To model the temperature as a sinusoidal function, we can start with the general form of a sinusoidal function:

$T(t) = A \sin(B(t - C)) + D$

where:

• $A$ is the amplitude,
• $B$ relates to the period,
• $C$ is the phase shift,
• $D$ is the vertical shift.

Step 1: Determine the Parameters

1. Amplitude ($A$): The temperature varies between 73 and 97 degrees. Thus, the amplitude is half the difference between the maximum and minimum temperatures: $A = \frac{97 - 73}{2} = 12$

2. Vertical Shift ($D$): The average temperature $D$ is the midpoint of the maximum and minimum: $D = \frac{73 + 97}{2} = 85$

3. Period ($B$): The temperature cycle completes in 24 hours, so the period is 24 hours. The formula for $B$ is given by: $B = \frac{2\pi}{\text{period}} = \frac{2\pi}{24} = \frac{\pi}{12}$

4. Phase Shift ($C$): The average temperature occurs at 8 AM (8 hours after midnight). Since the sinusoidal function starts at its midpoint when $t = 0$, we need to adjust for the phase shift. The cosine function starts at its maximum, so we can write: $T(t) = 12 \cos\left(\frac{\pi}{12}(t - 8)\right) + 85$

Step 2: Set Up the Equation for 82 Degrees

We want to find $t$ when the temperature $T(t) = 82$: $82 = 12 \cos\left(\frac{\pi}{12}(t - 8)\right) + 85$

Step 3: Solve for $t$

1. Rearrange the equation: $12 \cos\left(\frac{\pi}{12}(t - 8)\right) = 82 - 85$ $12 \cos\left(\frac{\pi}{12}(t - 8)\right) = -3$ $\cos\left(\frac{\pi}{12}(t - 8)\right) = -\frac{1}{4}$

2. Solve for $\frac{\pi}{12}(t - 8)$: $\frac{\pi}{12}(t - 8) = \cos^{-1}\left(-\frac{1}{4}\right)$ The cosine inverse gives two angles in the range [0, 2π]: $\theta_1 = \cos^{-1}\left(-\frac{1}{4}\right) \quad \text{and} \quad \theta_2 = 2\pi - \cos^{-1}\left(-\frac{1}{4}\right)$

3. Calculate $t$: $t_1 = 12 \cdot \frac{\theta_1}{\pi} + 8 \quad \text{and} \quad t_2 = 12 \cdot \frac{\theta_2}{\pi} + 8$

4. Substitute to find the specific values for $t_1$ and $t_2$.

Step 4: Numerical Calculation

Using a calculator, we find:

$\theta_1 \approx 1.82 \quad \text{radians} \quad (\text{for } -\frac{1}{4})$

Now calculating $t_1$: $t_1 = 12 \cdot \frac{1.82}{\pi} + 8 \approx 12 \cdot 0.58 + 8 \approx 6.96 + 8 \approx 14.96 \text{ hours}$

Now for $t_2$: $\theta_2 \approx 4.46 \text{ radians}$ $t_2 = 12 \cdot \frac{4.46}{\pi} + 8 \approx 12 \cdot 1.42 + 8 \approx 17.06 + 8 \approx 25.06 \text{ hours} \quad (\text{not applicable as it's past 24 hours})$

Conclusion

Thus, the first occurrence of 82 degrees occurs at approximately:

$\boxed{14.96} \text{ hours after midnight}$

Do you want more details or have any questions?

Here are some follow-up questions you might consider:

1. How does changing the temperature range affect the sinusoidal model?
2. What would happen if the average temperature shifted to a different time?
3. Can this model be applied to other daily temperature patterns?
4. How would you graph this function over a 24-hour period?
5. What other factors could affect daily temperature variations?

Tip: Always verify the phase shift and amplitude when dealing with sinusoidal functions to ensure accurate modeling.