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

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Answer 1 · 2017-02-26T16:57:08

$(7pi)/10$

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

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

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

So, if $f(x)=sin((7pi)/10)$ and $f^-1(x)=sin^-1(f(x))$

then $f^-1(f(x))=sin^-1(sin((7pi)/10))=(7pi)/10$

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