Problem 132 - Algebra and Trigonometry

Question

The number of hours of daylight in Boston is given by $$ y=3 \sin \left[\frac{2 \pi}{365}(x-79)\right]+12 $$where \(x\) is the number of days after January \(l\). Within a year, when does Boston have 13.5 hours of daylight? Give your answer in days after January 1 and round to the nearest day.

Step-by-Step Solution

Answer

After doing the calculations, it is found that Boston will have 13.5 hours of daylight 112 days after January 1.

Step 1: Set Up Equation

First, to find the day where there are 13.5 hours of daylight, set up the equation so \[y = 3 \sin \left[\frac{2 \pi}{365}(x-79)\right]+12 = 13.5.\] After this, solve for \(x\).

Step 2: Isolate the Trigonometric Function

Next, subtract 12 from both sides of the equation. It results in \[3 \sin \left[\frac{2 \pi}{365}(x-79)\right] = 1.5.\] Then divide both sides by 3 to isolate the sine function, giving \[\sin \left[\frac{2 \pi}{365}(x-79)\right] = 0.5.\]

Step 3: Solve for \(x\)

To find \(x\), first find the arcsine (inverse sine) of 0.5, which is \(\frac{\pi}{6}\) (or 30 degrees). This means \[\frac{2 \pi}{365}(x-79) = \frac{\pi}{6}.\] Multiply through by \(\frac{365}{2 \pi}\) to remove the fraction in front of \(x\), resulting in \[x-79 = \frac{365}{12}.\] Then add 79 to both sides of the equation, yielding \[x = \frac{365}{12} + 79\].

Step 4: Final Calculation

Now, compute \[x = \frac{365}{12} + 79\], and round it to the nearest whole day.

Key Concepts

Trigonometric Functions

Trigonometric functions, such as sine, cosine, and tangent, are a cornerstone of mathematics, particularly in the study of triangles and the modeling of periodic phenomena. These functions relate the angles of a triangle to the lengths of its sides in trigonometry. In the context of periodic functions, like the number of daylight hours in Boston, they prove to be very effective.

For instance, the sine function has the form \( y = A \times \text{sin}(Bx + C) + D \), where \( A \) dictates the amplitude, \( B \) the frequency, \( C \) the phase shift, and \( D \) the vertical shift. In our exercise, the daylight function \( y = 3 \text{sin}\big[\frac{2 \text{pi}}{365}(x - 79)\big] + 12 \) features these components that affect its shape and hence the model of daylight variation in Boston over the year.

To solve problems like the given one, the first step often involves manipulating the function's form to isolate the basic trigonometric function—in this case, sine. Then, we can apply appropriate methods to solve for the unknown variable, such as inverse operations or reference angles.

Inverse Trigonometric Functions

When solving trigonometric equations, we sometimes need to work backwards from the function's value to an angle. This is where inverse trigonometric functions come into play. They are the 'undoing' of the trigonometric functions—whereas trigonometric functions give us a ratio from an angle, the inverse provides us an angle from a ratio.

The primary inverse trigonometric functions include arcsine (\( \text{sin}^{-1}\)), arccosine (\( \text{cos}^{-1}\)), and arctangent (\( \text{tan}^{-1}\)). In this textbook problem, to find the value of \(x\) when the daylight is 13.5 hours, one part of the solution process necessitates using the arcsine function. We calculate \(x\) by finding the arcsine of 0.5 because the sine function we initially isolated equaled to 0.5. The angle corresponding to this value is \(\frac{\text{pi}}{6}\), or 30 degrees.

It is critical to remember that since trigonometric functions are periodic, there may be multiple angles that correspond to the same sine, cosine, or tangent value. Hence, when determining the inverse, we often refer to the principal value—the value of the angle within the function's principal domain.

Periodic Functions

Periodic functions have the unique characteristic of repeating their values at regular intervals, known as periods. The textbook exercise showcases a practical application of such a function by modeling the hours of daylight throughout the year. The sinusoidal function used in the problem displays the core attribute of periodicity, making it an apt model for these naturally cyclical events.

The period of a sinusoidal function is the length of one complete cycle. For the sine function in the equation, which is \( y = 3 \text{sin}\big[\frac{2 \text{pi}}{365}(x - 79)\big] + 12 \), the period is determined by the coefficient \( \frac{2 \text{pi}}{365} \), resulting from the number of days in a year. This period specifies how often the cycle repeats, and in this case, it completes once every 365 days.

To find when Boston has 13.5 hours of daylight, which is an event that occurs periodically, we solve for \(x\) within the confines of this periodic nature. Precedence is given to accuracy and understanding the relationship between the model's parameters and the real-world phenomena it represents to ensure the correct interpretation of the solution within the function's period.