Problem 47 - Algebra and Trigonometry with Analytic Geometry

Question

Find the period and sketch the graph of the equation. Show the asymptotes. $$ y=-\frac{1}{4} \csc \left(\frac{1}{2} x+\frac{\pi}{2}\right) $$

Step-by-Step Solution

Answer

The period is \(4\pi\); asymptotes are \(x = (2n-1)\pi\), \(n\) an integer.

Step 1: Identify the General Form of the Function

The function given is in the form of a cosecant function, which is the reciprocal of the sine function. The general form for \( ext{csc}(bx + c)\) is \( ext{csc}(bx + c)=-\frac{1}{\text{sin}(bx + c)}\). Here, the function is \(-\frac{1}{4} \csc(\frac{1}{2} x + \frac{\pi}{2})\).

Step 2: Determine the Period of the Function

The period of a cosecant function is determined by the coefficient of \('x'\) in the argument. For \(\csc(bx + c)\), the period is given by \(\frac{2\pi}{b}\). In this case, \('b' = \frac{1}{2}\). Substituting, the period is \(\frac{2\pi}{\frac{1}{2}} = 4\pi\).

Step 3: Find the Phase Shift

The phase shift is found by setting the argument of the cosecant function to zero. For \(\frac{1}{2}x + \frac{\pi}{2} = 0\), solve for \(x\). Thus, \(\frac{1}{2}x = -\frac{\pi}{2}\), giving \(x = -\pi\). This indicates the graph is shifted to the left by \(\pi\).

Step 4: Identify the Asymptotes

Asymptotes of the cosecant function occur where the sine function equals zero because the cosecant is the reciprocal of the sine. This happens at \(bx + c = n\pi\), where \(n\) is an integer. For \(\frac{1}{2}x + \frac{\pi}{2} = n\pi\), solving gives \(x = 2n\pi - \pi = (2n-1)\pi\).

Step 5: Sketch the Graph

Consider the amplitude \(-\frac{1}{4}\), which affects the maximum and minimum values of the cosecant curve. Plot the key points over one period \(4\pi\), indicating the vertical asymptotes derived in step 4. The curve will approach but never touch these asymptotes. Reflect this graph horizontally to include multiple periods if necessary.

Key Concepts

Period of Trigonometric Functions

The period of a trigonometric function is an essential characteristic that defines how frequently the function repeats its values. For sine and cosecant functions, the standard period is \(2\pi\). However, when the function includes a coefficient \(b\) in its argument \(bx + c\), the period becomes \(\frac{2\pi}{b}\).
In our example, we're given \(b = \frac{1}{2}\). By substituting this value into the formula, we calculate the period as \(\frac{2\pi}{\frac{1}{2}} = 4\pi\).

• The factor \(\frac{1}{2}\) in this case indicates that the cosecant function stretches horizontally, expanding the regular period.
• This means that the function will complete a full cycle of its behavior every \(4\pi\) units along the x-axis.

Understanding the adjustment in period is vital for graphing these functions correctly across intervals.

Phase Shift

Phase shift refers to horizontal movement of the graph along the x-axis. It shows how far along the x-axis the graph starts from zero. This is particularly important for shifted trigonometric functions.

To find the phase shift, take the argument \(bx + c\) and set it to zero. Solving \(\frac{1}{2}x + \frac{\pi}{2} = 0\) gives us the necessary condition for a phase shift.

• Solve for \(x\): \(\frac{1}{2}x = -\frac{\pi}{2}\), resulting in \(x = -\pi\).

This means the graph of our cosecant function shifts to the left by \(\pi\) units.

This leftward movement effectively repositions all the respective points and features of the function. Grasping phase shift is crucial when involved with graph transformations.

Vertical Asymptotes

Vertical asymptotes in trigonometric functions, especially for the cosecant function, occur where the related sine function equals zero. As the sine approaches zero, the reciprocal function, cosecant, heads toward infinity, thus creating asymptotes.

For the function \(\csc(bx + c)\), vertical asymptotes form at points where \(bx + c = n\pi\) for any integer \(n\).
Solving \(\frac{1}{2}x + \frac{\pi}{2} = n\pi\), we find \(x = 2n\pi - \pi = (2n-1)\pi\).

• These asymptotes divide the graph into distinct sections.
• Since the function cannot cross these lines, they significantly impact the shape of the graph.

Recognizing where these lines occur helps accurately sketch the graph in segments that avoid these vertical lines.

Graphing Trigonometric Functions

Graphing trigonometric functions such as the cosecant involves plotting its critical characteristics, like periods, phase shifts, and asymptotes. Each aspect contributes to the overall sketch and positioning of the function on the graph.

For the equation \(-\frac{1}{4} \csc\left(\frac{1}{2} x + \frac{\pi}{2}\right)\), the negative coefficient \(-\frac{1}{4}\) not only alters the amplitude, affecting how high and low the graph waves go, but also reflects the graph about the x-axis.

• Begin by marking previously calculated periods and shifts on the x-axis.
• Position asymptotes at \((2n-1)\pi\).
• Sketch the wave-like structure of the graph between them, being mindful of the vertical asymptotes which slice the wave into sections.

These strategies form repeating patterns throughout the graph, helping visualize the cosecant function over one or multiple cycles. Understanding how these elements interplay allows you to effectively display the function's behavior.