Question

If \(2\sin^{-1}x=\sin^{-1}k\), then \(k=\)

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

Hint:

For inverse trigonometric questions, assume \(\sin^{-1}x=\theta\), then convert the expression into normal trigonometric form.

Answer 1

The Correct Option is A

Concept: If \(\sin^{-1}x=\theta\), then: \[ x=\sin\theta \] Also, \[ \sin 2\theta=2\sin\theta\cos\theta \]

Step 1: Let \[ \sin^{-1}x=\theta \] Then, \[ x=\sin\theta \]

Step 2: Given: \[ 2\sin^{-1}x=\sin^{-1}k \] Using \(\sin^{-1}x=\theta\), we get: \[ 2\theta=\sin^{-1}k \]

Step 3: Take sine on both sides. \[ \sin(2\theta)=k \]

Step 4: Use the double-angle formula. \[ k=2\sin\theta\cos\theta \]

Step 5: Since \(\sin\theta=x\), \[ \cos\theta=\sqrt{1-\sin^2\theta} \] \[ \cos\theta=\sqrt{1-x^2} \]

Step 6: Substitute these values. \[ k=2x\sqrt{1-x^2} \] Therefore, \[ \boxed{2x\sqrt{1-x^2}} \]