Problem 51 - Precalculus: Functions and Graphs

Question

Refer to the graph of the equation on the specifled 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: \((-\frac{11\pi}{6}, -\frac{7\pi}{6})\), \((-\frac{\pi}{6}, \frac{\pi}{6})\), \((\frac{7\pi}{6}, \frac{11\pi}{6})\); (c) Remaining intervals.

Step 1: Understand the given function

The function given is the sine function, \( y = \sin x \) which ranges from \(-1\) to \(1\). We need to analyze this function over the interval \([-2\pi, 2\pi]\). The parameter \(a\) is \(\frac{1}{2}\). We will determine where \(y\) equals \(a\), where \(y > a\), and where \(y < a\).

Step 2: Determine when \(y = a\)

We need to find all values of \(x\) such that \( \sin x = \frac{1}{2} \) within the interval \([-2\pi, 2\pi]\). The sine function equals \(\frac{1}{2}\) at angles \(\frac{\pi}{6} + 2k\pi\) and \(\frac{5\pi}{6} + 2k\pi\). Considering the interval \([-2\pi, 2\pi]\), we calculate these values:- From \([-2\pi, 2\pi]\), \(\sin x = \frac{1}{2}\) at \(x = -\frac{11\pi}{6}, -\frac{7\pi}{6}, -\frac{\pi}{6}, \frac{\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}\).

Step 3: Determine when \(y > a\)

Now we need to find where \( \sin x > \frac{1}{2} \). This occurs in the intervals between the calculated values of \(x\) where \(y = \frac{1}{2}\), specifically:- Between \(-\frac{11\pi}{6}\) and \(-\frac{7\pi}{6}\), - Between \(-\frac{\pi}{6}\) and \(\frac{\pi}{6}\), - Between \(\frac{7\pi}{6}\) and \(\frac{11\pi}{6}\).

Step 4: Determine when \(y < a\)

The sine function is less than \(\frac{1}{2}\) in the remaining parts of the interval. These are:- Between \(-2\pi\) and \(-\frac{11\pi}{6}\),- Between \(-\frac{7\pi}{6}\) and \(-\frac{\pi}{6}\),- Between \(\frac{\pi}{6}\) and \(\frac{7\pi}{6}\),- Between \(\frac{11\pi}{6}\) and \(2\pi\).

Key Concepts

The Sine Function

The sine function, denoted as \( y = \sin x \), is a fundamental trigonometric function. It is periodic, which means it repeats its pattern over regular intervals. The sine function has a range from \(-1\) to \(1\). This means that for any angle \( x \), the value of \( \sin x \) will always lie between \(-1\) and \(1\).

The sine function is useful in various fields such as physics, engineering, and even in music theory due to its wave-like nature. Key points to remember about the sine function include:

• The sine function reaches its maximum value of \(1\) at \( \frac{\pi}{2} + 2k\pi \), where \( k \) is an integer.
• It reaches its minimum value of \(-1\) at \( \frac{3\pi}{2} + 2k\pi \).
• It crosses the x-axis, meaning \( \sin x = 0 \), at intervals of \( k\pi \).

The wave pattern of the sine function helps to understand how it behaves over any given interval.

Interval Analysis

Analyzing the sine function over a specific interval like \([-2\pi, 2\pi]\) involves checking the function's values across this range. Interval analysis helps in predicting the behavior of the function and determining its specific properties within certain bounds.

In the context of sine, interval analysis lets us determine where sine achieves certain values or falls within specified bands. For instance, knowing \( \sin x = \frac{1}{2} \), only specific angles satisfy this condition. Within the interval \([-2\pi, 2\pi]\), these angles are derived using the sine function's circular nature.

Key steps in interval analysis include:

• Identifying repetitive patterns or cycles within the interval because of the function's periodic nature.
• Solving equations such as \( \sin x = a \) for all potential \( x \) values in the specified range.
• Observing how the function value changes across half and full cycles of sine.

This process is crucial for systematic problem-solving in trigonometry.

Inequalities

When dealing with trigonometric functions, inequalities help identify where a function's value lies relative to a specific number. Here, looking at the inequality \( \sin x > \frac{1}{2} \), we're finding where the sine function's output exceeds \( \frac{1}{2} \) within the interval \([-2\pi, 2\pi]\).

For the given exercise, this occurs in certain arcs or sections between angles where the sine function equals \( \frac{1}{2} \). Specifically, these areas represent when sine increases past the threshold of \( \frac{1}{2} \):

• Between the points \(-\frac{11\pi}{6}\) to \(-\frac{7\pi}{6}\)
• Between points \(-\frac{\pi}{6}\) to \(\frac{\pi}{6}\)
• Between \(\frac{7\pi}{6}\) to \(\frac{11\pi}{6}\)

Inequalities allow us to map out regions of a graph where specific conditions hold true, crucial for understanding the larger behavior of functions across intervals.

Graph Interpretation

Graph interpretation is a pivotal skill in understanding functions like the sine function. It involves reading and understanding graphs to deduce information about the function's behavior. A graph of \( y = \sin x \) over the interval \([-2\pi, 2\pi]\) will show one complete cycle plus additional half cycles.

By analyzing this graph, we can visually confirm the solutions we find mathematically. Here's how:

• Identify where the curve reaches a specific horizontal line (like \( y = \frac{1}{2} \)) to determine the equality condition points.
• Note the intervals where the graph lies above or below this line. These sections correspond to inequalities (e.g., \( y > \frac{1}{2} \) or \( y < \frac{1}{2} \)).
• Observe periodic landmarks like peaks, valleys, and zero crossings, aiding comprehensive interpretation.

Using these interpretations, we validate our mathematical results and gain intuitive insights into the function's behavior.