Problem 32 - Algebra 2

Question

Solve each equation for \(0 \leq \theta<2 \pi\). $$ \sin \theta=-\sin \theta \cos \theta $$

Step-by-Step Solution

Answer

The solutions to the original equation \(\sin \theta=-\sin \theta \cos \theta\) for \(0 \leq \theta<2 \pi\) are \( \theta = 0, \pi\).

Step 1: Analyze and Simplify the Equation

Observe the equation \(\sin \theta=-\sin \theta \cos \theta \). We can simplify this equation by moving all terms to one side: \( \sin \theta + \sin \theta \cos \theta = 0 \). From here, factor out a \( \sin \theta \) to simplify further, giving: \( \sin \theta (1+\cos \theta ) = 0 \).

Step 2: Set Each Factor Equal to Zero

By setting each factor equal to zero, we get 2 separate equations: \( \sin \theta = 0 \) and \(1 + \cos \theta = 0 \). Solving \(\cos \theta = -1\) gives multiple possible solutions.

Step 3: Solve Each Equation Separately

The solutions of \( \sin \theta = 0 \) are \( \theta = 0, \pi \) since those are the angles for which the sine function is equal to zero while within our given range \(0 \leq \theta < 2\pi\). For the equation \(\cos \theta = -1\), the solution is \( \theta = \pi \) because cosine of \(\pi\) is -1.

Step 4: Gather all the Solutions

The solution to the equation \(\sin \theta=-\sin \theta \cos \theta \) for \(0 \leq \theta<2 \pi\) are found by combining the solutions from both equations, which gives us \( \theta = 0, \pi, \pi \). Here, we can see that \( \pi\) is repeated.

Key Concepts

Sine Function

The sine function, denoted as \( \sin \theta \), is an essential part of trigonometry. It represents the vertical component of a point on the unit circle, which is a circle with a radius of one. When given an angle \( \theta \), the sine function returns the y-coordinate of the corresponding point on the unit circle.

In our problem, we needed to solve the equation \( \sin \theta = 0 \). This means we're looking for angles where the sine of those angles is zero. Since the sine function corresponds to the y-coordinate, we know that \( \sin \theta = 0 \) when that point is on the x-axis of the unit circle.

• The angles that give \( \sin \theta = 0 \) within the range \(0 \leq \theta < 2\pi\) are \( \theta = 0 \) and \( \theta = \pi \).

Recognizing these angles is crucial, as it helps pinpoint the exact points where the waveform crosses the axis. These crossing points are often referred to as 'zeros' of the sine function.

Cosine Function

The cosine function, denoted as \( \cos \theta \), is another pillar of trigonometry. It gives the horizontal component of the same point on the unit circle mentioned for sine. When we evaluate \( \cos \theta \), we find the x-coordinate for the corresponding angle \( \theta \).

In this exercise, we encountered \( \cos \theta = -1 \). This means we're looking for the angle where the x-coordinate is -1. Only one angle within the standard range \(0 \leq \theta < 2\pi\) has a cosine of -1, and that is \( \theta = \pi \).

• Thus, \( \cos \pi = -1 \).

Understanding when the cosine function reaches specific values like -1 is an essential concept. It helps identify symmetry and cosine function characteristics, which are vital in solving trigonometric equations and graphing these functions.

Factorization

Factorization is a mathematical process used to break down expressions into simpler components that, when multiplied together, result in the original expression. In trigonometry, this skill is invaluable, especially when manipulating equations. The given equation \( \sin \theta + \sin \theta \cos \theta = 0 \) was made more manageable through factorization.

We factored \( \sin \theta \) out of the equation:

• \( \sin \theta (1 + \cos \theta) = 0 \).

This process allows us to consider each individual factor separately: either \( \sin \theta = 0 \) or \( 1 + \cos \theta = 0 \). By examining factors individually, we simplify complex problems and isolate solutions more easily.

Factorization is similar to breaking down a task into smaller, more manageable steps. It's a vital tool not just in algebra but is equally useful in trigonometry, as it helps solve equations efficiently while building deeper understanding of the trigonometric relationships at play.