Problem 45 - Precalculus Mathematics for Calculus

Question

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

\( \theta = 0 \) is the solution in the interval \([0, 2\pi)\).

Step 1: Recognize the formula to use

We need to use a double-angle identity for this expression. The double-angle formula for cosine is \( \cos(2\theta) = 2\cos^2(\theta) - 1 \). We will substitute this into the equation to simplify it.

Step 2: Substitute the double-angle formula

Substitute \( \cos(2\theta) = 2\cos^2(\theta) - 1 \) into the original equation \( \cos 2\theta + \cos \theta = 2 \). This gives us:\[2\cos^2(\theta) - 1 + \cos(\theta) = 2\].

Step 3: Rearrange into a quadratic equation

Combine like terms to transform the equation into a quadratic form:\[2\cos^2(\theta) + \cos(\theta) - 3 = 0\].

Step 4: Let \( x = \cos(\theta) \)

To solve, set \( x = \cos(\theta) \), which turns the equation into a standard quadratic:\[2x^2 + x - 3 = 0\].

Step 5: Solve the quadratic equation

Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2 \), \( b = 1 \), and \( c = -3 \):\[x = \frac{-1 \pm \sqrt{1^2 - 4\cdot2\cdot(-3)}}{4} = \frac{-1 \pm \sqrt{1 + 24}}{4} = \frac{-1 \pm 5}{4}\].This gives us two solutions: \( x = 1 \) and \( x = -\frac{3}{2} \).

Step 6: Discard non-feasible solution

Since \( x = \cos(\theta) \), and \( -1 \leq \cos(\theta) \leq 1 \), the solution \( x = -\frac{3}{2} \) is not feasible. Therefore, \( \cos(\theta) = 1 \) is the only solution.

Step 7: Find \( \theta \) that satisfies \( \cos(\theta) = 1 \)

Recognize that \( \cos(\theta) = 1 \) at \( \theta = 0 \) within the interval \([0, 2\pi)\). Hence, \( \theta = 0 \) is the solution.

Key Concepts

Double-Angle Formulas

Double-angle formulas are essential tools in trigonometry, particularly when simplifying expressions or solving equations involving trigonometric functions. The double-angle formula for cosine is: \[ \cos(2\theta) = 2\cos^2(\theta) - 1. \] This formula allows us to express the cosine of double the angle in terms of the square of the cosine of the angle itself. Such identities can help reduce complicated trigonometric equations to forms that are simpler to handle and solve.

• The double-angle formulas are derived from the angle addition formulas.
• They provide a way to reduce the degree of an angle, aiding in solving equations and proving other trigonometric identities.

In our original exercise, applying the double-angle formula enabled us to convert an expression involving \(\cos(2\theta)\) into one that solely involves \(\cos(\theta)\). This step was crucial in transforming the problem into a quadratic equation.

Quadratic Equations

Quadratic equations are foundational elements in algebra and appear frequently in various mathematical contexts, including trigonometry. A quadratic equation typically follows the form: \[ ax^2 + bx + c = 0, \] where \(a\), \(b\), and \(c\) are constants, and \(x\) is the variable. Solving quadratic equations can be achieved by factoring, completing the square, or using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. \]

• The discriminant \(b^2 - 4ac\) helps determine the nature of the roots.
• A positive discriminant indicates two distinct real solutions.
• A zero discriminant indicates one real solution.
• A negative discriminant indicates no real solutions.

In the exercise provided, converting the equation to the form \(2x^2 + x - 3 = 0\) allowed us to use the quadratic formula to identify potential solutions. It also highlighted the importance of checking the feasibility of solutions within the context, as trigonometric functions have constraints such as \(-1 \leq \cos(\theta) \leq 1\).

Cosine Function

The cosine function, denoted as \(\cos(\theta)\), is a fundamental trigonometric function that relates the angle \(\theta\) of a right triangle to the ratio of the adjacent side over the hypotenuse. The function has a range from -1 to 1. For any angle \(\theta\), the cosine function can be thought of as the horizontal coordinate of a point on the unit circle.

• The function is periodic with a period of \(2\pi\), which means that \(\cos(\theta + 2\pi) = \cos(\theta)\) for any \(\theta\).
• Some critical values include \(\cos(0) = 1\), \(\cos(\frac{\pi}{2}) = 0\), \(\cos(\pi) = -1\), and \(\cos(\frac{3\pi}{2}) = 0\).

In the exercise, identifying where \(\cos(\theta) = 1\) was key to finding the solution. Within the defined interval \([0, 2\pi)\), \(\cos(\theta) = 1\) occurs at \(\theta = 0\). Knowing the behavior and properties of the cosine function is essential for correctly identifying these solutions in the context of trigonometric equations.