Problem 38 - Precalculus with Calculus Previews

Question

Write the given expression as an algebraic expression in \(x\). $$ \cos \left(\sin ^{-1} x\right) $$

Step-by-Step Solution

Answer

The expression is \( \sqrt{1 - x^2} \).

Step 1: Understand the Problem

We need to express \( \cos(\sin^{-1}(x)) \) as an algebraic expression in terms of \( x \). The goal is to eliminate the trigonometric function and write it using an algebraic form.

Step 2: Use Definition of Inverse Sine

The function \( \sin^{-1}(x) \) gives us the angle \( \theta \) such that \( \sin(\theta) = x \). This means that we have an angle \( \theta \) with \( \sin(\theta) = x \).

Step 3: Apply Pythagorean Identity

From \( \sin(\theta) = x \), using the Pythagorean identity, we know that \( \cos(\theta) = \sqrt{1 - \sin^2(\theta)} \). Substitute \( \sin(\theta) = x \) to get \( \cos(\theta) = \sqrt{1 - x^2} \).

Step 4: Final Expression

Since \( \cos(\theta) = \cos(\sin^{-1}(x)) \), substitute \( \cos(\theta) = \sqrt{1 - x^2} \). Thus, \( \cos(\sin^{-1}(x)) = \sqrt{1 - x^2} \).

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are quite interesting because they allow you to find angles instead of side lengths in a right triangle. These functions include the inverse sine \( \sin^{-1}(x) \), inverse cosine \( \cos^{-1}(x) \), and inverse tangent \( \tan^{-1}(x) \). They are essential in dealing with trigonometric equations involving angles.

• In the exercise, you encountered \( \sin^{-1}(x) \), which represents an angle \( \theta \) such that \( \sin(\theta) = x \).
• Remember, the range of \( \sin^{-1}(x) \) is \([-\frac{\pi}{2}, \frac{\pi}{2}]\), meaning the angle is between \(-90^\circ\) and \(90^\circ\).

By setting \( \theta = \sin^{-1}(x) \), you're essentially solving for the angle whose sine is \( x \). This is pivotal in transforming trigonometric expressions into algebraic ones, making calculations simpler and symbol manipulation feasible.

Pythagorean Identity

The Pythagorean identity is an invaluable tool in trigonometry, helping to relate sine and cosine of an angle. It's based on the Pythagorean theorem and expressed as:\[\sin^2(\theta) + \cos^2(\theta) = 1\]When dealing with inverse trigonometric expressions, this identity can simplify your work significantly.

• In our case, having \( \sin(\theta) = x \) implies \( \sin^2(\theta) = x^2 \).
• This can be transformed using the identity to find \( \cos^2(\theta) = 1 - x^2 \).
• Consequently, \( \cos(\theta) = \sqrt{1 - x^2} \) as we only consider the principal square root due to angle restrictions of the inverse sine function.

Thus, using the Pythagorean identity, you can rewrite complex trigonometric expressions as simpler algebraic ones, facilitating easier handling in computations.

Algebraic Expressions

Algebraic expressions are combinations of variables, constants, and operations that can be simplified or transformed. They follow the rules of algebra to solve or simplify equations.

• In our exercise, \( \cos(\sin^{-1}(x)) \) was simplified to \( \sqrt{1 - x^2} \).
• This transformation from a trigonometric function to an algebraic expression makes computation much more straightforward, as it relies only on the variable \( x \).
• Algebraic expressions frequently use operations like addition, subtraction, multiplication, division, and taking roots, as seen in \( \sqrt{1 - x^2} \).

These transformations highlight the flexibility of algebra in dealing with mathematical expressions, enabling problem-solving across various mathematical domains. Understanding how to convert between forms is a key skill in mathematics.