Problem 45 - Precalculus: Functions and Graphs

Question

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

Step-by-Step Solution

Answer

Period: \(\pi\); Asymptotes: \(x = \frac{n\pi}{2} - \frac{\pi}{4}\).

Step 1: Identify the function

The given function is \(y = 2 \csc\left(2x + \frac{\pi}{2}\right)\). This is a transformation of the cosecant function, which is reciprocal to the sine function.

Step 2: Find the period of the base function

For \(y = \csc(x)\), the period is \(2\pi\). The period of \(\csc(kx)\) is calculated as \(\frac{2\pi}{k}\). Here, \(k = 2\), so the period is \(\frac{2\pi}{2} = \pi\).

Step 3: Determine phase shift and vertical stretch

The equation \(2x + \frac{\pi}{2}\) inside the cosecant function indicates a horizontal phase shift of \(-\frac{\pi}{4}\) to the right, calculated as \(-\frac{\text{horizontal shift}}{k}\). The number '2' outside introduces a vertical stretch, but it does not affect the period or asymptotes.

Step 4: Find the asymptotes

The function \(\csc(x)\) has vertical asymptotes where \(\sin(x) = 0\). Therefore, for \(y = 2\csc\left(2x + \frac{\pi}{2}\right)\), set \(\sin\left(2x + \frac{\pi}{2}\right) = 0\). Solve for \(2x + \frac{\pi}{2} = n\pi\), resulting in \(x = \frac{n\pi}{2} - \frac{\pi}{4}\) for integers \(n\).

Step 5: Sketch the graph

Sketch the period \(\pi\) from \(-\frac{\pi}{4}\) to \(\pi - \frac{\pi}{4}\). Within this interval, draw the vertical asymptotes at \(x = -\frac{\pi}{4}\) and \(x = \frac{3\pi}{4}\). Draw the graph of \(y = 2\csc(2x + \frac{\pi}{2})\), making sure it resembles a transformation of the basic \(\csc\) function, accounting for the vertical stretch and phase shift.

Key Concepts

Cosecant Function

The cosecant function, denoted as \( \csc(x) \), is the reciprocal of the sine function. This means, it is defined as \( \csc(x) = \frac{1}{\sin(x)} \), wherever the sine function is not zero. As a result, whenever sine touches zero, cosecant goes to infinity. This creates key features in the cosecant's graph: vertical asymptotes at points where \( \sin(x) = 0 \).
In its basic form, \( y = \csc(x) \) creates a series of U-shapes and upside-down U-shapes that repeat every \( 2\pi \), which is its natural period.
Because \( \csc(x) \) is undefined where \( \sin(x) = 0 \), its graph is broken into segments, each separated by vertical asymptotes. Understanding the basic properties of the cosecant function is crucial as transformations will simply shift or stretch these features.

Trigonometric Graph Transformations

Trigonometric functions can undergo transformations that alter their appearance on a graph without changing the fundamental nature of the function. These transformations include scaling, shifting, stretching, and reflecting.
For the function \( y = 2 \csc(2x + \frac{\pi}{2}) \):

• Amplitude/Vertical Stretch: The leading 2 outside the cosecant function means the graph is scaled vertically. Each point is multiplied by 2, making peaks and valleys twice as tall.
• Phase Shift: Inside the function, \( 2x + \frac{\pi}{2} \) indicates a horizontal shift. We compute it as \( -\frac{\frac{\pi}{2}}{2} = -\frac{\pi}{4} \), moving the graph \( \frac{\pi}{4} \) units to the right.
• Horizontal Stretch/Compression: The factor of 2 scaling the \( x \) changes the frequency, compressing its natural period down to \( \pi \) from \( 2\pi \).

This equation's final form is affected by these transformations, making it essential to account for each when sketching the graph.

Vertical Asymptotes

Vertical asymptotes occur in functions where the function approaches infinity as the variable approaches a specific value. In trigonometric functions like the cosecant, these asymptotes occur where the sine function equals zero.
For \( y = 2 \csc(2x + \frac{\pi}{2}) \), vertical asymptotes are critical. We set \( \sin(2x + \frac{\pi}{2}) = 0 \) to locate these asymptotes. Solving the equation \( 2x + \frac{\pi}{2} = n\pi \) results in solutions for \( x = \frac{n\pi}{2} - \frac{\pi}{4} \), where \( n \) is any integer.
This means the graph will have vertical asymptotes at points like \( x = -\frac{\pi}{4} \) and \( x = \frac{3\pi}{4} \), around which the graph of the cosecant function will come dangerously close but never actually touch.

Period of Trigonometric Functions

In trigonometry, the period of a function is the length of the smallest interval over which the function repeats itself. For most basic trigonometric functions like \( \sin(x) \), \( \cos(x) \), and \( \csc(x) \), the natural period is \( 2\pi \).
However, various transformations can alter this period. For example, in a function like \( y = \csc(kx) \), the period becomes \( \frac{2\pi}{k} \).
For our given function \( y = 2 \csc(2x + \frac{\pi}{2}) \), the value of \( k \) is 2. Therefore, the period becomes \( \frac{2\pi}{2} = \pi \). This means the entire wave pattern of the cosecant will repeat itself every \( \pi \) units along the \( x \)-axis.
Understanding the period is vital not only for graphing but also in applications requiring cyclic behavior emulation, like signal processing or seasonal predictions.