Problem 142 - Precalculus

Question

For the following exercises, find the exact value of the expression in terms of \(x\) with the help of a reference triangle. $$ \tan \left(\sin ^{-1}(x-1)\right) $$

Step-by-Step Solution

Answer

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

Step 1: Understand the Problem

We need to find the value of \( \tan(\sin^{-1}(x-1)) \) using a reference triangle. The expression involves applying trigonometric identities using a right triangle.

Step 2: Define the Variables in the Triangle

Since \( \sin^{-1}(x-1) \) is the angle whose sine is \( x-1 \), we set up a right triangle where the opposite side is \( x-1 \) and the hypotenuse is 1 (since sine is opposite over hypotenuse).

Step 3: Apply Pythagorean Theorem

Calculate the remaining side of the triangle (adjacent) using the Pythagorean theorem: \( \text{adjacent} = \sqrt{1^2 - (x-1)^2} = \sqrt{1 - (x-1)^2} \).

Step 4: Simplify the Expression

Simplify the expression for the adjacent side: \( \sqrt{1 - (x-1)^2} = \sqrt{1 - (x^2 - 2x + 1)} = \sqrt{2x - x^2} \).

Step 5: Calculate the Tangent

Now that we have the lengths of all three sides, tangent is opposite over adjacent: \( \tan(\sin^{-1}(x-1)) = \frac{x-1}{\sqrt{2x-x^2}} \).

Key Concepts

Reference Triangle

A reference triangle is used to translate inverse trigonometric expressions into an easier visualization form. This is especially handy when dealing with expressions like \( \tan(\sin^{-1}(x-1)) \). When you see \( \sin^{-1}(x-1) \), it helps to picture a right triangle where the angle has a sine value of \( x-1 \).
To construct this triangle:

• The angle \( \sin^{-1}(x-1) \) is represented at one of the corners.
• The opposite side, relative to this angle, is \( x-1 \).
• The hypotenuse is set to 1, as the sine function's ratio is opposite over hypotenuse: \( \sin(\theta) = \frac{x-1}{1} \).

By framing the problem into this triangle, it becomes clearer what you're solving for and how to apply trigonometric identities. With this visual aid, you're better equipped to understand and calculate related trigonometric functions like tangent.

Trigonometric Identities

Trigonometric identities are mathematical equations involving trigonometric functions that hold true for any angle. These are essential tools when working with trigonometric expressions and simplifying them.
In the expression \( \tan(\sin^{-1}(x-1)) \), we aim to find the tangent, which refers to the ratio of the opposite side to the adjacent side of a right triangle. Using the reference triangle:

• The opposite side is \( x-1 \).
• The adjacent side is computed with the help of other identities, specifically the Pythagorean identity in this context.

To further aid our calculation, remember that \( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \). By applying this identity, once you know both sides of the triangle, you can derive that \( \tan(\sin^{-1}(x-1)) = \frac{x-1}{\sqrt{2x-x^2}} \). These identities are powerful in revealing the relationships between different trigonometric functions.

Pythagorean Theorem

The Pythagorean Theorem is crucial for solving problems using reference triangles. It states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. Formally, it is expressed as \( a^2 + b^2 = c^2 \).
In our context, if you have a right triangle:

• The hypotenuse is 1.
• The opposite side is \( x-1 \).
• We want to find the length of the adjacent side.

Applying the Pythagorean Theorem, we solve for the adjacent side: \[1^2 = (x-1)^2 + \text{adjacent}^2\]Reorganizing and solving for the adjacent side gives us \[\text{adjacent} = \sqrt{1 - (x-1)^2} = \sqrt{2x - x^2}\]This solution gives the remaining side's length which is needed to calculate other trigonometric functions like tangent from \( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \). With this completed triangle, all relevant trigonometric functions can be evaluated accurately.