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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Evaluate ; \[ sin \bigg[ \frac{\pi}{3}-sin^{-1} \bigg( -\frac{1}{2} \bigg) \bigg] \]

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  • \(sin^{-1}(-\large\frac{1}{2})=-\large\frac{\pi}{6}\)
  • \(sin\large\frac{\pi}{2}=1\)
Given $sin \bigg(\large\frac{\pi}{3} - sin^{-1} ( -\large\frac{1}{2} ) \bigg)$:
We know that \(sin^{-1}(-\large\frac{1}{2})=-\large\frac{\pi}{6}\).
Therefore, $sin \bigg(\large\frac{\pi}{3} - sin^{-1} ( -\large\frac{1}{2} ) \bigg) = $ \( sin \bigg[ \large\frac{\pi}{3}- \bigg( -\large\frac{\pi}{6} \bigg) \bigg] = sin\large\frac{\pi}{2} =1\)
answered Feb 13, 2013 by rvidyagovindarajan_1
edited Mar 19, 2013 by rvidyagovindarajan_1
 

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