Question
If $$f\left( x \right) = {\sin ^{ - 1}}\left\{ {\frac{{\sqrt 3 }}{2}x - \frac{1}{2}\sqrt {1 - {x^2}} } \right\}, - \frac{1}{2} \leqslant x \leqslant 1,$$ 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 :
$${\sin ^{ - 1}}x - \frac{\pi }{6}$$
Solution :
Let $$x = \sin \theta .$$ Then $$f\left( x \right) = {\sin ^{ - 1}}\left\{ {\sin \left( {\theta - \frac{\pi }{6}} \right)} \right\}.$$
$$\eqalign{
& - \frac{1}{2} \leqslant x \leqslant 1 \cr
& \Rightarrow \,\, - \frac{1}{2} \leqslant \sin \theta \leqslant 1 \cr
& \Rightarrow \,\, - \frac{\pi }{6} \leqslant \theta \leqslant \frac{\pi }{2}. \cr} $$
So, $$\theta - \frac{\pi }{6}$$ is in the fourth or the first quadrant. Hence, $$f\left( x \right) = \theta - \frac{\pi }{6}.$$