How do you evaluate ${sin}^{- 1} (sin (5 \frac{π}{6}))$ $sin^-1(sin (5pi/6))$ ?

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Answer 1 · 2016-04-08T14:52:30

$150^{o}$ $150^o$

${sin}^{- 1}$ $sin^-1$ and $sin$ $sin$ cancel each other out. They are opposite functions - its like multiplying and dividing by the same number.

This just leaves $\frac{5 π}{6}$ $(5pi)/6$ , where $π$ $pi$ radians is equal to $180^{o}$ $180^o$ , in which case

$\frac{5 π}{6} = \frac{5 \cdot 180^{o}}{6} = \frac{900^{o}}{6} = 150^{o}$ $(5pi)/6 = (5*180^o)/6 = 900^o/6 = 150^o$

How do you evaluate sin−1(sin(5π6))sin^-1(sin (5pi/6))?