Problem 44 - Algebra and Trigonometry

Question

Rewrite the expression as an algebraic expression in \(x .\) \(\cos \left(\tan ^{-1} x\right)\)

Step-by-Step Solution

Answer

\(\cos(\tan^{-1}(x)) = \frac{1}{\sqrt{x^2 + 1}}\).

Step 1: Identify the Trigonometric Relation

First, note that \(\tan^{-1}(x)\) is the angle whose tangent is \(x\). If we represent this angle by \(\theta\), we have \(\tan(\theta) = x\). The task is to find \(\cos(\theta)\).

Step 2: Use the Pythagorean Identity

Recall the Pythagorean identity \( \tan^2(\theta) + 1 = \sec^2(\theta) \). Using \( \tan(\theta) = x \), we have \( x^2 + 1 = \sec^2(\theta) \).

Step 3: Express \(\sec(\theta)\) in Terms of \(x\)

The identity \( x^2 + 1 = \sec^2(\theta) \) implies that \( \sec(\theta) = \sqrt{x^2 + 1} \) since the secant function is the reciprocal of cosine, i.e., \( \sec(\theta) = \frac{1}{\cos(\theta)} \).

Step 4: Solve for \(\cos(\theta)\)

Since \( \sec(\theta) = \frac{1}{\cos(\theta)} \), we have \( \cos(\theta) = \frac{1}{\sqrt{x^2 + 1}} \). This gives us the expression for \(\cos(\tan^{-1}(x))\) in terms of \(x\).

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are a bridge between algebra and geometry. They allow us to find angles when we know the lengths of sides in a right-angled triangle. For instance, if you know the ratio of the opposite side to the adjacent side in a triangle, you can use the inverse tangent function, denoted as \( \tan^{-1}(x) \), to find the angle.
This function is essential as it helps us move from a ratio back to an angle. So, when you see \( \tan^{-1}(x) \), think of it as "the angle whose tangent is \( x \)." In our exercise, \( \tan^{-1}(x) \) helps us to define an angle \( \theta \) where \( \tan(\theta) = x \).
Understanding inverse trigonometric functions involves recognizing their domain and range. This ensures you apply them correctly in various problems. They are typically confined to principal values to make them single-valued and thus inverses in the true sense.

Algebraic Expression

Turning trigonometric expressions into algebraic expressions is a common task in math problems. It involves using known identities to express one type of mathematical expression in terms of another.
In this context, converting \( \cos(\tan^{-1}(x)) \) into an algebraic form requires us to substitute our original trigonometric expression using identities like the Pythagorean identity. By recognizing that \( \tan(\theta) = x \), you can find related expressions for other trigonometric functions using \( \theta \).
Ultimately, we seek to write these trigonometric expressions without explicitly involving the angle itself. This requires leveraging the trigonometric identities to eliminate the angle and express the function purely in terms of \( x \). This process is which we perform to achieve the goal of our given exercise.

Pythagorean Identity

The Pythagorean identity is a cornerstone in trigonometry. It states: \( \tan^2(\theta) + 1 = \sec^2(\theta) \). This identity is incredibly powerful because it connects the tangent and secant functions, which are derived from the basic sine and cosine functions.
By using the identity, we deduce that if \( \tan(\theta) = x \), then \( \tan^2(\theta) + 1 = \sec^2(\theta) \) converts to \( x^2 + 1 = \sec^2(\theta) \). From here, knowing \( \sec(\theta) \) is the reciprocal of \( \cos(\theta) \), we further solve for \( \cos(\theta) \).
Hence, the Pythagorean identity allows us to express \( \sec^2(\theta) \) as \( x^2 + 1 \), and eventually to find \( \cos(\theta) = \frac{1}{\sqrt{x^2 + 1}} \). This makes it possible to evaluate trigonometric functions beyond their initial context, extending their use in expressions where direct evaluation might not be evident.