In order to solve problems which require a sinusoidal model, it is necessary to A typical problem requiring a sinusoidal model is a relationship between time and some other data. We are given some information about data values that repeat over a certain interval or period of time. Scientists believe that the average temperatures at various places on Earth vary from cooler to warmer over thousands of years of gradual climate change. Suppose that at one place, the highest average temperature is 80º and the lowest is 60º. Also suppose that the time it takes to go from the high to low average is 20,000 years, and in the year 2000 the average temperature is at a high point of 80º. How can we use a sinusoidal expression to model this phenomenon? Let T(t) be the temperature in a given year t. We can model this situation using
where

A is the amplitude

f is the frequency which is

t is the year

C/2pf is the phase shift

D is a vertical displacement (from the temperature of 0º).
Sometimes this is written as
where

B = 2pf

C/B is the phase shift, which is 2000 years to the high point of the cosine function.
We first notice that the range of temperatures is 20º. D = 70° is the average temperature around which the highs and lows vary. The temperature varies 10º above and below this midpoint. The variation is given by A = 10º which is the amplitude or amount of temperature change above and below 70º. Second, the period of this temperature change is 40,000 years because that is the time it takes to go from 80º, then down to 60º, and finally back to (repeating) 80º. Therefore
.
We now have
We find C by noting that the value of this expression must be 80 when t = 2000. This means that
must equal 1 when t = 2000. This happens when
We could also set C/B = 2000 and solve for C, remembering that we have
Our final equation modeling this phenomenon is
This is a bit ugly, but it works! When we let t = 2000 we get
which was given as the average temperature in the year 2000. We could also rewrite this expression by factoring out the
which more clearly shows that at t = 2000, T(t) = 80. A graph is shown below using 80,000 < x < +80,000 counting by units of 20,000. A horizontal line is drawn at 70 indicating the midpoint of the average high and low temperatures. This calculator graph requires that the MODE be set for Radians. Notice that for year 2000 the average temperature is 80.
