Suppose that the length of daylight in Canberra is modelled by a sine or cosine function where the horizontal axis
shows the time in days after January 1st (e.g. x = 0 is January 1; x = 34 is February 4).
The longest day is 14.5 hours on December 22 (x = 355); the shortest day of the year is 9.5 hours on June 22 (x =
172). Calculate how many days in the year that daylight is above 13 hours.
We must first devise a viable equation.
Note that a =
14.5−9.5
2
= 2.5 and d =
14.5+9.5
2
= 12. Now, since the period is 365 days, we have

360
b
= 365 ⇒ b =
360
365
=
72
73
.

It is easier to use a cosine graph: since the maximum occurs at x = 355, we can then simply set c = −355.
Thus, we have y = 2.5 cos (
72
73
(x − 355)) + 12.

Solving 2.5 cos (
72
73
(x − 355)) + 12 = 13 yields cos (
72
73
(x − 355)) = 0.4.
Now, cos−1 0.4 ≈ 66.42° and so the other solution to cos θ = 0.4 is 360 − 66.42 ≈ 293.58°
Thus, we have: 72
73
(x − 355) = 66.42 ⇒ x ≈ 422.34;
72
73
(x − 355) = 293.58 ⇒ x ≈ 652.66.

Subtracting the period yields x ≈ 57.34, 287.66.
Accordingly, it is above 13 hours for the first 57 days and last 78 days of the year, so about 135 days in total.

The given problem is well on its way to modeling the number of days when daylight exceeds 13 hours in Canberra using a cosine function. Let's break down the steps for clarification:

### Step-by-step breakdown:

1. **Amplitude and Vertical Shift**:
   - **Amplitude**: The difference between the maximum and minimum daylight values is \( 14.5 - 9.5 = 5 \), so the amplitude \( a \) is:
     \[
     a = \frac{14.5 - 9.5}{2} = 2.5
     \]
   - **Vertical shift**: The average of the maximum and minimum values gives the vertical shift \( d \), which is:
     \[
     d = \frac{14.5 + 9.5}{2} = 12
     \]

2. **Period and Horizontal Stretch**:
   - Since the period of the function corresponds to 365 days, the relationship between the period and the cosine function’s coefficient \( b \) is:
     \[
     \frac{360}{b} = 365 \quad \Rightarrow \quad b = \frac{360}{365} = \frac{72}{73}
     \]

3. **Phase Shift**:
   - Since the maximum daylight occurs on December 22 (which corresponds to \( x = 355 \)), the phase shift is \( c = -355 \).

4. **The Function**:
   - The cosine function modeling the daylight \( y \) as a function of time \( x \) (in days after January 1) is:
     \[
     y = 2.5 \cos\left(\frac{72}{73}(x - 355)\right) + 12
     \]

5. **Solving for Daylight above 13 hours**:
   - To find when daylight is above 13 hours, we solve the equation:
     \[
     2.5 \cos\left(\frac{72}{73}(x - 355)\right) + 12 = 13
     \]
   - Simplifying:
     \[
     \cos\left(\frac{72}{73}(x - 355)\right) = \frac{13 - 12}{2.5} = 0.4
     \]

6. **Solving the Cosine Equation**:
   - The principal solution for \( \cos \theta = 0.4 \) is \( \theta \approx 66.42^\circ \).
   - The cosine function is periodic, so the second solution is \( 360^\circ - 66.42^\circ = 293.58^\circ \).
   
   Now, solve for \( x \):
   \[
   \frac{72}{73}(x - 355) = 66.42 \quad \Rightarrow \quad x \approx 422.34
   \]
   \[
   \frac{72}{73}(x - 355) = 293.58 \quad \Rightarrow \quad x \approx 652.66
   \]

7. **Converting the Solutions to Actual Days**:
   - Since \( x = 422.34 \) and \( x = 652.66 \) extend beyond the 365-day cycle, subtract 365 from these values:
     \[
     x_1 = 422.34 - 365 \approx 57.34 \quad \text{(around February 27)}
     \]
     \[
     x_2 = 652.66 - 365 \approx 287.66 \quad \text{(around October 14)}
     \]

8. **Result**:
   - Daylight exceeds 13 hours between days \( 57 \) and \( 287 \), which means:
     \[
     287 - 57 = 230 \quad \text{days total}
     \]

Thus, the number of days in the year that daylight is above 13 hours is approximately **230 days**.

---

Do you want any details or have any questions?

Here are 5 questions for further exploration:
1. What is the role of the amplitude in determining the range of daylight hours?
2. How does the period affect the shape of the cosine function for daylight modeling?
3. Can you explain why the phase shift is negative in this model?
4. How would the equation change if the shortest and longest days occurred on different dates?
5. How would the model be adjusted for a leap year with 366 days?

**Tip**: When solving trigonometric equations, remember to account for the periodic nature of sine and cosine functions, which often produce multiple solutions within a given cycle.

Suppose that the length of daylight in Canberra is modeled by a sine or cosine function where the horizontal axis shows the time in days after January 1st (e.g., x = 0 is January 1; x = 34 is February 4). The longest day is 14.5 hours on December 22 (x = 355); the shortest day of the year is 9.5 hours on June 22 (x = 172). Calculate how many days in the year that daylight is above 13 hours.

Explore how to model the length of daylight in Canberra using a cosine function. This example includes key trigonometric concepts such as amplitude, period, and phase shift. Learn how to calculate the number of days with more than 13 hours of daylight based on a trigonometric equation. Suitable for students in Grades 11-12.

Math Problem Statement

360 b = 365 ⇒ b = 360 365

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