Problem 100 - Precalculus with Limits

Question

In Exercises 93 - 104, use the trigonometric substitution tow rite the algebraic expression as a trigonometric function of \( \theta \), where \( 0 < \theta < \pi/2 \). \( \sqrt{x^2 + 100} \), \( x = 10 \tan \theta \)

Step-by-Step Solution

Answer

The expression \( \sqrt{x^2 + 100} \) can be written as a trigonometric function of \( \theta \) as \( 100 \sec \theta \) when \( x = 10 \tan \theta \).

Step 1. Substitution

Substitute \( x = 10 \tan \theta \) into the expression \( \sqrt{x^2 + 100} \). This will result in \( \sqrt{(10 \tan \theta)^2 + 100} \), which simplifies to \( \sqrt{100 \tan^2 \theta + 100} \).

Step 2. Factorization

Factor out the common factor 100 in the expression obtained in the previous step. This results in \( 100 \sqrt{\tan^2 \theta + 1} \).

Step 3. Recognize the Pythagorean trigonometric identity

The term under the square root \( \tan^2 \theta + 1 \) can be recognized as \( \sec^2 \theta \) because of the Pythagorean identity \( \tan^2 \theta + 1 = \sec^2 \theta \). Substitute this into the expression and we get \( 100 \sqrt{ \sec^2 \theta } \).

Step 4. Simplification

Simplify the square root of \( \sec^2 \theta \) which is \( |\sec \theta| \). Since the range for \( \theta \) given is \( 0 < \theta < \pi/2 \), and \( \sec \theta \) is positive in this range, \( |\sec \theta| \) simplifies to \( \sec \theta \). Hence, the final expression becomes \( 100 \sec \theta \).

Key Concepts

Pythagorean Trigonometric Identities

Understanding Pythagorean trigonometric identities is crucial for simplifying complex expressions in trigonometry. These identities are based on the Pythagorean theorem, which relates the sides of a right triangle. One of the fundamental identities is \( \tan^2 \theta + 1 = \sec^2 \theta \), which is derived from the right triangle where \( \tan \theta = \frac{\text{opposite}}{\text{adjacent}} \) and \( \sec \theta = \frac{1}{\cos \theta} \) or \( \frac{\text{hypotenuse}}{\text{adjacent}} \).

This identity is essential when working with algebraic expressions that involve tangent and secant functions. The benefit of using this identity is it allows for the simplification of square roots involving tangent functions by recognising that \( \tan^2 \theta + 1 \) is exactly \( \sec^2 \theta \), and thus, the square root of this expression is simply \( \sec \theta \), assuming \( \theta \) lies within the correct range—typically \( 0 < \theta < \frac{\pi}{2} \), where secant is positive.

By mastering these identities, students can transform daunting algebraic terms into more manageable trigonometric functions, streamlining the process of integration, differentiation, and solving equations.

Square Root Simplification

Square root simplification is a process that often involves identifying perfect squares within a radical expression and converting them back to algebraic or trigonometric forms without the radical. The main objective is to make the expression less complicated and easier to work with.

For instance, in the provided exercise, starting with \( \sqrt{100 \tan^2 \theta + 100} \) and factoring out the common 100 leads to \( 100\sqrt{\tan^2 \theta + 1} \). Recognizing the trigonometric identity allows us to further simplify \( \sqrt{\tan^2 \theta + 1} = \sqrt{\sec^2 \theta} \), which equals \( |\sec \theta| \).

In the context of square root simplification, it's important to consider the domain of the variable. In the given range of \( \theta \), \( |\sec \theta| \) simplifies to \( \sec \theta \) because secant is positive in the first quadrant. Simplification of square roots using such methods can significantly reduce the complexity of the problem, especially when incorporating trigonometric functions. Students should practice with similar problems to gain proficiency in identifying and applying the appropriate simplification techniques.

Algebraic Expression as Trigonometric Function

Converting an algebraic expression into a trigonometric function can be an insightful process, serving to illuminate hidden relationships in seemingly unrelated mathematical forms. This conversion is often essential in calculus, particularly in integrals where trigonometric substitution is applied to evaluate integrals involving radical expressions.

In our example, we begin with the algebraic expression \( \sqrt{x^2 + 100} \) and make a substitution with \( x = 10 \tan \theta \). Through this substitution, we are effectively translating the language of algebra into the language of trigonometry, which can then be manipulated using trigonometric identities to achieve a more elegant form.

By rewriting algebraic expressions as trigonometric functions, students can take advantage of the periodic and symmetric properties of trigonometric functions, which can simplify the process of solving equations. Furthermore, this trigonometric outlook frequently provides a new angle from which to approach a problem, opening up pathways to solutions that might not be evident from a purely algebraic perspective.

• Identify an appropriate trigonometric substitution that relates to the variables in the expression.
• Use trigonometric identities to simplify the resulting expression after the substitution.
• Investigate the domain of the trigonometric functions involved to ensure that the functions are defined and apply the appropriate signs.