Problem 52 - Algebra and Trigonometry with Analytic Geometry

Question

Find the period and sketch the graph of the equation. Show the asymptotes. $$ y=\sec \frac{\pi}{8} x $$

Step-by-Step Solution

Answer

The period is 16, with vertical asymptotes at \( x = 4 + 8n \).

Step 1: Understand the Function

The given function is \( y = \sec \frac{\pi}{8} x \). In general, a secant function \( y = \sec(bx) \) has the same period as the corresponding cosine function but with vertical asymptotes where the cosine function is zero.

Step 2: Determine the Period

The period of a basic trigonometric function \( y = \cos(kx) \) is given by \( \frac{2\pi}{|k|} \). For this function, we have \( k = \frac{\pi}{8} \), so the period is \( \frac{2\pi}{\frac{\pi}{8}} = \frac{2\pi \cdot 8}{\pi} = 16 \).

Step 3: Identify the Asymptotes

The vertical asymptotes for \( y = \sec(bx) \) occur where \( \cos(bx) = 0 \). Because \( \cos(bx) \) is zero at \( bx = \frac{\pi}{2} + n\pi \), the asymptotes occur at \( \frac{\pi}{8}x = \frac{\pi}{2} + n\pi \). Solving for \( x \), we get \( x = 4 + 8n \) for integer \( n \).

Step 4: Sketch the Graph

With a period of 16 and asymptotes at \( x = 4 + 8n \), we sketch one period of the secant function starting from \( x = 0 \) to \( x = 16 \). Draw the main curve of the function from the values near an asymptote at each half period interval, at \( x = 4 \) and \( x = 12 \). The graph resembles rolling 'U' (upward) and 'n' (downward) shapes with vertical lines where the asymptotes occur.

Key Concepts

Trigonometric Functions

Trigonometric functions are a key part of mathematics used to describe relationships in right-angled triangles and model periodic phenomena. The secant function, denoted as \( \sec(x) \), is one such trigonometric function. It is the reciprocal of the cosine function, meaning \( \sec(x) = \frac{1}{\cos(x)} \).

However, unlike \( \cos(x) \), which is defined for all real numbers, \( \sec(x) \) is undefined wherever \( \cos(x) = 0 \), because division by zero is undefined in mathematics. As such, there are points called vertical asymptotes in the graph of \( \sec(x) \).

Understanding secant and other trigonometric functions is useful in fields ranging from physics to engineering, where wave patterns and oscillations are modeled.

Periodicity

Periodicity is a fundamental property of trigonometric functions, referring to the regular intervals at which the function's values repeat. For standard trigonometric functions like cosine, sine, and secant, this repeating interval is known as the period.

For instance, the cosine function \( \cos(x) \) has a period of \( 2\pi \). This means that every \( 2\pi \), the function repeats its values. Similarly, in the function \( y = \sec(\frac{\pi}{8} x) \), periodicity is determined by the factor multiplying \( x \). Here, \( k = \frac{\pi}{8} \) stretches the period. To find the period, we use the formula \( \frac{2\pi}{|k|} \), which results in a period of 16 in this case.

Understanding periodicity is key to sketching accurate graphs of trigonometric functions and predicting their behavior over different intervals.

Vertical Asymptotes

Vertical asymptotes are lines corresponding to the values of \( x \) at which a function's value heads towards infinity, implying the function is undefined at these points. For a secant function \( y = \sec(bx) \), vertical asymptotes occur where the base cosine function \( \cos(bx) \) equals zero. This is because dividing by zero makes \( \sec(bx) \) undefined.

In the example \( y = \sec(\frac{\pi}{8} x) \), the asymptotes are located where \( \frac{\pi}{8}x = \frac{\pi}{2} + n\pi \). Solving for \( x \) gives \( x = 4 + 8n \), where \( n \) is any integer.

When sketching the graph, these vertical asymptotes should be clearly marked as they define the "breaks" in the curve of the secant function, indicating abrupt changes in the function's values. Knowing the positions of vertical asymptotes helps to correctly interpret the behavior and complete shape of trigonometric graphs.