Question
The formula $$2{\sin ^{ - 1}}x = {\sin ^{ - 1}}\left( {2x\sqrt {1 - {x^2}} } \right)$$ holds for
A.
$$x \in \left[ {0,1} \right]$$
B.
$$x \in \left[ { - \frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}} \right]$$
C.
$$x \in \left( { - 1,0} \right)$$
D.
$$x \in \left[ { - \frac{{\sqrt 3 }}{2},\frac{{\sqrt 3 }}{2}} \right]$$
Answer :
$$x \in \left[ { - \frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}} \right]$$
Solution :
Let $$x = \sin \theta .$$ Then $${\sin ^{ - 1}}\left( {2x\sqrt {1 - {x^2}} } \right) = {\sin ^{ - 1}}\left( {\sin 2\theta } \right) = 2\theta = 2{\sin ^{ - 1}}x$$ if $$ - \frac{\pi }{2} \leqslant 2\theta \leqslant \frac{\pi }{2},\,{\text{i}}{\text{.e}}{\text{.,}}\sin \left( { - \frac{\pi }{4}} \right) \leqslant \sin \theta \leqslant \sin \left( {\frac{\pi }{4}} \right),\,{\text{i}}{\text{.e}}{\text{.,}} - \frac{1}{{\sqrt 2 }} \leqslant x \leqslant \frac{1}{{\sqrt 2 }}.$$