Problem 45 - Algebra and Trigonometry with Analytic Geometry

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

The period is \(\pi\) and asymptotes occur at \(x = \frac{n\pi}{2} - \frac{\pi}{4}\).

Step 1: Identify the Function Type and Form

The given function is a cosecant function, specifically a scaled and shifted form of the standard cosecant function. The general form of a cosecant function is \(y = a \csc(bx - c)\). The function provided is \(y = 2 \csc\left(2x + \frac{\pi}{2}\right)\).

Step 2: Determine the Period of the Cosecant Function

The period of a general sine or cosecant function, \(y = a \csc(bx - c)\), is \(\frac{2\pi}{b}\). Here, \(b = 2\), so the period is \(\frac{2\pi}{2} = \pi\).

Step 3: Find the Phase Shift

The phase shift is determined by \(\frac{c}{b}\) in the equation \(y = a \csc(bx-c)\). Here, \(c = -\frac{\pi}{2}\) (note the sign change from the general form), so the phase shift is \(-\frac{\pi}{2}/2 = -\frac{\pi}{4}\). This means the graph is shifted to the left by \(\frac{\pi}{4}\).

Step 4: Identify the Vertical Stretch/Compression

The coefficient \(a = 2\) causes a vertical stretch of the cosecant function by a factor of 2. This means the \(y\)-values of the peaks and valleys of the graph are twice as far from the \(x\)-axis compared to a standard \(\csc(x)\) function.

Step 5: Determine the Asymptotes

The asymptotes of the \(\csc(x)\) function occur at the zeros of its corresponding \(\sin(x)\) function, because \(\csc(x) = \frac{1}{\sin(x)}\). Therefore, set the argument of \(\csc\) equal to multiples of \(\pi\), i.e., \(2x + \frac{\pi}{2} = n\pi\), where \(n\) is an integer. Thus, \(2x = n\pi - \frac{\pi}{2}\), giving the asymptotes at \(x = \frac{n\pi}{2} - \frac{\pi}{4}\).

Step 6: Sketch the Graph

Plot the asymptotes found in Step 5. Draw the basic shape of the \(y = 2\csc(2x)\) function, noting that the curve moves between the asymptotes peaking at \(2\) and \(-2\). Graphically adjust based on the phase shift found in Step 3.

Key Concepts

Period of Trigonometric Functions

In trigonometry, the period of a function is the interval over which it completes one full cycle before repeating itself. The concept of period is particularly important for sinusoidal functions like sine, cosine, and their reciprocals, such as cosecant.

For the general form of the cosecant function, which is given by \[ y = a \csc(bx - c) \]the period is determined by the coefficient \( b \). The formula for calculating the period of a cosecant function is \( \frac{2\pi}{b} \).

To find the period of our given function, \( y = 2 \csc(2x + \frac{\pi}{2}) \), we identify that \( b = 2 \), thus the period is:

• \( \frac{2\pi}{2} = \pi \)

This means that every \( \pi \) units along the x-axis, the function completes one full oscillation and starts to repeat. Understanding the period enables us to predict the behavior of the function over any interval. This knowledge is key when sketching graphs and solving related trigonometric problems.

Asymptotes in Trigonometry

Vertical asymptotes are critical features of trigonometric functions like the cosecant. These are essentially "invisible lines" that the function approaches but never actually touches or intersects.

For the cosecant function, asymptotes occur where its base function, sine, equals zero. This is due to the fact that \[ \csc(x) = \frac{1}{\sin(x)} \]and division by zero is undefined. Therefore, asymptotes exist at these points.

In the case of \( y = 2 \csc\left(2x + \frac{\pi}{2}\right) \),we set the argument of the cosecant to equal multiples of \( \pi \) to find the discontinuities:

• \( 2x + \frac{\pi}{2} = n\pi \) (where \( n \) is an integer)
• Solving for \( x \), we get: \( x = \frac{n\pi}{2} - \frac{\pi}{4} \)

These are the locations of vertical asymptotes for the graph. As you sketch the graph, remember to mark these limits, as the curve will arch upwards and downwards approaching each asymptote but never crossing it.

Phase Shift in Trigonometric Functions

Phase shift refers to the horizontal movement of a trigonometric graph along the x-axis. This shift is determined using the formula \[ \frac{c}{b} \]where \( c \) is the horizontal shift parameter in the function's argument and \( b \) is the coefficient of \( x \).

For the function given:\( y = 2 \csc\left(2x + \frac{\pi}{2}\right) \),we have:

• \( c = -\frac{\pi}{2} \) (taking the sign into account from the form \( y = a \csc(bx - c) \))
• \( b = 2 \)

Thus, the phase shift is:

• \( -\frac{\pi}{2} / 2 = -\frac{\pi}{4} \)

This indicates that the entire graph is shifted to the left by \( \frac{\pi}{4} \), as negative values of phase shift move the graph leftward. When drawing the graph, this phase shift alters the original starting point of the cycle, ensuring the correct alignment with the periodic continuation.