Problem 59 - Algebra and Trigonometry

Question

Phases of the Moon As the moon revolves around the earth, the side that faces the earth is usually just partially illuminated by the sun. The phases of the moon describe how much of the surface appears to be in sunlight. An astronomical measure of phase is given by the fraction \(F\) of the lunar disc that is lit. When the angle between the sun, earth, and moon is \(\theta\left(0 \leq \theta \leq 360^{\circ}\right),\) then $$ F=\frac{1}{2}(1-\cos \theta) $$ Determine the angles \(\theta\) that correspond to the following phases: (a) \(F=0\) (new moon) (b) \(F=0.25\) (a crescent moon) (c) \(F=0.5 \quad\) (first or last quarter) (d) \(F=1\) (full moon)

Step-by-Step Solution

Answer

(a) \(0^{\circ}\), (b) \(60^{\circ}\) and \(300^{\circ}\), (c) \(90^{\circ}\) and \(270^{\circ}\), (d) \(180^{\circ}\).

Step 1: Understand the Equation

The given formula for the fraction of the lunar disc that is lit is \( F=\frac{1}{2}(1-\cos \theta) \). This equation relates the phase of the moon, \( F \), to the angle \( \theta \) between the sun, earth, and moon.

Step 2: Substitute and Solve for Part (a)

For a new moon, \( F = 0 \). Substitute \( F = 0 \) into the equation: \( 0 = \frac{1}{2}(1-\cos \theta) \). Solve for \( \theta \):\[0 = \frac{1}{2}(1-\cos \theta)\]This simplifies to \( 1-\cos \theta = 0 \), which gives \( \cos \theta = 1 \). This occurs at \( \theta = 0^{\circ} \).

Step 3: Substitute and Solve for Part (b)

For a crescent moon, \( F = 0.25 \). Substitute \( F = 0.25 \) into the equation: \( 0.25 = \frac{1}{2}(1-\cos \theta) \). Solve for \( \theta \):\[0.25 = \frac{1}{2}(1-\cos \theta)\]This gives \( 1-\cos \theta = 0.5 \), leading to \( \cos \theta = 0.5 \). This occurs at \( \theta = 60^{\circ} \) and \( \theta = 300^{\circ} \).

Step 4: Substitute and Solve for Part (c)

For the first or last quarter, \( F = 0.5 \). Substitute \( F = 0.5 \) into the equation: \( 0.5 = \frac{1}{2}(1-\cos \theta) \). Solve for \( \theta \):\[0.5 = \frac{1}{2}(1-\cos \theta)\]This gives \( 1-\cos \theta = 1 \), which results in \( \cos \theta = 0 \). This occurs at \( \theta = 90^{\circ} \) and \( \theta = 270^{\circ} \).

Step 5: Substitute and Solve for Part (d)

For a full moon, \( F = 1 \). Substitute \( F = 1 \) into the equation: \( 1 = \frac{1}{2}(1-\cos \theta) \). Solve for \( \theta \):\[1 = \frac{1}{2}(1-\cos \theta)\]This simplifies to \( 1-\cos \theta = 2 \), leading to \( \cos \theta = -1 \). This occurs at \( \theta = 180^{\circ} \).

Key Concepts

Trigonometric Equations

Trigonometric equations are equations involving trigonometric functions like sine, cosine, and tangent. These functions play a vital role in understanding various phenomena in mathematics and physics, such as oscillations and wave patterns.
When solving trigonometric equations, we aim to find the values of the variable that satisfy the equation. For example, when solving for the moon phases, you deal with the cosine function. The equation given is:

• \( F = \frac{1}{2}(1-\cos \theta) \)

The task involves substituting the values of the lunar phase \( F \) and solving for \( \theta \). Each solution typically involves algebraic manipulation and sometimes recognizing known angles' trigonometric values.
This process can yield one or more solutions within a given range, such as 0 to 360 degrees when examining angles. Understanding the properties of trigonometric functions, like periodicity and symmetry, is crucial when solving these equations.

Cosine Function

The cosine function is one of the fundamental trigonometric functions, typically represented as \( \cos(\theta) \). It's related to the x-coordinate of a point on the unit circle.
The cosine of an angle in a right triangle is the ratio of the length of the adjacent side to the hypotenuse. In the context of the moon's phases, we use its trigonometric properties to track how the visible part of the moon varies with the angle \( \theta \).
Key characteristics of the cosine function:

• It has a range between -1 and 1.
• The function is periodic with a period of 360 degrees or \( 2\pi \) radians.
• It achieves a maximum value of 1 at \( 0^{\circ} \) and a minimum value of -1 at \( 180^{\circ} \).

These properties help determine key lunar phases, such as the new moon, when \( \cos(\theta) \) is 1, and the full moon, when \( \cos(\theta) \) becomes -1.

Angle Measurement

Angles can be measured in degrees or radians, and each unit has its applications in trigonometry and geometry. In this exercise, degrees are used because they are more intuitive when discussing angles in everyday circumstances.
When solving trigonometric equations related to the moon phases, it is essential to understand:

• The range given, often between 0 and 360 degrees, represents one complete revolution.
• Common angles such as \( 0^{\circ}, 90^{\circ}, 180^{\circ}, \) and \( 270^{\circ} \) are often used as they correspond to key positions on the unit circle.

Recognizing these angles and their cosine values speeds up solving trigonometric problems and enhances comprehension of periodic phenomena like moon phases. This understanding enables predicting which phases occur at specific angle values.

Lunar Phases

Lunar phases are the visible slices of the moon seen from Earth, caused by its orbit around us and its changing illuminated portion.
Understanding lunar phases is about identifying patterns. There are four main phases:

• New Moon: When the moon is between the Earth and the sun, making the visible side dark, \( F = 0 \).
• First Quarter: Half of the moon is visible, \( F = 0.5 \), occurring at \( 90^{\circ} \).
• Full Moon: The moon is opposite the sun, fully illuminated, \( F = 1 \), occurring at \( 180^{\circ} \).
• Last Quarter: Again, half-moon visibility but the opposite half, \( F = 0.5 \), occurs at \( 270^{\circ} \).

In between, the moon transitions through crescent and gibbous phases. Understanding these phases aids in astronomy and brings awareness to cosmic patterns shaping the night sky.