Problem 46 - Algebra and Trigonometry

Question

Involve trigonometric equations quadratic in form. Solve each equation on the interval \([0,2 \pi)\) $$ \cos ^{2} \theta-1=0 $$

Step-by-Step Solution

Answer

\(\theta = 0, \pi, 2 \pi\)

Step 1: Rewrite the Equation

We will first rewrite the equation to isolate the cosine function. Given \(\cos^2 \theta - 1 = 0\), it can be rearranged as \(\cos^2 \theta = 1\).

Step 2: Take square root

Next, we need to remove the square on the cosine function by taking the square root of both sides. This yields \(cos \theta = \sqrt{1}\) and \(cos \theta = -\sqrt{1}\). Hence, \(cos \theta = 1\) and \(cos \theta = -1\).

Step 3: Find the angle

The cosine function gives 1 at an angle of 0 and \(2 \pi\) and it gives -1 at an angle of \(\pi\) in the given interval [0, \(2 \pi\)]. Therefore, the solutions are \(\theta = 0, \pi, 2 \pi\).

Key Concepts

Understanding the Cosine Function

The cosine function is a fundamental part of trigonometry and is closely connected to the unit circle. When dealing with the cosine of an angle \( \theta \), it's important to remember:

• Cosine measures the horizontal distance from the origin to the point on the unit circle.
• Its values range between -1 and 1.

In trigonometric equations, finding where the cosine function takes certain values within an interval, such as \([0, 2\pi)\), is crucial. For example, \( \cos \theta = 1 \) corresponds to \( \theta = 0 \) or \( 2\pi \), while \( \cos \theta = -1 \) corresponds to \( \theta = \pi \). These key angles help us solve equations by pinpointing where the cosine reaches these values.

Transforming into Quadratic Form

Quadratic form refers to equations where terms are raised to the second power. This trigonometric equation, \( \cos^2 \theta - 1 = 0 \), resembles a typical quadratic expression. Here are the steps to handle it:

• The equation \( \cos^2 \theta - 1 = 0 \) is similar to a quadratic equation \( x^2 - 1 = 0 \), where \( x \) is the cosine of the angle.
• To solve, rearrange it to \( \cos^2 \theta = 1 \).

This transformation is akin to dealing with regular quadratic equations and sets the stage for further solutions, involving simple algebraic manipulation.

Using Interval Notation

Interval notation helps us define the range of possible solutions for trigonometric equations. In this context, the equation is solved within the interval \([0, 2\pi)\). Understanding this notation is key:

• Brackets \((\) and \([\) indicate if endpoints are included or excluded. Here, \([0, 2\pi)\) means 0 is included, but \(2\pi\) is not.
• This range represents a full rotation of a circle minus the endpoint, capturing all possible angles in the context of trigonometric problems.

By specifying where solutions should lie, interval notation narrows down potential answers to the exact required region on the unit circle.

Solving Equations with Cosine

Solving trigonometric equations often involves isolating the function and determining its possible values. For the given problem, the focus was on \( \cos^2 \theta = 1 \). The steps to solve such problems include:

• Take the square root to get \( \cos \theta = 1 \) or \( \cos \theta = -1 \) (since the square root of 1 is both 1 and -1).
• Identify angles within the specified interval \([0, 2\pi)\) that satisfy these equations.
• Solution: \( \theta = 0 \), \( \pi \), as at these points cosine attains values of 1 and -1.

Solving these equations requires a clear understanding of both algebraic manipulation and the behavior of the trigonometric function itself.