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 0.250.250, point, 25 lux of illumination (lux is the SI unit of illuminance). During a new moon, the moon provides 000 lux of illumination. The period of the lunar cycle is 29.5329.5329, point, 53 days long. The moon will be full on December 252525, 201520152015. Note that December 252525 is 777 days before January 111. Find the formula of the trigonometric function that models the illumination LLL of the moon ttt days after January 111, 201620162016. Define the function using radians.
Answers
Answer:
Functions using radiation on moon = white dust
Light and air
Answer:
L(t)=0.125(cos(
14.765
π
(t+7))+0.125
Explanation:
The expression for the trigonometric function is :
L(t) = A (cos (B(t - C)))+ D ----- equation (1)
where ;
A = \frac{max-min}{2}A=
2
max−min
A = \frac{0.25-0}{2}A=
2
0.25−0
A = 0.125
D = \frac{0+.025}{2}
2
0+.025
D = 0.125
Period of the lunar cycle = 29.53
Then;
\frac{2 \pi}{B} = 29.53
B
2π
=29.53
29.53 \ \ B = 2 \pi29.53 B=2π
B = \frac{2 \pi}{29.53}B=
29.53
2π
B = \frac{\pi}{29.53}B=
29.53
π
Also; we known that December 25 is 7 days before January 1.
Then L(-7) = 0.025
Plugging all the values into trigonometric function ; we have:
\begin{gathered}0.125 ( cos ( \frac{\pi}{14.765}((-7)-C)))+0.125 = 0.25 \\ \\ \\ ( cos ( \frac{\pi}{14.765}((-7)-C))) = \frac{0.25-0.125}{0.125}\end{gathered}
0.125(cos(
14.765
π
((−7)−C)))+0.125=0.25
(cos(
14.765
π
((−7)−C)))=
0.125
0.25−0.125
( cos ( \frac{\pi}{14.765}((-7)-C))) = 1(cos(
14.765
π
((−7)−C)))=1
( \frac{\pi}{14.765}((-7)-C))= cos^{-1} (1)(
14.765
π
((−7)−C))=cos
−1
(1)
$$}((-7)-C))=0$$
$$C= -7$$
$$L(t) = 0.125 (cos (\frac{\pi}{14.765}(t-(-7))) + 0.125$$
$$L(t) = 0.125 (cos (\frac{\pi}{14.765}(t+7)) + 0.125$$