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

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Answer 1 · 2016-05-12T12:38:02

$\frac{π}{6}$ $pi/6$ .

$sin and {sin}^{- 1}$ $sin and sin^(-1)$ are each inverse of the other.

So, applied in succession, both ${sin sin}^{- 1} and {sin}^{- 1} sin$ $sin sin^(-1) and sin^(-1) sin$ return the operand.

${sin sin}^{- 1} x = x and {sin sin}^{- 1} y = y$ $sin sin^(-1) x = x and sin sin^(-1) y = y$

How do you simplify sin−1(sin(16)(π))sin^-1(sin (1/6)(pi))?