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 $\sin^{-1} x = \theta$. This implies $x = \sin \theta$.
2. Rewrite the given function: \[ \sin^{-1} \left( \frac{\sqrt{3}}{2} x + \frac{1}{2} \sqrt{1-x^2} \right) \] Substitute $x = \sin \theta$. Since $x = \sin \theta$, then $\sqrt{1-x^2} = \sqrt{1-\sin^2 \theta} = \sqrt{\cos^2 \theta} = \cos \theta$ (assuming $\cos \theta \ge 0$). \[ = \sin^{-1} \left( \frac{\sqrt{3}}{2} \sin \theta + \frac{1}{2} \cos \theta \right) \]
3. Apply the angle addition formula for sine, $\sin(A+B) = \sin A \cos B + \cos A \sin B$. Recognize that $\frac{\sqrt{3}}{2} = \cos \frac{\pi}{6}$ and $\frac{1}{2} = \sin \frac{\pi}{6}$. \[ = \sin^{-1} \left( \sin \theta \cos \frac{\pi}{6} + \cos \theta \sin \frac{\pi}{6} \right) \] \[ = \sin^{-1} \left( \sin \left( \theta + \frac{\pi}{6} \right) \right) \] The inverse sine function $\sin^{-1}(\sin y)$ simplifies to $y$ for $-\frac{\pi}{2} \le y \le \frac{\pi}{2}$. Assuming $\theta + \frac{\pi}{6}$ is within this range: \[ = \theta + \frac{\pi}{6} \] Substitute back $\theta = \sin^{-1} x$: \[ = \sin^{-1} x + \frac{\pi}{6} \] Therefore, the correct answer is (2) $\frac{\pi}{6} + \sin^{-1} x$.