Problem 69 - Algebra 2

Question

Graph each function in the interval from 0 to 2\(\pi\). $$ y=\csc (-\theta) $$

Step-by-Step Solution

Answer

After identifying the characteristics of the function \(\csc(-\theta)\), it was determined to be undefined at \(0, \pi, 2\pi\) within the given range. Thus, the graph of the function will have vertical asymptotes at these points. The graph is reflective about the y-axis and is in the form of a hyperbola with points at \( \pi / 2, 5\pi /2\) for our interval.

Step 1: Identify characteristics of cosecant function

Firstly, it is important to remember that \(\csc(\theta)\) is equivalent to \(1/\sin(\theta)\). \(\csc(\theta)\) has a period of \(2\pi\). Since we are dealing with \(\csc(-\theta)\), there will be a reflection about the y-axis.

Step 2: Identify Undefined points in interval [0, 2\(\pi\)]

The \(\sin\) function has zeros where the \(\csc\) function will be undefined. So, identifying \(x: \sin(x) = 0\) in the interval [0,2\(\pi\)] is necessary. The \(\sin\) function is zero when \(\theta = 0, \pi, 2\pi\). Hence, \(\csc(-\theta)\) will be undefined at these points.

Step 3: Plot graph of \(\csc(-\theta)\) function

You can start by marking the values where \(\csc(-\theta)\) is undefined, i.e., the vertical asymptotes, on the graph at \(0, \pi, 2\pi\). Then you can mark points where \(\csc(-\theta) = \pm 1\), which is when \(\sin(-\theta) = \pm 1\). When \(\sin(-\theta) = 1\), \(-\theta = \pi / 2 + 2n\pi\), and when \(\sin(-\theta) = -1\), \(-\theta = 3\pi / 2 + 2n\pi\). This happens at \( \pi / 2, 5\pi /2\) for our interval. After these points are marked, one can draw the remaining parts of the graph which would be in the form of a hyperbola.

Key Concepts

Graphing Trigonometric Functions

Understanding how to graph trigonometric functions like the cosecant (\(\csc\)) function helps you visualize their behavior across different intervals. In this case, you'll be working with the function \(y = \csc(-\theta)\). Here are the steps to consider:

• The cosecant function is the reciprocal of the sine function, meaning \(\csc(\theta) = 1/\sin(\theta)\).
• This implies that wherever the sine function crosses zero, the cosecant function will have vertical asymptotes.
• Because the function is \(\csc(-\theta)\), a reflection about the y-axis is involved. This is because negating \(\theta\) reflects the graph across the y-axis.

Using these concepts, you can plot the general shape by determining key points where the function is defined and use vertical asymptotes to guide your sketch.

Periodicity in Trigonometry

Periodicity is a core concept when dealing with trigonometric functions. It means these functions repeat their values in regular intervals. For the cosecant function, the period is \(2\pi\).

• For \(y = \csc(-\theta)\), the periodicity remains \(2\pi\), so the pattern repeats every \(2\pi\) units.
• The reflection over the y-axis does not affect the period but changes the look of each period by reflecting the values.

Understanding periodicity is crucial for correctly sketching these functions over any interval, particularly from \(0\) to \(2\pi\), as required for this exercise.

Vertical Asymptotes

Vertical asymptotes occur where the function becomes undefined. In the case of the \(\csc(\theta)\) function, they happen when \(\sin(\theta) = 0\), as division by zero is undefined.

• For \(\csc(-\theta)\), focus on the zeros of the sine function: \(\theta = 0, \pi, 2\pi\).
• These points translate into vertical asymptotes on the graph at \(0, \pi,\) and \(2\pi\).

Once these asymptotes are plotted, sketch the cosecant curve around them. It appears as arch-like curves extending towards infinity, and between asymptotes, they reverse direction, displaying the hyperbolic shape typical of cosecant functions.