How do you evaluate ${sin}^{- 1} (sin (\frac{7 π}{6}))$ $sin^-1(sin((7pi)/6))$ ?

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Answer 1 · 2016-07-15T02:59:12

#(7pi)/6

The answer follows from the definition of the inverse $f^{- 1}$ $f^(-1)$ of a

function f:

If $y = f (x), x = f^{- 1} (y), f^{- 1} f (x) = x and f f^{- 1} (y) = y$ $y = f(x) , x = f^(-1)(y), f^(-1)f(x)=x and ff^(-1)(y) = y$

Here, $f = sin and x = \frac{7 π}{6}$ $f =sin and x = (7pi)/6$ .

How do you evaluate sin^-1(sin((7pi)/6))sin−1(sin(7π6))sin^-1(sin((7pi)/6))?