Problem 54 - Thomas Calculus

Question

For Exercises \(51-54,\) solve for the angle \(\theta,\) where \(0 \leq \theta \leq 2 \pi\). $$\cos 2 \theta+\cos \theta=0$$

Step-by-Step Solution

Answer

\( \theta = \frac{\pi}{3}, \frac{5\pi}{3}, \pi \)

Step 1: Use Double Angle Identity

The given equation is \( \cos 2 \theta + \cos \theta = 0 \). We start by using the double angle identity for cosine: \( \cos 2\theta = 2\cos^2 \theta - 1 \). Substituting this into the equation gives: \[ 2\cos^2 \theta - 1 + \cos \theta = 0. \]

Step 2: Substitute and Rearrange

We substitute the identity into the equation from Step 1:\[ 2\cos^2 \theta + \cos \theta - 1 = 0. \]Rearrange terms to form a standard quadratic equation:\[ 2\cos^2 \theta + \cos \theta - 1 = 0. \]

Step 3: Solve the Quadratic Equation

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

Step 4: Compute the Discriminant

Calculate the discriminant, \( b^2 - 4ac \):\[ 1^2 - 4(2)(-1) = 1 + 8 = 9. \]

Step 5: Apply Quadratic Formula

Substitute the values into the quadratic formula:\[ x = \frac{-1 \pm \sqrt{9}}{4} \]\[ x = \frac{-1 \pm 3}{4} \]This gives us two solutions: \( x = \frac{2}{4} = \frac{1}{2} \) and \( x = \frac{-4}{4} = -1. \)

Step 6: Determine Angles from Solutions

Recall that \( x \) represents \( \cos \theta \). Thus, \( \cos \theta = \frac{1}{2} \) or \( \cos \theta = -1 \).For \( \cos \theta = \frac{1}{2} \), \( \theta = \frac{\pi}{3}, \frac{5\pi}{3} \).For \( \cos \theta = -1 \), \( \theta = \pi \).

Step 7: Final Solution

The angles that satisfy \( \cos 2 \theta + \cos \theta = 0 \) in the range \( 0 \leq \theta \leq 2\pi \) are \( \theta = \frac{\pi}{3}, \frac{5\pi}{3}, \pi \).

Key Concepts

Double Angle Identity

In trigonometry, the double angle identity is a crucial tool that helps simplify expressions involving angles. It's an extension of basic trigonometric identities that allows you to rewrite expressions with double angles (such as \(2\theta\)) in terms of a single angle (\(\theta\)).
The identity for cosine, which we use here, is:

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

This identity is key for solving problems where angles need to be condensed or expanded. Applying it often simplifies the problem into a more manageable form.
When solving trigonometric equations, substituting double angle identities can make the equations linear or quadratic, which are easier to solve with standard algebraic techniques. In our exercise, by using this identity, we transformed the original equation into a quadratic form.

Quadratic Equation

A quadratic equation is a second-degree polynomial equation in a single variable, usually seen in the standard form \( ax^2 + bx + c = 0 \). Solving quadratic equations is a fundamental skill in algebra, with multiple methods available to find the roots (solutions). These methods include:

• Factoring (if applicable)
• Completing the square
• Using the quadratic formula

In our exercise, the equation \( 2\cos^2 \theta + \cos \theta - 1 = 0 \) is treated as a quadratic in terms of \( x = \cos \theta \), transforming it into:

• \( 2x^2 + x - 1 = 0 \)

To solve this, we used the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Solving for \( x \) gives us potential solutions for \( \cos \theta \). The quadratic formula is particularly handy when the equation doesn't easily factor, as in this example.

Cosine Function

The cosine function is a fundamental trigonometric function describing the relationship between the angle and the ratio of the adjacent side to the hypotenuse in a right triangle. It is periodic with a period of \(2\pi\), meaning it repeats its values over each complete cycle of \(2\pi\).Understanding the cosine function is essential, especially when solving equations involving angles. For instance, when we have \(\cos \theta = \frac{1}{2}\) and \(\cos \theta = -1\), we need to determine \(\theta\) values within the given range (\(0 \leq \theta \leq 2\pi\)) that satisfy these conditions.

• For \(\cos \theta = \frac{1}{2}\), \(\theta\) can be \(\frac{\pi}{3}\) and \(\frac{5\pi}{3}\).
• For \(\cos \theta = -1\), \(\theta\) is \(\pi\).

These solutions are found by identifying values of \(\theta\) that make the cosine function equal to the desired value, illustrating how we use the cosine function in coordinate geometry to solve trigonometric equations.