Question

If $f\left(x\right)={\sin }^{-1}\{\frac{\sqrt{3}}{2}x-\frac{1}{2}\sqrt{1-x^2}\},-\frac{1}{2}\le x\le 1,$ then $f\left(x\right)$ 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

Answer 1

Answer: (B)

${\sin }^{-1}x-\frac{\pi }{6}$

$f\left(x\right)={\sin }^{-1}\{\frac{\sqrt{3}}{2}x-\frac{1}{2}\sqrt{1-x^2}\}$
Substitute $x=\sin \theta$
$f\left(x\right)={\sin }^{-1}\{\frac{\sqrt{3}}{2}\sin \theta -\frac{1}{2}\cos \theta \}$
$f\left(x\right)={\sin }^{-1}\left\{\sin (\theta -\frac{\pi }{6})\right\}$
$=\theta -\frac{\pi }{6}$

$={\sin }^{-1}x-\frac{\pi }{6}$