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 · 2018-02-28T10:46:58

$\frac{π}{6}$ $pi/6$

Explanation:

We know that
${sin}^{- 1} (sin x) = x, \forall x \in [- \frac{π}{2}, \frac{π}{2}]$ $color(red)(sin^-1(sinx)=x,AAx in[-pi/2,pi/2])$
${sin}^{- 1} (\frac{1}{2}) = {sin}^{- 1} (sin (\frac{π}{6})) = \frac{π}{6}, w h e r e, \frac{π}{6} \in [- \frac{π}{2}, \frac{π}{2}]$ $sin^-1(1/2)=sin^-1(sin(pi/6))=pi/6, where, pi/6in[-pi/2,pi/2]$
So, ${sin}^{- 1} (\frac{1}{2}) = \frac{π}{6}$ $sin^-1(1/2)=pi/6$

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

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