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प्रश्न
Write the value of \[\cos^{- 1} \left( - \frac{1}{2} \right) + 2 \sin^{- 1} \left( \frac{1}{2} \right)\] .
उत्तर
For any x ∈ [−1, 1], cos−1x represents an angle in [0, \[\pi]\] whose cosine is x.
∴ \[\cos^{- 1} \left( - \frac{1}{2} \right)\] =any angle in [0, \[\pi\]] whose cosine is \[- \frac{1}{2}\] .
\[\Rightarrow \cos^{- 1} \left( - \frac{1}{2} \right) = \frac{2\pi}{3}\]
Similarly,
\[\sin^{- 1} \left( \frac{1}{2} \right)\] = an angle in \[\left[ - \frac{\pi}{2}, \frac{\pi}{2} \right]\] whose sine is \[\frac{1}{2}\] .
\[\Rightarrow \sin^{- 1} \left( \frac{1}{2} \right) = \frac{\pi}{6}\]
∴ \[\cos^{- 1} \left( - \frac{1}{2} \right) + 2 \sin^{- 1} \left( \frac{1}{2} \right)\] =
\[\frac{2\pi}{3} + 2\left( \frac{\pi}{6} \right) = \frac{4\pi + 2\pi}{6} = \pi\]
Hence,
\[\cos^{- 1} \left( - \frac{1}{2} \right) + 2 \sin^{- 1} \left( \frac{1}{2} \right) = \pi\] .
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