Problem 29 - Precalculus: Functions and Graphs

Question

Write the expression as an algebraic expression in \(x\) for \(x>0\). $$\cos \left(\frac{1}{2} \arccos x\right)$$

Step-by-Step Solution

Answer

The expression simplifies to \( \sqrt{\frac{1 + x}{2}} \).

Step 1: Understanding the Problem

We need to express \( \cos \left(\frac{1}{2} \arccos x\right) \) in terms of \( x \), given that \( x > 0 \). This involves using trigonometric identities and potentially inverse function properties to simplify the expression.

Step 2: Use of Trigonometric Identity

The expression \( \cos \left(\frac{1}{2} \arccos x\right) \) can be simplified using a trigonometric identity for half-angles: \( \cos \frac{\theta}{2} = \sqrt{\frac{1 + \cos \theta}{2}} \). In our case, \( \theta = \arccos x \).

Step 3: Substitute \(\theta\) with \(\arccos x\)

Substituting \( \theta \) in the half-angle identity gives us: \( \cos \left(\frac{1}{2} \arccos x\right) = \sqrt{\frac{1 + \cos(\arccos x)}{2}} \).

Step 4: Evaluate \(\cos(\arccos x)\)

Since \( \cos(\arccos x) = x \), we can plug this into the previous expression: \( \cos \left(\frac{1}{2} \arccos x\right) = \sqrt{\frac{1 + x}{2}} \).

Step 5: Final Expression

Thus, the algebraic expression for \( \cos \left(\frac{1}{2} \arccos x\right) \) in terms of \( x \) is \( \sqrt{\frac{1 + x}{2}} \).

Key Concepts

Algebraic Expression

An algebraic expression is a combination of variables, numbers, and mathematical operations. In the exercise, we express the function \(\cos \left(\frac{1}{2} \arccos x\right)\) in terms of purely algebraic components. Our goal is to represent the trigonometric function in a form that involves only the variable \(x\).
To achieve this, it often involves utilizing identities and simplifications that translate the original function into a simpler form.

• This process is critical in solving trigonometric equations and transforming them into solvable algebraic forms.
• Thus, in this exercise, by determining that \( \cos \left(\frac{1}{2} \arccos x \right)\) simplifies to \( \sqrt{\frac{1 + x}{2}} \), we've effectively replaced the trigonometric expression with an algebraic expression involving \(x\).

Understanding this transition is essential because it allows for the analysis and solving of problems without the direct complexities of trigonometric functions.

Inverse Trigonometric Functions

Inverse trigonometric functions allow us to determine the angle that corresponds to a given trigonometric value. In the original exercise, \(\arccos x\) represents the angle whose cosine is \(x\).
This function is critical in converting back and forth between angles and their trigonometric values:

• It acts like a bridge between an angle and its cosine value, effectively "reversing" the cosine function.
• For instance, if \(\arccos x = \theta\), then \(\cos \theta = x\).

These functions are bounded by their principal ranges to provide single-valued outputs. Specifically, for \(\arccos\), the range is \([0, \pi]\).
This characteristic allows us to understand what half of this angle would look like when translating it into a half-angle identity. In the problem, recognizing that \( \cos(\arccos x) = x \) simplifies significant parts of the algebra.

Half-Angle Identities

Half-angle identities are powerful tools in trigonometry that express a trigonometric function of half of a given angle in terms of the full angle's trigonometric functions.
In this problem, the identity \(\cos \frac{\theta}{2} = \sqrt{\frac{1 + \cos \theta}{2}}\) is used. With this, we can transform \(\cos \left(\frac{1}{2} \arccos x\right)\) into an algebraic expression.

• This identity helps simplify expressions where the angle is halved, converting them into something solvable with simpler arithmetic.
• The identity takes advantage of the primary trigonometric functions and their foundations in geometry to reveal simpler patterns.

Utilizing \(\theta = \arccos x\), the half-angle identity effortlessly allows \(\theta = \arccos x\), morphing the trigonometric problem into \(\sqrt{\frac{1 + x}{2}}\).
This method maximizes understanding of the relationships between these trigonometric expressions and usual algebraic forms.