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The daily low temperature in Guangzhou, China, varies over time in a periodic way that can be modeled by a trigonometric function. The period of change is exactly 1 year. The temperature peaks around July 26 at 78^(@)F, and has its minimum half a year later at 49^(@)F. Assuming a year is exactly 365 days, July 26 is (206)/(365) of a year after January 1 . Find the formula of the trigonometric function that models the daily low temperature T in Guangzhou t years after January 1,2015 . Define the function using radians. T(t)=◻

The daily low temperature in Guangzhou, China, varies over time in a periodic way that can be modeled by a trigonometric function.\newlineThe period of change is exactly 11 year. The temperature peaks around July 2626 at 78∘F 78^{\circ} \mathrm{F} , and has its minimum half a year later at 49∘F 49^{\circ} \mathrm{F} . Assuming a year is exactly 365365 days, July 2626 is 206365 \frac{206}{365} of a year after January 11 .\newlineFind the formula of the trigonometric function that models the daily low temperature T T in Guangzhou t t years after January 11,20152015 . Define the function using radians.\newlineT(t)=□ T(t)=\square

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Q. The daily low temperature in Guangzhou, China, varies over time in a periodic way that can be modeled by a trigonometric function.\newlineThe period of change is exactly 11 year. The temperature peaks around July 2626 at 78∘F 78^{\circ} \mathrm{F} , and has its minimum half a year later at 49∘F 49^{\circ} \mathrm{F} . Assuming a year is exactly 365365 days, July 2626 is 206365 \frac{206}{365} of a year after January 11 .\newlineFind the formula of the trigonometric function that models the daily low temperature T T in Guangzhou t t years after January 11,20152015 . Define the function using radians.\newlineT(t)=□ T(t)=\square
  1. Calculate Average Temperature: The average temperature is the midpoint between the maximum and minimum temperatures. Calculate the average: (78 ∘F+49 ∘F)/2(78\,^\circ\mathrm{F} + 49\,^\circ\mathrm{F}) / 2.
  2. Calculate Amplitude: The amplitude is half the distance between the maximum and minimum temperatures. Calculate the amplitude: (78°F−49°F)/2(78\degree\text{F} - 49\degree\text{F}) / 2.
  3. Calculate Angular Frequency: The period of the function is 11 year, which is 365365 days. To find the angular frequency, use the formula ω=2πperiod\omega = \frac{2\pi}{\text{period}}. Calculate ω\omega: 2π365.\frac{2\pi}{365}.
  4. Calculate Phase Shift: The function peaks at July 2626, which is (206)/(365)(206)/(365) of a year after January 11. To find the phase shift, use the formula φ=ω×(days after January 1)\varphi = \omega \times (\text{days after January 1}). Calculate φ\varphi: 2π×(206)/(365)2\pi \times (206)/(365).
  5. Model Temperature Function: The trigonometric function that models the temperature is of the form T(t)=A⋅cos⁡(ωt+φ)+DT(t) = A \cdot \cos(\omega t + \varphi) + D, where AA is the amplitude, ω\omega is the angular frequency, φ\varphi is the phase shift, and DD is the average temperature. Substitute the values calculated in the previous steps.
  6. Adjust Phase Shift: The function is T(t)=14.5×cos⁡(2π365t+2π×206365)+63.5T(t) = 14.5 \times \cos\left(\frac{2\pi}{365}t + \frac{2\pi \times 206}{365}\right) + 63.5. However, we need to adjust the phase shift because the cosine function peaks at 00, and we want it to peak at 206365\frac{206}{365} of a year. To do this, we subtract the phase shift from π\pi, not add it.

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