Question

The moon's distance from Earth varies in a periodic way that can be modeled by a trigonometric function.

When the moon is at its perigee (closest point to Earth), it's about 363,000 km away. When it's at its
apogee (farthest point from Earth), it's about 406,000 km away. The moon's apogees occur 27.3 days
apart. The moon will reach its apogee on January 2, 2016.
Find the formula of the trigonometric function that models the distance D between Earth and the moont
days after January 1, 2016. Define the function using radians.

Answer 1

D(t) = 21500 cos (2pi/27.3(t-1)) + 384500

Explanation:

Answer 2

`d = 384,500\,km + (21,500\,km)\cdot \cos \left[(2\pi)/(27.3)\cdot (t-1) \right]`

Explanation:

The trigonometric model of the distance between Earth and the Moon is:

`d = A + \Delta A \cdot \cos (\omega\cdot t + \phi)`

Where:

`A` - Apogee, measured in kilometers.

`\Delta A` - Amplitude, measured in kilometers.

`\omega` - Angular frequency, measured in radians.

`\phi` - Phase angle, measured in radians.

`t` - Time, measured in days.

The required information are derived below:

`A = (406,000\,km+363,000\,km)/(2)`

`A = 384,500\,km`

`\Delta A = (406,000\,km-363,000\,km)/(2)`

`\Delta A = 21,500\,km`

`\omega = (2\pi)/(27.3)`

`\phi = -(2\pi)/(27.3)`

The expression is:

`d = 384,500\,km + (21,500\,km)\cdot \cos \left[(2\pi)/(27.3)\cdot (t-1) \right]`