360 b = 365 ⇒ b = 360 365

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.

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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.