Problem 57 - Precalculus: Functions and Graphs

Question

Find the solutions of the equation that are in the interval \([0,2 \pi)\). $$\cot \alpha+\tan \alpha=\csc \alpha \sec \alpha$$

Step-by-Step Solution

Answer

The solutions in the interval \([0, 2\pi)\) are \(\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}\).

Step 1: Rewrite trigonometric functions in terms of sine and cosine

First, express each trigonometric function using sine and cosine. Recall that \( \cot \alpha = \frac{\cos \alpha}{\sin \alpha} \), \( \tan \alpha = \frac{\sin \alpha}{\cos \alpha} \), \( \csc \alpha = \frac{1}{\sin \alpha} \), and \( \sec \alpha = \frac{1}{\cos \alpha} \). The equation becomes:\[ \frac{\cos \alpha}{\sin \alpha} + \frac{\sin \alpha}{\cos \alpha} = \frac{1}{\sin \alpha \cos \alpha} \]

Step 2: Simplify the left-hand side

Use a common denominator to combine the fractions on the left-hand side:\[ \frac{\cos^2 \alpha + \sin^2 \alpha}{\sin \alpha \cos \alpha} \]Remember that \( \cos^2 \alpha + \sin^2 \alpha = 1 \). Thus, the expression simplifies to:\[ \frac{1}{\sin \alpha \cos \alpha} \]

Step 3: Recognize equivalent expressions

Notice that the left-hand side \( \frac{1}{\sin \alpha \cos \alpha} \) is identical to the right-hand side \( \frac{1}{\sin \alpha \cos \alpha} \). This implies that the equation holds true for values of \( \alpha \) where the denominators do not become zero, i.e., where \( \sin \alpha eq 0 \) and \( \cos \alpha eq 0 \).

Step 4: Determine the solutions within the interval

Since the equation holds true everywhere that the trigonometric functions involved are defined, identify exclusions due to the undefined terms. \( \sin \alpha eq 0 \) implies \( \alpha eq n\pi \). \( \cos \alpha eq 0 \) implies \( \alpha eq \frac{(2n+1)\pi}{2} \). Within the interval \([0, 2\pi)\), this leaves:- \( \alpha = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} \).

Step 5: Verify the solutions

Substitute the solution values back into the original equation to ensure they are valid:- For \( \alpha = \frac{\pi}{4} \) and \( \alpha = \frac{5\pi}{4} \): \[ \cot \alpha + \tan \alpha = \csc \alpha \sec \alpha = 2 \]- For \( \alpha = \frac{3\pi}{4} \) and \( \alpha = \frac{7\pi}{4} \): \[ \cot \alpha + \tan \alpha = \csc \alpha \sec \alpha = -2 \]All calculations confirm the equation holds.

Key Concepts

Trigonometric Identities

Trigonometric identities are fundamental relationships among the trigonometric functions. They allow one to express one function in terms of others, which is useful for simplifying equations. In this exercise, we rewrote several trigonometric functions using these identities, converting them into sine and cosine forms:

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

Expressing these functions in terms of sine and cosine helps in simplifying the given equation, allowing us to identify equivalent expressions on both sides of the equation. Knowing these identities is crucial for solving complex trigonometric problems. Always remember, they can simplify your work and help you recognize solutions easily.

Sine and Cosine Functions

Sine and cosine are the primary functions in trigonometry, from which other functions derive. Understanding how they work individually and together is key to solving problems involving trigonometric equations. In our solution, we heavily relied on:

• The identity \( \sin^2 \alpha + \cos^2 \alpha = 1 \), a fundamental formula in trigonometry.
• The behavior of these functions within the interval \([0, 2\pi)\).

The equation structured as \( \frac{1}{\sin \alpha \cos \alpha} \) is solved by recognizing that this form corresponds to the equation's right-hand side. This simplifies finding when equations hold true without evaluating more complex expressions. Mastery of these functions, and their identities, makes complex equations more approachable and less daunting.

Interval Notation

Interval notation is a straightforward way of expressing a set of solutions within a given range. In trigonometry, this is particularly helpful for specifying angles that solve an equation. For this exercise, we are interested in the interval \([0, 2\pi)\), which represents a full rotation without repeating the start boundary.

• The brackets "[" and ")" indicate that 0 is included in the interval but \(2\pi\) is not.
• The solutions that lie within this interval are: \( \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} \).

These points reflect angles where the original equation is satisfied without the denominator becoming zero. Understanding interval notation helps you convey where your solutions are valid, a crucial skill in trigonometry for precise communication of results. Always ensure you know which boundaries are included or excluded with the correct use of brackets.