How do you evaluate ${sin}^{- 1} (sin (\frac{π}{6}))$ $sin^-1(sin(pi/6))$ without a calculator?

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Answer 1 · 2016-08-26T04:23:30

$\frac{π}{6}$ $pi/6$

Use $f^{- 1} f (x) = x$ $f^(-1)f(x) = x$ that is valid for any function y = f(x) that is

bijective in an (infinitesimal) $\in$ $in$ - neighborhood $(x - \in, x + \in)$ $(x-in, x+in)$ of x.

Here, it is ${sin}^{- 1} sin (\frac{π}{6}) = \frac{π}{6}$ $sin^(-1)sin(pi/6)=pi/6$ .

It is important to note that $sin (\frac{π}{6}) = sin (\frac{13}{6} π) = \frac{1}{2}$ $sin (pi/6) = sin (13/6pi)=1/2$ and, while

${sin}^{- 1} (\frac{1}{2})$ $sin^(-1)(1/2)$ = the conventional principal value $\frac{π}{6}$ $pi/6$ ,,

${sin}^{- 1} sin (\frac{π}{6}) = \frac{π}{6}$ $sin^(-1)sin(pi/6)=pi/6$ and

${sin}^{- 1} sin (\frac{13}{6} π) = 13 \frac{π}{6}$ $sin^(-1)sin(13/6pi)=13pi/6$ .

How do you evaluate sin^-1(sin(pi/6))sin−1(sin(π6))sin^-1(sin(pi/6)) without a calculator?