Problem 45 - Algebra and Trigonometry

Question

\(43-52\) a Use a Double- or Half-Angle Formula to solve the equation in the interval \([0,2 \pi) .\) \(\cos 2 \theta+\cos \theta=2\)

Step-by-Step Solution

Answer

The solution is \( \theta = 0 \).

Step 1: Rearrange the Equation

The given equation is \(\cos 2\theta + \cos \theta = 2\). First, we'll rearrange it slightly for clarity: \(\cos 2\theta = 2 - \cos \theta\). We'll use a double-angle formula here.

Step 2: Apply the Double-Angle Formula

The double-angle formula for cosine is \( \cos 2\theta = 2 \cos^2 \theta - 1 \). Substitute this in the equation: \[ 2\cos^2 \theta - 1 + \cos \theta = 2 \] Simplify the equation.

Step 3: Simplify and Rearrange

Rearrange and simplify the equation: \[ 2\cos^2 \theta + \cos \theta - 3 = 0 \] This is a quadratic equation in terms of \( \cos \theta \).

Step 4: Solve the Quadratic Equation

Let \( x = \cos \theta \). The equation becomes: \[ 2x^2 + x - 3 = 0 \] Use the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \(a = 2\), \(b = 1\), \(c = -3\).

Step 5: Calculate Roots Using the Quadratic Formula

Calculate the discriminant: \( \Delta = 1^2 - 4 \cdot 2 \cdot (-3) = 1 + 24 = 25 \)Then, find the roots: \[ x = \frac{-1 \pm \sqrt{25}}{4} = \frac{-1 \pm 5}{4} \] This gives solutions, \( x = 1 \) and \( x = -\frac{3}{2} \).

Step 6: Determine Valid Solutions for \( \cos \theta \)

Since \( \cos \theta = x \), and \( -1 \leq \cos \theta \leq 1 \), \( x = -\frac{3}{2} \) is invalid. So, \( \cos \theta = 1 \).The valid solution is when \( \cos \theta = 1 \).

Step 7: Solve for \( \theta \)

The cosine function \( \cos \theta = 1 \) only at one point in the interval \( [0, 2\pi) \), which is \( \theta = 0 \).

Key Concepts

Double-Angle Formulas

Trigonometric identities are a powerful tool in solving equations involving angles. One such identity is the double-angle formula for cosine. It provides an expression for \(\cos 2\theta\) in terms of \(\cos \theta\). The formula is given as:

• \(\cos 2\theta = 2 \cos^2 \theta - 1\)

This transforms a trigonometric equation into a more manageable form.
For our problem, replacing \(\cos 2\theta\) with \(2 \cos^2 \theta - 1\) converts the equation into a quadratic form.
The double-angle formula helps in simplifying complex trigonometric expressions and is especially useful when the equation involves functions of angles in terms of their double values. It's crucial to understand and apply this efficiently when solving trigonometric problems.

Solving Quadratic Equations

Once we transform our trigonometric equation using the double-angle identity, we often encounter a quadratic form. The equation in our example becomes \(2 \cos^2 \theta + \cos \theta - 3 = 0\), which is now a quadratic equation in terms of \(\cos \theta\).

• To solve this type of quadratic equation, we often use the quadratic formula:
• \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

In this case, \(a = 2\), \(b = 1\), and \(c = -3\).
This leads us to determine the discriminant \(\Delta = b^2 - 4ac\) which reveals information about the nature of roots.
If \(\Delta \geq 0\), the equation has real solutions. It's important to assess these values and check if they are valid within the context of the original trigonometric function.
Solving quadratic equations is a critical step in many mathematical processes, allowing us to find roots that correspond to solutions in trigonometric problems.

Cosine Function

The cosine function, \(\cos \theta\), is one of the fundamental trigonometric functions. Understanding its properties is essential for solving equations involving cosine.

• The range of \(\cos \theta\) is between \(-1\) and \(1\).
• It is periodic with a period of \(2\pi\).
• Within one complete cycle \([0, 2\pi)\), it starts at \(1\), decreases to \(-1\), and then returns to \(1\).

In our solution, we found \(\cos \theta = 1\), which occurs only at \(\theta = 0\) within the given interval.
It's crucial to always check calculated values of a cosine function against its known limits. This ensures that every solution found is valid and applicable.
Thus, understanding the cosine function's behavior assists in narrowing down possible solutions to the given trigonometric problem.