How do you evaluate ${sin}^{- 1} (sin (\frac{π}{6}))$ $sin^-1(sin(pi/6))$ ?

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How do you evaluate ${sin}^{- 1} (sin (\frac{π}{6}))$ $sin^-1(sin(pi/6))$ ?

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Answer 1 · 2016-10-19T21:37:56

$\frac{π}{6}$ $pi/6$

Explanation:

The following is a pretty good definition for any inverse function:

Let $f (x)$ $f(x)$ be a function that has an inverse and let $f^{- 1} (x)$ $f^-1(x)$ be that inverse, then $f^{- 1} (f (x)) = x and f (f^{- 1} (x)) = x$ $f^-1(f(x)) = x and f(f^-1(x)) = x$

Therefore, the inverse sine "undoes" what the sine does.

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How do you evaluate ${sin}^{- 1} (sin (\frac{π}{6}))$ $sin^-1(sin(pi/6))$ ?

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How do you evaluate ${sin}^{- 1} (sin (\frac{π}{6}))$ $sin^-1(sin(pi/6))$ ?