Problem 41 - Precalculus: Functions and Graphs

Question

Find the period and sketch the graph of the equation. Show the asymptotes. $$y=\csc 2 x$$

Step-by-Step Solution

Answer

The period of \( y = \csc(2x) \) is \( \pi \) with asymptotes at \( x = \frac{k\pi}{2} \) for integer \( k \).

Step 1: Identify the Base Function

The function given is a transformation of the base cosecant function, given by \( y = \csc x \). We have the function \( y = \csc(2x) \), which involves a horizontal compression of \( y = \csc x \).

Step 2: Determine the Period

The period of the base cosecant function \( y = \csc x \) is \( 2\pi \). Since the function is \( y = \csc(2x) \), the period is adjusted by dividing the base period by the coefficient of \( x \), giving a new period of \( \frac{2\pi}{2} = \pi \).

Step 3: Find Critical Points for One Period

The significant points for one period of the cosecant function include the vertical asymptotes occurring at multiples of \( \frac{\pi}{2} \). For \( y = \csc(2x) \), the vertical asymptotes are at \( 2x = k\pi \), or \( x = \frac{k\pi}{2} \), where \( k \) is an integer.

Step 4: Sketch the Graph Including Asymptotes

Plot the vertical asymptotes at every \( x = \frac{k\pi}{2} \) on the graph. The function \( y = \csc(2x) \) approaches \( \pm \infty \) as \( x \) approaches the asymptotes from the left and right within each period.

Key Concepts

Graphing Trigonometric Functions

Graphing trigonometric functions like the cosecant function can sometimes feel a bit tricky. However, once you get the hang of the process, it becomes much clearer. The cosecant function, denoted as \( y = \csc x \), is the reciprocal of the sine function \( y = \sin x \). When you sketch the graph of a cosecant function, it's important to remember that it derives its behavior from the sine curve. When the sine function is zero, the cosecant will have vertical asymptotes because you cannot divide by zero. This results in portions of the graph that go off towards infinity. Therefore, a good start when graphing the cosecant function is to plot a few cycles of the sine function first. Identify where the sine curve equals zero — that's where you'll place the vertical asymptotes for the cosecant graph. Between these asymptotes, plot the curve segments that approach infinity, inversely following the arcs of the sine curve.

Trigonometric Function Transformation

Transformations are like steps we take to modify the parent trigonometric function into a new one. With the equation \( y = \csc(2x) \), the parent function is \( y = \csc x \), and the main transformation involves the horizontal stretching or compressing of the graph. In our case, the function is horizontally compressed. Here's how it works: the equation \( y = \csc(bx) \) means you adjust the period of the base cosecant function by dividing it by \( |b| \). If \( b > 1 \), like with our \( y = \csc(2x) \), the graph compresses; it squeezes the graph by reducing the length over which one cycle completes. So, in short, the period of the function becomes shorter as compared to \( y = \csc x \). The compressions don't affect the vertical asymptotes outside adjusting the spacing between them, as the zeros of the sine function it's based on reoccur more frequently within the x-axis interval.

Period and Asymptotes

The period and the asymptotes effectively define the behavior and shape of trigonometric functions like the cosecant function. For \( y = \csc x \), the period is the length it takes for the function to start over its pattern, which is \( 2\pi \) for the basic cosecant function. But with \( y = \csc(2x) \), we find that the period is halved to \( \pi \) as we divide \( 2\pi \) by the coefficient \( 2 \) in \( 2x \). As we make a full turn on the x-axis, the pattern of the function repeats every \( \pi \) units now. Asymptotes are where we find the function shooting off to infinity. They happen at the point's sine equals zero, specifically multiples of \( \pi \) for \( y = \csc x \). Because of the compression in \( y = \csc(2x) \), these critical asymptote points are more frequent; they occur at every \( \frac{k\pi}{2} \), where \( k \) is an integer, as we established their occurrence by solving \( 2x = k\pi \). This means every half a cycle (\( \frac{\pi}{2} \)), we will find these asymptotes.