Problem 2 - Algebra 2

Question

Find all solutions of each equation for the given interval. \(2 \sin ^{2} \theta-1=0 ; 90^{\circ}<\theta<270^{\circ}\)

Step-by-Step Solution

Answer

Solutions are \(\theta = 135^{\circ}\) and \(\theta = 225^{\circ}\).

Step 1: Rearrange the Equation

Start with the equation \(2 \sin^2 \theta - 1 = 0\). Add 1 to both sides to isolate the term with the sine function: \(2 \sin^2 \theta = 1\).

Step 2: Isolate \(\sin \theta\)

Divide both sides by 2 to solve for \(\sin^2 \theta\): \(\sin^2 \theta = \frac{1}{2}\).

Step 3: Solve for \(\sin \theta\)

Take the square root of both sides: \(\sin \theta = \pm \frac{1}{\sqrt{2}}\). Recall that \(\sin \theta = \frac{1}{\sqrt{2}}\) is equivalent to \(\sin \theta = \frac{\sqrt{2}}{2}\). Thus, we have \(\sin \theta = \pm \frac{\sqrt{2}}{2}\).

Step 4: Determine Possible Angles for \(\sin \theta = \pm \frac{\sqrt{2}}{2}\)

The general solutions for \(\sin \theta = \frac{\sqrt{2}}{2}\) are \(\theta = 45^{\circ} + 360^{\circ}k\) or \(\theta = 135^{\circ} + 360^{\circ}k\), and for \(\sin \theta = -\frac{\sqrt{2}}{2}\) they are \(\theta = 225^{\circ} + 360^{\circ}k\) or \(\theta = 315^{\circ} + 360^{\circ}k\), where \(k\) is an integer.

Step 5: Find Angles within the Given Interval

Considering the interval \(90^{\circ} < \theta < 270^{\circ}\), the angles that fall within this range are \(\theta = 135^{\circ}\) and \(\theta = 225^{\circ}\). These solutions correspond to \(\sin \theta = \frac{\sqrt{2}}{2}\) and \(\sin \theta = -\frac{\sqrt{2}}{2}\), respectively.

Key Concepts

Interval Notation

Interval notation is a math method used to describe a set of numbers between two endpoints. It's helpful for succinctly capturing the range in which a variable, like an angle in a trigonometric equation, can exist.

There are two main types of intervals:

• **Closed Interval**: Includes its endpoints, denoted by square brackets: \([a, b]\) means "from \(a\) to \(b\), inclusive."
• **Open Interval**: Excludes its endpoints, denoted by parentheses: \((a, b)\) means "from \(a\) to \(b\), not including \(a\) or \(b\)."

In our exercise, we have the interval \(90^{\circ} < \theta < 270^{\circ}\), which can be represented in interval notation as \((90^{\circ}, 270^{\circ})\). This indicates the angles we consider must be greater than \(90^{\circ}\) and less than \(270^{\circ}\), explicitly excluding the endpoints themselves.

Understanding how to interpret and use interval notation is crucial for solving trigonometric equations, as it defines the span of viable solutions.

Sine Function

The sine function is a fundamental concept in trigonometry. It represents the ratio of the length of the opposite side to the hypotenuse in a right triangle when considering an angle.

Characteristics of the sine function include:

• **Range**: The sine function outputs values between -1 and 1 inclusive. Every angle in the unit circle contributes to generating these values.
• **Periodicity**: Sine is periodic with a period of \(360^{\circ}\) (or \(2\pi\) radians). This means its pattern repeats every \(360^{\circ}\).

In our problem, solving \(2 \sin^2 \theta - 1 = 0\) led us to the values \(\sin \theta = \pm \frac{\sqrt{2}}{2}\). Knowing when these sine values occur helps us determine solutions to trigonometric equations within a specific interval. For example:

• \(\sin \theta = \frac{\sqrt{2}}{2}\) at angles such as \(45^{\circ}\), \(135^{\circ}\), etc.
• \(\sin \theta = -\frac{\sqrt{2}}{2}\) at angles like \(225^{\circ}\), \(315^{\circ}\), etc.

Mastering the sine function is key for navigating through trigonometric problems effectively.

Angle Solutions

When tasked with finding angle solutions for a trigonometric equation, like in our exercise, it's important to follow a systematic approach. This involves identifying possible angles that satisfy the equation, within the given interval.

To solve \(2 \sin^2 \theta - 1 = 0\):

1. Rearrange the equation to isolate the sine function: this gives \(\sin \theta = \pm \frac{\sqrt{2}}{2}\).
2. Identify general solutions for these sine values, knowing their common angles from the unit circle.
3. Filter these angles to find which fall within the interval \((90^{\circ}, 270^{\circ})\). This is crucial for getting the correct angle solutions.

In our problem, the angles that solve the equation in the desired interval are \(\theta = 135^{\circ}\) and \(\theta = 225^{\circ}\). These are derived from the known specific angles where sine has values of \(\frac{\sqrt{2}}{2}\) and \(-\frac{\sqrt{2}}{2}\) and applying the interval constraint to filter them properly.

Angle solutions are thus crucial for ensuring all conditions of the trigonometric problem are met, giving you the correct and complete answer.