Problem 33 - Precalculus with Calculus Previews

Question

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

Step-by-Step Solution

Answer

\( \frac{x}{\sqrt{x^2 + 1}} \)

Step 1: Understand the Inverse Trigonometric Function

The expression involves \( \tan^{-1} x \). This function represents the angle \( \theta \) such that \( \tan \theta = x \).

Step 2: Visualize with a Right Triangle

To find \( \sin(\tan^{-1} x) \), imagine a right triangle where \( \theta = \tan^{-1} x \). In this triangle, the opposite side is \( x \) and the adjacent side is 1, because \( \tan \theta = \frac{x}{1} \). The hypotenuse can be found using the Pythagorean theorem.

Step 3: Use Pythagorean Theorem to Find the Hypotenuse

Apply the Pythagorean theorem: hypotenuse \( h = \sqrt{x^2 + 1} \). This is because \( h^2 = x^2 + 1^2 = x^2 + 1 \), thus \( h = \sqrt{x^2 + 1} \).

Step 4: Determine \( \sin \theta \)

Since \( \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} \), we have \( \sin \theta = \frac{x}{\sqrt{x^2 + 1}} \), where the opposite side is \( x \) and the hypotenuse is \( \sqrt{x^2 + 1} \).

Step 5: Write the Final Algebraic Expression

Thus, the expression \( \sin(\tan^{-1} x) \) becomes \( \frac{x}{\sqrt{x^2 + 1}} \).

Key Concepts

Right Triangle Visualization

When tackling inverse trigonometric functions, it often helps to visualize the problem with a right triangle. The function \( \tan^{-1} x \) tells us about an angle, \( \theta \), such that \( \tan \theta = x \). In triangle terms, this means we are dealing with a right triangle where the ratio of the opposite side to the adjacent side equals \( x \). To make this tangible, consider:

• Opposite Side: This side is expressed by the value \( x \).
• Adjacent Side: This side, relating to the base of our triangle, is equal to 1.

Imagine placing this scenario within a triangle: The angle \( \theta \) lies between the opposite side of length \( x \) and the adjacent side of length 1. This method transforms an abstract concept into a clear picture, aiding our understanding of how to proceed further.

Pythagorean Theorem

The Pythagorean theorem is elemental when working with right triangles, especially in deriving unknown lengths. Here, we need to find the hypotenuse of our right triangle since it is essential for calculating the sine of \( \theta \) later. The theorem states:

• The square of the hypotenuse is equal to the sum of the squares of the other two sides.

For our triangle: Hypotenuse squared equals \( x^2 + 1^2 \).
Thus, hypotenuse \( h = \sqrt{x^2 + 1} \).
This formula allows us to conclude the hypotenuse's measure, ensuring all sides of our triangle are accounted for and making us ready to find the sine of the angle \( \theta \).

Algebraic Expression Conversion

The final goal in this problem is to express the trigonometric function algebraically in terms of \( x \). With the right triangle visualization and the hypotenuse clearly defined, we can now find \( \sin \theta \). Recall:

• Sine Function: Sine of an angle equals the ratio of the opposite side to the hypotenuse.

Substituting the known values:

• Opposite Side: \( x \)
• Hypotenuse: \( \sqrt{x^2 + 1} \)

Thus, \( \sin \theta = \frac{x}{\sqrt{x^2 + 1}} \).
This conversion succinctly transforms the inverse trigonometric expression \( \sin(\tan^{-1} x) \) into a straightforward algebraic expression, \( \frac{x}{\sqrt{x^2 + 1}} \), which is easier to use in further calculations or applications.