Problem 41 - Precalculus: Functions and Graphs

Question

Refer to the graph of \(y=\sin x\) or \(y=\cos x\) to find the exact values of \(x\) in the interval \([0,4 \pi]\) that satisfy the equation. $$\sin x=\frac{1}{2}$$

Step-by-Step Solution

Answer

The solutions are \( x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{13\pi}{6}, \text{ and } \frac{17\pi}{6} \).

Step 1: Understanding the equation

The problem requires finding the values of \( x \) in the interval \( [0, 4\pi] \) where the equation \( \sin x = \frac{1}{2} \) is true. The sine function achieves a value of \( \frac{1}{2} \) at angles that correspond to the reference angles of \( 30^\circ \) or \( \frac{\pi}{6} \), and also in the second quadrant at \( 150^\circ \) or \( \frac{5\pi}{6} \).

Step 2: Finding the reference angles

The reference angles where \( \sin x = \frac{1}{2} \) are \( x = \frac{\pi}{6} \) and \( x = \frac{5\pi}{6} \) within one complete cycle \([0, 2\pi]\). Due to the periodic nature of the sine function, these solutions repeat every \( 2\pi \).

Step 3: Utilizing periodicity

To find all solutions in the given interval \([0, 4\pi]\), consider the periodicity of the sine function. The sine function has a period of \( 2\pi \), meaning the solutions \( x = \frac{\pi}{6} \) and \( x = \frac{5\pi}{6} \) repeat every \( 2\pi \).

Step 4: Calculating additional solutions

To calculate additional solutions within \([0, 4\pi]\), add \( 2\pi \) (i.e., \( 2\pi\) and \( 4\pi \) for this interval) to each initial solution:- For \( x = \frac{\pi}{6} \): - \( x = \frac{\pi}{6} \) - \( x = \frac{\pi}{6} + 2\pi = \frac{\pi}{6} + \frac{12\pi}{6} = \frac{13\pi}{6} \)- For \( x = \frac{5\pi}{6} \): - \( x = \frac{5\pi}{6} \) - \( x = \frac{5\pi}{6} + 2\pi = \frac{5\pi}{6} + \frac{12\pi}{6} = \frac{17\pi}{6} \)

Step 5: Listing all solutions

The exact values of \( x \) in the interval \([0, 4\pi]\) that satisfy \( \sin x = \frac{1}{2} \) are: \( x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{13\pi}{6}, \text{ and } \frac{17\pi}{6} \).

Key Concepts

Sine Function

The sine function, often denoted as \( y = \sin x \), is a fundamental concept in trigonometry, representing the y-coordinate of a point on the unit circle as an angle \( x \) is measured in radians. The sine function has the following characteristics:

• Range: The values of \( \sin x \) lie between \(-1\) and \(1\).
• Periodicity: The sine function is periodic with a period of \( 2\pi \), meaning it repeats its values in regular intervals.
• Symmetry: It is an odd function, meaning that \( \sin(-x) = -\sin(x) \). This symmetry plays a crucial role in solving trigonometric equations like \( \sin x = \frac{1}{2} \).

Understanding the sine function's properties helps solve equations where you seek specific values for \( x \) that yield a particular sine value, such as \( \frac{1}{2} \).

Periodicity

One of the critical features of the sine function is its periodicity. This characteristic means that the sine function repeats its pattern every \( 2\pi \) radians. Therefore, if a particular value of \( x \) satisfies an equation like \( \sin x = \frac{1}{2} \) at some point in the interval, it will continue to satisfy the equation at every \( 2\pi \) interval thereafter. For example:

• The solution \( x = \frac{\pi}{6} \) means that within the interval \([0, 2\pi]\), \( \sin \left( \frac{\pi}{6} \right) = \frac{1}{2} \).
• Due to periodicity, adding \( 2\pi \) to \( x = \frac{\pi}{6} \) gives another solution: \( x = \frac{\pi}{6} + 2\pi = \frac{13\pi}{6} \).

This periodic property ensures we can find all solutions within the specified interval \((0, 4\pi]\).

Reference Angles

Reference angles are a critical concept in understanding sine values on the unit circle. They allow us to find angle equivalents in different quadrants with the same sine values. The reference angle is always the smallest angle subsection meant to reach the x-axis. It simplifies solving trigonometric equations by reducing them to well-known angle values such as \(30^\circ\) which converts to \(\frac{\pi}{6}\) in radians. When solving \( \sin x = \frac{1}{2} \), we're essentially looking for angles with reference angles like:

• \(x = \frac{\pi}{6}\) (in the first quadrant)
• \(x = \frac{5\pi}{6}\) (in the second quadrant)

These reference angles allow identifying all the possible values where \( \sin x = \frac{1}{2} \) in multiple occurrences along the unit circle's path. Using these reference angles and the periodic nature of sine, we map out all solutions within the interval.

Exact Values

When solving trigonometric equations like \( \sin x = \frac{1}{2} \), finding the exact values for \( x \) involves using precise trigonometric values associated with fundamental angles on the unit circle. These values can be achieved through:

• Memorization: Knowing that \( \sin(30^\circ) = \frac{1}{2} \) or \( \sin \left( \frac{\pi}{6} \right) = \frac{1}{2} \).
• Reference angles in radians: Recognizing that \( \sin \left( \frac{5\pi}{6} \right) = \frac{1}{2} \) exploits knowledge of the unit circle's symmetry.

For the exercise, the specific exact values of \( x \) that satisfy \( \sin x = \frac{1}{2} \) within the interval \([0, 4\pi]\) are:

• \(x = \frac{\pi}{6}\)
• \(x = \frac{5\pi}{6}\)
• \(x = \frac{13\pi}{6}\)
• \(x = \frac{17\pi}{6}\)

These values represent all angles within the interval where the sine attains its value of \( \frac{1}{2} \). The calculation of these values incorporates both the unit circle concepts and periodic behavior of the sine function.