Problem 51 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 51-54: Refer to the graph of the equation on the specified interval. Find all values of \(x\) such that for the real number \(a\), (a) \(y=a\), (b) \(y>a\), and (c) \(y

Step-by-Step Solution

Answer

(a) \( x = -\frac{11\pi}{6}, -\frac{7\pi}{6}, -\frac{\pi}{6}, \frac{\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6} \). (b) Intervals: \(\left(-\frac{11\pi}{6}, -\pi\right), \left(-\frac{\pi}{6}, \frac{\pi}{6}\right), \left(\pi, \frac{11\pi}{6}\right) \). (c) Intervals: \(\left(-2\pi, -\frac{11\pi}{6}\right), \left(-\frac{7\pi}{6}, -\frac{\pi}{6}\right), \left(\frac{\pi}{6}, \frac{7\pi}{6}\right), \left(\frac{11\pi}{6}, 2\pi\right) \).

Step 1: Understanding the Function and Interval

We are working with the function \( y = \sin x \) over the interval \([-2\pi, 2\pi]\). The sine function is periodic with a period of \(2\pi\) and oscillates between -1 and 1.

Step 2: Solving for y = a

Set \( y = a = \frac{1}{2} \) and solve for \( x \). Solve \( \sin x = \frac{1}{2} \). The solutions are \( x = \frac{\pi}{6} + 2n\pi \) and \( x = \frac{5\pi}{6} + 2n\pi \), where \( n \) is an integer.

Step 3: Find Specific Solutions in Given Interval for y = a

Substitute integer values for \( n \) to find solutions within \([-2\pi, 2\pi]\). \( x = -\frac{11\pi}{6}, -\frac{7\pi}{6}, -\frac{\pi}{6}, \frac{\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6} \) are the valid solutions within the interval.

Step 4: Finding x when y > a

We need \( \sin x > \frac{1}{2} \). Thus, solve the inequalities: \( x \in (\frac{\pi}{6}, \frac{5\pi}{6}) + 2n\pi \). Identify intervals using \( n \) values where this condition is satisfied in \([-2\pi, 2\pi]\). Solutions are: \( \left(-\frac{11\pi}{6}, -\pi\right), \left(-\frac{\pi}{6}, \frac{\pi}{6}\right), \left(\pi, \frac{11\pi}{6}\right) \).

Step 5: Finding x when y < a

We need \( \sin x < \frac{1}{2} \). Solve the inequalities: \( x \in \left(-2\pi, -\frac{11\pi}{6}\right), \left(-\frac{7\pi}{6}, -\frac{\pi}{6}\right), \left(\frac{\pi}{6}, \frac{7\pi}{6}\right), \left(\frac{11\pi}{6}, 2\pi\right) \).

Key Concepts

Sine Function

The sine function is a fundamental concept in trigonometry. It's often denoted as \( \sin x \) and is a function that maps the angle \( x \) (measured in radians) to a real number between -1 and 1. Understanding how this function behaves is key to solving trigonometric equations. It adds a wavelike motion to its graph, reminiscent of cycles. The sine wave is symmetric about the origin, and it is this symmetry that often makes it useful in a variety of mathematical and engineering applications.

• The sine function reaches its maximum value of 1 at \( \frac{\pi}{2} \) and its minimum value of -1 at \( \frac{3\pi}{2} \).
• At \( x = 0, \pi, \text{ and } 2\pi \), the function takes the value of 0, simplifying many calculations.

In the given problem, we explore the behaviors of the sine function across the interval \([-2\pi, 2\pi]\). The task includes determining where the sine equals, exceeds, or falls below a specific level: \( a = \frac{1}{2} \). Each scenario reflects various properties of the sine wave used in practical applications.

Inequalities

Inequalities play a crucial role when solving trigonometric problems like the one presented here. In this context, dealing with inequalities involves understanding where the sine function is greater than or less than the designated level of \( a = \frac{1}{2} \).For example, when working with the inequality \( \sin x > \frac{1}{2} \), we are interested in finding the range of \( x \) that makes the sine function's value above \( \frac{1}{2} \). This involves techniques such as interval analysis:

• To solve \( \sin x > \frac{1}{2} \), we find parts within the interval where this condition holds, like \( (\frac{\pi}{6}, \frac{5\pi}{6}) + 2n\pi \).
• On the other hand, \( \sin x < \frac{1}{2} \) requires identifying ranges where the sine is less than \( \frac{1}{2} \), such as in segments like \( \left(-2\pi, -\frac{11\pi}{6}\right) \).

These inequalities help isolate sections where the sine's behavior aligns with the condition, thereby allowing us to solve the posed problem over the specified interval. Mastery of these techniques is essential for handling more complex trigonometric expressions.

Periodic Functions

The sine function is a classic example of a periodic function, meaning it repeats its values in a consistent cycle. For the sine function, the cycle or period is \( 2\pi \). Understanding periodicity is vital since it explains the repetitive oscillation between -1 and 1—and therefore, knowing these patterns can simplify predicting future values of the function without direct calculation.

• The concept of periodicity implies that any solution for \( x \) is not unique within its period. In fact, \( \sin x \) returns the same value every \( 2\pi \) interval.
• This property helps determine the infinite number of solutions to equations involving sine, where each specific solution can be generated by adding \( 2n\pi \) to a known solution—where \( n \) is any integer.

In exercises like this, the periodic nature aids in listing all possible values of \( x \) that satisfy specific conditions within a set interval. This understanding is not only a cornerstone in trigonometry but also in applications beyond, such as signal processing and harmonic analysis.