrees celcius.

The hottest day occured exactly half a year later with an average daily temperature of 27 degrees celcius.

 

Take into account that 2016 was a leap year.

 

I already figured out this:

Sinusoidal line = 8

Amplitude = 19

Period = 366

 

The question is: Provide the final function in the form of: G(t) = a+bcos(c(t-d))

 

With t= 0 = January 1^st 2016.

Here d = horizontal shift.

 

8+19cos(2pi/366(t-

 

Can someone explain to me why the starting point can be found by = 31 + 1/2 · 366= 214 ?

Answer 1

using slightly different letter system

Temperature =

T= acos(b(t+c))+d

d=8= midline = average temperature

period =366,

b=2pi/period = 2pi/366 = pi/183

c = phase shift= 214

T = 19cos(pi/183)(t-183 )) +8

c = the number that makes cosine of the angle =-1 at time t= 0

with c= 183, cos(pi/183)(0-183) = cospi = -1

that would be true if t=0 meant Feb. 1

but the problem says t=0 is Jan. 1, 31 days earlier

so add 31 to 183 = 214 = the horizontal shift