Problem 34 - Precalculus with Calculus Previews

Question

Write the given 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: Understand the Trigonometric Identity

In this exercise, we're dealing with an inverse tangent function, \(\tan^{-1}(x)\), and we want to express \(\cos(\tan^{-1}(x))\) in terms of \(x\). To do this, we'll use the identity that relates the tangent and cosine: \(\cos(\theta) = \frac{1}{\sqrt{1 + \tan^2(\theta)}}\) for any angle \(\theta\).

Step 2: Set up the Right Triangle

Consider a right triangle where \(\theta = \tan^{-1}(x)\), so that \(\tan(\theta) = x\). This implies the opposite side is \(x\) and the adjacent side is \(1\) in a right triangle (since the tangent of an angle is opposite/adjacent).

Step 3: Calculate the Hypotenuse

Using the Pythagorean theorem, the hypotenuse \(h\) can be found as \( h = \sqrt{x^2 + 1^2} = \sqrt{x^2 + 1} \).

Step 4: Find the Cosine using Triangle Sides

With the triangle set, \(\cos(\theta)\) is the adjacent side divided by the hypotenuse. Therefore, \( \cos(\theta) = \frac{1}{\sqrt{x^2 + 1}} \). This gives us \(\cos(\tan^{-1}(x)) = \frac{1}{\sqrt{x^2 + 1}}\).

Key Concepts

Trigonometric Identities

Trigonometric identities are essential tools in mathematics, especially when dealing with calculus and trigonometry problems. These identities relate the trigonometric functions to each other and to complex numbers. A classic example is the Pythagorean identity:

• \( \sin^2(\theta) + \cos^2(\theta) = 1 \)

This identity shows a fundamental relationship between sine and cosine functions. They hold true for any angle \(\theta\) and are derived from the unit circle concept in trigonometry.

For our exercise, another key identity is used: \(\cos(\theta) = \frac{1}{\sqrt{1 + \tan^2(\theta)}}\). This identity helps convert a trigonometric expression involving an inverse tangent function, \(\tan^{-1}(x)\), into a purely algebraic expression using basic arithmetic and roots. Understanding and memorizing these identities can simplify complex problems significantly, turning them into more manageable algebraic computations.

Right Triangles

A right triangle is a significant type of triangle in geometry, characterized by one angle of exactly 90 degrees. It consists of three sides: the opposite, adjacent, and hypotenuse. Knowing the proportions of these sides helps solve trigonometric functions using right triangles.

In trigonometry, the tangent function is defined as the ratio of the opposite side to the adjacent side in a right triangle. In our exercise, the angle \(\theta = \tan^{-1}(x)\), gives an opposite side of \(x\), and an adjacent side of 1. This setup reflects how the tangent ratio represents angles in real-life problems, aiding in solving angles or distances.

• The opposite side is the side across from the angle \(\theta\).
• The adjacent side is the side next to the angle \(\theta\), other than the hypotenuse.

Using these definitions, right triangles become an intuitive way to understand inverse trigonometric functions.

Pythagorean Theorem

The Pythagorean theorem is a fundamental principle in mathematics, especially when dealing with right triangles. It states that the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. The formula is:

• \(a^2 + b^2 = c^2\)

Where \(a\) and \(b\) are the lengths of the legs, and \(c\) is the hypotenuse.

In the context of the current exercise, this theorem allows us to find the hypotenuse when \(\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = x\). This sets the opposite side as \(x\), the adjacent side as 1, and using the Pythagorean theorem:

• The hypotenuse is \(\sqrt{x^2 + 1^2} = \sqrt{x^2 + 1}\).

Thus, the Pythagorean theorem simplifies the process of finding the trigonometric functions' values within the context of right triangles, converting trigonometric expressions to algebraic values.