How do you evaluate ${sin}^{- 1} (sin (\frac{7 π}{10}))$ $sin^-1(sin((7pi)/10))$ ?

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Answer 1 · 2017-02-26T16:57:08

$\frac{7 π}{10}$ $(7pi)/10$

Recall that ${sin}^{- 1} (x)$ $sin^-1(x)$ is the inverse of $sin (x)$ $sin(x)$

Recall that a composition of inverse functions returns the independent variable.

$f^{- 1} (f (x)) = x$ $f^-1(f(x))=x$

So, if $f (x) = sin (\frac{7 π}{10})$ $f(x)=sin((7pi)/10)$ and $f^{- 1} (x) = {sin}^{- 1} (f (x))$ $f^-1(x)=sin^-1(f(x))$

then $f^{- 1} (f (x)) = {sin}^{- 1} (sin (\frac{7 π}{10})) = \frac{7 π}{10}$ $f^-1(f(x))=sin^-1(sin((7pi)/10))=(7pi)/10$

How do you evaluate sin^-1(sin((7pi)/10))sin−1(sin(7π10))sin^-1(sin((7pi)/10))?