To determine the period of the light from the moon given the function [tex]\( L(t) = 0.25 - \sin \left( \frac{2 \pi (t - 2)}{28.5} \right) \)[/tex], we need to analyze the sinusoidal component of the function.
The general form of a sinusoidal function is:
[tex]\[ \sin \left( \frac{2 \pi t}{P} \right) \][/tex]
where [tex]\( P \)[/tex] represents the period of the function. The period is the length of time it takes for the sine function to complete one full cycle.
Given the function in the problem:
[tex]\[ L(t) = 0.25 - \sin \left( \frac{2 \pi (t - 2)}{28.5} \right) \][/tex]
we see that it involves a sine function of the form:
[tex]\[ \sin \left( \frac{2 \pi (t - 2)}{28.5} \right) \][/tex]
To identify the period [tex]\( P \)[/tex], we compare this to the standard form [tex]\( \sin \left( \frac{2 \pi t}{P} \right) \)[/tex]. Notice that here, [tex]\( t \)[/tex] has been transformed to [tex]\( t - 2 \)[/tex], but this shift does not affect the period. The term to focus on is [tex]\( \frac{2 \pi (t - 2)}{28.5} \)[/tex].
The period [tex]\( P \)[/tex] is determined by the coefficient of [tex]\( t \)[/tex], which in this transformed function is [tex]\( \frac{2 \pi}{28.5} \)[/tex].
To find the period [tex]\( P \)[/tex], we set up the equation:
[tex]\[ \frac{2 \pi}{P} = \frac{2 \pi}{28.5} \][/tex]
By comparing these, we can see that [tex]\( P \)[/tex]:
[tex]\[ P = 28.5 \][/tex]
So, the period of the light from the moon is [tex]\( \boxed{28.5} \)[/tex] days.
