Problem 46 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 39-62: Find the solutions of the equation that are in the interval \([0,2 \pi\) ). $$ \sec \beta \csc \beta=2 \csc \beta $$

Step-by-Step Solution

Answer

The solutions are \( \beta = \frac{\pi}{3} \) and \( \beta = \frac{5\pi}{3} \).

Step 1: Analyze the Equation

The given equation is \( \sec \beta \csc \beta = 2 \csc \beta \). Our goal is to find values of \( \beta \) within the interval \( [0, 2\pi) \) that satisfy this equation.

Step 2: Simplify the Equation

First, notice that we can divide both sides of the equation by \( \csc \beta \) as long as \( \csc \beta eq 0 \). This simplifies the equation to \( \sec \beta = 2 \).

Step 3: Convert to Cosine

Recall that \( \sec \beta = \frac{1}{\cos \beta} \). So \( \sec \beta = 2 \) becomes \( \frac{1}{\cos \beta} = 2 \). Solving for \( \cos \beta \), we have \( \cos \beta = \frac{1}{2} \).

Step 4: Solve for \( \beta \)

We need to find \( \beta \) such that \( \cos \beta = \frac{1}{2} \) within the interval \([0, 2\pi)\). The cosine function equals \( \frac{1}{2} \) at \( \beta = \frac{\pi}{3} \) and \( \beta = \frac{5\pi}{3} \) within this interval.

Key Concepts

Secant Function

The secant function, denoted as \( \sec \beta \), is one of the trigonometric functions that can be derived from the cosine. It is defined as the reciprocal of the cosine function:

• \( \sec \beta = \frac{1}{\cos \beta} \)

This means that wherever the cosine equals zero, the secant will be undefined, because division by zero is not possible. Naturally, like other trigonometric functions, secant is periodic and repeats its values in regular intervals. In this context, understanding how to handle secant when it appears in trigonometric equations is crucial, as it can often be simplified by utilizing its relationship with cosine. In the original exercise, by equating \( \sec \beta = 2 \), we simplify the equation to find the solutions using principles from cosine.

Cosecant Function

Similar to the secant function, the cosecant function is the reciprocal of another trigonometric function: the sine. Denoted as \( \csc \beta \), it can be expressed as:

• \( \csc \beta = \frac{1}{\sin \beta} \)

The cosecant function becomes undefined whenever the sine equals zero. Because of this, trigonometric equations involving cosecant often need simplification or manipulation to handle points where these undefined values occur, as seen in the initial problem. In the step-by-step solution of the original exercise, \( \csc \beta \) was canceled during the process of solving the equation, highlighting a common technique where restrictions of the divisors themselves are considered to avoid division by zero.

Cosine

The cosine function, denoted as \( \cos \beta \), is a fundamental trigonometric function that describes the x-coordinate of a point on the unit circle for a given angle \( \beta \). It varies between -1 and 1 over a complete cycle, giving it a periodic interval of \( 2\pi \). This periodic behavior is essential when solving trigonometric equations, as it allows for multiple solutions within any given interval, such as \([0, 2\pi)\). In the original problem, simplifying \( \sec \beta = 2 \) to \( \cos \beta = \frac{1}{2} \) is a critical step that uses the relationship between secant and cosine. Because cosine value \( \frac{1}{2} \) is known to correspond to specific angles, solving for \( \beta \) within the desired interval becomes a manageable task.

Intervals in Trigonometry

Intervals in trigonometry define the range of angles under consideration. For periodic functions like sine, cosine, and their reciprocals, understanding intervals is crucial as it determines the number of possible solutions to an equation. Specifically for this problem, we consider the interval \([0, 2\pi)\), which encompasses a full cycle of trigonometric functions, capturing all distinct angle possibilities within one complete revolution of the unit circle. Within any given interval, certain values yield repeated function outputs, allowing us to identify which angles \( \beta \) satisfy the equation \( \sec \beta = 2 \), in this case, the angles \( \frac{\pi}{3} \) and \( \frac{5\pi}{3} \), each occurring once per cycle. This underscores the utility of comprehending interval constraints when addressing trigonometric equations.