Problem 1 - Algebra 2 and Trigonometry

Question

Isaiah said that if the equation \(\cos 2 x+2 \cos ^{2} x=2\) is divided by \(2,\) an equivalent equation is \(\cos x+\cos ^{2} x=1 .\) Do you agree with Isaiah? Explain why or why not.

Step-by-Step Solution

Answer

No, Isaiah's equation is incorrect; division was not done properly.

Step 1: Understand the Given Equation

We are given the equation \(\cos 2x + 2 \cos^2 x = 2\). Isaiah claims that dividing this equation by 2 gives an equivalent equation \(\cos x + \cos^2 x = 1\). We need to verify this claim by performing the division and checking the equality.

Step 2: Divide the Equation by 2

Dividing every term of the original equation \(\cos 2x + 2 \cos^2 x = 2\) by 2, we get: \[\frac{\cos 2x}{2} + \frac{2 \cos^2 x}{2} = \frac{2}{2}\] Simplifying, we obtain: \(\frac{1}{2} \cos 2x + \cos^2 x = 1\).

Step 3: Check for Equivalence

Let's compare the derived equation \(\frac{1}{2} \cos 2x + \cos^2 x = 1\) with Isaiah's claimed equation \(\cos x + \cos^2 x = 1\). These are not the same, since the terms involving \(\cos 2x\) and \(\cos x\) are different. This indicates that Isaiah divided incorrectly.

Key Concepts

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying expressions or equations to work towards a solution. In the problem at hand, we are tasked with simplifying and comparing trigonometric equations. The original equation is \(\cos 2x + 2 \cos^2 x = 2\), and the goal is to examine if dividing this equation by 2 results in an equivalent form.To perform this manipulation correctly, each term in the equation must be divided by 2. This means:

• The term \(\cos 2x\) is divided by 2, giving \(\frac{1}{2} \cos 2x\).
• The term \(2 \cos^2 x\) is divided by 2, simplifying to \(\cos^2 x\).
• The number 2 on the right side of the equation becomes 1.

This step-by-step approach helps ensure that each component of the equation is dealt with accurately, preserving the equality's logic. This kind of detailed algebraic manipulation is essential to solving and understanding equations correctly by simplifying terms without introducing errors.

Cosine Function

The cosine function is a fundamental trigonometric function that appears frequently in mathematical equations. It is often denoted as \(\cos\). The function maps angles to the ratios of the adjacent side over the hypotenuse in a right triangle.

For a deeper understanding, consider the cosine of two angles, \(\cos 2x\) and \(\cos x\). These involve angles that are multiples or transformations of the basic angle \(x\). The identity for double angles tells us that \(\cos 2x = 2 \cos^2 x - 1\) or \(\cos 2x = \cos^2 x - \sin^2 x\). Such identities are valuable when simplifying expressions or during algebraic manipulation as they link simpler trigonometric expressions directly to more complex ones.
Understanding how these functions behave and transform when multiplied, added, or simplified is crucial, especially when checking the equivalence of the equations, as in the given exercise. In our scenario, misinterpreting these functions leads to errors, as seen with Isaiah's mistake.

Equation Equivalence

Equation equivalence concerns whether two equations represent the same relationship or expression. To check equivalence, both sides must simplify to the same values or expressions under every circumstance or domain of discourse.Isaiah's claim about the equivalent equation involved dividing the entire equation by 2. However, when comparing \(\frac{1}{2} \cos 2x + \cos^2 x = 1\) and \(\cos x + \cos^2 x = 1\), the differences become clear:

• The term \(\cos 2x\) differs fundamentally from \(\cos x\).
• The equations do not hold equivalence because angles or cosine values don't simply scale down directly.

Incorrectly assuming equivalence, as seen here, can lead to wrong conclusions in solution steps. This highlights the importance of thoroughly checking each operation performed on an equation—to ensure no alteration of the relationship originally expressed by the equation. The lesson is that dividing through each term is one of several accurate methods for maintaining equation equivalence, but it must be executed carefully.