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The day length in Manila, Philippines, varies over time in a periodic way that can be modeled by a trigonometric function. Assume the length of the year (which is the period of change) is exactly 365 days long. The shortest day of the year is December 21 , and it's 675.85 minutes long. Manila's longest day is 779.60 minutes long. Note that December 21 is 11 days before January 1. Find the formula of the trigonometric function that models the length L of the day t days after January 1 . Define the function using radians. L(t)=◻ What is the day length on the People Power Anniversary (February 25, which is 55 days after January 1) in Manila? Round your answer, if necessary, to two decimal places. minutes

The day length in Manila, Philippines, varies over time in a periodic way that can be modeled by a trigonometric function.\newlineAssume the length of the year (which is the period of change) is exactly 365365 days long. The shortest day of the year is December 2121 , and it's 675675.8585 minutes long. Manila's longest day is 779779.6060 minutes long. Note that December 2121 is 1111 days before January 11 .\newlineFind the formula of the trigonometric function that models the length L L of the day t t days after January 11 . Define the function using radians.\newlineL(t)=□ L(t)=\square \newlineWhat is the day length on the People Power Anniversary (February 2525 , which is 5555 days after January 11) in Manila? Round your answer, if necessary, to two decimal places.\newlineminutes

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Q. The day length in Manila, Philippines, varies over time in a periodic way that can be modeled by a trigonometric function.\newlineAssume the length of the year (which is the period of change) is exactly 365365 days long. The shortest day of the year is December 2121 , and it's 675675.8585 minutes long. Manila's longest day is 779779.6060 minutes long. Note that December 2121 is 1111 days before January 11 .\newlineFind the formula of the trigonometric function that models the length L L of the day t t days after January 11 . Define the function using radians.\newlineL(t)=□ L(t)=\square \newlineWhat is the day length on the People Power Anniversary (February 2525 , which is 5555 days after January 11) in Manila? Round your answer, if necessary, to two decimal places.\newlineminutes
  1. Determine Amplitude: Determine the amplitude of the trigonometric function.\newlineThe amplitude is half the difference between the longest and shortest day lengths.\newlineAmplitude = (Longest day length−Shortest day length)/2(\text{Longest day length} - \text{Shortest day length}) / 2\newlineAmplitude = (779.60−675.85)/2(779.60 - 675.85) / 2\newlineAmplitude = 103.75/2103.75 / 2\newlineAmplitude = 51.87551.875 minutes
  2. Determine Vertical Shift: Determine the vertical shift of the trigonometric function.\newlineThe vertical shift is the average of the longest and shortest day lengths.\newlineVertical shift = (Longest day length+Shortest day length)/2(\text{Longest day length} + \text{Shortest day length}) / 2\newlineVertical shift = (779.60+675.85)/2(779.60 + 675.85) / 2\newlineVertical shift = 1455.45/21455.45 / 2\newlineVertical shift = 727.725727.725 minutes
  3. Determine Period: Determine the period of the trigonometric function.\newlineThe period is the length of the year, which is given as 365365 days.\newlinePeriod =365= 365 days
  4. Convert to Radians: Convert the period into radians.\newlineSince the period is one year and a full cycle in radians is 2π2\pi, the conversion factor is 2π2\pi radians per 365365 days.\newlineConversion factor = 2π365\frac{2\pi}{365}
  5. Determine Horizontal Shift: Determine the horizontal shift of the trigonometric function.\newlineThe shortest day is on December 2121, which is 1111 days before January 11. Therefore, the horizontal shift is 1111 days to the right.\newlineHorizontal shift = 1111 days
  6. Write Trigonometric Function: Write the trigonometric function using the values found.\newlineThe general form of a sinusoidal function is L(t)=A⋅cos⁡(B(t−C))+DL(t) = A \cdot \cos(B(t - C)) + D, where:\newline- AA is the amplitude,\newline- BB is the frequency (related to the period by B=2πPeriodB = \frac{2\pi}{\text{Period}}),\newline- CC is the horizontal shift, and\newline- DD is the vertical shift.\newlineUsing the values found:\newline- A=51.875A = 51.875,\newline- B=2π365B = \frac{2\pi}{365},\newline- C=11C = 11,\newline- D=727.725D = 727.725.\newlineThe function is AA00.
  7. Calculate Day Length: Calculate the day length on February 2525, which is 5555 days after January 11.\newlinePlug t=55t = 55 into the function L(t)L(t).\newlineL(55)=51.875×cos⁡(2π365(55−11))+727.725L(55) = 51.875 \times \cos\left(\frac{2\pi}{365}(55 - 11)\right) + 727.725\newlineL(55)=51.875×cos⁡(2π365(44))+727.725L(55) = 51.875 \times \cos\left(\frac{2\pi}{365}(44)\right) + 727.725\newlineNow calculate the cosine value and then the entire expression.
  8. Perform Calculation: Perform the calculation for the cosine value and the entire expression.\newlineL(55)=51.875×cos⁡(2π365×44)+727.725L(55) = 51.875 \times \cos\left(\frac{2\pi}{365} \times 44\right) + 727.725\newlineFirst, calculate the argument of the cosine function:\newline(2π365)×44≈0.7596\left(\frac{2\pi}{365}\right) \times 44 \approx 0.7596\newlineNow, calculate the cosine of this value and then multiply by the amplitude:\newlinecos⁡(0.7596)≈0.7314\cos(0.7596) \approx 0.7314 (using a calculator)\newline51.875×0.7314≈37.95151.875 \times 0.7314 \approx 37.951\newlineFinally, add the vertical shift:\newlineL(55)≈37.951+727.725L(55) \approx 37.951 + 727.725\newlineL(55)≈765.676L(55) \approx 765.676 minutes\newlineRound to two decimal places:\newlineL(55)≈765.68L(55) \approx 765.68 minutes

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