Problem 24 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 23-30: Write the expression as an algebraic expression in \(x\) for \(x>0\). $$ \tan (\arccos x) $$

Step-by-Step Solution

Answer

\( \tan(\arccos x) = \frac{\sqrt{1-x^2}}{x} \)

Step 1: Understanding the Problem

The exercise asks us to write the expression \( \tan(\arccos x) \) as an algebraic expression. Essentially, this means we need to express the tangent of an angle whose cosine is \( x \), using only \( x \).

Step 2: Visualizing with a Right Triangle

Consider a right triangle where \( \theta = \arccos x \). In this triangle, \( \cos \theta = x \), which can be represented as the ratio of the adjacent side over the hypotenuse: \( \frac{adjacent}{hypotenuse} = \frac{x}{1} \).

Step 3: Determining the Opposite Side

Applying the Pythagorean theorem, the side opposite \( \theta \) is calculated as: \( opposite = \sqrt{1^2 - x^2} = \sqrt{1 - x^2} \).

Step 4: Express \( \tan(\arccos x) \) in Terms of \(x\)

The tangent function is the ratio of the opposite side to the adjacent side in a right triangle. Therefore, \( \tan(\arccos x) = \frac{\sqrt{1-x^2}}{x} \).

Step 5: Conclusion

The expression \( \tan(\arccos x) \) simplifies to \( \frac{\sqrt{1-x^2}}{x} \) using algebraic manipulation and trigonometric identities.

Key Concepts

Algebraic Expressions

Algebraic expressions are mathematical phrases that contain numbers, variables, and operational symbols. They represent quantities and relationships in mathematics. Converting expressions with trigonometric functions into algebraic forms can greatly simplify solving equations and understanding mathematical concepts.
In the problem, we seek to express the trigonometric expression \( \tan(\arccos x) \) as a plain algebraic expression in terms of \( x \). Our goal is to eliminate the trigonometric functions and represent the entire operation just using the variable \( x \) and basic operations.
Here's how it's done:

• Identify the trigonometric function to be converted (e.g., \( \tan(\arccos x) \)).
• Use known relationships and identities to express these functions in terms of algebraic operations.
• Simplify the resulting expression, if possible, to ensure it only contains the variable \( x \).

The conversion helps in cases where further algebraic manipulation is required, and it assists in integrating trigonometric expressions with other algebraic expressions in complex equations.

Inverse Trigonometric Functions

Inverse trigonometric functions are used to obtain an angle from the known value of a trigonometric function. This is opposite to how we normally use trigonometric functions, where we calculate the function's value given an angle.
For instance, \( \arccos(x) \) returns the angle whose cosine is \( x \). The value of \( \arccos(x) \) ranges from 0 to \( \pi \), providing a principal value for the cosine function's inverse.
As we solve the exercise, \( \theta = \arccos(x) \) is an angle such that \( \cos(\theta) = x \). We find this angle and use right triangle properties to determine other trigonometric values related to \( \theta \). This helps in transforming the expression \( \tan(\arccos x) \) into an algebraic form.
When dealing with inverse trigonometric functions:

• Remember they provide angles as outputs.
• Use them to work backwards from a trigonometric function's value to an angle.
• Always consider their range, as it indicates the angles that are possible outputs.

Understanding these functions allows transforming and simplifying trigonometric expressions, an essential skill in many areas of mathematics.

Tangent and Cosine Relationship

The relationship between tangent and cosine is pivotal when converting trigonometric expressions, especially involving inverse functions. The tangent of an angle in a right triangle is defined as the ratio between the length of the opposite side and the adjacent side.
From the context of the given problem, if \( \theta = \arccos(x) \), it means we're dealing with an angle whose cosine is \( x \). For a right triangle where \( \cos(\theta) = \frac{x}{1} \), the adjacent side has length \( x \), and the hypotenuse is \( 1 \).
Using the Pythagorean theorem: \[ \text{opposite} = \sqrt{1^2 - x^2} = \sqrt{1-x^2} \] The tangent function simplifies to:
\[ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{\sqrt{1-x^2}}{x} \] This shows how tangent can be expressed using cosine and the lengths derived from it.

• Tangent involves the relationship between two specific sides of a triangle.
• The cosine provides a starting point for determining these side lengths.
• Understanding both functions and their interrelation helps simplify expressions and solve equations.

Recognizing how tangent can be calculated from given trigonometric inputs gives flexibility in solving trigonometric equations using geometric and algebraic techniques.