Question

Considering the principal values of the inverse trigonometric functions, $\sin^{-1} \left( \frac{\sqrt{3}}{2} x + \frac{1}{2} \sqrt{1-x^2} \right)$, $-\frac{1}{2}<x<\frac{1}{\sqrt{2}}$, is equal to

• A. $\frac{\pi}{4} + \sin^{-1} x$
• B. $\frac{\pi}{6} + \sin^{-1} x$
• C. $\frac{-5\pi}{6} - \sin^{-1} x$
• D. $\frac{5\pi}{6} - \sin^{-1} x$

Hint:

Use the angle addition formula for inverse trigonometric functions.

Answer 1

The Correct Option is B

1. Let $\sin^{-1} x = \theta$: \[ x = \sin \theta \]
2. Express the given function: \[ \sin^{-1} \left( \frac{\sqrt{3}}{2} x + \frac{1}{2} \sqrt{1-x^2} \right) \] \[ = \sin^{-1} \left( \frac{\sqrt{3}}{2} \sin \theta + \frac{1}{2} \cos \theta \right) \]
3. Use the angle addition formula: \[ = \sin^{-1} \left( \sin \theta \cos \frac{\pi}{6} + \cos \theta \sin \frac{\pi}{6} \right) \] \[ = \sin^{-1} \left( \sin \left( \theta + \frac{\pi}{6} \right) \right) \] \[ = \theta + \frac{\pi}{6} \] \[ = \sin^{-1} x + \frac{\pi}{6} \] Therefore, the correct answer is (2) $\frac{\pi}{6} + \sin^{-1} x$.