Question

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

• A. ${\sin }^{-1}\left(\frac{1}{2}\right)-{\sin }^{-1}\left(x\right)$
• B. ${\sin }^{-1}x-\frac{\pi }{6}$
• C. ${\sin }^{-1}x+\frac{\pi }{6}$
• D. none of these

Answer 1

Answer: (B)

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

$f\left(x\right)={\sin }^{-1}\left(\sin \right({60}^0\left)\cos \right({\cos }^{-1}\left(x\right))-\cos ({60}^0\left)\sin \right({\cos }^{-1}\left(x\right)\left)\right)$

$={\sin }^{-1}\left(\sin \right({60}^0-{\cos }^{-1}\left(x\right))$

$=\sin \left({60}^0\right)-{\cos }^{-1}\left(x\right)$

$=\frac{\pi }{3}-\left(\frac{\pi }{2}\right)+{\sin }^{-1}\left(x\right)$

$={\sin }^{-1}\left(x\right)-\frac{\pi }{6}$