If theta in left[-frac{pi}{2}, frac{pi}{2}rig | vedclass

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Question

(A) Given that $	heta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}ight]$.We know that $\sin 	heta = \cos \left(\frac{\pi}{2} - 	hetaight)$.Therefore

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$\frac{\pi}{2}-	heta$

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(A) Given that $	heta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}ight]$.We know that $\sin 	heta = \cos \left(\frac{\pi}{2} - 	hetaight)$.Therefore,$\cos^{-1}(\sin 	heta) = \cos^{-1} \left[ \cos \left(\frac{\pi}{2} - 	hetaight) ight]$.Since $	heta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}ight]$,it follows that $-	heta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}ight]$,and thus $\left(\frac{\pi}{2} - 	hetaight) \in [0, \pi]$.The range of the principal value branch of $\cos^{-1} x$ is $[0, \pi]$.Since $\left(\frac{\pi}{2} - 	hetaight) \in [0, \pi]$,we have $\cos^{-1} \left[ \cos \left(\frac{\pi}{2} - 	hetaight) ight] = \frac{\pi}{2} - 	heta$.

If $\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$,then $\cos^{-1}(\sin \theta)$ is equal to

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