Question

The equation \[ 2 \cos^{-1} x = \sin^{-1} \left( 2 \sqrt{1 - x^2} \right) \] is valid for all values of \(x\) satisfying:

• A. \( -1 \leq x \leq 1 \)
• B. \( 0 \leq x \leq 1 \)
• C. \( \frac{1}{\sqrt{2}} \leq x \leq 1 \)
• D. \( 0 \leq x \leq \frac{1}{\sqrt{2}} \)

Hint:

For inverse trigonometric functions, ensure that the domain and range of each function align with the constraints of the given equation.

Answer 1

The Correct Option is C

Given the equation: \[ 2 \cos^{-1} x = \sin^{-1} \left( 2 \sqrt{1 - x^2} \right) \] Let \( y = \cos^{-1} x \), then \( \cos y = x \). Thus, we have: \[ 2y = \sin^{-1} \left( 2 \sqrt{1 - \cos^2 y} \right) \] Simplifying this equation gives: \[ 2 \cos^{-1} x = \sin^{-1} \left( 2 \sqrt{1 - x^2} \right) \] Now, the equation holds valid for the values of \( x \) such that: \[ 0 \leq \cos^{-1} x \leq \frac{\pi}{2} \] This corresponds to the interval \( \frac{1}{\sqrt{2}} \leq x \leq 1 \).