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Question
35 - 40. evaluate.
- y = sin^{-1}(\frac{\sqrt{3}}{2})
Step1: Recall the definition of inverse sine function
The inverse sine function \(y = \sin^{-1}(x)\) has a range \([-\frac{\pi}{2},\frac{\pi}{2}]\) and \(\sin^{-1}(\sin\theta)=\theta\) when \(\theta\in[-\frac{\pi}{2},\frac{\pi}{2}]\). We know that \(\sin(\frac{\pi}{3})=\frac{\sqrt{3}}{2}\) and \(\frac{\pi}{3}\in[-\frac{\pi}{2},\frac{\pi}{2}]\).
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\(\frac{\pi}{3}\)