Question

If \( |x| \leq 1 \), then \( \sin \left( 2 \sin^{-1} x + \cos^{-1} x \right) \) is equal to:

• A. \( \sqrt{1 - x^2} \)
• B. \( \frac{1}{\sqrt{1 - x^2}} \)
• C. \( x^2 \)
• D. \( x \)
• E. \( \frac{x}{2} \)

Hint:

Use the identities \( \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \) and \( \cos(2A) = 1 - 2 \sin^2 A \) to simplify trigonometric expressions involving inverse functions.

Answer 1

The Correct Option is A

We are given the expression \( \sin \left( 2 \sin^{-1} x + \cos^{-1} x \right) \), and we need to simplify it. Let \( \theta = \sin^{-1} x \). 
Then \( \sin \theta = x \), and the angle \( \theta \) lies within the range \( \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \). Also, recall that: \[ \cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x = \frac{\pi}{2} - \theta \] Now, we can rewrite the expression as: \[ \sin \left( 2 \theta + \left( \frac{\pi}{2} - \theta \right) \right) \] This simplifies to: \[ \sin \left( \theta + \frac{\pi}{2} \right) \] Using the trigonometric identity \( \sin \left( \theta + \frac{\pi}{2} \right) = \cos \theta \), we have: \[ \sin \left( 2 \sin^{-1} x + \cos^{-1} x \right) = \cos \theta \] Since \( \sin \theta = x \), we use the Pythagorean identity \( \cos^2 \theta = 1 - \sin^2 \theta \) to find: \[ \cos \theta = \sqrt{1 - x^2} \] Thus, the value of \( \sin \left( 2 \sin^{-1} x + \cos^{-1} x \right) \) is \( \sqrt{1 - x^2} \). 
Thus, the correct answer is option (A), \( \sqrt{1 - x^2} \).