when t = T, the argument must be 2 pi so
cos ( 2 pi t/T + phase )
= cos (2 pi t/365 + phase)
the low point of the cosine function is when the angle inside is pi radians because cos(pi) = -1
so
2 pi (20)/365 + phase = pi
.11 pi + phase = pi
phase = .89 pi
so we have
cos ( 2 pi t/365 + .89 pi)
or
cos (.017 t + 2.8)
so when t = 20, then 2 pi t/365
Trying to decypher something, and am getting fouled up with the radians definition of periodicity.
Problem: Average temperatures vary over a one year period, or 365 days. Highest temperature = 5 C, lowest = -37 C. Lowest temperature is on day = 20, highest temperature is on day = 200
Write the cosine approximation of the equation for each day as T(d)= A+B(cos(C(d) + D).
OK. My solution, so far.
So the range = 5 --37 = 42.
Amplitude = B = 42/2 = 21
D = -20. (The low point is at day 20)
A = (5+-37)/2 = -32/2 = -16
But I can't logic out C properly. Period HAS to = 365, which is 2*pi radians, but then how do you get C from that?
5 answers
when t = T, the argument must be 2 pi so
cos ( 2 pi t/T + phase )
= cos (2 pi t/365 + phase)
the low point of the cosine function is when the angle inside is pi radians because cos(pi) = -1
so
2 pi (20)/365 + phase = pi
.11 pi + phase = pi
phase = .89 pi
so we have
cos ( 2 pi t/365 + .89 pi)
or
cos (.017 t + 2.8)
cos ( 2 pi t/T + phase )
= cos (2 pi t/365 + phase)
the low point of the cosine function is when the angle inside is pi radians because cos(pi) = -1
so
2 pi (20)/365 + phase = pi
.11 pi + phase = pi
phase = .89 pi
so we have
cos ( 2 pi t/365 + .89 pi)
or
cos (.017 t + 2.8)
That .89 pi or 2.8 radians is your D by the way
So I was throwing my "logic" off by assuming that I needed to subtract off the 20 days offset as "D", right?
Instead of 2.8 (or 2.796)
Instead of 2.8 (or 2.796)
And thank you very much, for your time and effort answering the above question.