Using Sine Functions to Model Real-World Problems

This tutorial explains how to use sine functions to model data arising from real-world situations. Given information about a phenomenon that varies periodically, we model it using a trigonometric function.

General Form of Sine and Cosine Functions

A sinusoidal model can be written in the form

or equivalently,

The sine and cosine functions oscillate between a maximum and a minimum value. Let \( M \) be the maximum value and \( m \) be the minimum value of \( f(x) \). Assuming that \( A > 0 \), the following hold:

The maximum value occurs when \( \sin(bx + c) = 1 \), giving

The minimum value occurs when \( \sin(bx + c) = -1 \), giving

Solving the system \( M = A + D \) and \( m = -A + D \) yields:

These formulas are essential for modeling data with sine functions.

Problem 1

Find \( A \), \( b \), \( c \), and \( D \) so that the function

satisfies the following conditions:

• The maximum value is 7
• The minimum value is 3
• \( f(0) = 7 \)
• The period is \( \frac{2\pi}{3} \)
• \( A, b, c > 0 \) and \( c < 2\pi \)

Solution:

• Using the maximum and minimum values formulas given above: \[ A = \frac{7 - 3}{2} = 2, \qquad D = \frac{7 + 3}{2} = 5 \]
• The period of \( f(x) = A\sin(bx + c) + D \) is given by \[ \text{Period} = \frac{2\pi}{b} = \frac{2\pi}{3} \] Solve for \( b \) to obtain \[ b = 3 \]
• Substituting \( A \), \( b \), and \( D \) gives \[ f(x) = 2\sin(3x + c) + 5 \]
• Using \( f(0) = 7 \): \[ 2\sin(c) + 5 = 7 \Rightarrow \sin(c) = 1 \]
• The solution is \[ c = \frac{\pi}{2} + 2k\pi \] Since \( 0 < c < 2\pi \), we choose \[ c = \frac{\pi}{2} \]

Final model:

Problem 2: Modeling Daylight Hours

In a certain city, the number of daylight hours \( H(t) \) at time \( t \) (in days) is modeled by

Here, \( t = 0 \) corresponds to January 1st. The maximum daylight (15 hours) occurs on June 21st, which is day 171. The minimum number of daylight hours is 11. Find \( A \), \( c \), and \( D \).

Solution:

• From the maximum and minimum values: \[ A = \frac{15 - 11}{2} = 2, \qquad D = \frac{15 + 11}{2} = 13 \]
• The model becomes: \[ H(t) = 2\sin\!\left(\frac{2\pi}{365}t + c\right) + 13 \]
• Using \( H(171) = 15 \): \[ 2\sin\!\left(\frac{2\pi}{365}\cdot171 + c\right) + 13 = 15 \] which gives \[ \sin\!\left(\frac{2\pi}{365}\cdot171 + c\right) = 1 \]
• Hence, \[ \frac{2\pi}{365}\cdot171 + c = \frac{\pi}{2} + 2k\pi \] Choosing \( k = 0 \): \[ c = \frac{\pi}{2} - \frac{2\pi}{365}\cdot171 \approx -1.37 \]

Final daylight model:

Further Reading

Trigonometry Tutorials and Practice Problems