Q. If $f(x) = \sin^{- 1}\left\{ \frac{\sqrt{3}}{2}x - \frac{1}{2}\sqrt{1 - x^{2}} \right\},\ - \frac{1}{2} \leq x \leq 1$, then $f(x)$ is equal to
A) $\sin^{- 1}{\frac{1}{2} - \sin^{- 1}x}$
B) $\sin^{- 1}{x - \frac{\pi}{6}}$
C) $\sin^{- 1}{x + \frac{\pi}{6}}$
D) None of these
Explanation
(b)
Put $x = \sin\theta$, we get
$\ \ \ \ \ \ f(x) = \sin^{- 1}\left\{ \sin\left( \theta - \frac{\pi}{6} \right) \right\}$
For, $- \frac{1}{2} \leq x \leq 1$
$\Rightarrow \ \ - \frac{1}{2} \leq \sin{\theta \leq 1}$
$\Rightarrow \ - \frac{\pi}{6} \leq \theta \leq \frac{\pi}{2}$
$\Rightarrow \ \ \theta = \frac{\pi}{6}$
is in the fourth on the first quadrant
$\therefore\ \ \ \ f(x) = \theta - \frac{\pi}{6} = \sin^{- 1}{x - \frac{\pi}{6}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.