Question

Let f (x) = $e^{{\cos }^{-1}\left\{\sin \left(x+\mathrm{\pi }/3\right)\right\}}$. Then, f (8π/9) =
(a) e^5π/18
(b) e^13π/18
(c) e^−2π/18
(d) none of these

Answer 1

(b) e^13π/18

Given: $f\left(x\right)=e^{{\cos }^{-1}\left\{\sin \left(x+\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$3$}\right.\right)\right\}}$

Then,
$$\begin{gathered} f\left(\raisebox{1ex}{$8\pi $}\!\left/ \!\raisebox{-1ex}{$9$}\right.\right)=e^{{\cos }^{-1}\left\{\sin \left(\raisebox{1ex}{$8\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$9$}\right.+\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\right)\right\}} \\ =e^{{\cos }^{-1}\left\{\sin \left(\raisebox{1ex}{$11\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$9$}\right.\right)\right\}} \\ =e^{{\cos }^{-1}\left\{\cos \left(\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.+\raisebox{1ex}{$13\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$18$}\right.\right)\right\}}\left[\because \cos \left(\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.+\theta \right)=\sin \theta \right] \\ =e^{{\cos }^{-1}\left\{\cos \left(\raisebox{1ex}{$13\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$18$}\right.\right)\right\}} \\ =e^{\raisebox{1ex}{$13\pi $}\!\left/ \!\raisebox{-1ex}{$18$}\right.} \end{gathered}$$