How do you evaluate $sin^-1(sin((9pi)/10))$ ?

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Answer 1 · 2016-07-12T02:06:52

$(9pi)/10$

Successive operations with f and its inverse $f^(-1)$ , either way, on

an operand, returns the operand as the result.

Explicitly, if y = f(x), then $f^(-1)f(x)=x and f f^(-1)y=y$ .

Here, $f = sin, f^(-1)=sin^(-1)$ and the operand is $(9pi)/10$ .

So, $sin^(-1)sin ((9pi)/10) = (9pi)/10$ .

How do you evaluate sin^-1(sin((9pi)/10))sin^-1(sin((9pi)/10))?