Problem 111 - Algebra and Trigonometry

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 for \(\sin \left(2 \sin ^{-1} x\right)\) is \(2x\sqrt{1 - x^2}\).

Step 1: Interpret the Inverse Sine Function

The function \(\sin^{-1} x\) gives us the angle whose sine is \(x\). Let's call this angle \(A\). That is, \(A = \sin^{-1} x\). Therefore, the function \(\sin \left(2 \sin ^{-1} x\right)\) becomes \(\sin(2A)\).

Step 2: Draw the Right Triangle and Calculate Cos(A)

Let's draw a right triangle with angle \(A\), where \(A = \sin^{-1} x\). From the definition of sine function, we can infer that the opposite side is \(x\) and the hypotenuse is \(1\). Now, using the Pythagorean theorem, we can calculate the length of the adjacent side (which is \(\sqrt{1 - x^2}\)), and consequently, \(\cos A = \frac{\sqrt{1 - x^2}}{1} = \sqrt{1 - x^2}\).

Step 3: Apply the Double Angle Identity for Sine

To rewrite \(\sin(2A)\) in algebraic terms, we use the double-angle identity for sine, that is, \(\sin(2A) = 2\sin(A)\cos(A)\). But we already know that \(\sin(A) = x\) and \(\cos(A) = \sqrt{1 - x^2}\). Replacing these values in, we obtain: \(\sin(2A) = 2x\sqrt{1 - x^2}\).