How do you evaluate ${sin}^{- 1} (- \frac{1}{2})$ $sin^-1 (-1/2)$ ?

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How do you evaluate ${sin}^{- 1} (- \frac{1}{2})$ $sin^-1 (-1/2)$ ?

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Answer 1 · 2016-07-10T22:43:56

$- \frac{π}{6} and \frac{7 π}{6}$ $- pi/6 and (7pi)/6$

Explanation:

Trig table, and unit circle give -->
$sin x = - \frac{1}{2}$ $sin x = - 1/2$ --> arc $x = - \frac{π}{6}$ $x = -pi/6$ and arc $x = π + \frac{π}{6} = \frac{7 π}{6}$ $x = pi + pi/6 = (7pi)/6$
$A n s w e r s : - \frac{π}{6}, and \frac{7 π}{6}$ $Answers: -pi/6, and (7pi)/6$

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How do you evaluate ${sin}^{- 1} (- \frac{1}{2})$ $sin^-1 (-1/2)$ ?

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