Problem 47 - Algebra 2

Question

Graph each function in the interval from 0 to 2\(\pi\) $$ y=\sec \frac{1}{4} \theta $$

Step-by-Step Solution

Answer

The graph of \(y = \sec(\frac{1}{4}\theta)\) from 0 to \(2\pi\) will be a wave-like pattern similar to the graph of the cosine function but with curves extending from 1 to positive infinity or from -1 to negative infinity rather than oscillating between 1 and -1. It will have a period of \(8\pi\) due to the \(\frac{1}{4}\) stretch in the function.

Step 1: Understand the Cosine Function

Start with an understanding of how a cosine function looks like. The function \(cos(\theta)\) is a wave-like function that oscillates between -1 and 1. At \(\theta = 0\) and \(\theta = 2\pi\), \(cos(\theta) = 1\).recognize that secant function is the reciprocal of the cosine function, i.e., \(sec(x) = 1/cos(x)\).

Step 2: Graph the Reciprocal Function

In order to graph \(y = \sec(\frac{1}{4}\theta)\), one has to take the reciprocal of each \(cos(\frac{1}{4}\theta)\) value. When the cosine function is equal to 1 or -1, the secant function will also be 1 or -1. However, as cosine of an angle approaches 0, secant of that angle approaches infinity or negative infinity resulting in vertical asymptotes in the secant function where the cosine function crosses the x-axis.

Step 3: Apply the Horizontal Stretch

Then, apply the horizontal stretch to the secant function. This means each point on the graph of the secant function will be moved to the right by a factor of 4. This transforms the original function \(y = sec(\theta)\) to \(y = sec(\frac{1}{4}\theta)\). The period of the secant function will increase from \(2\pi\) to \(8\pi\) due to this horizontal stretch.

Key Concepts

Secant Function

The secant function is a trigonometric concept that is closely related to the cosine function. Essentially, the secant function is the reciprocal of the cosine function. This means for any angle \( \theta \), the secant of that angle, expressed as \( \sec(\theta) \), is calculated as \( 1/\cos(\theta) \). This relationship is pivotal in understanding how the secant function behaves.

• When \( \cos(\theta) = 1 \), \( \sec(\theta) \) also equals 1, as you are dividing 1 by 1.
• Similarly, when \( \cos(\theta) = -1 \), \( \sec(\theta) = -1 \).
• As \( \cos(\theta) \) approaches zero, \( \sec(\theta) \) becomes undefined, resulting in a vertical asymptote at those points.

This unique behavior of vertical asymptotes occurs because division by zero is undefined in mathematics, which causes the secant graph to spike upward or downward near these values. Recognizing this can help in graphing and understanding how the secant function behaves in comparison to its cosine counterpart.

Cosine Function

The cosine function is one of the foundational trigonometric functions. It features a wave-like pattern that oscillates between -1 and 1. This characteristic oscillation is evident in the function \( y = \cos(\theta) \), where \( \theta \) measures the angle in radians.

• At \( \theta = 0 \) and \( \theta = 2\pi \), the cosine value is equal to 1.
• At \( \theta = \pi \), the cosine value drops down to -1.
• The cosine function smoothly transitions between these extremes, delivering a continuous wave.

Understanding this basic pattern is crucial for grasping the function's reciprocals like the secant function. Given that the cosine function serves as the denominator in \( \sec(\theta) = 1/\cos(\theta) \), its behavior directly influences the features of the secant function, such as its vertical asymptotes.

Graphing Trigonometric Functions

Graphing trigonometric functions like the secant and cosine may seem intimidating, but following systematic steps can simplify the process. For the function \( y = \sec(\frac{1}{4} \theta) \), the initial approach is to consider its base cosine function \( \cos(\frac{1}{4} \theta) \).Begin by sketching the cosine graph over one typical period from 0 to \( 2\pi \).

• The graph will visibly oscillate between -1 and 1, crossing the x-axis at \( \theta = \pi/2 \) and \( \theta = 3\pi/2 \), where vertical asymptotes will appear for the secant graph.
• These crossover points shift in the secant graph because the reciprocal at zero becomes undefined, stretching toward infinity.
• With the introduction of \( \frac{1}{4} \theta \), there is a horizontal stretch, effectively increasing the period of the function from \( 2\pi \) to \( 8\pi \).

Adjust your graphing points to accommodate this stretching, showing how each peak or trough is repositioned along the x-axis. This careful manipulation demonstrates how reciprocal and transformations influence the graphical outputs.