Problem 111 - Precalculus Essentials

Question

Use a right triangle to write \(\sin \left(2 \sin ^{-1} x\right)\) as an algebraic expression. Assume that \(x\) is positive and in the domain of the given inverse trigonometric function.

Step-by-Step Solution

Answer

The algebraic expression of \(\sin \left(2 \sin ^{-1} x\right)\) is \(2x\sqrt{1 - x^2}\).

Step 1: Set Up the Right Triangle

Firstly, set \(x=\sin \theta\), where \(\theta\) is an angle in a right triangle. This means \(\theta = \sin^{-1} x\). Consider a right triangle where opposite side is \(x\), adjacent side is \(\sqrt{1 - x^2}\), and hypotenuse is \(1\). So, from the right triangle we have \(\cos\theta = \sqrt{1 - x^2}\).

Step 2: Apply the Double Angle Identity

Using the double angle identity \(\sin 2\theta = 2\sin \theta \cos \theta\), replace the \(\theta\) with \(\sin ^{-1} x\) and substitute \(\sin \theta\) as \(x\) and \(\cos\theta\) as \(\sqrt{1 - x^2}\) that we found in the first step. So the expression becomes \(\sin 2\sin^{-1} x = 2x\sqrt{1 - x^2}\).

Step 3: Simplifying the Expression

The expression \(\sin 2\sin^{-1} x = 2x\sqrt{1 - x^2}\) is the final answer.