Problem 4 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 1-38: Find all solutions of the equation. $$ \cot \alpha=-\frac{1}{\sqrt{3}} $$

Step-by-Step Solution

Answer

The solutions are \( \alpha = \frac{2\pi}{3} + n\pi \) and \( \alpha = \frac{5\pi}{3} + n\pi \).

Step 1: Understand the Cotangent Function

The cotangent function, \( \cot \alpha \), is the reciprocal of the tangent function: \( \cot \alpha = \frac{1}{\tan \alpha} \). Given \( \cot \alpha = -\frac{1}{\sqrt{3}} \), we have \( \tan \alpha = -\sqrt{3} \).

Step 2: Identify Reference Angles

Find an angle \( \alpha \) such that \( \tan \alpha = -\sqrt{3} \). We know that \( \tan \frac{\pi}{3} = \sqrt{3} \), so the reference angles where the tangent is negative are in the second and fourth quadrants.

Step 3: Determine General Solutions

Since tangent is negative in the second and fourth quadrants: 1. The second quadrant angle: \( \alpha = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \). 2. The fourth quadrant angle: \( \alpha = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3} \).

Step 4: Write General Solution Using Periodicity

The general solution for \( \alpha \) uses the period of the tangent function, which is \( \pi \). So, the solutions are \( \alpha = \frac{2\pi}{3} + n\pi \) and \( \alpha = \frac{5\pi}{3} + n\pi \), where \( n \) is any integer.

Key Concepts

Cotangent Function

The cotangent function, represented as \( \cot \alpha \), is an important trigonometric function that you might encounter during solving trigonometric equations. It is defined as the reciprocal of the tangent function, which means:

• \( \cot \alpha = \frac{1}{\tan \alpha} \)

This implies that if you know the value of \( \tan \alpha \), you can calculate \( \cot \alpha \) by simply taking the reciprocal. Likewise, if you know \( \cot \alpha \), you can find \( \tan \alpha \) by taking the reciprocal. For example, if \( \cot \alpha = -\frac{1}{\sqrt{3}} \), then \( \tan \alpha = -\sqrt{3} \). Understanding the cotangent function's relationship to the tangent function is key to solving trigonometric equations involving \( \cot \).

Reference Angles

Reference angles are a powerful tool in trigonometry for determining the angles that will give the same sine, cosine, or tangent values. When solving trigonometric equations like \( \tan \alpha = -\sqrt{3} \), recognizing the reference angle can help in finding specific solutions.

• First, determine where the tangent function has a value of \( \sqrt{3} \). This happens at \( \frac{\pi}{3} \).
• Since \( -\sqrt{3} \) indicates a negative tangent, we consider the quadrants where the tangent is negative: the second and fourth quadrants.

In this situation, the reference angles help us quickly find the angles \( \alpha \) for which \( \tan \alpha = -\sqrt{3} \), specifically \( \frac{2\pi}{3} \) and \( \frac{5\pi}{3} \). These angles use the reference angle \( \frac{\pi}{3} \) but are adapted to the quadrants where the tangent is negative.

General Solution

In trigonometry, solving an equation means finding all angles that satisfy the equation within a given domain. Because trigonometric functions are periodic, they repeat their values at specific intervals known as the period.

• For the tangent and cotangent functions, the period is \( \pi \), meaning they repeat every \( \pi \) radians.

To find the general solution to \( \tan \alpha = -\sqrt{3} \), you should account for all possible angles \( \alpha \) that satisfy the equation. Start with specific solutions, such as \( \frac{2\pi}{3} \) and \( \frac{5\pi}{3} \), and then incorporate the function's periodicity by adding multiples of the period:

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

Here, \( n \) represents any integer, allowing for both positive and negative shifts of \( \pi \) radians, providing the infinite set of solutions to the equation.

Tangent Function

The tangent function, \( \tan \alpha \), is another central player in trigonometry, known for its distinct properties:

• It is defined as the ratio of the opposite side to the adjacent side in a right-angled triangle.
• Tangent values are positive in the first and third quadrants and negative in the second and fourth quadrants.
• Its period is \( \pi \), meaning the function repeats its pattern every \( \pi \) radians.

When solving our equation where \( \tan \alpha = -\sqrt{3} \), recognizing the behavior of tangent across its cycle helps identify which quadrants and angles yield negative values. Remember that at \( \frac{\pi}{3} \), the tangent is \( \sqrt{3} \), but since we want \( -\sqrt{3} \), we focus on areas where \( \tan \) shifts sign. This meticulous approach illuminates the values of \( \alpha \) needed for solutions and enhances understanding of trigonometric functions.