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Real-World Applications of Trigonometric Functions, Page 2

Real-World Application Using the Sine Function

The table below shows the number of daylight hours on the fifteenth of every month in Center City, where t = 1 represents January 15th, t = 2 represents February 15th, t = 3 represents March 15th, and so on.

Month
   t  
Hours of Daylight
Jan
1
10.67
Feb
2
11.45
Mar
3
11.97
Apr
4
12.75
May
5
13.42
June
6
13.89
July
7
13.67
Aug
8
13.09
Sept
9
12.34
Oct
10
11.62
Nov
11
10.92
Dec
12
10.53


Write a function that models the number of hours of daylight for Center City. Then use your model to determine the number of hours of daylight on May 30th.

Procedure:

1. Find the values of A, h

2. Compute c by substituting one of the coordinate pairs into the function.

3. Use the function to solve the problem.

Solution:

The value of A is one-half the difference between the most daylight (13.89 hours) and the least daylight (10.53 hours).

A equals (13.89 - 10.53) divided by 2 equals 1.68

The value of h is one-half the sum of the greatest value and the least value.

h = (13.89 + 10.53)/2 = 12.21

Since you know the period is 12 months, you can find k.

12 = 2pi/k, so k = pi/6

Substitute these values into the general form of the sinusoidal function.

f(t) = Asin(kt - c) + h, or f(t) = 1.68sin(pi/6 t - c) + 12.21

To compute c, substitute one of the coordinate pairs, (t, y), into the function. Use the pair (1, 10.67) to substitute in t = 1 and y = 10.67.

10.67 = 1.68 sin((pi/6)(1) - c) + 12.21 

Subtract 12.21 from both sides:

-1.54 = 1.68 sin(pi/6 - c)

Divide both sides by 1.68:

-.9167 = sin (pi/6 - c) or sine inverse of -.9167 equals pi/6 - c or -1.16 = pi/6 - c

To continue solving for c, remember that π is equal to approximately 3.14.

-c = -1.16 - .524, so c = 1.68 

The function f(t) = 1.68sin(pi/6 t - 1.68) + 12.21 is a model for the hours of daylight in Center City.

Next, you will determine the number of hours of daylight in Center City on May 30th using the function above.