If theta = sin^{-1}[sin(-600^circ)],then one | vedclass

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(A) Given $	heta = \sin^{-1}[\sin(-600^\circ)]$. Since $\sin(-	heta) = -\sin(	heta)$,we have $	heta = \sin^{-1}[-\sin(600^\circ)]$. Now,express $6

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$\frac{\pi}{3}$

Vedclass · Answer

(A) Given $	heta = \sin^{-1}[\sin(-600^\circ)]$. Since $\sin(-	heta) = -\sin(	heta)$,we have $	heta = \sin^{-1}[-\sin(600^\circ)]$. Now,express $600^\circ$ as a multiple of $180^\circ$ or $360^\circ$: $600^\circ = 360^\circ + 240^\circ$. So,$\sin(600^\circ) = \sin(360^\circ + 240^\circ) = \sin(240^\circ)$. Also,$\sin(240^\circ) = \sin(180^\circ + 60^\circ) = -\sin(60^\circ)$. Substituting this back: $	heta = \sin^{-1}[-(-\sin(60^\circ))] = \sin^{-1}[\sin(60^\circ)]$. Since $60^\circ$ is in the principal value branch $[-\frac{\pi}{2}, \frac{\pi}{2}]$,we have $	heta = 60^\circ = \frac{\pi}{3}$.

If $\theta = \sin^{-1}[\sin(-600^\circ)]$,then one of the possible values of $\theta$ is

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