Example 4 from Precalculus, 7.6 Modeling with Trigonometric Functions

Modeling an Equation and Sketching a Sinusoidal Graph to Fit Criteria

The average monthly temperatures for a small town in Oregon are given in Table 3. Find a sinusoidal function of the form $y = A \sin (B t - C) + D$ that fits the data (round to the nearest tenth) and sketch the graph.

Month	Temperature, $^{o} F$
January	42.5
February	44.5
March	48.5
April	52.5
May	58
June	63
July	68.5
August	69
September	64.5
October	55.5
November	46.5
December	43.5

Table 3

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Solution
Recall that amplitude is found using the formula

$A = \frac{largest value - smallest value}{2}$

Thus, the amplitude is

$\begin{array}{l} | A | = \frac{69 - 42.5}{2} \\ = 13.25 \end{array}$

The data covers a period of 12 months, so $\frac{2 π}{B} = 12$ which gives $B = \frac{2 π}{12} = \frac{π}{6} .$
The vertical shift is found using the following equation.

$D = \frac{highest value + lowest value}{2}$

Thus, the vertical shift is

$\begin{array}{l} \begin{array}{l} D = \frac{69 + 42.5}{2} \\ = 55.8 \end{array} \end{array}$

So far, we have the equation $y = 13.3 \sin (\frac{π}{6} x - C) + 55.8.$

To find the horizontal shift, we input the $x$ and $y$ values for the first month and solve for $C .$

$\begin{array}{l} 42.5 = 13.3 \sin (\frac{π}{6} (1) - C) + 55.8 \\ - 13.3 = 13.3 \sin (\frac{π}{6} - C) \\ - 1 = \sin (\frac{π}{6} - C) & \sin θ = - 1 \to θ = - \frac{π}{2} \\ \frac{π}{6} - C = - \frac{π}{2} \\ \frac{π}{6} + \frac{π}{2} = C \\ = \frac{2 π}{3} \end{array}$

We have the equation $y = 13.3 \sin (\frac{π}{6} x - \frac{2 π}{3}) + 55.8.$ See the graph in Figure 8.

Figure 8

How did it go?

Month

Temperature,

^{o} F

January

42.5

February

44.5

March

48.5

April

52.5

May

June

July

68.5

August

September

64.5

October

55.5

November

46.5

December

43.5

Solution

Solution