Dear student, y = sin-1cos2π3use sin-1x+cos-1x =π2 ⇒sin-1x =π2-cos-1xy = π2-cos-1cos2π3range of cos-1x ∈0,π then cos-1cos2π3 =2π3y = π2-2π3 = 3π-4π6=-π6 answer Hope this information will clear your doubts If you have any more doubts just ask here on the forum and our experts will try to help out as soon as possible Regards

Q7

Dear student,

y = \sin^{- 1} \cos \frac{2 π}{3} u s e \sin^{- 1} x + \cos^{- 1} x = \frac{π}{2} \Rightarrow \sin^{- 1} x = \frac{π}{2} - \cos^{- 1} x y = \frac{π}{2} - \cos^{- 1} \cos \frac{2 π}{3} r a n g e o f \cos^{- 1} x \in [0, π] t h e n \cos^{- 1} \cos \frac{2 π}{3} = \frac{2 π}{3} y = \frac{π}{2} - \frac{2 π}{3} = \frac{3 π - 4 π}{6} = \frac{- π}{6} a n s w e r

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