Question

If θ = sin⁻¹ {sin (−600°)}, then one of the possible values of θ is
(a) $\frac{\mathrm{\pi }}{3}$

(b) $\frac{\mathrm{\pi }}{2}$

(c) $\frac{2\mathrm{\pi }}{3}$

(d) $-\frac{2\mathrm{\pi }}{3}$

Answer 1

(a) $\frac{\mathrm{\pi }}{3}$

We know
${\sin }^{-1}\left(\sin x\right)=x$

Now,

$$\begin{gathered} \theta ={\sin }^{-1}\left\{\sin \left(-{600}^{\circ }\right)\right\} \\ ={\sin }^{-1}\left\{\sin \left({720}^{\circ }-{600}^{\circ }\right)\right\} \\ ={\sin }^{-1}\left\{\sin \left({120}^{\circ }\right)\right\} \\ ={\sin }^{-1}\left\{\sin \left({180}^{\circ }-{120}^{\circ }\right)\right\}\left[\because \sin x=\sin \left(\mathrm{\pi }-x\right)\right] \\ ={\sin }^{-1}\left(\sin {60}^{\circ }\right) \\ ={60}^{\circ } \end{gathered}$$