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Find the values of \[ sin^{-1} \bigg( sin \frac{2\pi}{3} \bigg) \]

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  • $sin(\pi - x) = sinx$
  • The range of the principal value of $\sin^{-1}x$ is $\left [ -\frac{\pi}{2}, \frac{\pi}{2} \right ]$
Given $sin^{-1} \bigg( sin \frac{2\pi}{3} \bigg)$
We know that $sin(\pi - x) = sinx \Rightarrow sin \frac{2\pi}{3} = sin \frac{\pi}{3}$
The given expression becomes $sin^{-1} sin \frac{\pi}{3}$ which is $= \frac{\pi}{3}$
answered Feb 23, 2013 by thanvigandhi_1
edited Mar 14, 2013 by balaji.thirumalai

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