Tutorial on how to use sine functions to model data. Given data and information about a certain situation, we model it in the form
f(x) = A sin (b x + c) + D
or
f(x) = A cos (b x + c) + D
The sine and cosine functions are known to vary between a maximum value and a minimum value. Let M be the maximum and m be the minimum values of f(x) defined above. If we assume that A is positive, then as x varies, the maximum value M of f(x) occurs when sin(Bx + C)=1, hence
M = A + D
The minimum m of f(x) occurs when sin(Bx + C)= 1, hence
m = A + D
Hence the system of equations M = A + D and m = A + D may be solved for A and D in terms of M and m as follows:
A = [ M  n ] / 2
and
D = [ M + m ] / 2
The above relationships between A, D, M and m may be used to model data using sine functions.
Problem 1: Find A, b, c and D so that function f defined by
f(x) = A sin(b x + c) + D
has the following properties: the maximum value of f(x) is 7 and f(0) = 7, the minimum value of f(x) is 3, the period of the graph of function f is equal to 2 pi/3. A, b and c are positive and c is less than 2 pi.
Solution to Problem 1:

Using the formula for A and D above, gives
A = [ 7  3 ] / 2 = 2
and
D = [ 7 + 3 ] / 2 = 5

Since the period is given, parameter b may be found using the formula for the period in terms of b
period = 2 pi / b = 2 pi / 3

Solve for b to find
b = 3

We now write a formula for f(x) using the values of A, b and D already found.
f(x) = 2 sin(3 x + c) + 5

We now use the fact that f(0) = 7 to find c.
f(0) = 2 sin(3 (0) + c) + 5 = 7

The above may be written as follows.
sin(c) = 1

Solve the above trigonmetric function to obtain.
c = pi / 2 + k(2pi), where k is an integer.

c must be positive and less than 2pi, hence
c = pi / 2

Finally f is given by
f(x) = 2 sin(3 x + pi/2) + 5
Problem 2: In a certain city, the number of daylight hours H(t) at a time t of the year is given by
H(t) = A sin [ (2pi/365) t + c ] + D
where t = 0 corresponds to January 1st. The maximum number of daylight hours occurs on June 21 st (or 171 days after January 1st) and is equal to 15 hours. The minimum number of daylight hours is equal to 11 hours. Find A, c and D.
Solution to Problem 2:

Using the formula for A and D above, gives
A = [ 15  11 ] / 2 = 2
and
D = [ 15 + 11 ] / 2 = 13

We now write a formula for H(t) using the values of A and D already found.
H(t) = 2 sin [ (2pi/365) t + c ] + 13

We now use the fact that H(171) = 15 to find c.
H(171) = 2 sin( (2pi/365) * 171 + c ) + 13 = 15

The above may be written as follows.
sin( (2pi/365) * 171 + c ) = 1

Solve the above trigonometric function to obtain.
(2pi/365) * 171 + c = pi / 2 + k (2 pi), where k is an integer.

Since there is no condition on c, we can choose k = 0 above and solve for c to obtain
c = pi / 2  (2pi/365) * 171 = 1.37 (approximated to 2 decimal places)

H(t) is given by
H(t) = 2 sin [ (2pi/365) t  1.37 ] + 13
More references and links on
Trigonometry Tutorials and Problems for Self Tests 