How do you simplify ${sin}^{- 1} (0.5)$ $sin^-1(0.5)$ ?

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Answer 1 · 2016-04-15T17:34:53

${sin}^{- 1} (0.5) = \frac{π}{6} + 2 π n, \frac{5 π}{6} + 2 π n$ $sin^-1 (0.5) =pi/6+2pin, (5pi)/6 +2pin$

Sin is positive in Quadrants 1 and 2 and all its coterminals

${sin}^{- 1} (0.5) = \frac{π}{6} + 2 π n, \frac{5 π}{6} + 2 π n$ $sin^-1 (0.5) =pi/6+2pin, (5pi)/6 +2pin$

How do you simplify sin−1(0.5)sin^-1(0.5)?