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The moon's illumination changes in a periodic way that can be modeled by a trigonometric function.
On the night of a full moon, the moon provides about 0.25 lux of illumination (lux is the SI unit of illuminance). During a new moon, the moon provides 0 lux of illumination. The period of the lunar cycle is 29.53 days long. The moon will be full on December 
25,2015. Note that December 25 is 7 days before January 1.
Find the formula of the trigonometric function that models the illumination 
L of the moon 
t days after January 1,2016 . Define the function using radians.

L(t)=◻

The moon's illumination changes in a periodic way that can be modeled by a trigonometric function.\newlineOn the night of a full moon, the moon provides about 00.2525 lux of illumination (lux is the SI unit of illuminance). During a new moon, the moon provides 00 lux of illumination. The period of the lunar cycle is 2929.5353 days long. The moon will be full on December 2525, 20152015. Note that December 2525 is 77 days before January 11 .\newlineFind the formula of the trigonometric function that models the illumination L L of the moon t t days after January 11,20162016 . Define the function using radians.\newlineL(t)= L(t)=\square

Full solution

Q. The moon's illumination changes in a periodic way that can be modeled by a trigonometric function.\newlineOn the night of a full moon, the moon provides about 00.2525 lux of illumination (lux is the SI unit of illuminance). During a new moon, the moon provides 00 lux of illumination. The period of the lunar cycle is 2929.5353 days long. The moon will be full on December 2525, 20152015. Note that December 2525 is 77 days before January 11 .\newlineFind the formula of the trigonometric function that models the illumination L L of the moon t t days after January 11,20162016 . Define the function using radians.\newlineL(t)= L(t)=\square
  1. Determine Amplitude: Determine the amplitude of the trigonometric function.\newlineThe amplitude AA is half the difference between the maximum and minimum values of the function. Since the maximum illumination is 0.250.25 lux (full moon) and the minimum is 00 lux (new moon), the amplitude is:\newlineA=0.2502=0.252=0.125A = \frac{0.25 - 0}{2} = \frac{0.25}{2} = 0.125 lux.
  2. Determine Period: Determine the period of the trigonometric function.\newlineThe period TT of the lunar cycle is given as 29.5329.53 days. In terms of radians, the period of a sine or cosine function is 2π2\pi, so we need to find the value that will stretch or compress the function to fit the lunar cycle. The period in the function will be:\newlineT=2πBT = \frac{2\pi}{B}, where BB is the horizontal stretch factor.\newlineTo find BB, we rearrange the equation: B=2πTB = \frac{2\pi}{T}.\newlineB=2π29.53B = \frac{2\pi}{29.53}.
  3. Calculate Stretch Factor: Calculate the horizontal stretch factor BB.B=2π29.530.2124B = \frac{2\pi}{29.53} \approx 0.2124 radians per day.
  4. Determine Shift: Determine the horizontal shift (phase shift) of the function.\newlineSince the full moon is on December 2525, 20152015, and we are considering tt days after January 11, 20162016, the phase shift (DD) will be the number of days from the full moon to January 11. This is 77 days since December 2525 is 77 days before January 11. The function will be at its maximum 77 days before t=0t = 0, so the phase shift is 7-7 days.
  5. Determine Vertical Shift: Determine the vertical shift of the function.\newlineThe vertical shift CC is the average of the maximum and minimum values of the function. Since the maximum is 0.250.25 lux and the minimum is 00 lux, the vertical shift is:\newlineC=0.25+02=0.252=0.125C = \frac{0.25 + 0}{2} = \frac{0.25}{2} = 0.125 lux.
  6. Write Function: Write the trigonometric function.\newlineWe can use a cosine function because it starts at its maximum value, which corresponds to the full moon. The general form of the cosine function is:\newlineL(t)=Acos(B(tD))+CL(t) = A \cdot \cos(B(t - D)) + C.\newlineSubstituting the values we have:\newlineA=0.125A = 0.125 lux,\newlineB0.2124B \approx 0.2124 radians per day,\newlineD=7D = -7 days,\newlineC=0.125C = 0.125 lux.\newlineL(t)=0.125cos(0.2124(t+7))+0.125L(t) = 0.125 \cdot \cos(0.2124(t + 7)) + 0.125.

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