Considering the principal values of the inver | vedclass

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(B) Let $y = \sin ^{-1}\left(\frac{\sqrt{3}}{2} x+\frac{1}{2} \sqrt{1-x^2}ight)$.Given $-\frac{1}{2} < x < \frac{1}{\sqrt{2}}$,let $x = \sin 	heta$

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$\frac{\pi}{6}+\sin ^{-1} x$

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(B) Let $y = \sin ^{-1}\left(\frac{\sqrt{3}}{2} x+\frac{1}{2} \sqrt{1-x^2}ight)$.Given $-\frac{1}{2} Since $-\frac{1}{2} Substituting $x = \sin 	heta$ into the expression:$y = \sin ^{-1}\left(\frac{\sqrt{3}}{2} \sin 	heta + \frac{1}{2} \cos 	hetaight)$$y = \sin ^{-1}\left(\sin \frac{\pi}{3} \sin 	heta + \cos \frac{\pi}{3} \cos 	hetaight)$ is incorrect; use $\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$ and $\cos \frac{\pi}{3} = \frac{1}{2}$.$y = \sin ^{-1}\left(\sin 	heta \cos \frac{\pi}{6} + \cos 	heta \sin \frac{\pi}{6}ight)$$y = \sin ^{-1}\left(\sin \left(	heta + \frac{\pi}{6}ight)ight)$.Since $-\frac{\pi}{6} Since the range of $\sin ^{-1}$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$ and $0 Thus,$y = \sin ^{-1} x + \frac{\pi}{6}$.

Considering the principal values of the inverse trigonometric functions,$\sin ^{-1}\left(\frac{\sqrt{3}}{2} x+\frac{1}{2} \sqrt{1-x^2}\right)$,for $-\frac{1}{2} < x < \frac{1}{\sqrt{2}}$,is equal to:

Similar Questions

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