Problem 25 - Precalculus: Functions and Graphs

Question

Find all solutions of the equation. $$\sec ^{2} \alpha-4=0$$

Step-by-Step Solution

Answer

Solutions are \( \alpha = \frac{\pi}{3} + 2n\pi, \frac{5\pi}{3} + 2n\pi, \frac{2\pi}{3} + 2n\pi, \frac{4\pi}{3} + 2n\pi \).

Step 1: Set the Equation to Zero

The given equation is \( \sec^2 \alpha - 4 = 0 \). First, we need to make this equation equal to zero to find the solutions.

Step 2: Isolate \( \sec^2 \alpha \)

From the equation \( \sec^2 \alpha - 4 = 0 \), add 4 to both sides of the equation to get \( \sec^2 \alpha = 4 \).

Step 3: Solve for \( \sec \alpha \)

Taking the square root of both sides, we obtain two possible solutions for \( \sec \alpha \): \( \sec \alpha = \pm 2 \).

Step 4: Convert to \( \cos \alpha \)

Recall the reciprocal identity \( \sec \alpha = \frac{1}{\cos \alpha} \). Thus, \( \sec \alpha = 2 \) corresponds to \( \cos \alpha = \frac{1}{2} \) and \( \sec \alpha = -2 \) corresponds to \( \cos \alpha = -\frac{1}{2} \).

Step 5: Determine \( \alpha \) for \( \cos \alpha = \frac{1}{2} \)

From trigonometric values, \( \cos \alpha = \frac{1}{2} \) occurs at \( \alpha = \frac{\pi}{3} \) and \( \alpha = \frac{5\pi}{3} \), and their general solutions are \( \alpha = \frac{\pi}{3} + 2n\pi \) and \( \alpha = \frac{5\pi}{3} + 2n\pi \) where \( n \) is any integer.

Step 6: Determine \( \alpha \) for \( \cos \alpha = -\frac{1}{2} \)

For \( \cos \alpha = -\frac{1}{2} \), the corresponding angles are \( \alpha = \frac{2\pi}{3} \) and \( \alpha = \frac{4\pi}{3} \). The general solutions are \( \alpha = \frac{2\pi}{3} + 2n\pi \) and \( \alpha = \frac{4\pi}{3} + 2n\pi \) where \( n \) is any integer.

Key Concepts

Secant Function

The secant function, denoted as \( \sec \alpha \), is a trigonometric function closely related to the cosine function. Specifically, it is the reciprocal of cosine, meaning \( \sec \alpha = \frac{1}{\cos \alpha} \). This means that when the cosine of an angle \( \alpha \) is known, the secant can be easily derived by taking the reciprocal.

The secant function is not as commonly used as sine, cosine, or tangent in everyday applications, but it plays an important role in calculus and higher-level mathematics. One notable feature of the secant function is that it is undefined whenever the cosine function is zero, which occurs at angles like \( \frac{\pi}{2} \), \( \frac{3\pi}{2} \), etc. This is because you cannot divide by zero.

Understanding the secant function is crucial for solving equations where it appears, particularly when simplifying expressions or converting to more familiar trigonometric functions.

Reciprocal Trigonometric Identities

Reciprocal identities are an essential part of trigonometry that help in expressing one trigonometric function in terms of another by using their reciprocal relationships. These identities simplify many equations and are fundamental when working with various trigonometric equations.

For example, some critical reciprocal identities include:

• \( \sec \alpha = \frac{1}{\cos \alpha} \)
• \( \csc \alpha = \frac{1}{\sin \alpha} \)
• \( \cot \alpha = \frac{1}{\tan \alpha} \)

These identities are particularly useful when you need to convert a less familiar function into a more familiar one, such as converting secant to cosine like we did in the original exercise. By using reciprocal identities, complex trigonometric equations can often be simplified, which makes solving them more manageable.

General Solutions

In trigonometry, finding general solutions to equations gives us all possible angle measures that satisfy the given equation. This is important because trigonometric functions are periodic, meaning they repeat their values at regular intervals.

For instance, in the exercise, we identified configurations for \( \cos \alpha = \frac{1}{2} \) as \( \alpha = \frac{\pi}{3} \) and \( \alpha = \frac{5\pi}{3} \). The general solutions for these are expressed as:

• \( \alpha = \frac{\pi}{3} + 2n\pi \)
• \( \alpha = \frac{5\pi}{3} + 2n\pi \)

where \( n \) is an integer. Here, \( 2n\pi \) accounts for the periodic nature of cosine, as every full circle \( 2\pi \) leads back to the original cosine value.

This approach not only provides specific solutions but also encompasses patterns for all other possible angles that fit the equation.

Trigonometric Values

Trigonometric values are specific values of trigonometric functions for common angle measures such as \( 0, \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}, \frac{\pi}{2}, \ldots \). Knowing these values is essential when solving trigonometric equations since they are often the key to finding specific solutions.

For example, in the original problem, knowledge of classic trigonometric values assists in identifying the angles \( \alpha \) that satisfy \( \cos \alpha = \frac{1}{2} \) or \( \cos \alpha = -\frac{1}{2} \). Some common trigonometric value pairs to remember include:

• \( \cos \frac{\pi}{3} = \frac{1}{2} \)
• \( \cos \frac{2\pi}{3} = -\frac{1}{2} \)
• \( \cos \frac{5\pi}{3} = \frac{1}{2} \)
• \( \cos \frac{4\pi}{3} = -\frac{1}{2} \)

Remembering these values can enhance problem-solving speed and accuracy in trigonometric exercises, making the process more efficient overall.