Question

The principal value of ${\sin }^{-1}\left[\sin \left(\raisebox{1ex}{$2\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\right)\right]$ is

• A. $\raisebox{1ex}{$-2\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$
• B. $\raisebox{1ex}{$2\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$
• C. $\raisebox{1ex}{$4\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$
• D. None of these

Answer 1

The correct option is D

None of these

Explanation of the correct option

Let ${\sin }^{-1}\left[\sin \left(\raisebox{1ex}{$2\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\right)\right]=\theta$

$\sin \theta =\sin \left(\raisebox{1ex}{$2\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\right)$

The principal value of $\sin x$ lies between $\raisebox{1ex}{$-\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ and $\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

$\sin \theta =\sin \left(\mathrm{\pi }-\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\right)=\sin \left(\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\right)$

The principal value of ${\sin }^{-1}\left[\sin \left(\raisebox{1ex}{$2\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\right)\right]$ is $\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$

Hence the correct option is option(d) i.e. None of these.