Question

Prove that: \( 3 \sin^{-1} x = \sin^{-1} (3x - 4x^3) \), \( x \in \left[ -\frac{1}{2}, \frac{1}{2} \right] \).

Hint:

To prove trigonometric identities involving inverse trigonometric functions, use standard trigonometric identities and substitute appropriate values.

Answer 1

Step 1: Start with the given identity: \[ 3 \sin^{-1} x = \sin^{-1} (3x - 4x^3) \] Step 2: Recall the identity for \( \sin 3\theta \): \[ \sin 3\theta = 3 \sin \theta - 4 \sin^3 \theta \] Set \( \theta = \sin^{-1} x \), so \( \sin \theta = x \), and the identity becomes: \[ \sin 3\theta = 3x - 4x^3 \] Step 3: Since \( 3 \sin^{-1} x = \sin^{-1} (3x - 4x^3) \), we see that both sides are equal, proving the identity. Thus, the equation holds.