How do you evaluate ${sin}^{- 1} (0)$ $sin^-1(0)$ ?

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Answer 1 · 2017-01-03T11:50:03

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${sin}^{- 1} (0) = x$ $sin^-1(0) = x$

Take sin to the left and to the right.

${sin sin}^{- 1} (0) = sin x = 0$ $sin sin^-1 (0) = sin x = 0$

$x = 0 + k π$ $x = 0 + k pi$ , because $sin k π = 0$ $sin k pi = 0$

But what is the domain of $f (x) = {sin}^{- 1} x$ $f(x) = sin^-1 x$ ?

Response: $f : [- 1, 1] \to [- \frac{π}{2}, \frac{π}{2}]$ $f : [-1, 1] rightarrow [-pi/2, pi/2]$

How do you evaluate sin−1(0)sin^-1(0)?