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प्रश्न
Write the principal value of `sin^-1(-1/2)`
उत्तर
Let `y=sin^-1(-1/2)`
Then,
\[\sin{y} = - \frac{1}{2} = \sin\left( - \frac{\pi}{6} \right)\]
\[y = - \frac{\pi}{6} \in \left[ - \frac{\pi}{2}, \frac{\pi}{2} \right]\]
Here,
\[\left[ - \frac{\pi}{2}, \frac{\pi}{2} \right]\] is the range of the principal value branch of the inverse sine function.
∴ \[\sin^{- 1} \left( - \frac{1}{2} \right) = - \frac{\pi}{6}\]
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