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sin inverse[root of 1+x+root of 1-x/2]

Dear Student, y = \sin^{- 1} [\frac{\sqrt{1 + x} + \sqrt{1 - x}}{2}] Putting x = \cos t = \sin^{- 1} [\frac{\sqrt{1 + cost} + \sqrt{1 - cost}}{2}] = \sin^{- 1} [\frac{\sqrt{1 + 2 \cos^{2} \frac{t}{2} - 1} + \sqrt{1 - 1 + 2 \sin^{2} \frac{t}{2}}}{2}] = \sin^{- 1} [\frac{\sqrt{2} \cos \frac{t}{2} + \sqrt{2} \sin \frac{t}{2}}{2}] = \sin^{- 1} [\frac{1}{\sqrt{2}} \times \cos \frac{t}{2} + \frac{1}{\sqrt{2}} \times \sin \frac{t}{2}] = \sin^{- 1} [\sin 45 \times \cos \frac{t}{2} + \cos 45 \times \sin \frac{t}{2}] = \sin^{- 1} [\sin (45 + \frac{t}{2})] = 45 + \frac{t}{2} Regards

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